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https://news.ycombinator.com/item?id=16535358
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Show HN: Convert screenshots of equations to LaTeX 456 points by slbenfica on Mar 7, 2018 | hide | past | web | favorite | 98 comments
Immediately made me think of a question on the state of handwriting to LaTeX from a few years ago, and all the massive challenges involved:https://tex.stackexchange.com/questions/1443/what-is-the-sta...Under API... you're already doing handwriting? This is uh, nontrivial work to say the least. Really impressive.The endorsements are a nice touch. :)Made me really curious how far the system goes, what cases break it.Oh... nevermind. You have a PDF of examples here: https://docs.mathpix.comIt's honing in on equations without getting distracted by nearby Hanzi or Cyrillic, or even pictures of dogs. Wow.I keep going back to dig through your resources and getting more impressed.EDIT: I guess my only constructive criticism is that you should brag more. I like a simple landing page, but I think you've earned a short list of examples of corner cases you tackle well, if the whole API is packed into that free app, because they're really impressive.
The perfection in those examples makes me suspect that they are cherry-picked or part of the training data. Especially the handwritten text is not always clear and could reasonably be interpreted differently. I'd expect a machine-learning model to get at least some things wrong some of the time.If I wanted to use this in an application, I'd definitely want to see some accuracy figures on validation data as well as a few failure cases to see whether the output remains reasonable even when it is wrong.
> the handwritten text is not always clear and could reasonably be interpreted differentlyDigital pen input contains more info than the resulting bitmap; strokes are lost while rasterizing.That info was the reason how old devices were able to reliably recognize characters written by a stylus. It worked well even on prehistoric hardware, such as 16MHz CPU + 128 kB RAM in the first Palm PDA.
In the OCR world this is known as offline OCR (OCR on the bitmap) vs online OCR (strokes information).Offline is way harder than online.
The examples are actually very simple compared to a lot of crazy stuff Mathpix can recognize so it's an honest representation of its capabilities. Mathpix is built for perfection because 99% isn't good enough.
This is really awesome OP! Thank you for sharing :)One note I should make: it was not entirely clear (to me) upon a cursory view of the website, that the purpose of mathpix was to convert handwritten text into LaTeX. For some reason (maybe my coffee hasn't kicked in yet) I thought this was strictly intended to take screenshots of equations on an existing pdf document or a website etc and that will be converted to LaTeX.My thought at that point was "I wonder if they could do this for handwritten text" and then I looked at the docs and facepalmed..
"Screenshot" is a very odd word to use in this context.
How? "Screenshot" is used correctly by the OP.For reference: screenshot (verb) an image of the data displayed on the screen of a computer or mobile device.
Their usage does not match that definition. When recognising writing, they are recognising an image of the page, not an image of what was on the screen at some other point.Also, your definition describes a noun but claims it's a verb.
The promotional material shows the application interpreting a screenshot of an equation, not handwriting.I also meant noun, not verb. Thanks.
I see absolutely nothing wrong with the usage here.
Screen grab?
Photograph or picture, surely?
You should reread OPs paragraph. Also if you go straight to the website it shows you the desktop app first which is using screenshots from whatever to turn it into Latex.
Stupid question, how well would this work on a PDF of a latex document?This would be great for blind people, as pdfed latex is extremely non-accessable, and I have to email authors of papers to get the original latex from them, which is often lost.
As someone who writes with Latex, thanks for pointing out this problem, I hadn't considered it.
> This would be great for blind people, as pdfed latex is extremely non-accessable, and I have to email authors of papers to get the original latex from them, which is often lost.Many years ago I "translated" course materials into a form which was accessible to a blind grad student. It was a really interesting job and taught me a lot about accessibility.I was effectively doing latex, but without all the leading \ characters. It made learning latex comparatively easy.What interface do you use to read equations? Screen reader speaking the straight latex, or do you have some Middleware to make it more digestible when listened to?
I'm not blind, but I have a blind PhD. He seems most happy with just straight latex. I suspect this is because it strikes a good balance between requiring little work for him, or others, as it usually already exists.
Isn't that the whole point of this utility? Or am I missing something?
Well I was hoping it could take a while page, rather than require me to cut out the equations.
InftyReader [1] does this. Its results are good but definitely not perfect. Unfortunately it is seriously expensive, at US $400 per computer. I may be naive here, but is there anything preventing running regular OCR over a page, and then feeding whatever it couldn't deal with into this? Sure, the plumbing for this is probably missing, but it sounds more like a matter of picking up the shovel rather than inventing something. I've done some work cleaning up documents for use by a blind student. OCR starts to fail really badly when math is involved. Common errors like O versus 0, any accents or additions to characters, small formulas embedded in sentences, graphs with captions, etc. Can all throw things off considerably.Best case I could copy and paste paragraphs at a time from a PDF of the textbook (with copy protection removed). Worst case I was retyping or fixing every few words in a sentence.I was working on this from about 2011-2013. Advances with image processing and machine learning have been significant since then, so there may be much better software available now.If anyone has ideas or packages they'd recommend, I'd be interested. It's not a stupid question. You need equation detection + equation OCR.Mathpix only does the equation OCR part.I've worked on this (for a PDF to HTML application), mail is in profile if you're interested. Im curious if the developers are fans of The Big Bang Theory TV series? They were using a smart phone app...and of course was less useful due to it being fiction... Wow!What kind of sorcery is this!?Is this using deep learning or "regular" OpenCV or similar?I would assume it's a highly tuned deep learning algo, but I'm not knowledgeable enough to distinguish a deep learning algo from a pile of rocks...Edit: Aha, someone already asked this and got an answer. Suggestion: instead of making me download a pdf to see examples of what the results look like, maybe put them on the page directly. You can have a couple. Then put the details in the pdf.Great software otherwise This is fantastic!!Bug report: it appears that multiline summation subscripts are not recognized correctly. For example, Eq. 8 of [1]. These are often created using \substack as part of amsmath.Awesome tool! Good catch!!! We're working on it, should be fixed by April 1st I assume you just got a lot of installs from India, because the large publishing houses contract out many re-typesetting jobs that are basically to take scans of technical texts and convert them back into LaTeX.I strongly suggest you talk to the publishers about integrating your tech into their TeX. Want a math-ish PDF and some LaTeX source for training on possible edge cases? Think I might get someone (or something) to read my dissertation this way... What's your dissertation about? yes please! nico@mathpix.com This is awesome work. What's the process of vision -> latex conversion? We published some research on one approach:http://lstm.seas.harvard.edu/latex/Here's how to do it with OpenNMT/PyTorch: I am curious to know as well. Yup! Please!! I really appeciate the testimonials. All software deserves personality and a soul like this! Any way you could make this available outside the Mac App store? Apple seems to have decided I did something horrible and unforgivable by moving to a different country after creating an account, thus making it impossible for me to use the store. Good to know, thanks for posting this. We'll post the dmg file on our landing page this week or next. Just a reminder, I'm still eagerly awaiting this-- I check your site everyday! :) Awesome! Is the plan for it to be free forever, or what might the pricing look like? Maybe you would consider open sourcing the model?Also it would be nice of some info on the process. Does work entirely locally, or is images uploaded to the cloud? from the site:$0.005 per request, first 1000 requests are free
Where does it say that? All I see is the (free) MacOS App.
It's at the bottom of https://mathpix.com/api.html
We used their API to make a simple screenshot2latex tool (select screen region -> puts latex formula in clipboard). From my experience it still fails on a couple of fairly common things like:- \mathcal letters (recognized as non-mathcal)- long equations (not recognized at all)- multi-line equations (not recognized at all)The screenshot2latex tool: https://github.com/rmst/screenshot2latex/blob/master/scripts...
All three points have seen big improvements in the last week, especially the first too, check again
That's aweson OP. I can't imagine the number of times I've wished for something like this.Coming from a grad student who hates writing equations in latex. I will probably try this out.
Is there any chance of a Linux client?
They have an api. You can write your own.
I was looking for an API that provides math OCR. Great, going to integrate it into our app soon :-) Let me know if you want to add us to your "trusted by" section.
sure! send the link to support@mathpix.com
Impressive. Any particular reason that you're using \left\{ \begin{array} ... \end{array} \right instead of \begin{cases} ... \end{cases} ?
No particular reason, we were thinking to start returning cases syntax soon!
Mathematicians use operator overloading all the time. It would be nice to have a tool that explains to me what an equation actually means in a given context.
You are talking about the semantics of an equation while this tool is already satisfying when understanding correctly the syntax (in LaTeX).There are actually a number of ongoing research projects to establish standards of semantical mathematical representations. Probably one of the best funded running projects (budget ~10MEUR) which has a work package on this topic is http://opendreamkit.org/ . Work is going on at https://mathhub.info/ from my knowledge. I would like to provide a deep link but the site seems to be in a broken state. Apparently people are working on it right in the moment.
I’ve long had this dream of building a system that would allow for symbolic algebra manipulations, where the representation knows about the semantics. The idea is to replace long and tedious pen-and-paper symbolic calculations, while still having direct control over each step of the calculation. In looking for something like this I’ve found specialized proof verifiers, which are doing something different. At the other end of the spectrum there’s Mathematica, which does symbolic manipulation but doesn’t understand enough semantics.Are these projects aiming at something like what I’m describing? Or are they more about something else like verifying proofs?
Well, in principle a CAS like Mathematica (or in principle every mature alternative) allows you to implement such a workflow (i.e. naming mappings or an algebra on your objects). I think what Wolfram is concentrating on since a decade is accumulating "knowledge" into their system, i.e. exactly this kind of semantic information we talk about. However, as an end user I don't really see that, it seems to be more well-groomed behind Wolfram alpha.On the other hand, there are these research projects which however seem to concentrate on standards rather then actually accumulating semantic knowledge.I would love to see an adoption of hypertext and semantic mathematical notation in scientific papers. Instead of writing $E=m c^2$ in a (LaTeX) paper, we would instead define the symbols machine readably with a code like set E = physics/Energy set m = physics/Mass set c = physics/constants/speed-of-light I have never seen actual scientific papers which do this kind of stuff, i.e. which are machine readable.
Isn't that usually what the document is for? :)The verbalization if most LaTeX commands can help learn to read the equations. Sometimes.
This is insanely impressive. Great work. Wish tools like this existed when I was still in school...almost makes me want to go back and do some more math :)
I tried this on some of my (fairly neat) handwritten physics notes and it was mostly pretty impressive. Failures: a lot of lowercase deltas became 8s and it had no idea what to make of hbar (converting it to n, pi, k, or most often refusing to convert the equation entirely). A fair number of little typos, but you'd want to check all its work anyway.
It's not reliable at complex handwritten math yet, it does do printed advanced math pretty reliably, as well as simple handwritten math.
It didn't recognize hbar as 1?
Looks like it could be really useful. Have you considered not requiring billing info to try the API?
we do it to protect against spamming... send me a note at nico@mathpix.com if you want to try the API
The Newton quip is a missed opportunity to make a reference to Diamond, his favorite dog who burned up all his papers...
This is gold, great job. The one clear bug that I've noticed thus far is that it seems to correctly identify \hat{} but not finish it correctly when it also has an index, so that: \hat { y _ { i } } erroneously becomes: \hat { y } _ { i }
This changes everything concerning my university life for me. I always was on the verge of doing everything digitally, but it was always cumbersome to type out (La)Tex by hand and convert handwritten notes to digital versions.Thank you so much!
This is something I would have definitely used if I were still a student, although I'm not a fan of it being a Chrome extension. I'm still curious enough to test it out.
There's also a macOS app.
Very cool! I had this exact idea of a service a few years back, but was one of those things I never found the time or motivation to actually do. The results seem very nice.
This is great! it is very useful for writing papers. I tried it with some of my equations! It actually improved it by adding more approrpiate braces than I had.
The example on the front page thinks the nu is a v. Still cool, though.
Read the text. The example on the front page contains both a \nu (frequency) and a v (velocity), and the tool distinguishes between them correctly.
Gr8! Would love a way to apply this to PDF and web eqns in Windows.
I maintain an idea notebook in latex via Lyx and presently, while reading a math PDF I have to paste a screenshot in paint, clip there an equation, and paste the image in lyx. Very clumsy.
Can you send me an email? I might have something like that working soon.
Cool! Any plans to port to Windows and Linux also?
Just tried it out and it worked great. Nice work!
Add code generator to complete the product. ;-p
I'm sorry if it's off the topic. The testimonials really cracked me up. “ If I had known about Mathpix earlier, perhaps I would have had enough time to work out the Grand Unified Theory. ” - Albert Einstein
My favorite is Alan Turings quote:> Mathpix's AI definitely passes THIS Turing test!
me to lol, really clever way to get attention.
This looks neat. Also, the Testimonials section made me laugh. Well done sir.
So I guess I can expect to see more low quality mathematical typesetting. The technology is great, but I don't see how it will help with high quality typesetting.I can type an equation into LaTeX more quickly that I can photograph it and then go in and manually correct all the spacing issues. And there are spacing issues. The examples PDF has things that just look horrible. No small spacing or negative spacing to space out things like matrices and integrals etc. If I'm going to manually tweak it anyway I might as well do it manually from the start. Typing it up is something that gets quicker with practice like everything else.But what I can actually see happening is people not tweaking the output manually. Either train yourself to use TeX properly, or let someone do it for you.
You may be correct, but your comment comes off sounding a bit elitist.Typesetting does not always need to be absolutely perfect, sometimes it just needs to work.For someone who isn't 100% fluent in TeX, making small edits is much easier than having to look up all the syntax and symbol names required.
It's not elitist at all. Anyone can learn TeX.
> Either train yourself to use TeX properly, or let someone do it for youI will let you do it for me. I usually work late at night just before the deadline. What’s your phone number?
Is that a serious offer? I charge 50GBP per hour. Still interested?
If you can match \$0.005 per request, I'll hire you.
I think the point is that for 50 GBP you will get a much higher quality typesetting than spending 1 GBP on the API.As.usual, it depends on the use case. If you are producing a textbook or something that requires high quality typesetting, you will probably pay for it.If not, you can use a tool like this one but you will have to accept certain margin of error.
"Either train yourself to use TeX properly, or let someone do it for you."The same complaint could be raised about using a biro (vs calligraphy), using a printing press (vs handwriting) or using a wysiwyg editor (vs the old word processing paradigm).In my view, this sort of accessibility is a great advance.
I disagree. Technology should either make things easier or better. Making things easier but worse is not progress. Your analogies are incorrect.Calligraphy is not handwriting. It's an art form and very slow to execute.The printing press isn't just an "easier handwriting". It's actually harder to use. But it's better. It's closer to calligraphy than handwriting, in fact.I'm not sure what your argument about WYSIWYG text editors is. In my experience they produce absolutely horrible results because, ironically, they require the user to know more about typesetting than TeX does if high quality output is desired.
thanks for your comments, can you share what you want your Latex to look like? how do you want to space out matrices and integrals? we're working on the Latex formatting so we can improve this
I would have put a small space (\,) between the upright "det" and the matrix. An integral sign is usually followed by a negative small space (\!) and the "dx" is separated by a small space (\,).Trouble is I'm not sure if these are absolute rules. I'm sure you can find pathological cases where the spacing will be wrong. Maybe the best thing would be to allow the user to modify the spacing easily after the automatic version is made.I trust you guys have read the TeXBook? Knuth dedicates two chapters to subtleties of mathematical typesetting.
Do you have a guide that you can point to for the correct typesetting in LaTeX?
The best treatment is in the original manual for TeX: The TeXBook by Donald Knuth.
Applications are open for YC Summer 2019
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2019-03-26 20:37:36
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{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5918060541152954, "perplexity": 1245.782927638604}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912206016.98/warc/CC-MAIN-20190326200359-20190326222359-00012.warc.gz"}
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https://www.physicsforums.com/threads/what-is-inductance-i-need-a-physical-meaning-and-not-equations.387136/
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# What is inductance? i need a physical meaning and not equations!
1. Mar 16, 2010
### hisham.i
what is inductance?...i need a physical meaning and not equations!
2. Mar 16, 2010
### collinsmark
Re: inductance
There's more than one way to think about inductance. Here is one such way, as applied to an inductor in a circuit.
Inductors tend to resist change in current. It takes work to get current flowing through an inductor in the first place. Once current is flowing through an inductor, the inductor tends to keep that same amount of current flowing through itself.
The emf (aka Voltage) across the terminals of an inductor are a result of this. The inductor produces an emf in such a way that the overall circuit tends to resist any change in current flowing through the inductor.
I know you didn't want any equations, but I'm going to give you one anyway. The induced Voltage, emf, across an inductor is:
$$V_{emf} = L \frac{di}{dt}$$
where L is the inductor's inductance, and di/dt is the rate of change of current.
So the magnitude of which a component resists change in current is called its inductance.
As a mechanical analogy, think of Newton's laws. An object in motion tends to stay in motion, unless acted upon by an outside force. The amount which a body tends to resist changes in motion is proportional to its mass. Inductors do the same thing, except replace motion with current. The amount that it resists changes in current is proportional to its inductance.
(btw., mutual inductance adds a whole different dimension to this picture, and is a property essential to transformers, power supplies, etc. But I'll leave that for another day.)
3. Mar 16, 2010
### Bob S
Re: inductance
Here is a physics explanation. In an inductor there is a magnetic field. The magnetic field is proportional to the inductor current. The magnetic field represents stored energy. The amount of stored magnetic energy is proportional to the square of the magnetic field** times its volume. The magnetic field (energy) can be in air, ferrite, laminated iron, etc.
** divided by the permeability of the material.
Bob S
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2018-12-16 00:44:53
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https://socratic.org/questions/how-do-you-write-0-00769-in-scientific-notation
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How do you write 0.00769 in scientific notation?
$0.00769 = 7.69 \times {10}^{-} 3$
$0.00769 = 7.69 \times {10}^{-} 3$
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2019-10-20 11:19:11
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https://sarahjanewoods.com/docs/archive.php?id=21f6bb-why-is-toluene-aromatic
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#### Sarah Jane
Life Coach for Women
Empowering women to appreciate themselves and find the balance, confidence and freedom to live a life they love
# why is toluene aromatic
When there is a single substituent on a benzene ring and the … While aromatic compounds are best represented by a continuous electron density evenly distributed around the aromatic core, the alternating single and double bonds that are commonly drawn are very useful when predicting the reactivity of aromatic compounds. By using this site, you agree to our, Previously Toluene was produced commercially from -, Present day the method used to produce Toluene is by separating the aromatic, The extractive distillation involves the use of a solvent for liquid-liquid extraction. 's' : ''}}. As a member, you'll also get unlimited access to over 83,000 The higher the octane number, the more compression the fuel can withstand before detonating. The most widely practiced example of this … 3D Structures. d. Solvent extraction - The most important requirement for a solvent in a solvent extraction process is that it should have a high selectivity for aromatics against non- aromatics, that a two-phase system is obtained in a reasonable range of temperature, and that the phases should separate within a reasonable range of temperature; of course it must also be non- corrosive, non- reactive and thermally stable. In contrast, fuels with lower octane numbers are ideal for diesel engines. Benzene and toluene are organic compounds. The extractive distillation involves the use of a solvent for liquid-liquid extraction. The color of the toluene solution of C60 is bright purple. Give any common name for each of the following compounds. Toluene is toxic but is sometimes used as a recreational inhalent and has the potential of causing severe neurological harm. As the solvent in some types of paint thinner, permanent markers, contact cement and certain types of glue, toluene is sometimes used as a recreational inhalant[7] and has the potential of causing severe neurological harm. The reaction preserves the pi system of electrons and therefore the aromatic character of the benzene ring. 7- Ceresana. "Toluene â Study: Market, Analysis, Trends - Ceresana". Create an account to start this course today. For substituted benzene rings where the substituent contains less than six carbons, the alkyl chain is added as a prefix with the ending changed to -yl. Inhalation of toluene in low to moderate levels can cause tiredness, confusion, weakness, drunken-type actions, memory loss, nausea, loss of appetite, hearing loss, and colour vision loss. For substituted benzene rings where the substituent contains more than six carbons, the benzene ring is noted by using a phenyl prefix on the alkane name. Methyl side groups present in toluene makes toluene metabolized differently than benzene.. Toluene is well absorbed through the lung, with an alveolar retention of 40 to 80 per cent of an inhaled dose. Toluene reacts as a normal aromatic hydrocarbon in electrophilic aromatic substitution. Plus, get practice tests, quizzes, and personalized coaching to help you electrophilic aromatic substitution (second step) The gain in stabilization attendant on regeneration of the aromatic ring is sufficiently advantageous that this, rather than combination of the cation with $$\ce{Y}^\ominus$$, normally is the favored course of reaction. When there are multiple substituents, ring atoms are numbered to minimize the numbers assigned to the substituted positions. [5] Toluene is used as a jet fuel surrogate for its content of aromatic compounds. This is in contrast to oxygenated octane boosters like ethanol or, Toluene - additive for racing fuels, fuel octane booster. Due to this connected network of $\pi$ bonds, the rings are planar, unlike the boat or table structures typical of cycloalkanes. - Definition & Example, What is Formaldehyde? Toluene Molecule - Ball and Stick Configuration, Toluene Molecule using Jsmol see -- For Toluene is produced from specially selected fractions of petroleum, rich in naphthenes, by catalytic reforming (also known as hydroforming). I agree to the Aromatic compounds, originally named because of their fragrant properties, are unsaturated hydrocarbon ring structures that exhibit special properties, including unusual stability, due to their aromaticity. The extent of toluene substitution has not been determined. Most amounts of toluene (in the form of benzene-toluene-xylene mixtures) are used in the blending of petrol and it also occurs as a by-product of styrene manufacture. Toluene is well absorbed through the lung, with an alveolar retention of 40 to 80 per cent of an inhaled dose. Toluene is also used as a component in gasoline and jet fuel because it boosts octane ratings, or the measurement of how much pressure gasoline can endure before igniting. Aromatic compounds are ring structures with unusual stability due to delocalized pi electron density that is shared between all of the carbon atoms in the ring. Benzoic acid and benzaldehyde are produced commercially by partial oxidation of toluene with oxygen. Toluene at 100% can be used as a fuel for both two-stroke and four-stroke engines; however, due to the density of the fuel and other factors, the fuel does not vaporize easily unless preheated to 70 °C (158 °F). Honda solved this problem in their Formula One cars by routing the fuel lines through a heat exchanger, drawing energy from the water in the cooling system to heat the fuel. Study.com has thousands of articles about every The hydrogen tends to reduce coke formation and thus maintain catalytic activity. Toluene is produced during the process of making gasoline and other fuels from crude oil, in making coke from coal, and as a by-product in the manufacture of styrene. Odor. Toluene is mainly used as a precursor to benzene via hydrodealkylation: The second ranked application involves its disproportionation to a mixture of benzene and xylene.[19]. Log in or sign up to add this lesson to a Custom Course. 36 kcal/mol). The extent of toluene substitution has not been determined. For the same reason, toluene is also used as a component in jet fuel. This involves catalytic dehydrogenation in the presence of hydrogen (which reduces coke formation) to yield a mixture of aromatic hydrocarbons, chiefly toluene. Toluene incurs no fuel excise tax, while other fuels are taxed at more than 40%, providing a greater profit margin for fuel suppliers. 1) Toluene reacts about 25 times faster than benzene under identical conditions.
### Sarah Jane WoodsLife Coach for Women
Sarah Jane is an NLP practitioner who believes that when we nourish our energy, our lives transform as we flourish.
We live with less fear, worry, doubt and anxiety and find the confidence to be ourselves every day.
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2021-01-16 18:05:28
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http://4chanarchives.cu.cc/board/sci/16
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Archived threads in /sci/ - Science & Math - 16. page
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What does /sci/ think of this
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>>8522563
xKCD is cucked
>>
>>8522568
Mathematics is an attempt to understand and organize the world around us. This comes from an innate desire to learn as much about life as we can, I.E. applied philosophy.
>>
I think the "math is derived from philosophy" argument is flawed. Philosophy depends on logical and axiomatic foundations that are more in the domain of mathematicians than philosophers nowadays (though historically there's a lot of overlap).
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I thought I should make a new one.
>>
I guess I'll start.
How do I figure out substitutions and all that stuff while doing integration?
Everytime I see a new problem I can't for the life of me figure out what to do to solve it but then I see the solution and it makes sense.But there is no way I could predict the substitutions and all the tweaking stuff.
Tl;Dr I suck at integration and need to learn it ASAP.
Pls tell me how to.
>>
Best way to study math? How much repeating concepts until they stick properly?
>>
What is even the meaning of these in the Laplace notation? Let's that I have
y'= -2x+4y
With initial y(0)=1
y'+2x-4y=0
L{y'}+2L{x}-4L{y}=L{0}
I understand everything up to here, but then
[sY(s)-y(0)]+2X(s)-4Y(s)=0
What is it with the Y(s) X(s) things? I know how to Laplace simple stuff and the results are always s's, but where are these giant X and Y even coming from?
Why are there no large-scale, well-funded efforts to significantly remediate the biological ageing process through scientific and technological means?
There seem to only be a handful of small charities with a few researchers and annual budgets in the low single digit millions of dollars; and a few researchers at a smattering of universities.
It doesn't seem to be considered a very big deal; why is this, given that everyone in the world will suffer as a result of ageing and there seems to be no in-principle reason why we could not do an awful lot more to...
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>>8521115
R&D is always at the back end of concerns, it seems, even when that concern is ultimately foremost for the selfish (i.e. rational) organism.
Still, if we accept the OP's frame, then this is a deeply encouraging fact together with our capacity to destroy ourselves. You see, I've reached a misanthropic conclusion recently, which is that the species ought to simple cease to exist, rather than to allow these arbitrary life extensions to continue unabated, even to the point of a game-changing biological...
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>>8521148
Edgy.
>>
Why do you keep making this thread?
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These are facts:
Women are more fertile during the 15-25 year range.
Males feel more attracted to the 15-25 year range, independely of their age.
A 30 year old women has a 70% risk of being infertile.
How long until feminism is crushed by hard biology facts?
>>
>>8528230
>A 30 year old women has a 70% risk of being infertile.
lol?
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>>8528253
http://www.natureworldnews.com/articles/4462/20131016/delaying-pregnancy-cause-infertility-miscarriages-obstetricians.htm
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>>8528260
says nothing about 70% and My personal view,
"as someone who's been a consultant obstetrician for 20 years, is that people should try to start having their family by 37 because after that things become more complex and it's harder to get pregnant"
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>be me
>taking a calculus final
>kid sitting behind me is hacking and coughing and sneezing every few seconds, without covering his mouth
>this continues for the entire two hours
>be me two days later
>starting to feel a soreness in my throat
Fuuuuuuuckk meeeeeeeeeeeeeeeeeeee
Are there any 100% guaranteed ways to prevent a cold this early, while the symptoms are *just*...
>>
>>
If you're showing symptoms you've probably been infected for a while. So I'd say no.
Source: I got a 90% in intro to bio, so I basically have a complete working understanding of the human body.
>>
zicam cold remedy
It's like antibiotics for colds, it's amazing.
zicam + tylenol can get rid of body aches and that general "sick" feeling almost completely
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Couldn't dark matter just be cold neutrinos?
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>>8527862
With that mass? Also how the hell do you define temperature for a single particle?
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>>8527866
by its kinetic energy and/or momentum?
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>>8527873
I didnt mean that there is no way to do it, I was genuinely asking, is there a known way to achieve this?
Also, I think neutrinos have too low mass to explain dark matter and why would they have to be cold?
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Are you blind?
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>>8527854
I wasnt until i saw this thread
>>
>>
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What would happen if you traveled back in time and killed your grandfather? If he dies yuo dont even exist ever, so you cannot kill him.
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>>8527774
You can't travel back in time.
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>>8527774
Different timelines i guess
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>>8527778
t. time traveler shill
OP, would you like the red pill or the blue one?
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can anyone with an above average iq who has the money and is willing to work hard capable of getting a PhD?
>>
yes
really you only need two of them
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>>8527667
>above average iq
So, around 45% of the world is above average IQ, OP
>money
Well duh, how will you pay for the PhD?
>willing to work hard
You obviously do this and everyone does it as well but it depends on how fast you can learn is the actual difference
>>
>>8527762
what are you getting at?
what I was really asking in the OP i guess is how special do you have to be to earn a PhD?
Are the Feynman lectures on physics a good place to start for someone who has studies all the hs physics courses but wants to brush up his skills and maybel learn something new? Or are they outdated? And if so, what would you recommend?
>>
they deal with stuff that hasn't changed since they were written so it's OK. not outdated
i dont like the way it is structured tho
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>>8527631
>all the hs physics courses
"All"? So you took algebra based physics, then took honors algebra freshman physics, and then took AP premed watered down physics?
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>>8527655
You assume I live in USA, I don't. Your pre uni school system sucks, I know.
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So I failed Calc 1 last year in College because I was a piece of shit that didn't go to class and missed exams. I knew the actual material since I took Calc 1 in highschool, but now my college won't let me reapply for Calc 1 unless I retake the placement test. Does anyone have some good pre-calc/calc1 books for review?
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>>8527626
lol
patrickjmt.com has a fuck ton of video lectures for (self admitted) pieces of shit like you
beyond that, if you're dead set on conventional textbooks, go to bookzz.org and do a keyword search
Stewart's books are serviceable, assuming you need calc but aren't a math major
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>>8527626
Lang's Basic Mathemaics
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>>8527626
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I want to become programming literate enough to A) land a comfy job hacking at Blockchain/Machine learning meme tech and B) eventually hack together my own projects in the future as a hobby. What is the best way to go about achieving these goals? Would the education path (Community College then 4 year school) be the best way, or self learning and getting some sort of credential?
I don't care enough to learn everything there is to learn about computers... just enough to hack shit together and know when I need other people to pick up my slack.
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>>8527522
>comfy job hacking at Blockchain/Machine learning meme tech
lol
>>
>>8527541
I was just coming in to chuckle at this too, gr8 laffs OP
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>>8527522
If you think albegra/statistics is easy then go learn by yourself
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So I saw a thread like that this week and I think it was actually a good idea.
ITT: We share scientific facts. Can be related to any /sci/ area.
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OP is a faggot. (Source: >>8527516)
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>>8527527
Kek
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niggers are inferior
source: ayn rnd
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Could Heisenberg's Uncertainty Principle be resolved if we had better tools to measure electrons?
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>>8527495
But then we wouldn't have free will asshole
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>>8527495
Literature says no
I say yes
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>>8527495
I can't even imagine what "better tools" would be able to do that
There are quantum mechanical reasons for the inability to measure electrons, if I recall correctly the foremost explanation is that at a quantum level the electron reciprocates with its observer somehow, which changes its momentum or position.
The double slit experiment is really fascinating and illustrates this better than I can.
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Alright, fellows. I'm in a deep one. I need to find the turning points (points where $\dot{r} = 0$ ) of pic related.
The result should be given by the equation in pic, but the one I'm getting has minus at the second term in the second parenthesis.
Can anybody tell me where am I wrong?
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>>8527447
Also, to use the initial conditions (r0, v0) to express the constants arising from the integrals of motion.
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>>8527447
>has minus
I meant to say that it has plus in the second term of the second parenthesis.
>>
>>8527447
bampo
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Can someone explain to me how the sum of an infinite number of zeros can equal not zero.
I understand that we can do a lot of cool things by assuming that it doesn't sum to zero, but it seems apparent that the limit of pic related would be zero.
>>
it is zero.
it's the sequence:{0,0,0,0,0...} which converges to zero
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>>8527339
Why wouldn't it be zero? What is happening in this board? Is /B/ raiding us?
>>
>>8527348
>>8527371
Doesn't integration require the sum of an infinite number of zeros not to equal zero, though. If you imagine integration as a Riemann Sum as the area of the box approaches zero, then you're just summing a bunch of zeros together and getting a non-zero number.
Or, since I'm more familiar with statistics, we know that, given a continuous distribution, the probability of getting any one number in particular is 0%, but the probability of...
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I find it so strange that no one is talking about this, this might seriously be a beginning to a new era of mankind.
Can we connect logical thinking to our reward center? Would you do it?
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>>8527293
>"Science Finds God?"
>Spinoza
My Gott.*sniff*
>>
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post rare slavojs
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>>8527557
its dat boj o shit waddup
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Suppose that there exists infinite series that converges conditionally in classic space R of real numbers. Riemann series theorem states that I can rearrange elements of this series to get what ever real number I want.
If we look at R as algebraic structure, what is the main property that gives us this theorem?
What does group theory says about using operator infinite number of time?
>>
group structures don't allow for infinitely applied operators
it's a purely sequential property
>>
>>8526979
>If we look at R as algebraic structure, what is the main property that gives us this theorem?
The least upper bound property, and how any bounded closed set is compact. These properties form the basis for the proof that R is complete.
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>>8527439
but that's a topological property,
>>8526986
this anon is right
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Monty Hall is psychological bullshit
After your first decision, the host will ALWAYS open a goat door. There will ALWAYS be one goat and one car remaining. You have no impact on this outcome no matter which door you pick. It will always end with the same result. The first decision has zero relevance.
The first decision exists purely to act as a mind game to make you consider additional variables in your second decision which ultimately hold no meaning. Monty wants you to hold a sentimental value towards one door over another, to frame the decision as 'stay'...
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>>8526737
but each time you switch doors the average amount approaches 2/3rds of the time, this can be demonstrated experimentally. http://math.ucsd.edu/~crypto/cgi-bin/MontyKnows/
Surely if you deny the abstract theory, you at least support empirical evidence?
>>
>>8526743
>>
>>8526743
You're missing the point anon
That experiment is still with framing the decision as "stay" or "switch", which is dependent on whatever initial decision you made
But your initial decision is meaningless. The outcome is always identical (one goat door opened). It serves no purpose for setting up the second decision besides framing it to be about some arbitrary attachment
Run the exact same experiment as "pick door 1 or 2" rather than "stay or switch" which...
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I'm a scientist, specializing in criminology. AMA
>>
why do you make AMA here instead of reddit?
>>
Why can't I get a gf?
>>
>>8526635
Funny way to say psychologist
Anyway, i hope you get them pedos with the cooperation with the Gods of the Internet (NSA,FBI,CIA)
Also pls don't delet my chinesecartoons tho
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Lets say someone shot a giant space laser at the yellowstone caldera and burned a whole through the rock into the magma chamber, could it set off the volcano?
>>
That's a solid maybe.
>>
>>8526499
Why would it have to be a space laser?
Why not just dump a bunch of thermite down it?
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>>8526506
down what?
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Depression is not a mental disorder, it's a rational response to having a shit life
>>
So how do You convince your depressed mate that?
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>>8526413
Then it's not depression.
Depression is being chronically sad when you should not be. Being sad when your life is shit is not depression.
When you're successful, have a loving gf you bone every night, a nice house and car, and a good career and are so sad you can't get out of bed and want to blow your brains out, that's depression.
>>
>>8526418
But You are stating generic things. There can be other things in that Said persons Life. Things they want to be different, which they deem as depression, since they havent been true to Themselves about what they seek to change. For that Said person: he could be gay, only keeping that gf for his image to friends and family, while going to his job, to Pay for shit he Dont want. Like the House etc.
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"What" exactly are we?
>>
>>8526390
thoughts, according to Daykart
>>
>>8526390
Autistic
>>
>>8526390
woooooah #weed
i fucking love science
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Solve for x.
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>>8525844
ur a faggit
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>>8525844
>Solve for x.
Step 1: Type all that into wolfram
Step 2: hit enter
Works every time.
>>
>>8525844
try plugging in 1 or -1 first
Hey /sci/
I'm looking to start learning math from ground up up until college level.
Anyone has any books they recommend? Preferably in order, I don't mind sitting hours on end studying.
Thanks a bunch.
>>
literally khan
>>
>>8525390
So just hours of videos and interactive problems?
I'd prefer books and pdfs with clear explanations if that's possible because I don't have a Computer, but I do have quite a large pdf reader.
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>>8525384
Wanting to do the same as you OP. I have been using Pauls's online notes and would recommend using those to you, but as for books i have no idea so also interested in seeing responses to this.
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What is the difference between category theory, group theory, and set theory?
If they aren't fucking synonyms I'm going to beat up a math professor (I would have said shoot up a math class but I don't want the FBI breaking down my door. Now that I've said that let me say I plan to neither beat up a math professor nor shoot up a math class.) Because that shit is ridiculous.
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>>8525356
>cs major detected
>>
category theory studies objects and arrows between them
group theory studies symmetries
set theory studies sets
>>
They are completely different.
That being said your professor probably just wanted you to shut up and stop asking pedantic questions so he could teach the class and go home.
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mean?
>>
It means "I hate white people"
>>
>>8527598
more like....i hate European science..because....I hate white people
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>>8527587
relocating modern science to India
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>It seems like Elon is the only one attempting to leave Earth before 2020.
Will it take us another 200 years to send humans outside of our solar system?
>>
>>8527547
honesty it's pretty easy to send someone out of the solar system. The "problems" are.
Bringing them back.
Whats worth sending someone out there.
Why cant a sensor robot go
muh safety
In the early days of jets, rockets, and space flight the guys that were flying were daredevils who got pushed to the limit. No reason not to be on the moon right now.
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>>8527547
There's nothing worthwhile outside of earth you cockmuncher.
>>
WE
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Do photons exist inside an electron, well, or in any fermion, lepton, that can emit, absorb this package of energy?
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Photons are electrons
>>
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>>8527328
Yes; this is why electrons have lower energy levels as they get closer to the nucleus. Once the photons excape from the inside of the electrons, the electron loses mass, and therefrom momentum. Since it loses momentum, it also loses kinetic engergy, resulting in the n=1 energy level being the lowest. glad i could help
>>
>>8527341
oh wow, but about that loses mass, aren't photons massless, particles without mass, what do you mean by that when a photon leaves and electron, electron loses mass?
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If 2+3×4=14 then how come 2+3×4÷2=8? Shouldn't it be 7?
>>
this is the worst thread i have ever seen
>>
2+3x4÷2≠(2+3x4)÷2
>>
>>8527199
I just want to know. I was never very good at math.
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2017-01-17 15:11:07
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http://wiki.seeedstudio.com/Xadow_Barometer/
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edit
The module can measure the air pressure and temperature of the current environment, and then you can convert this signal to information of altitude. If you are planning to use Xadow to do a wearable device and measure the altitude when you climb the mountain, then the module will be your perfect choice.
## Specification¶
• Working Voltage: 1.8 ~ 3.3v
• Pressure Range: 300 ~ 1100hPa(+9000m ~ 4500m above sea level)
• Low Power
• Control Mode: I2C (address 0x77)
• Operating Temperature：-40 ~ +85 °C
• Dimensions: 25.43mm x 20.35mm
## Demonstration¶
The demo will show you how to get the air pressure and temperature of the current environment.
Hardware Installation
Note
When connect Xadow Barometer to Xadow Main Board, you should concern about the connection direction. The connection method is that the unfilled corner of one Xadow module need to connect to the right angle of another module(see four corners of each Xadow module).
• Xadow Barometer is sharing the library with Grove - Barometer Sensor. You can download directly the library from Github.
• Now you can open the code and upload it to Xadow Main Board. If you have not installed the Xadow Driver, You can learn the operation steps by referring to the getting started in wiki main page .
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 /* * Get pressure, altitude, and temperature from the BMP085. * Serial.print it out at 9600 baud to serial monitor. */ #include "Barometer.h" #include float temperature; float pressure; float atm; float altitude; Barometer myBarometer; void setup(){ Serial.begin(9600); myBarometer.init(); } void loop() { temperature = myBarometer.bmp085GetTemperature(myBarometer.bmp085ReadUT()); //Get the temperature, bmp085ReadUT MUST be called first pressure = myBarometer.bmp085GetPressure(myBarometer.bmp085ReadUP());//Get the temperature altitude = myBarometer.calcAltitude(pressure); //Uncompensated calculation - in Meters atm = pressure / 101325; Serial.print("Temperature: "); Serial.print(temperature, 2); //display 2 decimal places Serial.println("deg C"); Serial.print("Pressure: "); Serial.print(pressure, 0); //whole number only. Serial.println(" Pa"); Serial.print("Ralated Atmosphere: "); Serial.println(atm, 4); //display 4 decimal places Serial.print("Altitude: "); Serial.print(altitude, 2); //display 2 decimal places Serial.println(" m"); Serial.println(); delay(1000); //wait a second and get values again. }
• Open the serial monitor to receive the sensor's data including temperature, barometric pressure value, relative atmosphere pressure and altitude.
• Here is a reference graph plotting out the relationship between altitude above sea level and barometric pressure.
## Tech Support¶
Please submit any technical issue into our forum.
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2019-09-21 08:50:46
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http://www.physicsforums.com/showthread.php?t=608834
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Implicit Euler Integration
I'm trying to write a code to implement he backwards Euler method to integrate the equation of motion. The sticking point seems to be that the acceleration is due to drag, and thus is dependent on the new position and velocity.
I understand the method to be:
$v_{i+1}=v_{i}+a_{i+1}δ$
$x_{i+1}=x_{i}+v_{i+1}δ$
With only the current conditions I can’t evaluate $a_{i+1}$ and am stuck.
Any help on how the implicit methods work would be really appreciated.
I’ve considered using $a_{i+1}= a_{i}+ (a_{i}-a_{i-1}){δ}$ but then that’s not really an implicit method is it? – you’re simply reusing the acceleration from last time around.
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The backwards Euler method is iterative. The first guess is usually the forward Euler method, so the right-hand side is determined using the information at time "i". You now have a first estimate of the properties at time "i+1". You can now use these estimates in the right-hand side to get a better estimate of the properties at time "i+1". so for your example, first use a and v at time "i", then you have v and x at time "i+1". Now calculate a at "i+1" and recalculate v an x, but using your new estimate of a and v at "i+1": first iteration: $v^1_{i+1}=v_{i}+a_{i}\Delta t$ $x^1_{i+1}=x_{i}+v_{i}\Delta t$ calculate $a_{i+1}=f(v^1_{x+1})\Delta t$ second iteration: $v^2_{i+1}=v_{i}+a^1_{i+1}\Delta t$ $x^2_{i+1}=x_{i}+v^1_{i+1}\Delta t$
Recognitions:
Quote by RH10 I’ve considered using $a_{i+1}= a_{i}+ (a_{i}-a_{i-1}){δ}$
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2013-05-18 18:35:41
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https://www.quizover.com/physics-k12/section/constant-acceleration-force-is-applied-in-the-opposite-direction
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# 2.6 One dimensional motion with constant acceleration (Page 5/6)
Page 5 / 6
## Interpretation of equations of motion
One dimensional motion felicitates simplified paradigm for interpreting equations of motion. Description of motion in one dimension involves mostly the issue of “magnitude” and only one aspect of direction. The only possible issue of direction here is that the body undergoing motion in one dimension may reverse its direction during the course of motion. This means that the body may either keep moving in the direction of initial velocity or may start moving in the opposite direction of the initial velocity at certain point of time during the motion. This depends on the relative direction of initial velocity and acceleration. Thus, there are two paradigms :
• Constant force is applied in the direction of initial velocity.
• Constant force is applied in the opposite direction of initial velocity.
Irrespective of the above possibilities, one fundamental attribute of motion in one dimension is that all parameters defining motion i.e initial velocity, final velocity and acceleration act along a straight line.
## Constant acceleration (force) is applied in the direction of velocity
The magnitude of velocity increases by the magnitude of acceleration at the end of every second (unit time interval). In this case, final velocity at any time instant is greater than velocity at an earlier instant. The motion is not only in one dimension i.e. linear , but also unidirectional. Take the example of a ball released (initial velocity is zero) at a certain height ‘h’ from the surface. The velocity of the ball increases by the magnitude of ‘g’ at the end of every second. If the body has traveled for 3 seconds, then the velocity after 3 seconds is 3g (v= 0 + 3 x g = 3g m/s).
In this case, all parameters defining motion i.e initial velocity, final velocity and acceleration not only act along a straight line, but also in the same direction. As a consequence, displacement is always increasing during the motion like distance. This fact results in one of the interesting aspect of the motion that magnitude of displacement is equal to distance. For this reason, average speed is also equal to the magnitude of average velocity.
$\begin{array}{l}s=|x|\end{array}$
and
$\begin{array}{l}\frac{\Delta s}{\Delta t}=|\frac{\Delta x}{\Delta t}|\end{array}$
## Constant acceleration (force) is applied in the opposite direction of velocity
The magnitude of velocity decreases by the magnitude of acceleration at the end of every second (unit time interval). In this case, final velocity at any time instant is either less than velocity at an earlier instant or has reversed its direction. The motion is in one dimension i.e. linear, but may be unidirectional or bidirectional. Take the example of a ball thrown (initial velocity is ,say, 30 m/s) vertically from the surface. The velocity of the ball decreases by the magnitude of ‘g’ at the end of every second. If the body has traveled for 3 seconds, then the velocity after 3 seconds is 30 - 3g = 0 (assume g = 10 $\phantom{\rule{2pt}{0ex}}m/{s}^{2}$ ).
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
n=a+b/T² find the linear express
Quiklyyy
Moment of inertia of a bar in terms of perpendicular axis theorem
How should i know when to add/subtract the velocities and when to use the Pythagoras theorem?
Centre of mass of two uniform rods of same length but made of different materials and kept at L-shape meeting point is origin of coordinate
A balloon is released from the ground which rises vertically up with acceleration 1.4m/sec^2.a ball is released from the balloon 20 second after the balloon has left the ground. The maximum height reached by the ball from the ground is
work done by frictional force formula
Torque
Why are we takingspherical surface area in case of solid sphere
In all situatuons, what can I generalize?
the body travels the distance of d=( 14+- 0.2)m in t=( 4.0 +- 0.3) s calculate it's velocity with error limit find Percentage error
Explain it ?Fy=?sN?mg=0?N=mg?s
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2018-02-24 19:42:56
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https://owenduffy.net/blog/?p=22446
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# On testing coax cable loss with an analyser / VNA – part 1
A recent online video provides instruction on how to measure loss of a section of coax cable, loss to mean Matched Line Loss, $$MLL=\frac{P_{in}}{P_{out}}$$ when the cable is terminated in its characteristic impedance Zo, and which can be expressed in db as $$MLL=10 log_{10} \frac{P_{in}}{P_{out}}$$. Note that MLL in dB is ALWAYS a +ve value for a passive DUT such as this.
There is nothing new in the method, it appears in lots of analyser user manuals, and has a built in assist in many analysers.
The video deals with the case of an antenna analyser that has a ‘measure cable loss’ function and using a VNA. Lets use the VNA graphic as it shows more detail of what is happening.
Above is the video’s graphic for the case. The narration says to use the dB magnitude of s11 or ReturnLoss as equivalents. They aren’t equivalent (a hammy Sammy thing), $$ReturnLoss=-20 log_{10}|s11|$$ or $$ReturnLoss_{dB}=-s11mag_{dB}$$ (both wrt the VNA reference impedance).
The video states that $$MLL=\frac{50}{length} ReturnLoss_{dB} \text{ dB/100′}$$.
I am going to use metric length units in this example, the same formula applies but the units are metres.
Let’s calculate the expected values of such a test on a 10m section of Belden 8267 (RG213) at 3.5MHz. We will use TLLC for the calculation of both MLL and MLL’ calculated from half ReturnLoss.
Above is the complete output from TLLC, we are interested in the highlighted quantities.
MLL is given by “Line loss (matched)” as 0.105dB (for 10m).
ReturnLoss is given by “RL(50)in” as 0.146dB, so half ReturnLoss is 0.073dB for 10m.
This gives MLL’ (based on half ReturnLoss) of about two thirds cable MLL.
Why is that?
What does it say for the measurement technique?
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2021-11-30 13:08:19
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http://time.to.pullthepl.ug/post/2014/02/01/xv6-on-mac-os-x-now-easy/
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# xv6 on Mac OS X, now easy
Xv6, if you haven’t heard about it, is a reimplementation of Version 6 Unix in ANSI C for x86 processors for an operating systems class at MIT. When I first came across it a while back I wanted to play with it, but unfortunately it’s rather harder to get working under OS X because xv6 uses ELF for its executables while OS X uses COFF Mach-O (update: I’m not sure how I horked that one up. I must have either been thinking of COFF for some reason, or read something else that claimed OS X used COFF, or I don’t even know what. Fixed, in any case.). You have to build a cross compiler to get xv6 building under OS X. Unfortunately that isn’t a very straightforward process, and I wasn’t real interested in devoting tons of time getting it to all work right. I had found instructions for getting xv6 to build on OS X using MacPorts (and other instructions too), but never had any luck with the assembler actually building anything.
The other day, though, I bumped into xv6 again and decided to take a look once more. Someone out there was trying to build xv6 with clang somehow, and it got me thinking about the cross compiler thing again. Googling a little, I saw that someone made just the homebrew formulas for the cross compilers that I needed, so I downloaded those, put them in with my homebrew formulas, and ran brew install i386-elf-gcc and brew install i386-elf-gdb. I had already installed qemu and the prerequisites mentioned on that gist about trying to build xv6 with clang, so after making some little changes to the Makefile (seen here, between lines 32 and 78) xv6 finally built and started up for me.
Of course, xv6 doesn’t actually do much. I’ve been dinking around with it some though (because obviously what I need is another project), adding some simple utilities and whatnot, that I’ve tossed up on github. So far all I’ve done is add some simple versions of basic utilities, but I’m exploring. Even if it doesn’t do much, it’s pretty interesting.
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2021-05-19 01:42:06
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http://crypto.stackexchange.com/questions/15305/hkdf-vs-tls-prf-which-of-the-one-is-better?answertab=active
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# HKDF vs TLS PRF. Which of the one is better? [closed]
Which of the one (HKDF or TLS1.1+ PRF) is more secure and why?
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## closed as unclear what you're asking by e-sushi, D.W., DrLecter, figlesquidge, GillesMar 31 at 14:43
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.
Better, in what sense? More secure, in what sense? For what purposes? Under what threat model? This question is too vague as it stands; I suggest you edit it to make it more specific. – D.W. Mar 30 at 18:23
It is unclear if you wanted to compare TLS 1.1 PRF or TLS 1.2 PRF. Different TLS versions have different PRFs. Assuming you meant TLS 1.1 PRF although you linked TLS 1.2 RFC.
## TLS 1.1 PRF
Short: HKDF is commonly a better choice than TLS 1.1 PRF, but not always.
Consider these aspects:
• HKDF is a generic construct.
• HKDF is extract and expand.
• TLS1.1 PRF is just extract or expand.
• TLS1.1 PRF is intended for TLS protocol.
• HKDF can be used with hash functions chosen by the user.
• TLS1.1 PRF uses MD5 and SHA-1 hashes. Thus, it does not allow using stronger hash functions.
• TLS1.1 PRF uses two hash functions which causes extra complexity and complicates security analysis.
• TLS1.2 defined new PRF function to use. (So for purposes of TLS, TLS1.1 PRF has gotten deprecated).
TLS1.1 PRF has been analyzed in Hash function combiners in TLS and SSL. Based on this analysis, HKDF, when used with a strong hash function is likely stronger than TLS1.1 PRF.
## TLS1.2 PRF
Comparing HKDF with TLS 1.2's PRF.
NIST SP 800-108 Recommendation for Key Derivation Functions defines various ways to construct Key Derivation Function: - Counter Mode - Feedback Mode - Double-Pipeline Iteration Mode
For NIST, all these approaches are equally acceptable. The Feedback Mode matches HKDF's Expand phase. The Double-Pipeline Iteration Mode matches TLS 1.2 PRF. Thus we could assume, that TLS 1.2 PRF could be considered to be similar to HKDF's Expand phase if used with the same hash function underlying the hmac.
Purpose of HKDF's extract phase is quoted from RFC 5869:
In many applications, the input keying material is not necessarily
distributed uniformly, and the attacker may have some partial
knowledge about it (for example, a Diffie-Hellman value computed by a
key exchange protocol) or even partial control of it (as in some
entropy-gathering applications). Thus, the goal of the "extract"
stage is to "concentrate" the possibly dispersed entropy of the input
keying material into a short, but cryptographically strong,
pseudorandom key. In some applications, the input may already be a
good pseudorandom key; in these cases, the "extract" stage is not
necessary, and the "expand" part can be used alone.
For applications where extract phase is beneficial (distribution of input keying material is not uniform), HKDF is certainly the way to go.
For other uses, where expand phase is used alone, any one of these two (or in some cases counter mode) from NIST SP 800-108 could be used. (Note: If you use TLS 1.2 PRF outside context of TLS protocol, maybe better define the function in terms of NIST SP 800-108, instead of TLS 1.2 PRF from the RFC).
Short summary:
• If you need extract use HKDF.
• If you need expand only, use functions from NIST SP 800-108.
• Generally you should not use TLS 1.1 PRF except for compatibility with TLS protocol.
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Note that the NIST SP 800-108 specs are more or less used to cover all secure KDF's that are allowed. I've created generic implementations of them and tested them against the test vectors. Those test vectors are not covering everything and are already multiple MB's for each KDF construction that is defined. Specifying the practical security level of NIST SP 800-108 is a moot point; there are too many variables. Heck, the KDF's don't even have official names let alone ASN.1 definitions or OID's. – owlstead Apr 24 at 11:13
So you can validate your implementation (say, HSM) to be FIPS compliant, but please don't try to find 2 vendors that implement exactly the same KDF :( Reporting this to NIST resulted in a clear "MEH" from NIST – owlstead Apr 24 at 11:16
@owlstead: I almost agree on your points. However, 2 vendors implementing exactly the same KDF are occasionally found, but this is a result of them implementing support for the same protocol(s) using KDFs. BTW, there are also some CAVP unofficial KDF test vectors available from NIST, for a preliminary compliance check of implementation. Furthermore, if you have a product in being validated in CAVP (a part of FIPS 140-2), you get to choose a lot of options for KDF, to specifically pinpoint how to test your implementation. – user4982 Apr 25 at 4:18
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2014-11-01 13:24:12
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https://code.tutsplus.com/tutorials/counting-word-frequency-in-a-file-using-python--cms-25965
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# Counting Word Frequency in a File Using Python
Difficulty:IntermediateLength:ShortLanguages:
Ever wondered about a quick way to tell what some document is focusing on? What is its main topic? Let me give you this simple trick. List the unique words mentioned in the document, and then check how many times each word has been mentioned (frequency). This way would give you an indication of what the document is mainly about. But that wouldn't work easily manually, so we need some automated process, don't we?
Yes, an automated process will make this much easier. Let's see how we can list the different unique words in a text file and check the frequency of each word using Python.
## Test File
In this tutorial, we are going to use test.txt as our test file. Go ahead and download it, but don't open it! Let's make a small game. The text inside this test file is from one of my tutorials at Envato Tuts+. Based on the frequency of words, let's guess which of my tutorials this text was extracted from.
Let the game begin!
## Regular Expressions
Since we are going to apply a pattern in our game, we need to use regular expressions (regex). If "regular expressions" is a new term to you, this is a nice definition from Wikipedia:
A sequence of characters that define a search pattern, mainly for use in pattern matching with strings, or string matching, i.e. "find and replace"-like operations. The concept arose in the 1950s, when the American mathematician Stephen Kleene formalized the description of a regular language, and came into common use with the Unix text processing utilities ed, an editor, and grep, a filter
If you want to know more about regular expressions before moving ahead with this tutorial, you can see my other tutorial Regular Expressions In Python, and come back again to continue this tutorial.
## Building the Program
Let's work step by step on building this game. The first thing we want to do is to store the text file in a string variable.
Now, in order to make applying our regular expression easier, let's turn all the letters in our document into lower case letters, using the lower() function, as follows:
Let's write our regular expression that would return all the words with the number of characters in the range [3-15]. Starting from 3 will help in avoiding words that we may not be interested in counting their frequency like if, of, in, etc., and words having a length larger than 15 might not be correct words. The regular expression for such a pattern looks as follows:
\b is related to word boundary. For more information on the word boundary, you can check this tutorial.
The above regular expression can be written as follows:
Since we want to walk through multiple words in the document, we can use the findall function:
Return all non-overlapping matches of pattern in string, as a list of strings. The string is scanned left-to-right, and matches are returned in the order found. If one or more groups are present in the pattern, return a list of groups; this will be a list of tuples if the pattern has more than one group. Empty matches are included in the result unless they touch the beginning of another match.
At this point, we want to find the frequency of each word in the document. The suitable concept to use here is Python's Dictionaries, since we need key-value pairs, where key is the word, and the value represents the frequency words appeared in the document.
Assuming we have declared an empty dictionary frequency = { }, the above paragraph would look as follows:
We can now see our keys using:
Finally, in order to get the word and its frequency (number of times it appeared in the text file), we can do the following:
Let's put the program together in the next section, and see what the output looks like.
## Putting It All Together
Having discussed the program step by step, let's now see how the program looks:
If you run the program, you should get something like the following:
Let's come back to our game. Going through the word frequencies, what do you think the test file (with content from my other Python tutorial) was talking about?
(Hint: check the word with the maximum frequency).
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2021-05-10 19:42:36
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https://www.math-only-math.com/worksheet-on-expansion-of-a-plus-minus-b-whole-square-and-its-corollaries.html
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# Worksheet on Expansion of (a ± b)$$^{2}$$ and its Corollaries
Practice the questions given in the worksheet on expansion of (a ± b)$$^{2}$$ and its corollaries.
1. Expand the squares of the following:
(i) 4x + y
(ii) 5a + 3b
(iii) 2x + $$\frac{1}{x}$$
2. Expand the following:
(i) (x – 2y)$$^{2}$$
(ii) (3y – 2z)$$^{2}$$
(iii) (3x - $$\frac{1}{3x}$$)$$^{2}$$
3. Expand the squares of the following:
(i) $$\frac{3}{2}$$ + $$\frac{5}{3}$$n
(ii) 4x - $$\frac{3}{4}$$
(iii) a$$^{2}$$ + 1
(iv) a$$^{2}$$ - $$\frac{1}{a}$$
4. Simplify:
(i) (3x + 5y)$$^{2}$$ + (3x – 5y)$$^{2}$$
(ii) (a - $$\frac{1}{a}$$)$$^{2}$$ + (a + $$\frac{1}{a}$$)$$^{2}$$
(iii) (7x + 4y)$$^{2}$$ – (7x – 4y)$$^{2}$$
(iv) (x + $$\frac{1}{x}$$)$$^{2}$$ – (x - $$\frac{1}{x}$$)$$^{2}$$
5. If a + $$\frac{1}{a}$$ = 2, find the value of
(i) a$$^{2}$$ + $$\frac{1}{a^{2}}$$
(ii) a$$^{4}$$ + $$\frac{1}{a^{4}}$$
6. If x - $$\frac{1}{x}$$ = 3, find the value of
(i) x$$^{2}$$ + $$\frac{1}{x^{2}}$$
(ii) x$$^{4}$$ + $$\frac{1}{x^{4}}$$
7. If x + y = 7 and x – y = 3, evaluate
(i) xy
(ii) x$$^{2}$$ + y$$^{2}$$
8. If a + b = 8 and ab = 12, find
(i) a$$^{2}$$ + b$$^{2}$$
(ii) the difference between a and b.
9. If x$$^{2}$$ + 4x – 1 = 0 and x is positive then find the value of
(i) x + $$\frac{1}{x}$$
(ii) x$$^{2}$$ + $$\frac{1}{x^{2}}$$
10. (i) If a + $$\frac{1}{a}$$ = 5, find the difference between a and $$\frac{1}{a}$$.
(ii) m$$^{2}$$ + n$$^{2}$$ = 34 and mn = 10$$\frac{1}{2}$$, find the value of 2(m + n)$$^{2}$$ + (m – n)$$^{2}$$.
Answers for the worksheet on expansion of (a ± b)$$^{2}$$ and its corollaries are given below.
1.(i) 16x$$^{2}$$ + 8xy + y$$^{2}$$
(ii) 25a$$^{2}$$ + 30ab + 9b$$^{2}$$
(iii) 4x$$^{2}$$ + 4 + $$\frac{1}{x^{2}}$$
2. (i) x$$^{2}$$ – 4xy + 4y$$^{2}$$
(ii) 9y$$^{2}$$ – 12yz + 4z$$^{2}$$
(iii) 9x$$^{2}$$ – 2 + $$\frac{1}{9x^{2}}$$
3. (i) $$\frac{9}{4}$$ + 5n + $$\frac{25}{9}$$n$$^{2}$$
(ii) 16x$$^{2}$$ – 6x + $$\frac{9}{16}$$
(iii) a$$^{4}$$ + 2a$$^{2}$$ + 1
(iv) a$$^{4}$$ – 2a + $$\frac{1}{a^{2}}$$
4. (i) 18x$$^{2}$$ + 50y$$^{2}$$
(ii) 2a$$^{2}$$ + $$\frac{2}{a^{2}}$$
(iii) 112xy
(iv) 4
5. (i) 2
(ii) 2
6. (i) 11
(ii) 119
7. (i) 10
(ii) 29
8. (i) 40
(ii) 4
9. $$\sqrt{20}$$
(ii) 18
10. $$\sqrt{21}$$
(ii) 123
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2019-10-22 01:59:35
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http://mathhelpforum.com/trigonometry/207510-rewrite-expression-w-o-fractions-cscx-1-sinx.html
|
# Math Help - Rewrite as an expression w/o fractions: (cscx)/(1-sinx)
.
2. ## Re: Rewrite as an expression w/o fractions: (cscx)/(1-sinx)
Rewrite as an expression w/o fractions: (cscx)/(1-sinx)
$\frac{\csc{x}}{1-\sin{x}} = \sec^2{x}(\csc{x}+1)$
also ... please post problems in the body of the post, not the title.
3. ## Re: Rewrite as an expression w/o fractions: (cscx)/(1-sinx)
It would be good if you gave some indication that you had tried to do the problem yourself, too!
4. ## Re: Rewrite as an expression w/o fractions: (cscx)/(1-sinx)
I meant to say that I multiplied by the conjugate
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2016-02-12 21:03:08
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https://socratic.org/questions/55fce4da581e2a428193afe4
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# Question #3afe4
Sep 19, 2015
the minimum length of ladder is 5.406m
#### Explanation:
answer is very simple . the length of the ladder which is to be caluclated must be minimum .
so the length of the ladder = 3*cosec (x)+sec(x)
applying derrivation on both sides gives the minimum length of ladder required
0= -3cosec(x)cot(x)+sec(x)tan(x)
that implies
${\tan}^{3} \left(x\right) = 3$
x=55.7 degrees
the minimumlength of the ladder is 3cosec(55.7)+sec(55.7) (in degrees)
the minimum length of the ladder is 5.406m
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2021-01-21 11:51:35
|
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http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=ufa&paperid=387&option_lang=eng
|
RUS ENG JOURNALS PEOPLE ORGANISATIONS CONFERENCES SEMINARS VIDEO LIBRARY PACKAGE AMSBIB
General information Latest issue Archive Impact factor Search papers Search references RSS Latest issue Current issues Archive issues What is RSS
Ufimsk. Mat. Zh.: Year: Volume: Issue: Page: Find
Ufimsk. Mat. Zh., 2017, Volume 9, Issue 3, Pages 63–77 (Mi ufa387)
Adiabatic approximation in a resonance capture problem
L. A. Kalyakin
Institute of Mathematics, Ufa Scientific Center, RAS, Chernyshevsky str. 112, 450008, Ufa, Russia
Abstract: By means of the averaging method, we analyze two model problems on capture into resonance that leads us to the adiabatic approximation in the leading term in the asymptotics. The main aim is an approximate (by using a small parameter) description of the domain of capture into resonance. This domain is in the phase plane and it is formed by the initial points for the resonance solutions with an unboundedly increasing energy. The capture domain depends on an additional parameter involved in the equation. We show that the adiabatic approximation fails as the capture domain becomes narrow. In this case we have to modify substantially the averaging method. As a result, a system of nonlinear differential equations arises for the leading term in the asymptotics and this system is not always integrable.
Keywords: nonlinear oscillations, small parameter, asymptotics, capture into a resonance, adiabatic approximation.
Funding Agency Grant Number Russian Science Foundation 17-11-01004 The research is supported by the grant of Russian Science Foundation (project no. 17-11-01004).
Full text: PDF file (886 kB)
References: PDF file HTML file
English version:
Ufa Mathematical Journal, 2017, 9:3, 61–75 (PDF, 837 kB); https://doi.org/10.13108/2017-9-3-61
Bibliographic databases:
UDC: 517.956.226
MSC: 34D10, 34D20, 37J25, 34E13
Citation: L. A. Kalyakin, “Adiabatic approximation in a resonance capture problem”, Ufimsk. Mat. Zh., 9:3 (2017), 63–77; Ufa Math. J., 9:3 (2017), 61–75
Citation in format AMSBIB
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2019-09-22 09:00:19
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https://math.stackexchange.com/questions/2944167/infinite-sum-sk-lambda-2k-lambda2-3k-lambda3-dots
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Infinite sum $S(k) = \lambda + 2^k \lambda^2 + 3^k \lambda^3 + \dots$
I'm trying to find an expression for:
$$S(k) = \sum_{n=0}^{\infty} n^k \lambda^n$$ I have found a recursive expression for this where I first find $$S(0)$$, then use that to find $$S(1)$$. Then use those to find $$S(2)$$ and so on until I get to $$S(k)$$. For large $$k$$, I will need a computer for this. I want to know if there is a more efficient method, preferably a closed form solution I can calculate without a computer. Here is how my method works:
$$S(k) = \lambda + 2^k \lambda^2 + 3^k \lambda^3 + \dots$$ $$\frac{S(k)}{\lambda} = 1 + 2^k \lambda + 3^k \lambda^2 + \dots$$
Subtracting the first equation from the second equation we get:
$$S(k) \left(\frac{1-\lambda}{\lambda} \right) = (1^k-0^k)\lambda^0 + (2^k-1^k)\lambda^1+\dots$$ $$= \sum_{n=0}^{\infty} ((n+1)^k-n^k)\lambda^n$$
$$=\sum_{n=0}^\infty \sum_{r=0}^{k-1} {k \choose r}n^r\lambda^n$$ $$=\sum_{r=0}^{k-1}{k \choose r} \sum_{n=0}^\infty n^r \lambda^n$$ $$= \sum_{r=0}^{k-1} {k \choose r} S(r)$$
So, $$S(k) = \frac{\lambda}{1-\lambda} \sum_{r=0}^{k-1} {k \choose r} S(r)$$
We know $$S(0) = \frac{1}{1-\lambda}$$. Can use this to get $$S(1), S(2), \dots$$. But is there a better way?
• With $|\lambda|<1$ you get $S(k,\lambda) = \Phi(\lambda,-k,0)$ where $\Phi$ is the Lerch Transcendent – gammatester Oct 6 '18 at 7:59
• This is related to the Eulerian polynomials (see Identities section), which in turn has a nice combinatoric interpretation in terms of descent numbers of permutations. – Sangchul Lee Oct 6 '18 at 8:00
• This is the [polylogarithm function][1] $\operatorname{Li}_{-k}(\lambda)$. [1]: en.wikipedia.org/wiki/Polylogarithm – user10354138 Oct 6 '18 at 8:01
• This is the polylogarithm of order $-kk. Google may help. – MPW Oct 6 '18 at 8:01 • At least we have the following more explicit way of writing$S(k)\$, which also appears in the wikipedia link above: $$\sum_{i=0}^{\infty} i^k x^i = \frac{x \sum_{m=0}^{k-1} A(k,m)x^m}{(1-x)^{k+1}}, \qquad A(k,m)=\sum _{i=0}^{m}(-1)^{i}\binom{k+1}{i}(m+1-i)^k.$$ – Sangchul Lee Oct 6 '18 at 8:23
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2019-10-22 05:45:13
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https://studydaddy.com/question/eng-225-week-2-dqs
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QUESTION
# ENG 225 Week 2 DQs
In this paperwork of ENG 225 Week 2 Discussion Questions you will find the next information:
DQ 1: The text defines five types of actors: Impersonator, Personality, Star, Wild Card, and Character. Come up with examples of each type of actor (in addition to those identified in the text) and explain in detail why you think they fit each particular category.
DQ 2: Choose any film clip from Movieclips.com (preferably from a film with which you are familiar). After reviewing your chosen scene, explain how cinematography is used within that scene. How does the cinematography inform the setting? How does it inform the characters? What do the choices made by the cinematographer tell you about the mood at this point within the story?
• $5.19 ANSWER Tutor has posted answer for$5.19. See answer's preview
*** 225 **** 2 ***
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2017-12-15 04:13:10
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https://www.iacr.org/cryptodb/data/paper.php?pubkey=30759
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## CryptoDB
### Paper: Ranking Loss: Maximizing the Success Rate in Deep Learning Side-Channel Analysis
Authors: Gabriel Zaid , Univ Lyon, UJM-Saint-Etienne, CNRS Laboratoire Hubert Curien UMR 5516 F-42023, Saint-Etienne, France; Thales ITSEF, Toulouse, France Lilian Bossuet , Univ Lyon, UJM-Saint-Etienne, CNRS Laboratoire Hubert Curien UMR 5516 F-42023, Saint-Etienne, France François Dassance , Thales ITSEF, Toulouse, France Amaury Habrard , Univ Lyon, UJM-Saint-Etienne, CNRS Laboratoire Hubert Curien UMR 5516 F-42023, Saint-Etienne, France Alexandre Venelli , Thales ITSEF, Toulouse, France DOI: 10.46586/tches.v2021.i1.25-55 URL: https://tches.iacr.org/index.php/TCHES/article/view/8726 Search ePrint Search Google The side-channel community recently investigated a new approach, based on deep learning, to significantly improve profiled attacks against embedded systems. Compared to template attacks, deep learning techniques can deal with protected implementations, such as masking or desynchronization, without substantial preprocessing. However, important issues are still open. One challenging problem is to adapt the methods classically used in the machine learning field (e.g. loss function, performance metrics) to the specific side-channel context in order to obtain optimal results. We propose a new loss function derived from the learning to rank approach that helps preventing approximation and estimation errors, induced by the classical cross-entropy loss. We theoretically demonstrate that this new function, called Ranking Loss (RkL), maximizes the success rate by minimizing the ranking error of the secret key in comparison with all other hypotheses. The resulting model converges towards the optimal distinguisher when considering the mutual information between the secret and the leakage. Consequently, the approximation error is prevented. Furthermore, the estimation error, induced by the cross-entropy, is reduced by up to 23%. When the ranking loss is used, the convergence towards the best solution is up to 23% faster than a model using the cross-entropy loss function. We validate our theoretical propositions on public datasets.
##### BibTeX
@article{tches-2020-30759,
title={Ranking Loss: Maximizing the Success Rate in Deep Learning Side-Channel Analysis},
journal={IACR Transactions on Cryptographic Hardware and Embedded Systems},
publisher={Ruhr-Universität Bochum},
volume={2021, Issue 1},
pages={25-55},
url={https://tches.iacr.org/index.php/TCHES/article/view/8726},
doi={10.46586/tches.v2021.i1.25-55},
author={Gabriel Zaid and Lilian Bossuet and François Dassance and Amaury Habrard and Alexandre Venelli},
year=2020
}
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2022-05-20 14:27:59
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https://www.hackerearth.com/problem/algorithm/lazy-vs-active/
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Lazy-Active vs Strict
Tag(s):
## Medium, maths, prime factorization, sieve
Problem
Editorial
Analytics
Mike is a Lazy guy.He sleeps in lectures.Today the professor saw him sleeping. So he gave him an assignment.
As the prof. is a Strict guy ,he announced that he will deduct 10 marks if he fails to submit the assignment in the next lecture.
Given a number X ,He needs to find the probability that any divisor ( proper ) of X is a perfect square and a even number also.
NOTE :- Output of probability must be in N/D format ,where N and D are positive coprime numbers.
but if numerator is 0 then simply print 0.
Mike seeks for help from an Active guy who pays attention in the lecture.So now it's lazy and active vs Strict.
Can you help mike.
INPUT
• First line of input contains an integer T denoting the test cases.
• Following T lines, each line contains an integer X.
OUTPUT
• output contains T lines . each line contains the answer in described format.
CONSTRAINTS
• 1<=T<=40000
• 1<=x<=1000000
SAMPLE INPUT
2
2
900
SAMPLE OUTPUT
0
3/26
Explanation
In first test case for 2 - {1} is only proper divisor so it's answer is 0.
In second test case 900 , there are 26 proper divisors but only 3 are perfect square and even {4,36,100} so answer is 3/26
Time Limit: 1.0 sec(s) for each input file.
Memory Limit: 256 MB
Source Limit: 1024 KB
Marking Scheme: Marks are awarded when all the testcases pass.
Allowed Languages: C, C++, C++14, Clojure, C#, D, Erlang, F#, Go, Groovy, Haskell, Java, Java 8, JavaScript(Rhino), JavaScript(Node.js), Julia, Kotlin, Lisp, Lisp (SBCL), Lua, Objective-C, OCaml, Octave, Pascal, Perl, PHP, Python, Python 3, R(RScript), Racket, Ruby, Rust, Scala, Swift, Visual Basic
## CODE EDITOR
Initializing Code Editor...
## This Problem was Asked in
Challenge Name
MNIT-Codes Episode-01
OTHER PROBLEMS OF THIS CHALLENGE
• Math > Number Theory
• Math > Polygon
• Basic Programming > Implementation
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2018-01-23 15:32:10
|
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|
https://socratic.org/questions/how-do-you-solve-for-r-in-i-n-p-q-r
|
# How do you solve for R in i= (N-P)/( Q-R)?
May 13, 2016
Use the rule $\frac{a}{b} = \frac{m}{n} \to b \times m = n \times a$
#### Explanation:
$i \left(Q - R\right) = N - P$
$Q - R = \frac{N - P}{i}$
$- R = \frac{N - P}{i} - Q$
$R = - \frac{N - P}{i} + Q$
Hopefully this helps!
May 13, 2016
$R = \frac{- N + P + i Q}{i}$
#### Explanation:
The strategy is to get $R$ out of the denominator using multiplication, then isolate all terms including $R$, then factor $R$ out if necessary, and finally divide by the coefficient of $R$.
$i = \frac{N - P}{Q - R}$
Multiply each side by $Q - R$.
$\implies i \left(Q - R\right) = \frac{N - P}{Q - R} \left(Q - R\right) = N - P$
Apply the distributive property to the left hand side.
$\implies i Q - i R = N - P$
Subtract $i Q$ from each side.
$\implies i Q - i R - i Q = N - P - i Q$
$\implies - i R = N - P - i Q$
Divide each side by $- i$
$\implies \frac{- i R}{- i} = \frac{N - P - i Q}{- i}$
$\implies R = - \frac{N - P - i Q}{i} = \frac{- N + P + i Q}{i}$
May 14, 2016
$R = \frac{P - N}{i} + Q$ or $R = Q - \frac{N - P}{i}$
#### Explanation:
Here is another method. The biggest problem is that $R$ is in the denominator. However, there is only one term on each side of the equal sign., so we can simply invert the entire equation
$\frac{1}{i} = \frac{Q - R}{N - P}$
Multiply by $\left(N - P\right) \Rightarrow \frac{N - P}{i} = Q - R$
Now : EITHER..... Move $R$ to the left and the whole of the term on the left to the right, remembering to change the signs.
$R = Q - \frac{N - P}{i}$
OR: Move the Q to the left and then multiply through by -1 to make
-R into +R
$\frac{N - P}{i} = Q - R$
$\frac{N - P}{i} - Q = - R$
$\frac{P - N}{i} + Q = R$
|
2019-10-17 18:44:34
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|
https://zbmath.org/?q=ut%3AHadamard+matrix
|
## Found 2,717 Documents (Results 1–100)
100
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2022-10-03 15:43:25
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https://mathematica.stackexchange.com/questions/162070/chaplygin-sleigh-problem-plotting-the-trajectory-of-the-sleigh-in-the-xy-plane
|
# Chaplygin sleigh problem: plotting the trajectory of the sleigh in the xy-plane
The Chaplygin sleigh problem is very popular in nonholonomic mechanics (more details can be seen in Bloch, Nonholonomic Mechanics and Control, Springer, New York, 2015, section 1.7, pp 29-30),
The Chaplygin sleigh is a rigid body moving on two sliding posts and one knife edge
The equations are:
$\qquad v =\dot{x}\cos\theta+\dot{y}sin\theta,$
$\qquad \ddot{\theta} =\dot{\omega}=-\frac{ma}{1+ma^2}v\omega.$
where $v$ is the velocity in the direction of motion, and
$\qquad \dot{v}=a(\cos^2\theta+\sin^2\theta)\dot{\theta}^2=a\dot{\theta}^2=a\omega^2.$
In Mathematica, I can plot the trajectory in $vω$ space, but how can I plot the Chaplygin sleigh model trajectory in $xy$ space?
My code as follows (maybe something wrong):
Clear["Global*"]
m = 1; a = 1;
sol =
NDSolve[
{w'[t] == -v[t]*w[t]*m*a/(1 + m*a^2),
v'[t] == a*w[t]^2,
w[0] == 0.2, v[0] == 0.1},
{w, v}, {t, -4, 4}]
ParametricPlot[Evaluate[{w[t], v[t]} /. sol], {t, -4, 4}]
The result in the referenced text is:
• Are you missing a minus sign in the equation for w'[t]? Dec 16 '17 at 10:00
• yeah, thank you, there is some mistake.@aardvark2012
– Ben
Dec 16 '17 at 10:11
If you want to plot in $xy$ space, you need to solve for $x$ and $y$ first. It's not obvious to me what parameters and initial conditions are used for the reference figure, so I chose them casually:
m = 1; a = 1;
sol = NDSolve[{w'[t] == -v[t]*w[t]*m*a/(1 + m*a^2), v'[t] == a*w[t]^2, w[0] == 0.2,
v[0] == 0.1,
θ''[t] == w'[t], θ[0] == Pi, θ'[0] == 1,
x'[t] == v[t] Cos[θ@t], y'[t] == v[t] Sin[θ@t],
x[0] == 0, y[0] == 0}, {w, v, θ, x, y}, {t, -10, 6}];
ParametricPlot[Evaluate[{x[t], y[t]} /. sol], {t, -9, 6}]
`
• I appreciate your help, it enriched my knowledge on mathematica, thank you for your kindness and enthusiasm!
– Ben
Dec 16 '17 at 15:35
• 谢谢,才发现咱们都是中国人,向你学习! :) @xzczd
– Ben
Dec 17 '17 at 9:06
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2021-09-24 09:04:46
|
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|
https://or.stackexchange.com/questions/4635/min-cost-flow-with-per-edge-flow-conservation
|
Min-cost flow with per-edge flow conservation
I am trying to solve a linear program that is identical to a min-cost flow problem, except for a difference in the flow-conservation constraint.
Instead of the summed outgoing flow equaling the summed incoming flow for each node:
$$\sum_i f_{i,n} = \sum_j f_{n, j} \ \forall n$$ The flow of each outgoing edge should equal the summed incoming flow for each node:
$$\sum_i f_{i,n} = f_{n, m} \ \forall n, m$$
In other words: The incoming flow is replicated along all outgoing edges.
I wonder if:
1.) This problem has a particular name in the literature
2.) It has an optimal integral solution when all edge capacities are integral
• Are you sure about the definition of per-edge flow conservation? This would result in increasing the total flow while traversing the network from the origin to the destination. Suppose the incoming flow to a node is 100 and there are five outgoing edges from that node. Based on your constraint, the total outgoing flow would be 500. – Ehsan Aug 6 '20 at 6:18
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2021-04-18 23:46:53
|
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|
http://mathhelpforum.com/advanced-algebra/118735-spanning-set.html
|
# Math Help - Spanning set.
1. ## Spanning set.
a) Show that the set of vectors {u1 = <1 2 -1>, u2 = <-1 -1 0>, u3= <2 5 4>} spans R^3
b)Express the vector v = <1 3 -2> in terms of the spanning set in part a).
I already answered a), I just don't know how to answer b)
By expressing does that mean I should just include the vector v in the set of given vectors? I'm not really sure with this one. Please help me. thanks!
2. Originally Posted by cookiejar
a) Show that the set of vectors {u1 = <1 2 -1>, u2 = <-1 -1 0>, u3= <2 5 4>} spans R^3
b)Express the vector v = <1 3 -2> in terms of the spanning set in part a).
I already answered a), I just don't know how to answer b)
By expressing does that mean I should just include the vector v in the set of given vectors? I'm not really sure with this one. Please help me. thanks!
Reduce the augmented matrix $\begin{bmatrix}1 &-1 & 2 & 1\\ 2 & -1 & 5 & 3\\ -1 & 0 & 4 & -2\end{bmatrix}$ and you'll find the constants a,b,c such that $a\left<1,2,-1\right>+b\left<-1,-1,0\right>+c\left<2,5,4\right>=\left<1,3,-2\right>$
3. So, I'll just solve the constants a, b, c to answer b)? I didn't think of that... but, Thanks anyway!
|
2014-07-31 05:40:08
|
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|
http://www.physicsforums.com/showthread.php?p=3866764
|
## 3D spherical vs 2D radial waves
The Green's functions for a 3d wave are like δ(r - ct)/r -- so if you have static source at the origin that is turned on at t=0, you get an expanding ball around it of radius ct, with strength 1/r. If you look just at the XY plane, you see an expanding disc of value 1/r.
Similarly, if you turned the source on at t = 0 and off at t=1, you would get a an expanding spherical shell of radius ct and thickness c, or an expanding annulus in the XY plane.
However, for 2d waves, there is an "afterglow" -- I don't recall the exact Green function, but rather than a δ-function, it's a lograthmic or exponential decay, and there's something similar for 1-D waves -- e.g. If someone turns a light on for one second, you will still see lit for longer than 1s, though it will get dimmer as time goes on.
This is often discussed around Huygens' principle -- IIRC the principle is the "no afterglow" rule for 3D, which holds in odd dimensions > 1.
However, because of symmetry, the spherical wave equation satisfies the one-dimensional wave equation: (rV),tt = (rV),rr -- where r is the distance from the origin (√xx+yy+zz), V(r, t) is the amplitude a distance r from the origin, and t is the time.
So, then, shouldn't our point source at the origin - which is obviously spherically-symmetrical -- exhibit the afterglow,and therefore *not* give the simple constant 1/r dependence? E.g. Since there is an afterglow, the value at r is not just affected by the source at t=r/c, but also all previous times, leading to (apparently) V -> infinity as t-> infinity?
Similarly, it would seem th XY plane (or any plane through the origin) would satisfy a 2D wave equation, by symmetry...but I'm not so sure there.
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Recognitions: Science Advisor The wave equation in any spatial dimension reads (setting the phase speed of the wave to unity) $$(\partial_t^2-\Delta_D) \Phi(t,\vec{x})=0,$$ where $\Delta_D$ is the Laplace(-Beltrami) operator of Euclidean space in $D$ dimensions. For isotropic problems, we have $\Phi(t,\vec{x})=\Phi(t,r)$ with $r=|\vec{x}|$, and the equation reads $$\left [\partial_t^2 -\frac{1}{r^{D-1}}\frac{\partial}{\partial r} \left (r^{D-1} \frac{\partial}{\partial r} \right ) \right ] \Phi(t,r)=0.$$ This shows that the equations are different for 2 or 3 dimensions. The equation also holds for the Green's function of the wave operator (except at the origin), $$(\partial_t^2-\Delta_{D}) G(t,\vec{x})=\delta(t) \delta^{(D)}(\vec{x}).$$ This explains why the Green's functions are different for different space dimensions.
Recognitions:
Homework Help
Quote by jjustinn However, for 2d waves, there is an "afterglow" -- e.g. If someone turns a light on for one second, you will still see lit for longer than 1s, though it will get dimmer as time goes on. This is often discussed around Huygens' principle -- IIRC the principle is the "no afterglow" rule for 3D, which holds in odd dimensions > 1.
A light is in 3D, so why do you expect afterglow?
The afterglow from a doused incandescent lamp is because it takes a while to cool down.
You won't see it with a LED.
Quote by jjustinn However, because of symmetry, the spherical wave equation satisfies the one-dimensional wave equation: (rV),tt = (rV),rr -- where r is the distance from the origin (√xx+yy+zz), V(r, t) is the amplitude a distance r from the origin, and t is the time. So, then, shouldn't our point source at the origin - which is obviously spherically-symmetrical -- exhibit the afterglow,and therefore *not* give the simple constant 1/r dependence? E.g. Since there is an afterglow, the value at r is not just affected by the source at t=r/c, but also all previous times, leading to (apparently) V -> infinity as t-> infinity?
You've lost me. There's no afterglow in one or three dimensions, right? So why do you think there should be afterglow?
And the reason the spherical wave equation can be transformed into a 1-D equation does not follow from symmetry (or it would happen in all dimensions). Read http://bigbro.biophys.cornell.edu/~t..._Principle.htm.
## 3D spherical vs 2D radial waves
Quote by haruspex A light is in 3D, so why do you expect afterglow? The afterglow from a doused incandescent lamp is because it takes a while to cool down. You won't see it with a LED.
I didn't mean to imply that the was any observed evidence of the "afterglow" in 3D -- there's clearly not (to take the incandescent lamp, the source itself is what has a finite turn-off time).
You've lost me. There's no afterglow in one or three dimensions, right? So why do you think there should be afterglow?
For some reason, I recalled reading that there was an afterglow in 1D / that Huygens' principle held in 2n+1, n > 0 dimensions. If I'm recalling correctly, isn't the "afterglow" actually equal to the original impulse in 1D? E.g. a unit origin impulse at t=0 is felt at x for *all* t > x/c? If I'm retroactively hallucinating, forgive me.
And the reason the spherical wave equation can be transformed into a 1-D equation does not follow from symmetry (or it would happen in all dimensions).
I think that's what was confusing me -- why it doesn't happen in all dimensions. Ill have to check out the links you and vanhees posted, and see if those clear anything up.
Thanks,
Justin
Tags green functions, huygens’s principle, potential, wave eqaution
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2013-05-22 14:34:57
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https://de.mathworks.com/help/comm/ug/eye-diagram-analysis.html
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## Eye Diagram Analysis
In digital communications, an eye diagram provides a visual indication of how noise might impact system performance.
Use the Eye Diagram Scope block to examine the eye diagram of signals.
You can obtain the following measurements on an eye diagram:
• Amplitude Measurements
• Eye Amplitude
• Eye Crossing Amplitude
• Eye Crossing Percentage
• Eye Height
• Eye Level
• Eye SNR
• Quality Factor
• Vertical Eye Opening
• Time Measurements
• Deterministic Jitter
• Eye Crossing Time
• Eye Delay
• Eye Fall Time
• Eye Rise Time
• Eye Width
• Horizontal Eye Opening
• Peak-to-Peak Jitter
• Random Jitter
• RMS Jitter
• Total Jitter
Measurements assume that the eye diagram object has valid data. A valid eye diagram has two distinct eye crossing points and two distinct eye levels.
The deterministic jitter, horizontal eye opening, quality factor, random jitter, and vertical eye opening measurements utilize a dual-Dirac algorithm. Jitter is the deviation of a signal’s timing event from its intended (ideal) occurrence in time [1]. Jitter can be represented with a dual-Dirac model. A dual-Dirac model assumes that the jitter has two components: deterministic jitter (DJ) and random jitter (RJ). The DJ PDF comprises two delta functions, one at μL and one at μR. The RJ PDF is assumed to be Gaussian with zero mean and variance σ.
The Total Jitter (TJ) PDF is the convolution of these two PDFs, which is composed of two Gaussian curves with variance σ and mean values μL and μR.
The dual-Dirac model is described in [5] in more detail. The amplitude of the two Dirac functions may not be the same. In such a case, the analyze method estimates these amplitudes, ρL and ρR.
### Amplitude Measurements
You can use the vertical histogram to obtain a variety of amplitude measurements. For complex signals, measurements are done on both in-phase and the quadrature components, unless otherwise specified.
Note
For amplitude measurements, at least one bin per vertical histogram must reach 10 hits before the measurement is taken, ensuring higher accuracy.
#### Eye Amplitude (EyeAmplitude)
Eye Amplitude, measured in Amplitude Units (AU), is defined as the distance between two neighboring eye levels. For an NRZ signal, there are only two levels: the high level (level 1 in figure) and the low level (level 0 in figure). The eye amplitude is the difference of these two values.
#### Eye Crossing Amplitude (EyeCrossingLevel)
Eye crossing amplitudes are the amplitude levels at which the eye crossings occur, measured in Amplitude Units (AU). The analyze method calculates this value using the mean value of the vertical histogram at the crossing times [3].
The next figure shows the vertical histogram at the first eye crossing time.
#### Eye Crossing Percentage (EyeOpeningVer)
Eye Crossing Percentage is the location of the eye crossing levels as a percentage of the eye amplitude.
#### Eye Height (EyeHeight)
Eye Height, measured in Amplitude Units (AU), is defined as the 3σ distance between two neighboring eye levels.
For an NRZ signal, there are only two levels: the high level (level 1 in figure) and the low level (level 0 in figure). The eye height is the difference of the two 3σ points. The 3σ point is defined as the point that is three standard deviations away from the mean value of a PDF.
#### Eye Level (EyeLevel)
Eye Level is the amplitude level used to represent data bits, measured in Amplitude Units (AU).
For an ideal NRZ signal, there are two eye levels: +A and –A. The analyze method calculates eye levels by estimating the mean value of the vertical histogram in a window around the EyeDelay, which is also the 50% point between eye crossing times [3]. The width of this window is determined by the EyeLevelBoundary property of the eye measurement setup object.
The analyze method calculates the mean value of all the vertical histograms within the eye level boundaries. The mean vertical histograms show two distinct PDFs, one for each eye level.
#### Eye SNR (EyeSNR)
Eye signal-to-noise ratio is defined as the ratio of the eye amplitude to the sum of the standard deviations of the two eye levels. It can be expressed as:
SNR = $\frac{{L}_{1}-{L}_{0}}{{\sigma }_{1}+{\sigma }_{0}}$
where L1 and L0 represent eye level 1 and 0, respectively, and σ1 and σ2 are the standard deviation of eye level 1 and 0, respectively.
For an NRZ signal, eye level 1 corresponds to the high level, and the eye level 0 corresponds to low level.
#### Quality Factor (QualityFactor)
The analyze method calculates Quality Factor the same way as the eye SNR. However, instead of using the mean and standard deviation values of the vertical histogram for L1 and σ1, the analyze method uses the mean and standard deviation values estimated using the dual-Dirac method. For more detail, see dual-Dirac section in [2].
#### Vertical Eye Opening (EyeOpeningVer)
Vertical Eye Opening is defined as the vertical distance between two points on the vertical histogram at EyeDelay that corresponds to the BER value defined by the BERThreshold property of the eye measurement setup object. The analyze method calculates this measurement taking into account the random and deterministic components using a dual-Dirac model [5] (see the Dual Dirac Section). A typical BER value for the eye opening measurements is 10-12, which approximately corresponds to the 7σ point assuming a Gaussian distribution.
### Time Measurements
You can use the horizontal histogram of an eye diagram to obtain a variety of timing measurements.
Note
For time measurements, at least one bin per horizontal histogram must reach 10 hits before the measurement is taken.
#### Deterministic Jitter (JitterDeterministic)
Deterministic Jitter is the deterministic component of the jitter. You calculate it using the tail mean value, which is estimated using the dual-Dirac method as follows [5]:
DJ = μL — μR
where μL and μR are the mean values returned by the dual-Dirac algorithm.
#### Eye Crossing Time (EyeCrossingTime)
Eye crossing times are calculated as the mean of the horizontal histogram for each crossing point, around the reference amplitude level. This value is measured in seconds. The mean value of all the horizontal PDFs is calculated in a region defined by the CrossingBandWith property of the eye measurement setup object.
The region is from -Atotal* BW to +Atotal* BW, where Atotal is the total amplitude range of the eye diagram (i.e., A total = A maxAmin) and BW is the crossing band width.
Because this example assumes two symbols per trace, the average PDF in this region indicate there are two crossing points.
Note
When an eye crossing time measurement falls within the [-0.5/Fs, 0) seconds interval, the time measurement wraps to the end of the eye diagram, i.e., the measurement wraps by 2*Ts seconds (where Ts is the symbol time). For a complex signal case, the analyze method issues a warning if the crossing time measurement of the in-phase branch wraps while that of the quadrature branch does not (or vice versa).
To avoid the time-wrapping or a warning, add a half-symbol duration delay to the current value in the MeasurementDelay property of the eye diagram object. This additional delay repositions the eye in the approximate center of the scope.
#### Eye Delay (EyeDelay)
Eye Delay is the distance from the midpoint of the eye to the time origin, measured in seconds. The analyze method calculates this distance using the crossing time. For a symmetric signal, EyeDelay is also the best sampling point.
#### Eye Fall Time (EyeFallTime)
Eye Fall Time is the mean time between the high and low threshold values defined by the AmplitudeThreshold property of the eye measurement setup object. The fall time is calculated from 10% to 90% of the eye amplitude.
#### Eye Rise Time (EyeRiseTime)
Eye Rise Time is the mean time between the low and high threshold values defined by the AmplitudeThreshold property of the eye measurement setup object. The rise time is calculated from 10% to 90% of the eye amplitude.
#### Eye Width (EyeWidth)
Eye Width is the horizontal distance between two points that are three standard deviations (3σ ) from the mean eye crossing times, towards the center of the eye. The value for Eye Width measurements is seconds.
#### Horizontal Eye Opening (EyeOpeningHor)
Horizontal Eye Opening is the horizontal distance between two points on the horizontal histogram that correspond to the BER value defined by the BERThreshold property of the eye measurement setup object. The measurement is take at the amplitude value defined by the `ReferenceAmplitude` property of the eye measurement setup object. It is calculated taking into account the random and deterministic components using a dual-Dirac model [5] (see the Dual Dirac Section).
A typical BER value for the eye opening measurements is 10-12, which approximately corresponds to the 7σ point assuming a Gaussian distribution.
#### Peak-to-Peak Jitter (JitterP2P)
Peak-To-Peak Jitter is the difference between the extreme data points of the histogram.
#### Random Jitter (JitterRandom)
Random Jitter is defined as the Gaussian unbounded component of the jitter. The analyze method calculates it using the tail standard deviation estimated using the dual-Dirac method as follows [5]:
RJ = (QL + QR) * σ
where
`${Q}_{L}=\sqrt{2}*erf{c}^{-1}\left(\frac{2*BER}{{\rho }_{L}}\right)$`
and
`${Q}_{R}=\sqrt{2}*erf{c}^{-1}\left(\frac{2*BER}{{\rho }_{R}}\right)$`
BER is the bit error ratio at which the random jitter is calculated. It is defined with the BERThreshold property of the eye measurement setup object.
#### RMS Jitter (JitterRMS)
RMS Jitter is the standard deviation of the jitter calculated from the horizontal histogram.
#### Total Jitter (JitterTotal)
Total Jitter is the sum of the random jitter and the deterministic jitter [5].
## References
[1] Nelson Ou, et al, Models for the Design and Test of Gbps-Speed Serial Interconnects,IEEE Design & Test of Computers, pp. 302-313, July-August 2004.
[2] HP E4543A Q Factor and Eye Contours Application Software, Operating Manual, http://agilent.com
[3] Agilent 71501D Eye-Diagram Analysis, User’s Guide, http://www.agilent.com
[4] 4] Guy Foster, Measurement Brief: Examining Sampling Scope Jitter Histograms, White Paper, SyntheSys Research, Inc., July 2005.
[5] Jitter Analysis: The dual-Dirac Model, RJ/DJ, and Q-Scale, White Paper, Agilent Technologies, December 2004, http://www.agilent.com
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2021-09-27 01:37:43
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http://gate-exam.in/EC/Syllabus/Electronics-Communication-Engineering/Networks
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# Questions & Answers of Networks
#### Topics of Networks 97 Question(s) | Weightage 11 (Marks)
Consider a two-port network with the transmission matrix: $T=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$. If the network is reciprocal, then
An AC voltage source V = 10 sin(t) volts is applied to the following network. Assume that R1 = 3 kΩ, R2 = 6 kΩ and R3 = 9 kΩ, and that the diode is ideal.
RMS current Irms (in mA) through the diode is ________
In the circuit shown in the figure, the maximum power (in watt) delivered to the resistor R is __________
In the circuit shown below,VS is a constant voltage source and IL is a constant current load.
The value of IL that maximizes the power absorbed by the constant current load is
The switch has been in position 1 for a long time and abruptly changes to position 2 at t=0.
If time t is in seconds, the capacitor voltage VC (in volts) for t > 0 is given by
The figure shows an RLC circuit with a sinusoidal current source.
At resonance, the ratio $\left|{\mathbf{I}}_{\mathbf{L}}\right|/\left|{\mathbf{I}}_{\mathbf{R}}\right|$, i.e., the ratio of the magnitudes of the inductor current phasor and the resistor current phasor, is ________
The z-parameter matrix for the two-port network shown is
where the entries are in Ω. Suppose
Then the value of Rb (in Ω) equals ________
In the given circuit, each resistor has a value equal to 1 Ω.
What is the equivalent resistance across the terminals a and b ?
In the circuit shown in the figure, the magnitude of the current (in amperes) through R2 is ___
In the RLC circuit shown in the figure, the input voltage is given by
$v_i\left(t\right)=2\;\cos\left(200t\right)+4\;\sin\left(500t\right)$
The output voltage $v_0\left(t\right)$ is
Assume that the circuit in the figure has reached the steady state before time t=0 when the 3 Ω resistor suddenly burns out, resulting in an open circuit. The current i(t) (in ampere) at t=0+ is __________
In the figure shown, the current i (in ampere) is __________
The z-parameter matrix $\begin{bmatrix}z_{11}&z_{12}\\z_{21}&z_{22}\end{bmatrix}$ for the two-port network shown is
In the circuit shown, at resonance, the amplitude of the sinusoidal voltage (in Volts) across the capacitor is ______ .
In the network shown in the figure, all resistors are identical with R = 300 $\mathrm{\Omega }$. The resistance Rab (in $\mathrm{\Omega }$) of the network is __________.
In the given circuit, the values of $V_1$ and $V_2$ respectively are
The damping ratio of a series $RLC$ circuit can be expressed as
In the given circuit, the maximum power (in Watts) that can be transferred to the load RL is ____.
The voltage (VC) across the capacitor (in Volts) in the network shown is _______
In the circuit shown, the average value of the voltage Vab (in Volts) in steady state condition is _______.
The 2-port admittance matrix of the circuit shown is given by
An LC tank circuit consists of an ideal capacitor C connected in parallel with a coil of inductance L having an internal resistance R. The resonant frequency of the tank circuit is
In the circuit shown, the Norton equivalent resistance (in Ω) across terminals a-b is ___________.
For the circuit shown in the figure, the Thevenin equivalent voltage (in Volts) across terminals a-b is _____.
In the circuit shown, the voltage Vx (in Volts) is ____.
At very high frequencies, the peak output voltage Vo (in Volts) is_____.
In the circuit shown, the current I flowing through the 50 Ω resistor will be zero if the value of capacitor C (in μF) is _______.
The ABCD parameters of the following 2-port network are
For maximum power transfer between two cascaded sections of an electrical network, the relationship between the output impedance Z1 of the first section to the input impedance Z2 of the second section is
Consider the configuration shown in the figure which is a portion of a larger electrical network
For R=1 Ω and currents i1=2 A,i4=−1 A,i5= −4 A, which one of the following is TRUE?
A Y-network has resistances of 10Ω each in two of its arms, while the third arm has a resistance of 11 Ω. In the equivalent Δ-network, the lowest value (in Ω.) among the three resistances is ________
A 230 V rms source supplies power to two loads connected in parallel. The first load draws 10 kW at 0.8 leading power factor and the second one draws 10 kVA at 0.8 lagging power factor. The complex power delivered by the source is
A periodic variable x is shown in the figure as a function of time. The root-mean-square (rms) value of x is _________.
In the circuit shown in the figure, the value of capacitor C (in mF) needed to have critically damped response i(t) is________.
Norton’s theorem states that a complex network connected to a load can be replaced with an equivalent impedance
In the figure shown, the ideal switch has been open for a long time. If it is closed at t=0, then the magnitude of the current (in mA) through the 4 kΩ resistor at t= 0+ is _______.
In the h-parameter model of the 2-port network given in the figure shown, the value of h22 (in S) is ______ .
In the figure shown, the capacitor is initially uncharged. Which one of the following expressions describes the current I(t) (in mA) for t >0?
A series RC circuit is connected to a DC voltage source at time t = 0. The relation between the source voltage VS, the resistance R, the capacitance C, and the current i(t) is given below:
${V}_{s}=Ri\left(t\right)+\frac{1}{C}{\int }_{0}^{t}i\left(u\right).$
Which one of the following represents the current i(t)?
In the figure shown, the value of the current I (in Amperes) is __________.
Consider the building block called ‘Network N’ shown in the figure.
Let C = 100 μF and R = 10 kΩ.
Two such blocks are connected in cascade, as shown in the figure.
The transfer function $\frac{{V}_{3}\left(s\right)}{{V}_{1}\left(s\right)}$ of the cascaded network is
In the circuit shown in the figure, the value of node voltage V2 is
In the circuit shown in the figure, the angular frequency ω (in rad/s), at which the Norton equivalent impedance as seen from terminals b-b′ is purely resistive, is _________.
For the Y-network shown in the figure, the value of R1 (in Ω) in the equivalent Δ-network is ____.
The circuit shown in the figure represents a
The steady state output of the circuit shown in the figure is given by
$y\left(t\right)=A\left(\omega\right)\sin\left(\omega t+\phi\left(\omega\right)\right).$If the amplitude $\left|A\left(\omega \right)\right|=0.25$ , then the frequency $\omega$ is
In the circuit shown in the figure, the value of v0(t) (in Volts) for t→∞ is ______.
For the two-port network shown in the figure, the impedance (Z) matrix (in Ω) is
Consider a delta connection of resistors and its equivalent star connection as shown below. If all elements of the delta connection are scaled by a factor k, k> 0, the elements of the corresponding star equivalent will be scaled by a factor of
The transfer function $\frac{{V}_{2}\left(s\right)}{{V}_{1}\left(s\right)}$ of the circuit shown below is
A source ${v}_{s}\left(t\right)=Vcos100\pi t$ has an internal impedance of $\left(4+j3\right)\Omega$ . If a purely resistive load connected to this source has to extract the maximum power out of the source, its value in $\Omega$ should be
In the circuit shown below, if the source voltage VS = $100\angle 53.13°$ V then the Thevenin’s equivalent voltage in Volts as seen by the load resistance RL is
The following arrangement consists of an ideal transformer and an attenuator which attenuates by a factor of 0.8. An ac voltage VWX1 = 100V is applied across WX to get an open circuit voltage VYZ1 across YZ. Next, an ac voltage VYZ2 =100V is applied across YZ to get an open circuit voltage VWX2 across WX. Then, VYZ1 / VWX1, VWX2 / VYZ2 are respectively,
Two magnetically uncoupled inductive coils have Q factors q1 and q2 at the chosen operating frequency. Their respective resistances are R1 and R2. When connected in series, their effective Q factor at the same operating frequency is
Three capacitors C1, C2 and C3 whose values are 10μF, 5μF, and 2μF respectively, have breakdown voltages of 10V, 5V, and 2V respectively. For the interconnection shown below, the maximum safe voltage in Volts that can be applied across the combination, and the corresponding total charge in μC stored in the effective capacitance across the terminals are respectively,
Consider the following figure
The current IS in Amps in the voltage source, and voltage VS in Volts across the current source respectively, are
Consider the following figure
The current in the 1 $\Omega$ resistor in Amps is
In the following figure, C1 and C2 are ideal capacitors. C1 has been charged to 12 V before the ideal switch S is closed at t = 0. The current i(t) for all t is
The average power delivered to an impedance (4-j3)$\Omega$ by a current $5\mathrm{cos}\left(100\pi t+100\right)A$ is
The impedance looking into nodes 1 and 2 in the given circuit is
In the circuit shown below, the current through the inductor is
Assuming both the voltage sources are in phase, the value of R for which maximum power is transferred from circuit A to circuit B is
If VA-VB=6V, then VC-VD is
With 10 V dc connected at port A in the linear nonreciprocal two-port network shown below, the following were observed:
(i) 1Ω connected at port B draws a current of 3 A
(ii) 2.5Ω connected at port B draws a current of 2 A
With 10 V dc connected at port A, the current drawn by 7Ω connected at port B is
With 10 V dc connected at port A in the linear nonreciprocal two-port network shown below, the following were observed:
(i) 1Ω connected at port B draws a current of 3 A
(ii) 2.5Ω connected at port B draws a current of 2 A
For the same network, with 6 V dc connected at port A, 1Ω connected at port B draws 7/3 A. If 8 V dc is connected to port A, the open circuit voltage at port B is
In the circuit shown below, the Norton equivalent current in amperes with respect to the terminals P and Q is
In the circuit shown below, the value of RL such that the power transferred to RL is maximum is
The circuit shown below is driven by a sinusoidal input vi = Vp cos (t /RC). The steady state output vo is
In the circuit shown below, the network N is described by the following Y matrix:
$Y=\left[\begin{array}{cc}0.1\mathrm{S}& -0.01\mathrm{S}\\ 0.01\mathrm{S}& 0.1\mathrm{S}\end{array}\right]$. The voltage gain $\frac{{V}_{2}}{{V}_{1}}$ is
In the circuit shown below, the initial charge on the capacitor is 2.5 mC, with the voltage polarity as indicated. The switch is closed at time t=0. The current i(t) at a time t after the switch is closed is
In the circuit shown below, the current I is equal to
For the two-port network shown below, the short-circuit admittance parameter matrix is
For parallel RLC circuit, which one of the following statements is NOT correct?
If the scattering matrix [S] of a two port network is
$\left[\mathrm{s}\right]=\left[\begin{array}{cc}0.2\angle {0}^{o}& 0.9\angle {90}^{o}\\ 0.9\angle {90}^{o}& 0.1\angle {90}^{o}\end{array}\right]$
then the network is
In the circuit shown, the switch S is open for a long time and is closed at t=0. The current i(t) for t≥0+ is
The current I in the circuit shown is
In the circuit shown, the power supplied by the voltage source is
In the interconnection of ideal sources shown in the figure, it is known that the 60 V source is absorbing power
Which of the following can be the value of the current source I ?
If the transfer function of the following network is $\frac{{V}_{0}\left(s\right)}{{V}_{i}\left(s\right)}=\frac{1}{2+sCR}$,
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2018-10-22 18:46:37
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https://www.physicsforums.com/threads/eigenvalues-of-j_2-k_1-j_1-k_2.158172/
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Eigenvalues of J_2 + K_1; -J_1 + K_2
1. Feb 25, 2007
wandering.the.cosmos
Weinberg in volume 1 of his QFT text says we do not observe any non-zero eigenvalues of $A = J_2 + K_1; B = -J_1 + K_2$. He says the "problem" is that any nonzero eigenvalue leads to a continuum of eigenvalues, generated by performing a spatial rotation about the axis that leaves the standard vector invariant.
What are the observable consequences of non-zero eigenvalues of A and B?
2. Feb 27, 2007
alphaone
Very interesting question to which I unfortunately do not know an answer. But I agree that it is very unsatisfying to argue along the usual lines that no continuous degree of freedom is observed experimentally for massless paricles and then concluding that therefore A and B have to be 0. It would be much more satisfying if we could get this result out of theoretical physics without referring to experiment but unfortunately I have not heard of any such argument, so if anybody knows one I would be delighted to hear about it.
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2017-02-20 09:02:38
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https://studyqas.com/the-following-information-is-available-units-in-process/
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# The following information is available: Units in process, Dec. 1 (60 percent converted) 2,000 units
The following information is available: Units in process, Dec. 1 (60 percent converted) 2,000 units Units in process, Dec. 31 (50 percent converted) 1,000 units Units started during the month 7,500 units Materials are added at the beginning of the process. How many equivalent units in process for conversion were there in December
## Coates inc. experienced the following events in 2014, in its first year of operation: (1) received $20,000 ## Steven and Emily Campbell are planning to open a casual dining restaurant in downtown Akron, Ohio, and ## Roxy operates a dress shop in arlington, virginia. lisa, a maryland resident, comes in for a measurement ## Asap which statement is true? a. as the risk of an investment decreases, the opportunity of gains increases. ## Which tool or feature reduces speckling, dust particles, scratches, and minute light ## FIllmore Company began operations on Sept. 1 by purchasing$6000 of inventory and $600 of cleaning supplies. During the month, ## This Post Has 10 Comments 1. coggonp says: $p\%=\dfrac{p}{100}\\\\3\%=\dfrac{3}{100}=3:100=0.03$ 2. Bryson2148 says: Explanation: UNITS TO ACCOUNT FOR: Beginning Work in Process units 2,000 Add: Units Started in Process 7,500 Total Units to account for: 9,500 Equivalent Units: UNITS Conversion cost % Completion Units Units completed 8500100% 8,500 Ending Work in Process 100030% 300 Total Equivalent units 9500 8,800 3. s237200 says: 65% = 0.65 = 65/100 (65/100) / (5/5) = 13/20 13/20 is the simplest fraction achievable. I hope this helps, let me know if you have any questions! 4. nataliemakin7123 says: 2.01yrs Explanation: future value = present value * (1+r)^n where r = interest rate per period and n = number of periods => 1480 * (1+7.7%/2)^n = 1300 * (1+7.7%)^n => n = 2.01 year 5. kendarriuskj1990 says: 206/100 103/50 2 3/50 2 whole and 3/50 6. jenn8055 says:$915.71
Explanation:
Price of the bond is determined by calculating the present value of all cash flows.
We will use following formula in Excel
=nper(rate,pmt.-pv,fv)
rate = Interest rate = 12% x 6/12 = 6%
pmt = Coupon Payment=$1000 x 8.3% x 6/12=$41.50
pv = Price of first coupon = $813.04 fv = Face value =$1000
Placinf all the values in the formula
=nper(6%,41.50,-813.04,1000)= 16.00
Years to maturity=16/2=8 years
The years to maturity of second bond=8+3=11 years
price of second bond=-pv(rate,nper,pmt,fv)
rate = 6%
nper = 11 years x 2= 22
pmt =5.3% x $1000=$53
fv = $1000 Placing values in the formula =-pv(6%,22,53,1000)=$915.71
7. alexdonnelly2123 says:
$915.71 Explanation: In order to determine the second bond price we need to determine the number of years to maturity of the first bond using nper formula in excel. =nper(rate,pmt.-pv,fv) rate is the semiannual interest rate of 6% (12%*6/12) pmt is the semiannual interest=$1000*8.3%*6/12=$41.50 pv is the current price at$813.04
fv is the face value of $1000 =nper(6%,41.50,-813.04,1000)= 16.00 The years to maturity=16/2=8 years The years to maturity of second bond=8+3=11 years price of second bond=-pv(rate,nper,pmt,fv) rate is 6% nper is 11 years multiplied by 2= 22 pmt =5.3%*$1000=$53 fv is$1000
=-pv(6%,22,53,1000)=\$915.71
8. bmr12 says:
$p\%=\dfrac{p}{100}\\\\3\%=\dfrac{3}{100}=3:100=0.03$
9. gagedugan2019 says:
7,800 units
Explanation:
work in process December 1 = 2,000 units x 60% = 1,200 equivalent units
work in process December 31 = 1,000 units x 50% = 500 equivalent units
unis started during the month = 7,500
total units transferred out = 2,000 units + 7,500 units - 1,000 units = 8,500
equivalent conversion units = total units transferred out + ending WIP - beginning WIP = 8,500 units + 500 units - 1,200 units = 7,800
10. greeneashlynt says:
$p\%=\dfrac{p}{100}\\\\206\%=\dfrac{206}{100}=\dfrac{206:2}{100:2}=\dfrac{103}{50}=\dfrac{100+3}{50}=\dfrac{100}{50}+\dfrac{3}{50}=2\dfrac{3}{10}\\\\\boxed{206\%=\frac{103}{50}=2\frac{3}{50}}$
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2023-02-08 06:20:10
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http://gmatclub.com/forum/gmat-problem-solving-ps-140/index-300.html?sk=ku&sd=a
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Electronic Resource
Springer
Acta mechanica 100 (1993), S. 241-251
ISSN: 1619-6937
Source: Springer Online Journal Archives 1860-2000
Topics: Mechanical Engineering, Materials Science, Production Engineering, Mining and Metallurgy, Traffic Engineering, Precision Mechanics , Physics
Notes: Summary The study of two-dimensional, irrotational, inviscid, incompressible steady state motion generated by a polygonal and a smooth obstruction, is made in terms of the linearized theory. The bottom is represented in integral form using Fourier's double integral theorem. Then following Lamb [1], a linear free-surface profile is obtained for the supercritical and subcritical cases. The results are plotted and discussed for the two cases of the flow for different shapes of the bottom and different values of Froude number,F.
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• 2
Electronic Resource
Springer
Acta mechanica 29 (1978), S. 287-290
ISSN: 1619-6937
Source: Springer Online Journal Archives 1860-2000
Topics: Mechanical Engineering, Materials Science, Production Engineering, Mining and Metallurgy, Traffic Engineering, Precision Mechanics , Physics
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• 3
Electronic Resource
Springer
Acta mechanica 76 (1989), S. 243-252
ISSN: 1619-6937
Source: Springer Online Journal Archives 1860-2000
Topics: Mechanical Engineering, Materials Science, Production Engineering, Mining and Metallurgy, Traffic Engineering, Precision Mechanics , Physics
Notes: Summary The problem considered is a steady, two-dimensional, and irrotational flow of an incompressible and inviscid fluid under the action of gravity. The boundary of the flow domain consists of a free boundary part “C f ” and an analytic solid boundary part “C s ” and they meet in a common endpoint “0”. The flow is determined by the inverse velocity potential functionz=f(w) and the conjugate flow velocity $$u = \frac{{dw}}{{dz}}$$ is given by the reciprocal derivative $$u\left( w \right) = \frac{1}{{f'\left( w \right)}}$$ . Following the method suggested by Carter, we will be able to expressz(w) andu(w) as an infinite asymptotic expansion near the origin in powers ofw 1/2 andw 1/2 lnw. Upon imposing the flow conditions on these expansions we will obtain the expansion of the conjugate velocityu to terms of orderw 3/2 as well as the shape of free streamline. The importance of the results of this problem is that it gives simple, and more accurate, qualitative features of the local flow behaviour. Numerical examples have been considered and the results are given in quantitative diagrams to show the progress of the obtained higher approximation.
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• 4
Electronic Resource
Springer
Acta mechanica 78 (1989), S. 219-233
ISSN: 1619-6937
Source: Springer Online Journal Archives 1860-2000
Topics: Mechanical Engineering, Materials Science, Production Engineering, Mining and Metallurgy, Traffic Engineering, Precision Mechanics , Physics
Notes: Summary A theoretical study is made, of an irrotational, inviscid incompressible and steady flow over a two-dimensional trapezoidal obstacle; with disturbing height ε, lying on the bottom of the running stream, in terms of a linearised theory. Particular attention is given to two cases of the flow, the supercritical and the subcritical. The bottom is represented in integral form using Fourier's double-integral theorem. Following the method suggested by Thomson (1886) and Lamb (1932), we obtain a linearised free-surface profile in series form for the two cases of flow. The linearised solution obtained is based on the assumption that the height of the trapezoidal bump, ε, is small compared to the channel depth,h. The nature of the free-surface formed depends on whether the flow is subcritical or supercritical. The results are plotted for the two cases of the flow for different shapes of the bottom and different values of Froude number,F. The effect of the Froude number, the bottom height and the shape of the bottom are discussed.
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• 5
Electronic Resource
Springer
The European physical journal 1 (1986), S. 363-372
ISSN: 1434-6079
Keywords: 34.00 ; 34.80B ; 34.90
Source: Springer Online Journal Archives 1860-2000
Topics: Physics
Notes: Abstract Precision data on transient magnetic fields (TF) in ferromagnetic Gd and Fe were obtained for oxygen ions at velocities between 2 and 8v 0 (v 0=c/137) using the16O(3−) state as probe. At the high velocities the TF in Gd were found to be twice as large as in Fe which is in contrast to the generally observed scaling with the polarization density of the hosts. Degrees of polarization could be deduced from TF strengths using measured ion fractions of singly occupiedK-shell. The values derived were compared with theoretical estimates based on spin exchange scattering as a possible polarization mechanism.
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• 6
Electronic Resource
Springer
Boundary layer meteorology 62 (1993), S. 3-20
ISSN: 1573-1472
Source: Springer Online Journal Archives 1860-2000
Topics: Geosciences , Physics
Notes: Abstract As a result of several air quality model evaluation exercises involving a large number of source scenarios and types of models, it is becoming clear that the magnitudes of the uncertainties in model predictions are similar from one application to another. When considering continuous point sources and receptors at distances of about 0.1 km to 1 km downwind, the uncertainties in ground-level concentration predictions lead to typical mean biases of about ±20 to 40% and typical relative root-mean-square errors of about 60 to 80%. In fact, in two otherwise identical model applications at two independent sites, it is not unusual for the same model to overpredict by 50% at one site and underpredict by 50% at the second site. It is concluded that this fundamental level of model uncertainty is likely to exist due to data input errors and stochastic fluctuations, no matter how sophisticated a model becomes. The tracer studies that lead to these conclusions and have been considered in this study include: (1) tests of the Offshore and Coastal Dispersion (OCD) model at four coastal sites; (2) tests of the Hybrid Plume Dispersion Model (HPDM) at five power plants; (3) tests of a similarity model for near-surface point source releases at four sites; and (4) tests of 14 hazardous gas models at eight sites including six sets of experiments where dense gases were released.
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• 7
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Springer
Nonlinear dynamics 18 (1999), S. 25-50
ISSN: 1573-269X
Keywords: liquid sloshing modeling ; impact ; parametric resonance
Source: Springer Online Journal Archives 1860-2000
Topics: Mathematics
Notes: Abstract The parametric excitation of an elevated water tower experiencing sloshing hydro-dynamic impact is studied using the multiple scales method. The liquid sloshing mass is replaced by a mechanical model in the form of a simple pendulum experiencing impacts with the tank walls. The impact loads are modeled based on a phenomenological representation in the form of a power function with a higher exponent. In this case the system equations of motion include impact nonlinearities (selected to be of fifth power) and cubic structural geometric nonlinearities. When the first mode is parametrically excited the system exhibits hard nonlinear behavior and the impact loading reduced the response amplitude. On the other hand, when the second mode is parametrically excited, the impact loading results in complex response behavior characterized by multiple steady state solutions, where the response switches from soft to hard nonlinear characteristics. Under combined parametric resonance, the system possesses a single steady-state response in the absence and in the presence of impact. However, the system behaves like a soft system in the absence of impact and like a hard system in the presence of impact.
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• 8
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Springer
Czechoslovak journal of physics 36 (1986), S. 378-382
ISSN: 1572-9486
Source: Springer Online Journal Archives 1860-2000
Topics: Physics
Notes: Abstract Preliminary results of measurements of proton energy spectra and search for coincidentp, n pair emission following negative muon capture in12C,16O,27Al are given. A very clear signal for proton emission is found, which can be exploited in future detailed studies.
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• 9
Electronic Resource
[S.l.] : American Institute of Physics (AIP)
Journal of Applied Physics 75 (1994), S. 5625-5627
ISSN: 1089-7550
Source: AIP Digital Archive
Topics: Physics
Notes: Traditionally ferrites have been used in accelerators for tuning rf cavities and in nonreciprocal devices controlling the power flow in rf accelerating systems. Recently, the development of cavity tuners based on perpendicularly biased ferrites has shown good progress. Yttrium iron garnet (YIG) is gradually replacing the traditional Ni Zn ferrites. The use of conventional parallel-biased Ni Zn ferrites for varying the frequency of accelerating cavities has the disadvantage of high saturation magnetization (4πMs). This precludes practical operation in low magnetic loss regions. Different substitutions have been used with YIG to reduce its 4πMs, making it a practical candidate for perpendicular biasing operating in the saturation region. In addition, YIG is known for its low dielectric and magnetic losses. In this paper we give a short review of development in accelerator cavity tuners based on perpendicularly biased iron garnet ferrites. We use the operation of a 52 MHz stripline-based YIG tuner which we have tested at BNL as an example to demonstrate the advantages of using YIG in cavity tuners. We also discuss magnetic tuning techniques based on magnetostriction of Ni.
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• 10
Electronic Resource
[S.l.] : American Institute of Physics (AIP)
Journal of Applied Physics 67 (1990), S. 5505-5505
ISSN: 1089-7550
Source: AIP Digital Archive
Topics: Physics
Notes: This experimental investigation is concerned with the study of diffraction of magnetostatic waves (MSWs) by surface acoustic waves (SAWs) when both types of waves travel in the same yttrium iron garnet (YIG) film. The SAWs are used to set up a space and time modulation of the medium in which the MSW propagates. Because of their mostly isotropic propagation properties in the plane of the film, magnetostatic forward volume waves are employed in this investigation. In the configuration used, MSW is launched at different angles with respect to the acoustic beam. Different MSW microstrip transducers are used to detect the scattered and unscattered MSW signals. The frequency of the Bragg-diffracted MSW signal is shifted by an amount equal to the SAW frequency. The MSW microstrip transducers are 65-μm-wide and 15-μm-thick copper lines on an RT/duroid substrate of thickness of 5 mils and εr=10.5. A YIG film of 120 μm thickness is placed in contact with the microstrip structure described above. A dc magnetic field is applied normal to the film plane. The SAW is excited using bulk wave transducers (36° Y-cut lithium niobate) placed on plexiglass wedges to convert the bulk acoustic waves into SAW. Our experimental results are in good agreement with calculations based on Bragg's diffraction. Different device applications that employ this effect for spectrum analysis1 and frequency shifting2 have been proposed recently.
Type of Medium: Electronic Resource
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2020-01-27 22:10:28
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https://physics.stackexchange.com/questions/469540/why-is-pressure-volume-work-calculated-from-the-external-pressure-if-the-work-is
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Why is pressure-volume work calculated from the external pressure if the work is done by the gas? [duplicate]
Greeting everyone,
The formula for pressure-volume work I have been given in the textbook I am using to learn general chemistry (Chemistry.A.Molecular.Approach.Global.Edition.4th.Edition, Nivaldo J.Tro) is as follows:
w = -PΔV where P is the external pressure. Why is the external pressure used if the force is from the expanding gas, not from the external atmosphere?
N.B: The system is a gas trapped in cylinder between a dead end and a piston. We assume that the piston is massless. We assume that the pressure is constant.
• Think about if the external pressure was $0$. What would the gas be doing work on? – Aaron Stevens Mar 30 at 13:44
• @AaronStevens The piston. – Stooniel Schiffer Mar 30 at 14:00
• @StoonielSchiffer But you said the piston is massless. – Bob D Mar 30 at 14:15
• Yes, indeed. My mistake. Maybe I should learn about gases before diving further into the problem. – Stooniel Schiffer Mar 30 at 14:24
• @AaronStevens Unfortunately the possible duplicate you are indicating contains part of the answer but something is missing. But the way SE works makes not very appealing to add a new answer there. – GiorgioP Mar 30 at 15:58
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2019-12-08 10:07:35
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https://codereview.stackexchange.com/questions/224176/boolean-class-using-shared-referenced-memory
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# Boolean class using shared referenced memory
To improve my knowledge I am trying to make a bool class from scratch, or derivative of. I am using MinGW g++11 compiler on a Windows 7 laptop.
#include <iostream>
#include <bitset>
class Zeroth{
uint8_t* MEM = nullptr;
public:
const static uint8_t NullSet; // if data is declared sttic it is shared acros all instances, const means it can not be changed
const static uint8_t UnitSet;
/*
the idea is that instead copying integers i can define a constant shared memories [ generated on a fly or in a header ]
and then just store memory of the value that is pointed to
so if object is present: point to memory of UnitSet
if object isn't present: point to memory of NullSet
therefore i have 1 8 bit space instanced and deleted with every value, but all those spaces point to the same constant memories
*/
Zeroth():MEM(new uint8_t){ this->MEM = (uint8_t*)&Zeroth::NullSet; }
Zeroth(int VAL):MEM(new uint8_t) {
this->MEM = (uint8_t*)( VAL&0x1 ? &Zeroth::UnitSet : &Zeroth::NullSet );
}
~Zeroth(){ delete MEM; }
const bool isPresent() const {
return (int)&Zeroth::UnitSet == (int)&(*this->MEM);
}
const bool isNotPresent() const {
return (int)&Zeroth::NullSet == (int)&(*this->MEM);
}
friend std::ostream& operator<<( std::ostream& os, const Zeroth& data );
};
std::ostream& operator<<( std::ostream& os, const Zeroth &data ){
return os << ( data.isPresent() ? 'T' : 'F');
}
const uint8_t Zeroth::NullSet = 0;
const uint8_t Zeroth::UnitSet = 1;
int main(){
Zeroth Az(3);
std::cout << Az << std::endl;
std::cout << Az.isPresent() << std::endl;
std::cout << Az.isNotPresent() << std::endl;
Zeroth Bz;
std::cout << Bz << std::endl;
std::cout << Bz.isPresent() << std::endl;
std::cout << Bz.isNotPresent() << std::endl;
return 0;
}
And result is :
CD: F:\cppDMmods
Current directory: F:\cppDMmods
g++ -Wno-unused-variable -O2 F:\cppDMmods\test.cpp -o F:\cppDMmods\test.exe
Process started (PID=4296) >>>
<<< Process finished (PID=4296). (Exit code 0)
IF: "Exists" != "Exists" goto NOEXE
test.exe
Process started (PID=5516) >>>
T // Az printout
1 // Az "presence" :> aka true
0 // Az "negated presence" :> aka !true
F // Bz printout
0 // Bz "presence"
1 // Bz "negated presence"
<<< Process finished (PID=5516). (Exit code 0)
cmd /c del test.exe
Process started (PID=6124) >>>
<<< Process finished (PID=6124). (Exit code 0)
NO EXE FILE TO DELETE
I am worried that due to my presumed lack of knowledge there might be some memory leaks, massive future bugs and etc.
Any suggestions, recommendations and pointed out issues are welcomed.
• The size of a pointer to an uint8_t should be either 32bit or 64bit and is therefore heavier than if you'd use the value itself. – AlexV Jul 14 '19 at 21:32
• Are you talking about MEM member ? How can i declare an 34 bit ( by defining an uint64_t ? ) , and what are possible issues that could rise if i don't change the size of the pointer ? – Danilo Jul 14 '19 at 21:35
• The pointer has the right size, C++ will take care of this. What I was trying to say is that "instead copying integers i [...] just store memory [address] of the value" (from your code) increases the memory footprint (see yourself in this online example: cpp.sh/2xmm). – AlexV Jul 14 '19 at 21:49
• So , i am guessing the same thing arises if i try to store address of instead of pointer? Which makes sense since size of an address is tied to OS bit size... and since we push the elements on the registry the variable is destroyed once it exits the scope. So bool a = true expands to bool a = new bool(1) ? – Danilo Jul 14 '19 at 22:03
I'm sorry to be so straightforward: This is madness! Your code seems to indicate a sincere lack of several language features.
## Pointer to const
You define MEM to be of type uint8_t*, but it actually points to values of const uint8_t. To account for this difference you cast the const away in both constructors. If you'd ever (accidentally) write to the memory MEM points to, undefined behavior might ensue. So it should be at least of type uint8_t const *.
## Memory leak in constructor
As you have suspected, your code leaks memory. Both constructors assign new uint8_t to MEM before overwriting it directly afterwards. Now there is no way to reference the values created on the heap with new. Both allocations are not necessary since MEM always points to one of the two static variables. Therefore you should also not try to delete them once Zeroth goes out of scope.
Since we are at the constructor: why do you accept an 32bit integer as argument when all you care about is one bit and all your internal values are stored in an uint8_t? I also find it unintuitive to treat all the even numbers as false, and all the odd ones as true (this is basically a consequence of your bit mask). Often all values except 0 are treated as truthy values (exempt of this rule are exit codes, where 0 is usually seen as success).
## isPresent/isNotPresent
Those functions do not even compile on a 64bit platform, since the completely unnecessary cast to int narrows the bidwidth of the pointer type from 64bit to 32bit. Apart from that, there is a whole lot of other weird things going on here.
Let's try to break this down:
(int)&(*this->MEM)
this->MEM is a pointer. You then for whatever reason decided to dereference (*this->MEM) it to get the value it's pointing to. After that, you take the address of the value (&(*this->MEM)), which should be the same as this->MEM. And if all of this weren't enough, you then try to cram the address into an int ((int)&(*this->MEM)).
Same procedure for the value you try to compare it with:
(int)&Zeroth::UnitSet
You have the value Zeroth::UnitSet, take the address of it &Zeroth::UnitSet and then try to cram this address into a 32bit integer.
You get the same effect using
bool isPresent() const { return &Zeroth::UnitSet == this->MEM; }
bool isNotPresent() const { return &Zeroth::NullSet == this->MEM; }
without all this madness.
## Memory footprint
The internal state of your class is represented by a pointer to a const uint8_t. As I have already tried to tell you in the comments, the size of the pointer is 32bit or 64bit depending on which type of architecture your compiler tries to target. This 4 to 8 times larger than the 8bit you would need to store the value. If you stick with the value, there are far less possibilities to shoot yourself in the foot.
class ZerothVal {
public:
ZerothVal() = default; // see https://en.cppreference.com/w/cpp/language/default_constructor
ZerothVal(uint8_t VAL) {
val = (VAL & 0x1 ? ZerothVal::UnitSet : ZerothVal::NullSet);
}
bool isPresent() const { return val == ZerothVal::UnitSet; }
bool isNotPresent() const { return val == ZerothVal::NullSet; }
friend std::ostream& operator<<(std::ostream& os, const ZerothVal& data);
private:
const uint8_t NullSet = 0;
const uint8_t UnitSet = 1;
uint8_t val = ZerothVal::NullSet;
};
std::ostream& operator<<(std::ostream& os, const ZerothVal& data) {
return os << (data.isPresent() ? 'T' : 'F');
}
## Operator support
Since you'd like to implement some kind of bool value, I'd highly recommend implementing the Safe Bool Idiom. For you that would mean that you have to implement operator bool() for your class (see this SO post for an explanation).
• This is madness i tell you! Madness!! :D Don't apologise for straightforwardness and bluntness ...that is why i am here for. Operator overloading will come after i stop shooting myself in the foot. It does compile on 64 bit OS ( i am currently on it ), but this could be IDE /compiler difference. Why did you remove static ? Ok i get that all that memory , but the idea is to use shared members so each class don't have to initialise them ? Does const in this manner do the similar action ? – Danilo Jul 15 '19 at 11:53
• There are 32bit and 64bit versions of Cygwin, so it depends on which version you've got. I removed static because IMHO it does not buy you anything here. Until you have serious indications that there will be a problem in your application, you should stick with the simplest possible implementation. Then if there is a problem, measure carefully and then and only then look at the results and think about optimization. – AlexV Jul 15 '19 at 12:03
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2020-07-13 14:04:58
|
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|
https://msp.org/agt/2004/4-2/p03.xhtml
|
Volume 4, issue 2 (2004)
Recent Issues
The Journal About the Journal Subscriptions Editorial Board Editorial Interests Editorial Procedure Submission Guidelines Submission Page Author Index To Appear ISSN (electronic): 1472-2739 ISSN (print): 1472-2747
Peripheral separability and cusps of arithmetic hyperbolic orbifolds
D B McReynolds
Algebraic & Geometric Topology 4 (2004) 721–755
arXiv: math.GT/0409278
Abstract
For $X=ℝ$, $ℂ$, or $ℍ$, it is well known that cusp cross-sections of finite volume $X$–hyperbolic $\left(n+1\right)$–orbifolds are flat $n$–orbifolds or almost flat orbifolds modelled on the $\left(2n+1\right)$–dimensional Heisenberg group ${\mathfrak{N}}_{2n+1}$ or the $\left(4n+3\right)$–dimensional quaternionic Heisenberg group ${\mathfrak{N}}_{4n+3}\left(ℍ\right)$. We give a necessary and sufficient condition for such manifolds to be diffeomorphic to a cusp cross-section of an arithmetic $X$–hyperbolic $\left(n+1\right)$–orbifold.
A principal tool in the proof of this classification theorem is a subgroup separability result which may be of independent interest.
Keywords
Borel subgroup, cusp cross-section, hyperbolic space, nil manifold, subgroup separability.
Primary: 57M50
Secondary: 20G20
|
2017-12-14 15:11:53
|
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|
http://www.ck12.org/trigonometry/Inverses-of-Trigonometric-Functions/lesson/Inverses-of-Trigonometric-Functions/r4/
|
<meta http-equiv="refresh" content="1; url=/nojavascript/"> Inverses of Trigonometric Functions ( Read ) | Trigonometry | CK-12 Foundation
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# Inverses of Trigonometric Functions
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Your instructor gives you a trigonometric function, $f(x) = 3\sin (x) + 5$ , and asks you to find the inverse. You are all set to start manipulating the equation, when you realize that you don't know just how to do this. Your instructor suggests that you try finding the inverse through graphing instead.
Are you able to do this?
Keep reading, and by the end of this Concept, you'll be able to find the inverse of trig function and others through graphing instead of algebra.
### Guidance
In other Concepts, two different ways to find the inverse of a function were discussed: graphing and algebra. However, when finding the inverse of trig functions, it is easy to find the inverse of a trig function through graphing. Consider the graph of a sine function shown here:
In order to consider the inverse of this function, we need to restrict the domain so that we have a section of the graph that is one-to-one. If the domain of $f$ is restricted to $-\frac{\pi}{2} \le x \le \frac{\pi}{2}$ a new function $f(x) = \sin x, -\frac{\pi}{2} \le x \le \frac{\pi}{2}$ . is defined. This new function is one-to-one and takes on all the values that the function $f(x) = \sin x$ takes on. Since the restricted domain is smaller, $f(x) = \sin x, -\frac{\pi}{2} \le x \le \frac{\pi}{2}$ takes on all values once and only once.
The inverse of $f(x)$ is represented by the symbol $f^{-1}(x)$ , and $y = f^{-1}(x) \Leftrightarrow f(y) = x$ . The inverse of $\sin x, -\frac{\pi}{2} \le x \le \frac{\pi}{2}$ will be written as $\sin^{-1} x$ . or $\arcsin x$ .
$\begin{Bmatrix}y = \sin^{-1} x\\\quad or\\y = \arcsin x\end{Bmatrix} \Leftrightarrow \sin y = x$
In this Concept we will use both $\sin^{-1} x$ and $\arcsin x$ and both are read as “the inverse sine of $x$ ” or “the number between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ whose sine is $x$ .”
The graph of $y = \sin^{-1} x$ is obtained by applying the inverse reflection principle and reflecting the graph of $y=\sin x, -\frac{\pi}{2} \le x \le \frac{\pi}{2}$ in the line $y = x$ . The domain of $y = \sin x$ becomes the range of $y = \sin^{-1} x$ , and hence the range of $y = \sin x$ becomes the domain of $y = \sin^{-1} x$ .
Another way to view these graphs is to construct them on separate grids. If the domain of $y = \sin x$ is restricted to the interval $\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$ , the result is a restricted one-to one function. The inverse sine function $y = \sin^{-1} x$ is the inverse of the restricted section of the sine function.
The domain of $y = \sin x$ is $\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$ and the range is [-1, 1].
The restriction of $y = \sin x$ is a one-to-one function and it has an inverse that is shown below.
The domain of $y = \sin^{-1}$ is [-1, 1] and the range is $\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$ .
The inverse functions for cosine and tangent are defined by following the same process as was applied for the inverse sine function. However, in order to create one-to-one functions, different intervals are used. The cosine function is restricted to the interval $0 \le x \le \pi$ and the new function becomes $y = \cos x, 0 \le x \le \pi$ . The inverse reflection principle is then applied to this graph as it is reflected in the line $y = x$ The result is the graph of $y = \cos^{-1} x$ (also expressed as $y = \arccos x$ ).
Again, construct these graphs on separate grids to determine the domain and range. If the domain of $y = \cos x$ is restricted to the interval $[0, \pi]$ , the result is a restricted one-to one function. The inverse cosine function $y = \cos^{-1} x$ is the inverse of the restricted section of the cosine function.
The domain of $y = \cos x$ is $[0, \pi]$ and the range is [-1, 1].
The restriction of $y = \cos x$ is a one-to-one function and it has an inverse that is shown below.
The statements $y = \cos x$ and $x = \cos y$ are equivalent for $y-$ values in the restricted domain $[0, \pi]$ and $x-$ values between -1 and 1.
The domain of $y = \cos^{-1} x$ is [-1, 1] and the range is $[0, \pi]$ .
The tangent function is restricted to the interval $-\frac{\pi}{2} < x < \frac{\pi}{2}$ and the new function becomes $y = \tan x, -\frac{\pi}{2} < x < \frac{\pi}{2}$ . The inverse reflection principle is then applied to this graph as it is reflected in the line $y = x$ . The result is the graph of $y = \tan^{-1} x$ (also expressed as $y = \arctan x$ ).
Graphing the two functions separately will help us to determine the domain and range. If the domain of $y = \tan x$ is restricted to the interval $\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$ , the result is a restricted one-to one function. The inverse tangent function $y = \tan^{-1} x$ is the inverse of the restricted section of the tangent function.
The domain of $y = \tan x$ is $\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$ and the range is $[-\infty, \infty]$ .
The restriction of $y = \tan x$ is a one-to-one function and it has an inverse that is shown below.
The statements $y = \tan x$ and $x = \tan y$ are equivalent for $y-$ values in the restricted domain $\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$ and $x-$ values between -4 and +4.
The domain of $y = \tan^{-1} x$ is $[-\infty, \infty]$ and the range is $\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$ .
The above information can be readily used to evaluate inverse trigonometric functions without the use of a calculator. These calculations are done by applying the restricted domain functions to the unit circle. To summarize:
Restricted Domain Function Inverse Trigonometric Function Domain Range Quadrants
$y = \sin x$ $\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$ [-1, 1] 1 AND 4
$y = \arcsin x$
$y = \sin^{-1} x$
[-1, 1] $\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]$
$y = \cos x$ $[0, \pi]$ [-1, 1] 1 AND 2
$y = \arccos x$
$y = \cos^{-1} x$
[-1, 1] $[0, \pi]$
$y = \tan x$ $\left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$ $(-\infty, \infty)$ 1 AND 4
$y = \arctan x$
$y = \tan^{-1}x$
$(-\infty, \infty)$ $\left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )$
Now that the three trigonometric functions and their inverses have been summarized, let’s take a look at the graphs of these inverse trigonometric functions.
#### Example A
Establish an alternative domain that makes $y=sin(x)$ a one to one function.
Solution: Any number of possible solutions can be given, but the important point is that the function must pass the "horizontal line test" and the "vertical line test". This means that a horizontal line drawn through the graph will intersect the function in only one place, and a vertical line drawn through the graph will intersect the function in only one place.
For the sine curve, this means that the function can't "turn over" or "go in the other direction", since then it couldn't pass the horizontal line test. So any part of the function that starts at the bottom of the "y" values and stops at the top of the "y" values will work. (Any value that starts at the top of the "y" values and stops at the bottom of the "y" values will work as well.
In this example, you can see that the function starts at $\frac{\pi}{2}$ and stops at $\frac{3\pi}{2}$ .
#### Example B
Find the range of the function given in Example A
Solution: You can see that the function still has the same "y" range of values, since the function $y = \sin x$ moves up and down between -1 and 1. Therefore, the range is $-1 \le y \le 1$ .
#### Example C
Find the domain and range of the inverse of the function given in Example A.
Solution: Since the domain of the inverse function is the range of the original function and the range of the inverse function is the domain of the original function, you only have to take the "x" and "y" values of the original function and reverse them to get the domain and range range of the inverse function.
Therefore, the domain of $y = \sin^{-1} x$ as described in Example A is $-1 \le x \le 1$ and the range is $\frac{\pi}{2} \le y \le \frac{3\pi}{2}$ .
### Vocabulary
Inverse Function: An inverse function is a function that undoes another function.
### Guided Practice
1. Sketch a graph of $y = \frac{1}{2} \cos^{-1} (3x+1)$ . Sketch $y = \cos^{-1} x$ on the same set of axes and compare how the two differ.
2. Sketch a graph of $y = 3-\tan^{-1} (x-2)$ . Sketch $y = \tan^{-1} x$ on the same set of axes and compare how the two differ.
3. Graph $y = 2\sin^{-1}(2x)$
Solutions:
1.
$y = \frac{1}{2} \cos^{-1} (3x+1)$ is in blue and $y =\cos^{-1}(x)$ is in red. Notice that $y = \frac{1}{2} \cos^{-1}(3x+1)$ has half the amplitude and is shifted over -1. The 3 seems to narrow the graph.
2.
$y = 3-\tan^{-1} (x-2)$ is in blue and $y = \tan^{-1} x$ is in red. $y = 3-\tan^{-1} (x-2)$ is shifted up 3 and to the right 2 (as indicated by point $C$ , the “center”) and is flipped because of the $-\tan^{-1}$ .
3.
### Concept Problem Solution
To find the inverse of this function through graphing, first restrict the domain of the function so that it is one to one. A graph of $f(x) = 3\sin (x) + 5$ , restricted so that the domain is $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ looks like this:
If you apply the inverse reflection principle, you can see that the inverse of this function looks like this:
### Practice
1. Why does the domain of a trigonometric function have to be restricted in order to find its inverse function?
2. If the domain of $f(x)=\cos(x)$ is $[0,\pi]$ , what is the domain and range of $f^{-1}(x)$ ?.
3. If the domain of $g(x)=\sin(x)$ is $[-\frac{\pi}{2},\frac{\pi}{2}]$ , what is the domain and range of $g^{-1}(x)$ ?.
4. Establish an alternative domain that makes $y=\cos(x)$ a function.
5. What is the domain and range of the inverse of the function from the previous problem.
6. Establish an alternative domain that makes $y=\tan(x)$ a function.
7. What is the domain and range of the inverse of the function from the previous problem.
Sketch a graph of each function. Use the domains presented in this concept.
1. $y=2\sin^{-1}(3x-1)$
2. $y=-3+\cos^{-1}(2x)$
3. $y=1+2\tan^{-1}(x+2)$
4. $y=4\sin^{-1}(x-4)$
5. $y=2+\cos^{-1}(x+3)$
6. $y=1+\cos^{-1}(2x-3)$
7. $y=-3+\tan^{-1}(3x+1)$
8. $y=-1+2\sin^{-1}(x+5)$
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2014-10-31 08:05:02
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http://mathhelpforum.com/calculus/5570-analysis.html
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# Math Help - Analysis
1. ## Analysis
Have a question:
Suppose a, b can be any real number and for every e greater than 0, a is less than or equal to b +e. Prove that a is less than or equal to b.
It makes sense, just thinking about it, but how do you actually prove it?
MK
2. Say it is false, then a>b.
Then a-b>0. So let e=a-b.
See what happens when you use the e in the given.
3. I still don't see where the contradiction occurs.
4. You have,
a<b+e
then,
a-b<e
If,
a>b then, a-b>0
Say,
c=a-b
then,
c<e
c>0
thus
0<c<e
but for any positive integer c
we can choose a smaller e, i.e. e=c/2
thus,
0<c<e is not true for all e.
5. Originally Posted by MKLyon
I still don't see where the contradiction occurs.
Let e = a - b > 0 (the last step is by assumption of a > b.)
Thus b + e = b + (a - b) = a. (Which holds for the condition in the problem statement.)
Now, e is positive so there is some real number e' such that 0 < e' < e.
The problem statement says that a =< b + e' also, since any positive real number will do.
But e' < e implies b + e' < b + e.
Thus b + e' < a, contrary to the problem statement.
Thus a is not greater than b.
What can you do to show that a is not equal to b?
-Dan
6. We are given that for each e>0, a<b+e.
If it were the case that a>b then let e=a-b>0.
By the given a<b+e=b+(a-b)=a.
The above says a<a. That is the clear contradiction.
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2015-01-27 17:40:09
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https://stats.stackexchange.com/questions/424252/how-to-decide-the-boundary-of-a-2d-kd-tree
|
# How to decide the boundary of a 2d kd-tree?
People often cites this figure when talking about k-d tree
in the context of plotting, this is a $$10 \times 10$$ square, considering this particular dataset, the maximum of x axis is 9, the minimum is 2, [1,7] along the y axis.
is there a rule or algorithm to induce the $$10 \times 10$$ square based on the dataset?
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2020-04-03 02:54:49
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https://meangreenmath.com/category/news-clips/
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# Article on John Urschel
I enjoyed reading this article about John Urschel, a former professional football player who is now pursuing a Ph.D. in mathematics at MIT.
https://hmmdaily.com/2018/09/28/john-urschel-goes-pro/
# Why Should Physicists Study History?
I’ve always enjoyed reading about the history of both mathematics and physics, and so I really appreciated this perspective from Physics Today magazine about the importance of this field. One of many insightful paragraphs:
And a more human physics is a good thing. For starters, it makes physics more accessible, particularly for students. Many promising students drop out of the sciences because the material seems disembodied and disconnected from their lives. Science education researchers have found that those lost students “hungered—all of them—for information about how the various methods they were learning had come to be, why physicists and chemists understand nature the way they do, and what were the connections between what they were learning and the larger world.” Students can potentially lose the wonder and curiosity that drew them to science in the first place. Historical narratives naturally raise conceptual, philosophical, political, ethical, or social questions that show the importance of physics for the students’ own lives. A field in which people are acknowledged as people is much more appealing than one in which they are just calculating machines.
The whole article can be found here: https://physicstoday.scitation.org/doi/full/10.1063/PT.3.3235
# Richard Feynman’s Integral Trick
“I had learned to do integrals by various methods shown in a book that my high school physics teacher Mr. Bader had given me. [It] showed how to differentiate parameters under the integral sign — it’s a certain operation. It turns out that’s not taught very much in the universities; they don’t emphasize it. But I caught on how to use that method, and I used that one damn tool again and again. [If] guys at MIT or Princeton had trouble doing a certain integral, [then] I come along and try differentiating under the integral sign, and often it worked. So I got a great reputation for doing integrals, only because my box of tools was different from everybody else’s, and they had tried all their tools on it before giving the problem to me.” (Surely you’re Joking, Mr. Feynman!)
I read Surely You’re Joking, Mr. Feynman! dozens of times when I was a teenager, and I was always curious about exactly what this integration technique actually was. So I enjoyed reading this article about the Leibniz Integration Rule: https://medium.com/dialogue-and-discourse/richard-feynmans-integral-trick-e7afae85e25c
# Texas slide rule competitions
I got a kick out of reading this retrospective of Texas high school slide rule competitions… including a 1959 picture of Janis Joplin on her high school slide rule team and a 1980 Dallas Morning News article eulogizing the competition.
https://mikeyancey.com/uil-slide-rule-resources/
# Optimal wedding seat assignments
Mixed-integer linear programming at work: finding optimal seating assignments at a wedding when everyone doesn’t necessarily get along with everyone else.
https://blogs.sas.com/content/operations/2014/11/10/do-you-have-an-uncle-louie-optimal-wedding-seat-assignments/
# Powers Great and Small
I enjoyed this reflective piece from Math with Bad Drawings about determining whether $a^b$ or $b^a$ is larger. The final answer, involving the number $e$, was a complete surprise to me.
Short story: $e$ is the unique number so that $e^x > x^e$ for all positive $x$.
Powers Great and Small
# A Classical Math Problem Gets Pulled Into the Modern World
I enjoyed this article about how the solution of a pure mathematics problem from a century ago is finding an unlikely application now: https://www.quantamagazine.org/a-classical-math-problem-gets-pulled-into-the-modern-world-20180523\
From the introductory paragraphs:
Long before robots could run or cars could drive themselves, mathematicians contemplated a simple mathematical question. They figured it out, then laid it to rest — with no way of knowing that the object of their mathematical curiosity would feature in machines of the far-off future.
The future is now here. As a result of new work by Amir Ali Ahmadi and Anirudha Majumdar of Princeton University, a classical problem from pure mathematics is poised to provide iron-clad proof that drone aircraft and autonomous cars won’t crash into trees or veer into oncoming traffic.
“You get a complete 100-percent-provable guarantee that your system” will be collision-avoidant, said Georgina Hall, a final-year graduate student at Princeton who has collaborated with Ahmadi on the work.
The guarantee comes from an unlikely place — a mathematical problem known as “sum of squares.” The problem was posed in 1900 by the great mathematician David Hilbert. He asked whether certain types of equations could always be expressed as a sum of two separate terms, each raised to the power of 2.
Mathematicians settled Hilbert’s question within a few decades. Then, almost 90 years later, computer scientists and engineers discovered that this mathematical property — whether an equation can be expressed as a sum of squares — helps answer many real-world problems they’d like to solve.
“What I do uses a lot of classical math from the 19th century combined with very new computational math,” said Ahmadi.
# Peer-reviewed math and science journal for kids
I’m filing this away for future reference: a peer-reviewed journal for explaining advanced concepts in science and mathematics to kids… and the peer reviewers are the kids.
Journal: https://kids.frontiersin.org/
# Solutions to Exercises in Math Textbooks
I read a very thought-provoking blog post on the pros and cons of having answers in the back of math textbooks. The article and comments on the article are worth reading.
https://blogs.ams.org/bookends/2017/10/11/solutions-to-exercises-in-math-textbooks/
# 5 Ways to go Beyond Recitation
Most students will encounter recitation in a math class during their academic career. How can math professors make the experience more meaningful? MAA Teaching Tidbits blog has 5 ways educators can enhance the student experience during recitation.
1. Focus on getting students to do the work instead of doing it for them.
2. Incorporate group work into your sessions.
3. Get students to communicate what they understand to each other and to the class.
4. Have students relate mathematics to their own experiences.
5. Cultivate an environment where failure is ok and experimentation is encouraged.
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2019-05-23 00:38:57
|
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https://zbmath.org/?q=an%3A1039.62047
|
zbMATH — the first resource for mathematics
A new class of multivariate skew distributions with applications to Bayesian regression models. (English) Zbl 1039.62047
Can. J. Stat. 31, No. 2, 129-150 (2003); erratum ibid. 301-302 (2009).
Summary: The authors develop a new class of distributions by introducing skewness in multivariate elliptically symmetric distributions. The class, which is obtained by using transformation and conditioning, contains many standard families including the multivariate skew-normal and $$t$$ distributions. The authors obtain analytical forms of the densities and study distributional properties. They give practical applications in Bayesian regression models and results on the existence of the posterior distributions and moments under improper priors for the regression coefficients. They illustrate their methods using practical examples.
MSC:
62H05 Characterization and structure theory for multivariate probability distributions; copulas 62F15 Bayesian inference 62J05 Linear regression; mixed models 62H12 Estimation in multivariate analysis
Full Text:
References:
[1] Adcock, Asset Pricing and Portfolio Selection Based on the Multivariate Skew-Student Distribution (2002) · Zbl 1233.91112 [2] Aigner, Formulation and estimation of stochastic frontier production function model, Journal of Econometrics 12 pp 21– (1977) · Zbl 0366.90026 [3] Arnold, Hidden truncation models, Sankhyǎ, Series A 62 pp 23– (2000) [4] Arnold, Skewed multivariate models related to hidden truncation and/or selective reporting (with discussion), Test 11 pp 7– (2002) · Zbl 1033.62013 [5] Azzalini, Statistical applications of the multivariate skew normal distribution, Journal of the Royal Statistical Society Series B 61 pp 579– (1999) · Zbl 0924.62050 [6] Azzalini, The multivariate skew-normal distribution, Biometrika 8 pp 715– (1996) · Zbl 0885.62062 [7] Bernardo, Bayesian Theory. (1994) [8] Branco, Bayesian analysis of calibration problem under elliptical distributions, Journal of Statistical Planning and Inference 90 pp 69– (2000) · Zbl 1066.62040 [9] Branco, A general class of multivariate skew elliptical distributions, Journal of Multivariate Analysis 79 pp 99– (2001) · Zbl 0992.62047 [10] Chen, A new skewed link model for dichotomous quantal response data, Journal of the American Statistical Association 94 pp 1172– (1999) · Zbl 1072.62655 [11] Chib, Bayes prediction in regressions with elliptical errors, Journal of Econometrics 38 pp 349– (1988) · Zbl 0677.62030 [12] DiCiccio, Computing Bayes factors by combining simulation and asymptotic approximations, Journal of the American Statistical Association 92 pp 903– (1997) · Zbl 1050.62520 [13] Fang, Symmetric Multivariate and Related Distributions. (1990) [14] Fernandez, On Bayesian modeling of fat tails and skewness, Journal of the American Statistical Association 93 pp 359– (1998) [15] Gelfand, Bayesian model choice-Asymptotics and exact calculations, Journal of the Royal Statistical Society Series B 56 pp 501– (1994) · Zbl 0800.62170 [16] Geweke, Bayesian treatment of the independent Student-i linear model, Journal of Applied Econometrics 8 pp 519– (1993) [17] Huang, Foundations for Financial Economics. (1988) · Zbl 0677.90001 [18] Jones, Probability and Statistical Models with Applications pp 269– (2001) [19] Kelker, Distribution theory of spherical distributions and a location-scale parameter generalization, Sankhyà 32 pp 419– (1970) · Zbl 0223.60008 [20] Mardia, Measures of multivariate skewness and kurtosis with applications, Biometrika 57 pp 519– (1970) · Zbl 0214.46302 [21] Osiewalski, Robust Bayesian inference in elliptical regression models, Journal of Econometrics 57 pp 345– (1993) · Zbl 0776.62029 [22] Spiegelhalter, Bayesian measures of model complexity and fit (with discussion), Journal of the Royal Statistical Society Series B 64 pp 583– (2002) · Zbl 1067.62010 [23] Spiegelhalter, Bayesian Statistics 5: Proceedings of the 5th Valencia International Meeting Held in Alicante, June 5-9, 1994 pp 407– (1996) [24] Whittaker, A Course of Modern Analysis: An Introduction to the General Theory of Infinite Processes and of Analytic Functions. (1927) · JFM 53.0180.04 [25] Zellner, Bayesian and non-Bayesian analysis of the regression model with multivariate Student-t error term, Journal of the American Statistical Association 11 pp 400– (1976) · Zbl 0348.62026
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.
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2021-12-08 14:04:28
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https://dsp.stackexchange.com/questions/79360/should-i-normalize-fft-signal-with-z-score
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# Should i normalize FFT signal with z-score?
I am working with EEG data (time domain) in a machine learning task, where each input signal must be mapped to a class/frequency. I am using FFT in order to get data in frequency domain and make classification easier.
I'd like to know if it is necessary to normalize data after (or before) applying FFT with z-score normalization (subtract mean and divide by std) in order to center data with zero mean and unit variance.
I know that machine learning models perform a lot better with normalized data, but i don't know if this is the proper way of working with frequency data.
I can provide more details if needed, but i'm using default scipy.fft and sklearn.preprocessing.StandardScaler modules in python.
Probably not.
Applying Z-score to an FFT is problematic. The FFT is a complex signal and you need to define exactly how to normalize.
For example you could normalize the complex frequency domain signal directly. However that doesn't make much sense. Example: the FFT of a unit impulse $$\delta(n)$$ has a mean of 1 and a standard deviation of 0. If you time shift by one sample the mean of the FFT becomes 0 and the standard deviation is 1. I don't think you want minor time shifts make that dramatic a difference.
The next option would be to normalize the power spectrum. Z-score is not a great fit for that either: power spectra are always positive and if you take out the mean, it's not a power spectrum any more.
The 3rd option would be to normalize to log power/amplitude spectrum. That's at least mathematically plausible but I don't think the physics would warrant that either.
The best normalization is one that is based on the physical/physiological understanding of the features and the underlying mechanism. Blindly applying a z-score probably doesn't help much.
Example: if the phenomenon is related to absolute activity levels in the brain, than any normalization procedure must maintain the absolute level. If the phenomenon is related to the relative difference of two events/activities than the normalization should be using the "reference" activity or event.
Some options:
• EEG signals often have a lot of bias. Use a DC blocking filter to eliminate this
• Normalize to the same overall power
• Normalize to the power of some "background" activity
• Normalize to the power of a "reference frequency" band
• Stay fully calibrated. Make sure that your FFT represents the actual spectral density in the brain measured in $$V/\sqrt{Hz}$$
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2022-12-05 00:13:16
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http://www.math.princeton.edu/events/seminars/349/archive/2015-2016
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# Seminars & Events for PACM/Applied Mathematics Colloquium
September 21, 2015
4:30pm - 5:30pm
##### Reduced Order Models for the numerical modelling of complex systems
###### PACM/Applied Mathematics Colloquium
This talk will address the challenge of complexity in numerical simulations of complex physical problems.When the latter exhibit a multiscale and or a multiphysics nature, appropriate mathematical models and accurate numerical methods are required to catch the essential features of the manifold components of the physical solution. Often the associated numerical problem is so large that devising computational reduction techniques, and developing efficient parallel algorithms by exploiting a dimensional reduction paradigm, becomes necessary.
Speaker: Alfio Quarteroni , École Polytechnique Fédérale de Lausanne, Switzerland
Location:
Fine Hall 214
September 24, 2015
4:30pm - 5:30pm
##### Special Colloquium: Bayesian Inversion for Functions and Geometry
###### PACM/Applied Mathematics Colloquium
Please note special day (Thursday). Many problems in the physical sciences require the determination of an unknown function from a finite set of indirect measurements. Examples include oceanography, medical imaging, oil recovery, water resource management and weather forecasting. Furthermore there are numerous inverse problems where geometric characteristics, such as interfaces, are key unknown features of the overall inversion. Applications include the determination of layers and faults within subsurface formations, and the detection of unhealthy tissue in medical imaging. We describe a theoretical and computational Bayesian framework relevant to the solution of inverse problems for functions, and for geometric features. We formulate Bayes' theorem on separable Banach spaces, a conceptual approach which leads to a form of probabilistic well-posedness and also to new and efficient MCMC algorithms which exhibit order of magnitude speed-up over standard methodologies. Furthermore the approach can be developed to apply to geometric inverse problems, where the geometry is parameterized finite-dimensionally and, via the level-set method, to infinite-dimensional parameterizations. In the latter case this leads to a well-posedness that is difficult to achieve in classical level-set inversion, but which follows naturally in the probabilistic setting.
[1] A.M. Stuart. Inverse problems: a Bayesian perspective. Acta Numerica 19(2010) 451--559. http://homepages.warwick.ac.uk/~masdr/BOOKCHAPTERS/stuart15c.pdf
[2] M. Dashti and A.M. Stuart. The Bayesian approach to inverse problems.To appear in Handbook of Uncertainty Quantification, Springer, 2016. http://arxiv.org/abs/1302.6989
[3] S.L.Cotter, G.O.Roberts, A.M. Stuart and D. White, MCMC methods for functions: modifying old algorithms to make them faster. Statistical Science, 28 (2013) 424-446. http://homepages.warwick.ac.uk/~masdr/JOURNALPUBS/stuart103.pdf
[4] M.A. Iglesias, K. Lin, A.M. Stuart, "Well-Posed Bayesian Geometric Inverse Problems Arising in Subsurface Flow", Inverse Problems, 30 (2014) 114001. http://arxiv.org/abs/1401.5571
[5] M.A. Iglesias, Y. Lu, A.M. Stuart, "A level-set approach to Bayesian geometric inverse problems", submitted. http://arxiv.org/abs/1504.00313
Speaker: Andrew Stuart , Warwick University
Location:
Fine Hall 214
September 28, 2015
4:30pm - 5:30pm
##### The Resilience of the Perceptron
###### PACM/Applied Mathematics Colloquium
The most widely used optimization method in machine learning practice is the Perceptron Algorithm, also known as the Stochastic Gradient Method (SGM). This method has been used since the fifties to build statistical estimators, iteratively improving models by correcting errors observed on single data points. SGM is not only scalable, robust, and simple to implement, but achieves the state-of-the-art performance in many different domains. In contemporary systems, SGM powers enterprise analytics systems and is the workhorse tool used to train complex pattern-recognition systems in speech and vision.
Speaker: Ben Recht, University of California, Berkeley
Location:
Fine Hall 214
October 5, 2015
4:30pm - 5:30pm
##### An Applied Math Perspective on Climate Science, Turbulence, and Other Complex Systems
###### PACM/Applied Mathematics Colloquium
Speaker: Andrew Majda , New York University
Location:
Fine Hall 214
October 12, 2015
4:30pm - 5:30pm
##### Comparison Lemmas and Convex-Optimization-Based Signal Recovery
###### PACM/Applied Mathematics Colloquium
In the past couple of decades, non-smooth convex optimization has emerged as a powerful tool for the recovery of structured signals (sparse, low rank, finite constellation, etc.) from (possibly) noisy measurements in a variety of applications in statistics, signal processing, machine learning, communications, etc. I will describe a fairly general theory for how to determine the performance (minimum number of measurements, mean-square-error, probability-of-error, etc.) of such methods for certain measurement ensembles (Gaussian, Haar, quantized Gaussian, etc.).
Speaker: Babak Hassibi , Caltech
Location:
Fine Hall 214
October 19, 2015
4:30pm - 5:30pm
##### Applications of interlacing polynomials in graph theory
###### PACM/Applied Mathematics Colloquium
In this talk, I will discuss some recent applications of interlacing polynomials in graph theory. I will begin by discussing the more classical results concerning graph polynomials, including the real rootedness of the matching polynomial and the extension of Chudnovsky and Seymour to the independence polynomial of claw-free graphs. I will then discuss results using the method of interlacing polynomials'' developed by the speaker with Daniel Spielman and Nikhil Srivastava. In particular, I will mention recent results concerning the existence of Ramanujan graphs of all sizes and degrees as well as a recent improvement of the best known upper bound on the integrality gap in the asymmetric traveling salesman problem due to Anari and Oveis-Gharan.
Speaker: Adam Marcus , Princeton University
Location:
Fine Hall 214
October 26, 2015
4:30pm - 5:30pm
##### The master equation and the convergence problem in mean-field games
###### PACM/Applied Mathematics Colloquium
We discuss the convergence, as $N$ tends to infinity, of a system of $N$ coupled Hamilton-Jacobi equations, called the Nash system. This system arises in differential game theory. We describe the limit problem in terms of the so-called master equation", a kind of second order partial differential equation stated on the space of probability measures. Our first main result is the well-posedness of the master equation. To do so, we first show the existence and uniqueness of a solution to the mean field game system with common noise", which consists in a coupled system made of a backward stochastic Hamilton-Jacobi equation and a forward stochastic Kolmogorov equation and which plays the role of characteristics for the master equation.
Speaker: François Delarue , Université Nice-Sophia Antipolis
Location:
Fine Hall 214
November 9, 2015
4:30pm - 5:30pm
##### An Axiomatic Foundation for Non-Bayesian Learning in Networks
###### PACM/Applied Mathematics Colloquium
Rational learning postulates that individuals incorporate new information into their beliefs in a Bayesian fashion. Despite its theoretical appeal, this Bayesian learning framework has been criticized on the basis of placing unrealistic computational demands on the agents. Furthermore, experiments have shown that the way agents update their beliefs in networked settings is often inconsistent with predictions of Bayesian learning models. Motivated by these issues, A large body of literature has emerged that proposes a series of non-Bayesian updates that are often inspired by the linear (consensus) learning model of DeGroot. However, a systematic framework that captures behavioral deviations of such updates from Bayesian learning has been lacking.
Speaker: Ali Jadbabaie , Massachusetts Institute of Technology (MIT)
Location:
Fine Hall 214
November 16, 2015
4:30pm - 5:30pm
##### On the Complexity of Detecting Planted Solutions
###### PACM/Applied Mathematics Colloquium
Many combinatorial problems appear to be hard even when inputs are drawn from simple, natural distributions, e.g., SAT for random formulas, clique in random graphs etc. To understand their complexity, we consider random problems with planted solutions, e.g., planted k-SAT/k-CSP (clauses are drawn at random from those satisfying a fixed assignment), planted clique (a large clique is added to a random graph), planted partitions (edges of a random graph have different probabilities between and within parts of a fixed vertex partition). How many clauses/edges does one need to find (or detect) planted solutions? At the moment there are large gaps between the point at which the planted solution becomes unique (information-theoretically) and when it can be found efficiently (in time polynomial in the size of the input).
Speaker: Santosh Vempala , Georgia Institute of Technology
Location:
Fine Hall 214
November 30, 2015
4:30pm - 5:30pm
##### Dissipation at Maximal Rate
###### PACM/Applied Mathematics Colloquium
The lecture will present facets of the conjecture that the role of entropy is to maximize the rate of dissipation. BIO: Constantine Dafermos was born in Athens, in 1941. He received a diploma in Civil Engineering from the National Technical University, in 1964, and a Ph.D. in Mechanics from Johns Hopkins, in 1967. After teaching for three years at Cornell, he moved to Brown, in 1971, where he is currently a Professor of Applied Mathematics. His work lies at the interface between continuum mechanics and partial differential equations.
Speaker: Constantine M. Dafermos , Brown University
Location:
Fine Hall 214
December 7, 2015
4:30pm - 5:30pm
##### Faster Convex Optimization - Simulated Annealing with an Efficient Universal Barrier
###### PACM/Applied Mathematics Colloquium
Interior point methods and random walk approaches have been long considered disparate approaches for convex optimization. We show how simulated annealing, one of the most common random walk algorithms, is equivalent, in a certain sense, to the central path interior point algorithm applied to the entropic universal barrier function. Using this observation we improve the state of the art in polynomial time convex optimization in the membership-oracle model.
Speaker: Elad Hazan , Princeton University
Location:
Fine Hall 214
December 14, 2015
4:30pm - 5:30pm
##### Active subspaces: Emerging ideas for dimension reduction in parameter studies
###### PACM/Applied Mathematics Colloquium
Scientists and engineers use computer simulations to study relationships between a physical model's input parameters and its outputs. However, thorough parameter studies---e.g., constructing response surfaces, optimizing, or averaging---are challenging, if not impossible, when the simulation is expensive and the model has several inputs. To enable studies in these instances, the engineer may attempt to reduce the dimension of the model's input parameter space. Active subspaces are a set of dimension reduction tools that identify important directions in the parameter space. I will describe methods for discovering a model's active subspace and propose strategies for exploiting the reduced dimension to enable otherwise infeasible parameter studies.
Speaker: Paul Constantine , Colorado School of Mines
Location:
Fine Hall 214
February 8, 2016
4:30pm - 5:30pm
##### Julia Computing (with some random matrices)
###### PACM/Applied Mathematics Colloquium
This talk is designed to be of interest for a computer science and mathematics audience, I will combine my two “passions” by (1) Explaining why the Julia Computing language is so fast. Julia in some ways resembles Python, or R, or MATLAB, which makes Julia easy to use, but Julia is deeply architected very differently from these earlier languages, making Julia's modern approach fast, flexible, and productive. Julia is being used worldwide by scientists, engineers, in classes, and in industry for big data, financial applications, and the internet of things. (2) Showing how Julia has enabled research in one of my own favorite topics: random matrices. We will demonstrate several results that quantify and illuminate finite random matrix theory.
Speaker: Alan Edelman , Massachusetts Institute of Technology
Location:
Fine Hall 214
February 15, 2016
4:30pm - 5:30pm
##### Laplacian growth, sandpiles and scaling limits
###### PACM/Applied Mathematics Colloquium
How can repeating simple local operations lead to an intricate large scale structure? This phenomenon arises in several growth models originating in Physics: Internal diffusion limited aggregation (IDLA) and the Abelian sandpile. The first of these is closely related to free boundary problems for the Laplacian and an algebraic operation introduced by Diaconis and Fulton known as smash sum’’. These connections allow a precise description of large scale geometry, using a least action principle. The abelian sandpile, discovered independently by Statistical Physicists and Combinatorialists is harder to analyze, yet has recently yielded many of its secrets in works of Pegden, Smart and Levine.
Speaker: Yuval Peres , Microsoft Research & UC Berkeley
Location:
Fine Hall 214
February 22, 2016
4:30pm - 5:30pm
##### Computationally feasible greedy algorithms for neural nets
###### PACM/Applied Mathematics Colloquium
Previously, greedy algorithms have been shown to have good approximation and estimation properties for superpositions of a sigmoidal function or other ridge activation functions. In each step the parameters of a new sigmoid are fit to the residuals of the previous sigmoids.
Speaker: Andrew Barron , Yale University
Location:
Fine Hall 214
February 29, 2016
4:30pm - 5:30pm
##### Using Graphons to Model Large Networks
###### PACM/Applied Mathematics Colloquium
A fundamental question in the study of large networks is how we compare networks. Given two large networks, say FB and LinkedIn today, or alternatively, FB 5 year ago and FB today, how similar do we consider them to be? Another question is how to describe large networks in terms of a suitable class of (possibly non-parametric) models. Finally, how do we consistently estimate such a model from an observed graph? In this talk, I describe how the theory of graph limits and graphons developed in the last ten years can help to address these questions, reviewing both the classical theory for dense graphs and newer results for sparse graphs including power law graphs.
Speaker: Christian Borgs, Microsoft
Location:
Fine Hall 214
March 7, 2016
4:30pm - 5:30pm
##### Navigating SU(2) and Universal Quantum Gates
###### PACM/Applied Mathematics Colloquium
We discuss recent developments concerning number theoretic generators of SU(2) and their application to optimally efficient universal quantum gates.
Speaker: Peter Sarnak, Princeton University
Location:
Fine Hall 214
March 28, 2016
4:30pm - 5:30pm
##### Beyond Matrix Completion
###### PACM/Applied Mathematics Colloquium
Here we study some of the statistical and algorithmic problems that arise in recommendation systems. We will be interested in what happens when we move beyond the matrix setting, to work with higher order objects — namely, tensors. To what extent does inference over more complex objects yield better predictions, but at the expense of the running time? We will explore the computational vs. statistical tradeoffs for some basic problems about recovering approximately low rank tensors from few observations, and will show that our algorithms are nearly optimal among all polynomial time algorithms, under natural complexity-theoretic assumptions. This is based on joint work with Boaz Barak.
Speaker: Ankur Moitra, Massachusetts Institute of Technology
Location:
Fine Hall 214
April 4, 2016
4:30pm - 5:30pm
##### Coloring some perfect graphs
###### PACM/Applied Mathematics Colloquium
Perfect graphs are a class of graphs that behave particularly well with respect to coloring. In the 1960's Claude Berge made two conjectures about this class of graphs, that motivated a great deal of research, and by now they have both been solved. The following remained open however: design a combinatorial algorithm that produces an optimal coloring of a perfect graph. Recently, we were able to make progress on this question, and we will discuss it in this talk. Last year, in joint work with Lo, Maffray, Trotignon and Vuskovic we were able to construct such an algorithm under the additional assumption that the input graph is square-free (contains no induced four-cycle).
Speaker: Maria Chudnovsky, Princeton University
Location:
Fine Hall 214
April 11, 2016
4:30pm - 5:30pm
##### A 3D Code in the Human Genome
###### PACM/Applied Mathematics Colloquium
Stretched out from end-to-end, the human genome – a sequence of 3 billion chemical letters inscribed in a molecule called DNA – is over 2 meters long. Famously, short stretches of DNA fold into a double helix,
Speaker: Erez Lieberman Aiden , Baylor College of Medicine and Rice University
Location:
Fine Hall 214
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2017-11-23 09:01:05
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http://mathoverflow.net/feeds/question/100793
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Effective way of finding generators on the curve and the rank conjecture - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T14:58:45Z http://mathoverflow.net/feeds/question/100793 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100793/effective-way-of-finding-generators-on-the-curve-and-the-rank-conjecture Effective way of finding generators on the curve and the rank conjecture Shanmukha_Srinivasan 2012-06-27T17:49:11Z 2012-06-28T16:50:24Z <p>Hello everyone, </p> <p>I have never heard of a polynomial time running algorithm that finds the generators of elliptic curves efficiently. I do know that <a href="http://en.wikipedia.org/wiki/Nagell%E2%80%93Lutz_theorem" rel="nofollow">Nagell-Lutz theorem</a> is useful in computing the torsion part in $$E(\mathbb{Q})=\mathbb{Z}^{\phi} \oplus E_{\rm{Torsion }}(Q).$$ So what about the count of $\phi$ and the effective generation of points. </p> <p>I do know 2-descent , that doesn't not work perfectly always. Regarding the 3-descent , only some curves having CM are said to pass through it. But are there any latest advancements in the area of computing the generators of the curve, if so please give some references. </p> <p>I also read the rank conjecture, for expressing the rank of the underlying abelian group $E(\mathbb{Q})$ in terms of the order of vanishing of the taylor expansion of associated $L$-function. So I again got stuck. </p> <p>Why do one go for a rank, if one has a point that has infinite order ? . </p> <ol> <li><p>Rank 1-curves usually mean that they have a point of infinite order on them that can be used to generate all other points by successive chord and tangent methods.</p></li> <li><p>Rank 2- Curves have 2 such points of infinite order that can be used to generate all other points. </p></li> </ol> <p>So my question is why should one bother about $n$ points ( of infinite order ) if we have one point of infinite order. To express an analogous statement, suppose think that you are given a secret map that will lead you to some treasure. After the hard journey, you at-last said " Eureka" and you have found a " Machine X" . Its written on the Machine X, that it will produce dollar notes, as many as you want. So its limit is infinity. You can extract infinite number of dollar notes from the machine. You took that machine and packed it and again saw the secret-map. There are in fact some other markings on the map that will lead you to a place that contains the same Machine X. </p> <p>Do you again go to those places searching for another Machine-X if you have a Machine-X that will produce infinite amount of money. I think the analogy is clear. So if we already have a point that produces infinite points why to again bother about other points that give rise to infinite points. </p> <p>So if I am right, does the task of finding $\phi$ number of generators reduce to the task of finding one generator ? . That will make the above equation look like this $$E(\mathbb{Q})=\mathbb{Z} \oplus E_{\rm{Torsion }}(Q).$$</p> <p>( Which is nothing but the Rank-1 situation ).</p> <p>Thank you. </p> http://mathoverflow.net/questions/100793/effective-way-of-finding-generators-on-the-curve-and-the-rank-conjecture/100796#100796 Answer by Timo Keller for Effective way of finding generators on the curve and the rank conjecture Timo Keller 2012-06-27T18:08:11Z 2012-06-28T16:50:24Z <p>I am not sure I am understanding correctly what you are asking for, but if (1) $\mathrm{ord}_{s=1}L(E/K,s) = \mathrm{rk}E(K)$ is known <em>or</em> (2) if the Tate-Shafarevich group $III(E/K)$ is finite, there is an algorithm for calculating the Mordell-Weil group $E(K)$ of an elliptic curve $E$ over a number field $K$. Ask if I shall give you more details.</p> <p>Edit:</p> <p>For calculating the torsion, use that $E(K)[m] \hookrightarrow \tilde{E}(k_v)$ for $(v,m) = 1$ and $v$ a place of good reduction.</p> <p>For calculating the rank:</p> <ol> <li><p>$\mathrm{ord}_{s=1}L(E/K,s) = \mathrm{rk}E(K)$: There is an algorithm which computes $L^{(n)}(E/K,1)$ up to an arbitrary precision, so one can check if it does not vanish (one can <em>not</em> determine if it is equal to $0$. Now the algorithm is as follows: Search in parallel for a non-torsion point and calculate $L(E/K,1)$. If you find a point, move on to $L^{(2)}(E/K,1)$; else you will find after a finite time that $L(E/K,1) \neq 0$. Repeat.</p></li> <li><p>$III(E/K) = H^1(\mathcal{O}_K, \mathcal{E})$, $\mathcal{E}$ the Néron model of $E/K$, is finite: This is motivated by the analogy with the (finite) class group $H^1(\mathcal{O}_K, \mathbf{G}_m)$. See <a href="http://jmilne.org/math/Books/ectext0.pdf" rel="nofollow">http://jmilne.org/math/Books/ectext0.pdf</a> p. 126 for the algorithm.</p></li> </ol> http://mathoverflow.net/questions/100793/effective-way-of-finding-generators-on-the-curve-and-the-rank-conjecture/100843#100843 Answer by Jesper Petersen for Effective way of finding generators on the curve and the rank conjecture Jesper Petersen 2012-06-28T06:46:17Z 2012-06-28T06:46:17Z <p>Your question seems to suggest that you think that one point of infinite order is enough to generate the full set of solutions of a given curve. This is not always the case. If a curve has rank $n$ then exactly $n$ points of infinite order is needed to generate the full group.</p> <p>Think of a curve of rank, say, $3$. The full group is isomorphic to $\mathbb{Z}^3$, yet none of the points corresponding to $(\pm1,0,0)$, $(0,\pm1,0)$ or $(0,0,\pm1)$ all of infinite order can generate $\mathbb{Z}^3$.</p> http://mathoverflow.net/questions/100793/effective-way-of-finding-generators-on-the-curve-and-the-rank-conjecture/100870#100870 Answer by Joe Silverman for Effective way of finding generators on the curve and the rank conjecture Joe Silverman 2012-06-28T15:21:50Z 2012-06-28T15:21:50Z <p>We do not currently know an <em>effective</em> algorithm to compute the rank of an elliptic curve or to compute generators for its Mordell-Weil group. One can "do descents by day and search for points by night", and in practice, the process will stop. BTW, people do 3 and 4 (and maybe even 5) descents these days, and they're not restricted to CM curves. Finally, I'll mention that if $E(\mathbb{Q})$ has rank 1, then one can use Heegner points to find a rational point of infinite order, or one can use the value of $L'(E,1)$ (sort of as described in Keller's answer) to do a very efficient search. (This last algorithm is in a paper of mine). However, neither method is practical if the conductor $N$ of the curve is too large. Roughly, for both methods one needs to compute the value of a series that doesn't begin to converge until you take $O(\sqrt{N})$ terms.</p>
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2013-05-18 14:58:45
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https://learnche.org/3E4/Tutorial_5_-_2010
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# Tutorial 5 - 2010
Due date(s): 25 October 2010 (PDF) Tutorial questions Other instructions Hand-in at class.
<rst> <rst-options: 'toc' = False/> <rst-options: 'reset-figures' = False/>
.. rubric:: Tutorial objectives: solution of a single, nonlinear equation.
.. note:: All questions below will consider the following problem.
The heat of reaction for a certain reaction is given by :math:\Delta H_{r}^{0}(T)= -24097 -0.26 T+1.69\times 10^{-3}T^2 + {\displaystyle{\frac{1.5\times10^5}{T}}}\; cal/mol. Compute the temperature at which :math:\Delta H_{r}^{0}(T)= -23505 cal/mol.
Question 1 [1]
###### ===
Show the first 4 iterations of the bisection method to solve for :math:T, justifying your choice for the initial bracket.
For iteration 2, 3, and 4, please also report these two relative errors:
.. math::
\varepsilon_\text{rel,x}^{(k)} = \displaystyle \left|\frac{x_m^{(k)} - x_m^{(k-1)}}{x_m^{(k)}} \right| \qquad\qquad \varepsilon_\text{rel,f}^{(k)} = \displaystyle \left|\frac{f(x_m^{(k)}) - f(x_m^{(k-1)})}{f(x_m^{(k)})} \right|
Question 2 [1]
###### ===
1. . Derive a :math:g(x) = x function to use in the fixed-point algorithm.
2. . Show the first 3 iterations of using the fixed-point algorithm, starting with an initial guess of :math:T = 380 K.
3. . Will the fixed-point method converge for this problem, using your :math:g(x)?
Question 3 [1]
###### ==
1. . Write the Newton-Raphson iteration formula that you would use to solve this nonlinear equation.
2. . Apply 3 iterations of this formula, also starting from :math:T = 380 K, and calculate the error tolerances.
Question 4 [1]
###### ==
Comment on the 3 approaches used so far. Are your calculations what you would expect from each method?
Bonus question [1]
###### =======
A naive code for the bisection algorithm would evaluate the function :math:f(x) at the three points, :math:[x_\ell, x_m, x_u] in every iteration. Fewer function evaluations can be obtained though.
Write a function, in either MATLAB or Python, that implements the bisection method, that evaluates :math:f(x) as few times as possible. You should report the following 8 outputs in each iteration: :math:[x_\ell^{(k)},\, x_m^{(k)},\, x_u^{(k)},\, f(x_\ell^{(k)}),\, f(x_m^{(k)}),\, f(x_u^{(k)}),\, \varepsilon_\text{rel,x}^{(k)},\, \varepsilon_\text{rel,f}^{(k)}].
Use this code to find the solution to the above problem, within a tolerance of :math:\sqrt{\text{eps}} based on :math:\varepsilon_\text{rel,x}^{(k)}.
.. raw:: latex
\vspace{0.25cm} \hrule \begin{center}END\end{center} </rst>
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2022-12-01 20:54:34
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http://www.maplesoft.com/support/help/MapleSim/view.aspx?path=MapleSimControlDesignToolbox/whatsNew
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Control Design Toolbox - MapleSim Help
# Online Help
###### All Products Maple MapleSim
Home : Support : Online Help : MapleSim Toolboxes : MapleSim Control Design Toolbox : MapleSimControlDesignToolbox/previousReleases/6.4/whatsNew
Control Design Toolbox
New Features of the MapleSim Control Design Toolbox 6.4
Numerous improvements and additions have been made to the MapleSim Control Design toolbox, including:
• New command for PID controller automatic tuning
• Updates to existing PID tuning commands to choose the format of the returned controller parameters
• Updates to existing control design commands to support the DynamicSystems parameters option, which is new in Maple 18
• New commands for the automatic computation of design parameters, such as poles and weighting matrices, for state feedback control design
• New commands for system manipulation (for example, determining the closed-loop equations of common feedback system configurations) and simplification (for example, eliminating the structurally non-minimal states of a system)
>
PID automatic tuning
The PIDAuto command performs automatic tuning of a PID controller based on a single tuning parameter, Tc, which has the same order of magnitude as the desired time constant of the closed-loop. The time constant is proportional to the settling time of the system (first order approximation) and, therefore, provides a simple way to specify the desired closed-loop system response. This command can obtain the PID controller parameters for a wide range of plants, including unstable systems. The best-suited tuning method is automatically selected according to both the order and stability of the input system. For instance, for stable higher order systems (order greater than 2), Skogestad internal model control (SIMC) tuning rules are applied.
>
Design a PID controller for a system of order 7 with desired time constant Tc = 0.77.
> $\mathrm{sys1}≔\mathrm{TransferFunction}\left(\frac{\left(1-0.3\cdot s\right)\cdot \left(1+0.08\cdot s\right)}{\left(s+1\right)\cdot \left(2\cdot s+1\right)\cdot \left(0.4\cdot s+1\right)\cdot \left(0.2\cdot s+1\right)\cdot {\left(0.05\cdot s+1\right)}^{3}}\right):$
>
PIDAuto: Using Skogestad IMC tuning rules
${\mathrm{pid1}}{:=}{{\mathrm{Record}}}_{{\mathrm{packed}}}{}\left({\mathrm{Kp}}{=}{2.077922078}{,}{\mathrm{Ki}}{=}{0.6493506494}{,}{\mathrm{Kd}}{=}{1.558441558}{,}{\mathrm{Tf}}{=}{0.01558441558}\right)$ (1.1)
Return a DynamicSystems system object directly.
>
PIDAuto: Using Skogestad IMC tuning rules
$\left[\begin{array}{l}{\mathbf{Transfer Function}}\\ {\mathrm{continuous}}\\ {\mathrm{1 output\left(s\right); 1 input\left(s\right)}}\\ {\mathrm{inputvariable}}{=}\left[{\mathrm{u1}}{}\left({s}\right)\right]\\ {\mathrm{outputvariable}}{=}\left[{\mathrm{y1}}{}\left({s}\right)\right]\\ {{\mathrm{tf}}}_{{1}{,}{1}}{=}\frac{{1.590824759}{}{{s}}^{{2}}{+}{2.088041828}{}{s}{+}{0.6493506494}}{{0.01558441558}{}{{s}}^{{2}}{+}{s}}\end{array}\right$ (1.2)
Design a PID controller for an integrating process with desired time constant Tc = 1.0.
>
>
PIDAuto: Using PIDUnstable
${\mathrm{pid2}}{:=}{{\mathrm{Record}}}_{{\mathrm{packed}}}{}\left({\mathrm{Kp}}{=}{2.}{,}{\mathrm{Ki}}{=}{1.}{,}{\mathrm{Kd}}{=}{0.}{,}{\mathrm{Tf}}{=}{0}\right)$ (1.3)
Return a DynamicSystems system object directly.
>
PIDAuto: Using PIDUnstable
$\left[\begin{array}{l}{\mathbf{Transfer Function}}\\ {\mathrm{continuous}}\\ {\mathrm{1 output\left(s\right); 1 input\left(s\right)}}\\ {\mathrm{inputvariable}}{=}\left[{\mathrm{u1}}{}\left({s}\right)\right]\\ {\mathrm{outputvariable}}{=}\left[{\mathrm{y1}}{}\left({s}\right)\right]\\ {{\mathrm{tf}}}_{{1}{,}{1}}{=}\frac{{2.}{}{s}{+}{1}}{{s}}\end{array}\right$ (1.4)
PID tuning commands returned format
Existing PID tuning commands accept new options such as factored and returntype which are useful for specifying the way in which the controller parameters are returned. Use the option factored to return either the controller gains (proportional, integral, and derivative gain Kp, Ki, and Kd) or their factored version (proportional gain, integral time, and derivative time K, Ti, Td). Use the option returntype to return a record, a list or a system containing the controller parameters in either form or a DynamicSystems system object . The updated commands are CohenCoon, DominantPole, GainPhaseMargin, ZNFreq, and ZNTimeModified.
>
>
${\mathrm{fotd3}}{:=}\left[{1.}{,}{1.414213562}{,}{1.}\right]$ (2.1)
By default, the CohenCoon command returns a list containing the PID controller gains Kp, Ki, and Kd.
>
${\mathrm{pid_list}}{:=}\left[{2.152188309}{,}{1.079532192}{,}{0.7019962725}\right]$ (2.2)
Using the new options factored and returntype, the CohenCoon command can return a record containing the factored version of the PID controller gains.
>
${\mathrm{pid_rcrd}}{:=}{{\mathrm{Record}}}_{{\mathrm{packed}}}{}\left({K}{=}{2.152188309}{,}{\mathrm{Ti}}{=}{1.993630504}{,}{\mathrm{Td}}{=}{0.3261779044}\right)$ (2.3)
By default, the DominantPole command returns a sequence of lists containing the controller gains ( Kp and Ki in this case) and the closed-loop poles.
>
${\mathrm{pi_list}}{:=}\left[\frac{{47}}{{8}}{,}\frac{{91}}{{8}}\right]{,}\left[{-}\frac{{7}}{{8}}{,}{-}{3}{-}{2}{}{I}{,}{-}{3}{+}{2}{}{I}\right]$ (2.4)
Using the option returntype, the DominantPole command can return a DynamicSystems system object directly.
>
$\left[\begin{array}{l}{\mathbf{Transfer Function}}\\ {\mathrm{continuous}}\\ {\mathrm{1 output\left(s\right); 1 input\left(s\right)}}\\ {\mathrm{inputvariable}}{=}\left[{\mathrm{u1}}{}\left({s}\right)\right]\\ {\mathrm{outputvariable}}{=}\left[{\mathrm{y1}}{}\left({s}\right)\right]\\ {{\mathrm{tf}}}_{{1}{,}{1}}{=}\frac{{47}{}{s}{+}{91}}{{8}{}{s}}\end{array}\right$ (2.5)
Support of parameters option Existing ControlDesign commands that accept DynamicSystems system objects as input now accept the parameters option. You can use this option to define numeric values for symbolic parameters existing in the system object. This option is the same as the DynamicSystems parameters option. For examples of its use see the sections below.
State Feedback design parameters
Two new commands ComputePoles and ComputeQR help with the calculation of the pole locations for Ackermann's formula and the Q and R weighting matrices for LQR design.
ComputePoles
The Ackermann command calculates the state feedback gain required in order to place the closed-loop poles in the desired locations. The ComputePoles automatically computes these pole locations such that the closed-loop system response has a desired time constant, Tc.
> $\mathrm{sys4}≔\mathrm{StateSpace}\left(\mathrm{Matrix}\left(\left[\left[1,f,3\right],\left[g,4,-1\right],\left[2,5,7\right]\right]\right),\mathrm{Matrix}\left(\left[\left[3\right],\left[9\right],\left[5\right]\right]\right),\mathrm{Matrix}\left(\left[\left[1,0,0\right],\left[0,1,0\right],\left[0,0,1\right]\right]\right),\mathrm{Matrix}\left(3,1\right)\right):$
> $\mathrm{PrintSystem}\left(\mathrm{sys4}\right)$
$\left[\begin{array}{l}{\mathbf{State Space}}\\ {\mathrm{continuous}}\\ {\mathrm{3 output\left(s\right); 1 input\left(s\right); 3 state\left(s\right)}}\\ {\mathrm{inputvariable}}{=}\left[{\mathrm{u1}}{}\left({t}\right)\right]\\ {\mathrm{outputvariable}}{=}\left[{\mathrm{y1}}{}\left({t}\right){,}{\mathrm{y2}}{}\left({t}\right){,}{\mathrm{y3}}{}\left({t}\right)\right]\\ {\mathrm{statevariable}}{=}\left[{\mathrm{x1}}{}\left({t}\right){,}{\mathrm{x2}}{}\left({t}\right){,}{\mathrm{x3}}{}\left({t}\right)\right]\\ {\mathrm{a}}{=}\left[\begin{array}{ccc}{1}& {f}& {3}\\ {g}& {4}& {-1}\\ {2}& {5}& {7}\end{array}\right]\\ {\mathrm{b}}{=}\left[\begin{array}{c}{3}\\ {9}\\ {5}\end{array}\right]\\ {\mathrm{c}}{=}\left[\begin{array}{ccc}{1}& {0}& {0}\\ {0}& {1}& {0}\\ {0}& {0}& {1}\end{array}\right]\\ {\mathrm{d}}{=}\left[\begin{array}{c}{0}\\ {0}\\ {0}\end{array}\right]\end{array}\right$ (4.1.1)
>
>
>
${p}{:=}\left[{-}{1.124260354}{+}{3.158898890}{}{I}{,}{-}{1.124260354}{-}{3.158898890}{}{I}{,}{-}{10.}\right]$ (4.1.2)
>
${\mathrm{Kc}}{:=}\left[\begin{array}{ccc}{3.522839239}& {0.6091131254}& {1.639596973}\end{array}\right]$ (4.1.3)
ComputeQR
The LQR command calculates the linear quadratic state feedback regulator (LQR) gain given the weighting matrices Q and R. The ComputeQR command automatically computes these weighting matrices Q and R to obtain a closed-loop system with desired time constant, Tc.
>
${\mathrm{QRmat}}{:=}{\mathrm{Record}}{}\left({Q}{=}\left[\begin{array}{ccc}{1.}& {0.}& {0.}\\ {0.}& {1.}& {0.}\\ {0.}& {0.}& {1.}\end{array}\right]{,}{R}{=}\left[\begin{array}{c}{2.83828026835937}\end{array}\right]\right)$ (4.2.1)
>
${\mathrm{Kc}}{:=}\left[\begin{array}{ccc}{1.72323935395080}& {0.712987014076495}& {3.18235779686362}\end{array}\right]$ (4.2.2)
System manipulation and simplification
The new commands for system manipulation and simplification are ControllerObserver, PIDClosedLoop, StateFeedbackClosedLoop, and ReduceSystem.
ControllerObserver
The ControllerObserver command calculates the equations of the subsystem comprised of a state feedback controller and an observer. This command is useful, for example, to construct an LQG controller comprised of an LQR state feedback controller and a Kalman observer. Use the closedloop option to obtain the closed-loop equations of the system with state feedback and observer.
> $\mathrm{sys5}≔\mathrm{StateSpace}\left(\mathrm{Matrix}\left(\left[\left[0,1\right],\left[0,0\right]\right]\right),\mathrm{Matrix}\left(\left[\left[\mathrm{α}\right],\left[1\right]\right]\right),\mathrm{Matrix}\left(\left[\left[1,0\right]\right]\right),\mathrm{Matrix}\left(1,1\right)\right):$
> $\mathrm{PrintSystem}\left(\mathrm{sys5}\right)$
$\left[\begin{array}{l}{\mathbf{State Space}}\\ {\mathrm{continuous}}\\ {\mathrm{1 output\left(s\right); 1 input\left(s\right); 2 state\left(s\right)}}\\ {\mathrm{inputvariable}}{=}\left[{\mathrm{u1}}{}\left({t}\right)\right]\\ {\mathrm{outputvariable}}{=}\left[{\mathrm{y1}}{}\left({t}\right)\right]\\ {\mathrm{statevariable}}{=}\left[{\mathrm{x1}}{}\left({t}\right){,}{\mathrm{x2}}{}\left({t}\right)\right]\\ {\mathrm{a}}{=}\left[\begin{array}{cc}{0}& {1}\\ {0}& {0}\end{array}\right]\\ {\mathrm{b}}{=}\left[\begin{array}{c}{\mathrm{\alpha }}\\ {1}\end{array}\right]\\ {\mathrm{c}}{=}\left[\begin{array}{cc}{1}& {0}\end{array}\right]\\ {\mathrm{d}}{=}\left[\begin{array}{c}{0}\end{array}\right]\end{array}\right$ (5.1.1)
Define the numeric value for parameter $\mathrm{\alpha }$
>
Design the LQR controller
>
${W}{:=}{\mathrm{Record}}{}\left({Q}{=}\left[\begin{array}{cc}{1.}& {0.}\\ {0.}& {0.}\end{array}\right]{,}{R}{=}\left[\begin{array}{c}{1.27083333333333}\end{array}\right]\right)$ (5.1.2)
>
${\mathrm{Kc5}}{,}{\mathrm{Kr5}}{:=}\left[\begin{array}{cc}{0.887065525145487}& {0.444353124522434}\end{array}\right]{,}\left[\begin{array}{c}{0.887065525145487}\end{array}\right]$ (5.1.3)
Design the Kalman observer
> $\mathrm{G1}≔\mathrm{LinearAlgebra}:-\mathrm{IdentityMatrix}\left(\mathrm{sys5}:-\mathrm{statecount}\right):$
>
> $\mathrm{H1}≔\mathrm{Matrix}\left(\left[\left[1,0\right]\right]\right):$
> $\mathrm{R1}≔\mathrm{Matrix}\left(\left[\left[2\right]\right]\right):$
>
${\mathrm{Kobs}}{:=}\left[\begin{array}{c}{1.36803398874989}\\ {0.500000000000000}\end{array}\right]$ (5.1.4)
Obtain the LQG controller.
>
> $\mathrm{PrintSystem}\left(\mathrm{LQG}\right)$
$\left[\begin{array}{l}{\mathbf{State Space}}\\ {\mathrm{continuous}}\\ {\mathrm{1 output\left(s\right); 2 input\left(s\right); 2 state\left(s\right)}}\\ {\mathrm{inputvariable}}{=}\left[{\mathrm{y1_ref}}{}\left({t}\right){,}{\mathrm{y1_meas}}{}\left({t}\right)\right]\\ {\mathrm{outputvariable}}{=}\left[{\mathrm{u1}}{}\left({t}\right)\right]\\ {\mathrm{statevariable}}{=}\left[{\mathrm{x1_est}}{}\left({t}\right){,}{\mathrm{x2_est}}{}\left({t}\right)\right]\\ {\mathrm{a}}{=}\left[\begin{array}{cc}{-}{3.1421650390408695}& {0.1112937509551315}\\ {-}{1.3870655251454873}& {-}{0.44435312452243425}\end{array}\right]\\ {\mathrm{b}}{=}\left[\begin{array}{cc}{1.7741310502909748}& {1.368033988749895}\\ {0.8870655251454874}& {0.5}\end{array}\right]\\ {\mathrm{c}}{=}\left[\begin{array}{cc}{-}{0.8870655251454874}& {-}{0.44435312452243425}\end{array}\right]\\ {\mathrm{d}}{=}\left[\begin{array}{cc}{0.887065525145487421}& {0}\end{array}\right]\end{array}\right$ (5.1.5)
Obtain the LQG control system closed-loop equations.
>
> $\mathrm{PrintSystem}\left(\mathrm{clLQG}\right)$
$\left[\begin{array}{l}{\mathbf{State Space}}\\ {\mathrm{continuous}}\\ {\mathrm{1 output\left(s\right); 1 input\left(s\right); 4 state\left(s\right)}}\\ {\mathrm{inputvariable}}{=}\left[{\mathrm{y1_ref}}{}\left({t}\right)\right]\\ {\mathrm{outputvariable}}{=}\left[{\mathrm{y1}}{}\left({t}\right)\right]\\ {\mathrm{statevariable}}{=}\left[{\mathrm{x1}}{}\left({t}\right){,}{\mathrm{x2}}{}\left({t}\right){,}{\mathrm{x1_est}}{}\left({t}\right){,}{\mathrm{x2_est}}{}\left({t}\right)\right]\\ {\mathrm{a}}{=}\left[\begin{array}{cccc}{0}& {1}& {-1.77413105029097484}& {-0.888706249044868501}\\ {0}& {0}& {-0.887065525145487421}& {-0.444353124522434251}\\ {1.36803398874989490}& {0.}& {-3.14216503904086952}& {0.111293750955131499}\\ {0.500000000000000000}& {0.}& {-1.38706552514548731}& {-0.444353124522434251}\end{array}\right]\\ {\mathrm{b}}{=}\left[\begin{array}{c}{1.77413105029097484}\\ {0.887065525145487421}\\ {1.77413105029097484}\\ {0.887065525145487421}\end{array}\right]\\ {\mathrm{c}}{=}\left[\begin{array}{cccc}{1.0}& {0.0}& {0.0}& {0.0}\end{array}\right]\\ {\mathrm{d}}{=}\left[\begin{array}{c}{0.0}\end{array}\right]\end{array}\right$ (5.1.6)
PIDClosedLoop
The PIDClosedLoop command calculates the closed-loop system equations of a feedback system with the controller Gc located in the forward path before the plant Gp and inside the feedback loop. The feedback system has unity negative feedback. This is a basic feedback structure commonly used for P/PI/PID control, hence the name of the command.
>
>
>
> $\mathrm{PrintSystem}\left(\mathrm{cl}\right)$
$\left[\begin{array}{l}{\mathbf{Transfer Function}}\\ {\mathrm{continuous}}\\ {\mathrm{1 output\left(s\right); 1 input\left(s\right)}}\\ {\mathrm{inputvariable}}{=}\left[{\mathrm{y1_ref}}{}\left({s}\right)\right]\\ {\mathrm{outputvariable}}{=}\left[{\mathrm{y1}}{}\left({s}\right)\right]\\ {{\mathrm{tf}}}_{{1}{,}{1}}{=}\frac{{{\mathrm{\omega }}}^{{2}}{}{A}{}{\mathrm{kp}}{}{s}{+}{{\mathrm{\omega }}}^{{2}}{}{A}{}{\mathrm{ki}}}{{{s}}^{{3}}{+}{2}{}{\mathrm{\omega }}{}{\mathrm{\theta }}{}{{s}}^{{2}}{+}\left({{\mathrm{\omega }}}^{{2}}{}{A}{}{\mathrm{kp}}{+}{{\mathrm{\omega }}}^{{2}}\right){}{s}{+}{{\mathrm{\omega }}}^{{2}}{}{A}{}{\mathrm{ki}}}\end{array}\right$ (5.2.1)
>
> $\mathrm{PrintSystem}\left(\mathrm{clnum}\right)$
$\left[\begin{array}{l}{\mathbf{Transfer Function}}\\ {\mathrm{continuous}}\\ {\mathrm{1 output\left(s\right); 1 input\left(s\right)}}\\ {\mathrm{inputvariable}}{=}\left[{\mathrm{y1_ref}}{}\left({s}\right)\right]\\ {\mathrm{outputvariable}}{=}\left[{\mathrm{y1}}{}\left({s}\right)\right]\\ {{\mathrm{tf}}}_{{1}{,}{1}}{=}\frac{{0.7200000000}{}{s}{+}{0.1440000000}}{{{s}}^{{3}}{+}{4.800000000}{}{{s}}^{{2}}{+}{2.160000000}{}{s}{+}{0.1440000000}}\end{array}\right$ (5.2.2)
Use the PIDClosedLoop command to access the controller output node through the augment_output option.
>
> $\mathrm{PrintSystem}\left(\mathrm{claug}\right)$
$\left[\begin{array}{l}{\mathbf{Transfer Function}}\\ {\mathrm{continuous}}\\ {\mathrm{2 output\left(s\right); 1 input\left(s\right)}}\\ {\mathrm{inputvariable}}{=}\left[{\mathrm{y1_ref}}{}\left({s}\right)\right]\\ {\mathrm{outputvariable}}{=}\left[{\mathrm{y1}}{}\left({s}\right){,}{\mathrm{u1}}{}\left({s}\right)\right]\\ {{\mathrm{tf}}}_{{1}{,}{1}}{=}\frac{{{\mathrm{\omega }}}^{{2}}{}{A}{}{\mathrm{kp}}{}{s}{+}{{\mathrm{\omega }}}^{{2}}{}{A}{}{\mathrm{ki}}}{{{s}}^{{3}}{+}{2}{}{\mathrm{\omega }}{}{\mathrm{\theta }}{}{{s}}^{{2}}{+}\left({{\mathrm{\omega }}}^{{2}}{}{A}{}{\mathrm{kp}}{+}{{\mathrm{\omega }}}^{{2}}\right){}{s}{+}{{\mathrm{\omega }}}^{{2}}{}{A}{}{\mathrm{ki}}}\\ {{\mathrm{tf}}}_{{2}{,}{1}}{=}\frac{{\mathrm{kp}}{}{{s}}^{{3}}{+}\left({2}{}{\mathrm{kp}}{}{\mathrm{\omega }}{}{\mathrm{\theta }}{+}{\mathrm{ki}}\right){}{{s}}^{{2}}{+}\left({2}{}{\mathrm{ki}}{}{\mathrm{\omega }}{}{\mathrm{\theta }}{+}{\mathrm{kp}}{}{{\mathrm{\omega }}}^{{2}}\right){}{s}{+}{{\mathrm{\omega }}}^{{2}}{}{\mathrm{ki}}}{{{s}}^{{3}}{+}{2}{}{\mathrm{\omega }}{}{\mathrm{\theta }}{}{{s}}^{{2}}{+}\left({{\mathrm{\omega }}}^{{2}}{}{A}{}{\mathrm{kp}}{+}{{\mathrm{\omega }}}^{{2}}\right){}{s}{+}{{\mathrm{\omega }}}^{{2}}{}{A}{}{\mathrm{ki}}}\end{array}\right$ (5.2.3)
StateFeedbackClosedLoop
The StateFeedbackClosedLoop command calculates the closed-loop system equations of a plant with state feedback. This command requires the state feedback gain matrix Kc, but can also accept the feed-forward matrix Kr to support a state feedback structure with control law ${u}_{c}=-{K}_{c}.x+{K}_{r}.r.$
>
${\mathrm{Kc}}{,}{\mathrm{Kr}}{:=}\left[\begin{array}{ccc}{3.522839239}& {0.6091131254}& {1.639596973}\end{array}\right]{,}\left[\begin{array}{ccc}{2.54751899020608}& {0.536319787411807}& {-}{0.871519654544186}\end{array}\right]$ (5.3.1)
>
> $\mathrm{PrintSystem}\left(\mathrm{clsf}\right)$
$\left[\begin{array}{l}{\mathbf{State Space}}\\ {\mathrm{continuous}}\\ {\mathrm{3 output\left(s\right); 3 input\left(s\right); 3 state\left(s\right)}}\\ {\mathrm{inputvariable}}{=}\left[{\mathrm{y1_ref}}{}\left({t}\right){,}{\mathrm{y2_ref}}{}\left({t}\right){,}{\mathrm{y3_ref}}{}\left({t}\right)\right]\\ {\mathrm{outputvariable}}{=}\left[{\mathrm{y1}}{}\left({t}\right){,}{\mathrm{y2}}{}\left({t}\right){,}{\mathrm{y3}}{}\left({t}\right)\right]\\ {\mathrm{statevariable}}{=}\left[{\mathrm{x1}}{}\left({t}\right){,}{\mathrm{x2}}{}\left({t}\right){,}{\mathrm{x3}}{}\left({t}\right)\right]\\ {\mathrm{a}}{=}\left[\begin{array}{ccc}{-}{9.568517717}& {0.17266062380000013}& {-}{1.918790919}\\ {-}{31.705553151}& {-}{1.4820181286}& {-}{15.756372757}\\ {-}{15.614196195000002}& {1.9544343730000002}& {-}{1.1979848650000005}\end{array}\right]\\ {\mathrm{b}}{=}\left[\begin{array}{ccc}{7.642556970618246}& {1.6089593622354208}& {-}{2.614558963632558}\\ {22.92767091185474}& {4.826878086706262}& {-}{7.843676890897674}\\ {12.737594951030411}& {2.681598937059035}& {-}{4.35759827272093}\end{array}\right]\\ {\mathrm{c}}{=}\left[\begin{array}{ccc}{1.0}& {0.0}& {0.0}\\ {0.0}& {1.0}& {0.0}\\ {0.0}& {0.0}& {1.0}\end{array}\right]\\ {\mathrm{d}}{=}\left[\begin{array}{ccc}{0.0}& {0.0}& {0.0}\\ {0.0}& {0.0}& {0.0}\\ {0.0}& {0.0}& {0.0}\end{array}\right]\end{array}\right$ (5.3.2)
ReduceSystem
The ReduceSystem command simplifies a state-space system by detecting and removing its structural unobservable and uncontrollable states. These struturally non-minimal states may occur, for example, as a result of extracting a subsystem from a subset of its inputs and/or outputs. The reduced state-space system is structurally minimal, that is, contains a subset of the original system states but preserves its state structure and variable names. The input/output response of the reduced system is equivalent to the transfer function of the original system.
>
> $\mathrm{PrintSystem}\left(\mathrm{sys6}\right)$
$\left[\begin{array}{l}{\mathbf{State Space}}\\ {\mathrm{continuous}}\\ {\mathrm{2 output\left(s\right); 3 input\left(s\right); 4 state\left(s\right)}}\\ {\mathrm{inputvariable}}{=}\left[{\mathrm{u1}}{}\left({t}\right){,}{\mathrm{u2}}{}\left({t}\right){,}{\mathrm{u3}}{}\left({t}\right)\right]\\ {\mathrm{outputvariable}}{=}\left[{\mathrm{y1}}{}\left({t}\right){,}{\mathrm{y2}}{}\left({t}\right)\right]\\ {\mathrm{statevariable}}{=}\left[{\mathrm{x1}}{}\left({t}\right){,}{\mathrm{x2}}{}\left({t}\right){,}{\mathrm{x3}}{}\left({t}\right){,}{\mathrm{x4}}{}\left({t}\right)\right]\\ {\mathrm{a}}{=}\left[\begin{array}{cccc}{\mathrm{\delta }}& {0}& {-}{\mathrm{\phi }}& {\mathrm{\sigma }}\\ {3}& {0}& {0}& {3}\\ {1}& {0}& {-1}& {1}\\ {0}& {0}& {0}& {0}\end{array}\right]\\ {\mathrm{b}}{=}\left[\begin{array}{ccc}{5}& {z}& {-1}\\ {1}& {3}& {-}{x}\\ {0}& {2}& {y}\\ {0}& {0}& {z}\end{array}\right]\\ {\mathrm{c}}{=}\left[\begin{array}{cccc}{1}& {0}& {3}& {5}\\ {-3}& {0}& {\mathrm{\sigma }}& {7}\end{array}\right]\\ {\mathrm{d}}{=}\left[\begin{array}{ccc}{1}& {0}& {0}\\ {0}& {1}& {1}\end{array}\right]\end{array}\right$ (5.4.1)
Get a subsystem with inputs u1(t) and u2(t). This subsystem is uncontrollable and unobservable.
>
${\mathrm{false}}{,}{\mathrm{false}}$ (5.4.2)
Get a structurally minimal system removing the non-minimal (structural uncontrollable and unobservable) states.
>
${\mathrm{true}}{,}{\mathrm{true}}$ (5.4.3)
> $\mathrm{PrintSystem}\left(\mathrm{redsys}\right)$
$\left[\begin{array}{l}{\mathbf{State Space}}\\ {\mathrm{continuous}}\\ {\mathrm{2 output\left(s\right); 2 input\left(s\right); 2 state\left(s\right)}}\\ {\mathrm{inputvariable}}{=}\left[{\mathrm{u1}}{}\left({t}\right){,}{\mathrm{u2}}{}\left({t}\right)\right]\\ {\mathrm{outputvariable}}{=}\left[{\mathrm{y1}}{}\left({t}\right){,}{\mathrm{y2}}{}\left({t}\right)\right]\\ {\mathrm{statevariable}}{=}\left[{\mathrm{x1}}{}\left({t}\right){,}{\mathrm{x3}}{}\left({t}\right)\right]\\ {\mathrm{a}}{=}\left[\begin{array}{cc}{\mathrm{\delta }}& {-}{\mathrm{\phi }}\\ {1}& {-1}\end{array}\right]\\ {\mathrm{b}}{=}\left[\begin{array}{cc}{5}& {z}\\ {0}& {2}\end{array}\right]\\ {\mathrm{c}}{=}\left[\begin{array}{cc}{1}& {3}\\ {-3}& {\mathrm{\sigma }}\end{array}\right]\\ {\mathrm{d}}{=}\left[\begin{array}{cc}{1}& {0}\\ {0}& {1}\end{array}\right]\end{array}\right$ (5.4.4)
Verify the input/output response is not affected by the removed states.
>
${\mathrm{true}}$ (5.4.5)
See Also
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2016-10-25 06:44:32
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https://dsp.stackexchange.com/questions/462/why-did-my-sine-wave-turn-into-a-square-wave-when-written-to-a-wav-file-in-octav/463
|
# Why did my sine wave turn into a square wave when written to a WAV file in Octave?
I am trying to use Octave to generate a pure sine wave. The code for the same is as follows:
x = 10.*sin(2*pi*(300/16000)*(0:1:400));
The sampling rate is 16000Hz, the sine wave is at 300 Hz. I write the above wave to a file using wavwrite like so:
wavwrite(x, 16000, 16, "temp.wav")
When I try to read it back into a variable, like so:
y = wavread('temp.wav');, I get square waves upon plotting y.
I have checked the sine wave and the period indicates a frequency of 300Hz.
How can a pure sine wave become a square wave on simply writing and reading? Or am I going wrong somewhere?
x is being clipped and that is why y looks like a square wave. When writing x to disk using wavwrite, the samples of x are stored in 16 bits Q15 fixed-point format. That means your data must be in range -1 to +1 (in principle +1 minus one lsb). Therefore, x must be normalized to be in this range before calling wavwrite in order to avoid clipping.
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2019-10-18 20:09:48
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{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7828097939491272, "perplexity": 1032.1490719049536}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986684425.36/warc/CC-MAIN-20191018181458-20191018204958-00378.warc.gz"}
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# Iowa ($$) vs. BYU new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Intern Joined: 19 Sep 2012 Posts: 25 Concentration: Marketing GMAT 1: 680 Q45 V38 GPA: 3.6 WE: Marketing (Advertising and PR) Followers: 0 Kudos [?]: 5 [0], given: 0 BYU or Iowa [#permalink] ### Show Tags 19 Feb 2013, 13:50 Given the two options what would you choose? BYU No scholarship (may come, but don't know anything yet) Tuition 10,600 per year. Ranked mid 30s Average Salary: 90,000 Iowa Full Tuition Scholarship (plus health insurance) Ranked mid 40s Average Salary: 80,000 I attended BYU for my undergrad. I'm a Mormon so the BYU honor code isn't an issue for me. I will be studying Marketing. Manager Joined: 29 Nov 2011 Posts: 114 Location: United States GMAT 1: 760 Q50 V45 Followers: 0 Kudos [?]: 22 [0], given: 11 Re: Iowa ($$$) vs. BYU [#permalink] ### Show Tags 20 Feb 2013, 11:46 I'd say go to BYU. The cost of attendance is low even without a scholarship, and salaries are higher which will make up for the additional cost pretty quickly. Intern Joined: 19 Sep 2012 Posts: 25 Concentration: Marketing GMAT 1: 680 Q45 V38 GPA: 3.6 WE: Marketing (Advertising and PR) Followers: 0 Kudos [?]: 5 [0], given: 0 Re: Iowa ($$) vs. BYU [#permalink] ### Show Tags 20 Feb 2013, 11:50 Thanks for the input. Manager Joined: 13 Aug 2009 Posts: 223 Location: United States GMAT 1: 780 Q51 V46 GMAT 2: 800 Q51 V51 GRE 1: 340 Q170 V170 Followers: 29 Kudos [?]: 135 [0], given: 81 Re: BYU or Iowa [#permalink] ### Show Tags 20 Feb 2013, 17:21 I would definitely choose Iowa, but that's only because I really, really love corn. In all seriousness, you're in a great position--congratulations on the admits and scholarship offer! At this stage, I wouldn't worry at all about the cost, since the two are ultimately so similar in your case, and an extra 20,000 is a rounding error when you think about it over the course of your career. And I wouldn't obsess over the average salary data or the rankings, either. You could argue that BYU maybe has a slightly higher profile nationally, but the differences in salary and rankings aren't all that huge. I'm not an expert on BYU or Iowa's business school, but I have a very positive impression of both places in terms of the quality of the teaching. I don't think you can go too wrong with either place, but I'm willing to wager that you'll encounter substantially different environments at the two schools, and there are definitely differences in the curriculum, culture, and types of firms that recruit on each campus. Honestly, I think you should focus on the qualitative differences, and don't worry so much about the data from US News or Businessweek. Go wherever you think you'll have the more enjoyable learning experience, and you'll be fine. But maybe it all comes down to whether you want a change of scenery, since you already did your undergrad at BYU? Have you been to Iowa City? It's roughly the same size as Provo, but the similarities pretty much end right there. I like Iowa City, and it's a great place if you want a good, solid, tight-knit, Midwestern-type community. But if you plan to stay in Utah or somewhere else out west, then BYU might make more sense. Sorry, I'm rambling. Good luck with the decision, and let us know what you decide to do! _________________ Helping students kick the GMAT in the nuts since 2002... www.gmatninja.com. Intern Joined: 19 Sep 2012 Posts: 25 Concentration: Marketing GMAT 1: 680 Q45 V38 GPA: 3.6 WE: Marketing (Advertising and PR) Followers: 0 Kudos [?]: 5 [0], given: 0 Re: BYU or Iowa [#permalink] ### Show Tags 20 Feb 2013, 17:29 Thanks for the insights. My only real concern with BYU is that it is the same school as my undergrad. I graduated from a different program in my undergrad so my network at BYU will still be mostly new, but does it look bad/weird to a potential employer to have both degrees from the same school? I have heard that this can be a negative, but I'm not sure how much of an impact it would really have. Manager Joined: 13 Aug 2009 Posts: 223 Location: United States GMAT 1: 780 Q51 V46 GMAT 2: 800 Q51 V51 GRE 1: 340 Q170 V170 Followers: 29 Kudos [?]: 135 [0], given: 81 Re: BYU or Iowa [#permalink] ### Show Tags 20 Feb 2013, 17:40 I don't think that attending the same institution is inherently a negative in the eyes of employers. If anything, there might be a few people out there who think that your loyalty is endearing. The only real issue is if your potential employers aren't familiar with BYU--or if they aren't particularly impressed by BYU for whatever silly reason. I think that might depend on the industry and the region of the country, to some degree. If you're at all interested in living in the Midwest or if you're interested in working in, say, agribusiness or insurance, you might as well diversify and go with Iowa. Otherwise? If you think you'd be happier at BYU, I really don't see much of a downside. And the skiing in Utah is way better than the skiing in Iowa. _________________ Helping students kick the GMAT in the nuts since 2002... www.gmatninja.com. Re: BYU or Iowa [#permalink] 20 Feb 2013, 17:40 Similar topics Replies Last post Similar Topics: 1 BYU with 16K vs GWU with 60K 7 30 Apr 2016, 22:06 4 Marriott (BYU) v/s Sauder (UBC, Canada) 11 10 Feb 2015, 22:19 5 Iowa Tippie vs Georgia Terry ($$$) 3 10 Feb 2015, 11:37
1 BYU vs Wisconsin 3 10 Mar 2013, 22:09
BYU or Iowa 0 20 Feb 2013, 17:40
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2016-05-24 18:37:39
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{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.29651689529418945, "perplexity": 3599.3498979304904}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-22/segments/1464049272823.52/warc/CC-MAIN-20160524002112-00047-ip-10-185-217-139.ec2.internal.warc.gz"}
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https://codegolf.stackexchange.com/questions/77221/create-your-own-gym-exercise-and-follow-along-irl
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# Create your own Gym Exercise and follow along irl
Let's do an exercise together, shall we? Simply follow the movements of the stickman. What stickman you ask? The one we are about to create!
Make a program that allows for a string-input only containing integers, and outputs the following nine stick-figures with an interval in between:
@ \@/ |@_ _@| @/ \@ |@| \@\ /@/
/|\ | | | | | | | |
/ \ / \ / \ / \ /|\ /|\ / \ / \ / \
1 2 3 4 5 6 7 8 9
This will result in an animation, which we can then follow along irl.
Here is an example of the output created when the input is "123245762":
Some rules:
• This is tagged , so shortest code in bytes wins
• The input is a string only containing [1-9] with a length potentially varying from 2 to 100 characters **
• The animation must be an endless loop
• The interval time must be 750 ms
• No duplicated adjacent integers are allowed in the input - this also includes the first and last integers of the input (see the seventh and eight test cases) **
** How it fails in case of invalid input is completely up to you, but it should be clear that it fails. It could throw an exception; simply return at the start; reboot your PC; delete it's own source-/compile-code; hack into the military and shoots a rocket to the compiling GPS-location. Your call. EDIT: It's not allowed to simply display a (correct / half) animation as failing input. It must be clear that something is wrong with the input for the failing testcases below. Thanks to @user81655 for bringing this to my attention, hence the edit.
Subrules:
• Please post the gif for one of the (non-failing) test cases below in your answer, which you can make very easy with the following program: screentogif.codeplex.com (Any other gif-creating program is also allowed of course.)
• Bonus points if you actually do the exercise alongside the animation irl when it's finished. ;)
Correct testcases:
1. 123245762
2. 65
3. 121234346565879879132418791576
Failing testcases:
1. 2 // Too few characters
2. 7282918274959292747383785189478174826894958127476192947512897571298593912374857471978269591928974518735891891723878 // Too much characters
3. 1232405762 // Containing invalid input (the 0)
4. 112212 // Fails because of the 11 and 22 present
5. 1232457621 // Fails because of the starting and leading 1
• IMO this is a borderline dupe of this – Peter Taylor Apr 6 '16 at 11:44
• BTW, making edits after a challenge is posted is discouraged. – Blue Apr 6 '16 at 19:17
• imo the input validation ruins the challenge. – FlipTack Feb 12 '17 at 19:36
• @FlipTack You're indeed right. I should have just kept with the challenge and assume all input is valid. This was my very first question here on PPCG. Bit too late to change it now, though. – Kevin Cruijssen Feb 13 '17 at 7:44
## Pyth, 114 bytes
.V0j@cv.Z"xÚí» À0 DW¹NM@+Üñ\">íÂA¸êÄÓw»3±2þ&'NövfAé8é~0,p'ÆìÞúr_'¥,d! YÈBíéqs"3tv@zb.d.75
Try it here
(No pausing, non-infinite)
• Hmm, it isn't possible to return/throw or loop infinitely in Pyth? Or it will take too much bytes/is too time-consuming to make? (Currently it works the same for all correct testcases, as well as all failing testcases. And it also adds in the output for 0.) Regardless, thanks for the entry. Definitely one of the most unreadable code-snippets for this entry is my guess. :) – Kevin Cruijssen Apr 6 '16 at 11:02
• The online interpreter doesn't flush IO during a sleep so it isn't noticed. You did say you wanted an infinite loop so I'm not sure what you're asking there. "How it fails is up to you" - I'm doing undefined behavior for this; the code might do anything (even though it SHOULD behave the same) Also I'm expecting this to be relatively readable compared to some possible entries – Blue Apr 6 '16 at 11:06
# SpecBAS - 387 bytes
1 DIM m$=" @"#13"/|\"#13"/ \","\@/"#13" |"#13"/ \","|@_"#13" |"#13"/ \","_@|"#13" |"#13"/ \","@/"#13" |"#13"/|\"," \@"#13" |"#13"/|\","|@|"#13" |"#13"/ \","\@\"#13" |"#13"/ \","/@/"#13" |"#13"/ \" 2 INPUT a$: o=0
3 IF LEN a$<2 OR LEN a$>100 THEN 10
4 FOR i=1 TO LEN a$5 n=VAL(a$(i))
6 IF n=0 OR n=o THEN 10
7 CLS : ?m\$(n): o=n: WAIT 750
8 NEXT i
9 GO TO 4
10 CLS : ?" @"#13"-O-"#13"/ \"#13"FAT"
Keeps looping until you press ESC. Failure to exercise properly (incorrect input - in this example a 0 as one of the steps) leads to fatness. The GIF loops, in the program it just stops at that point.
#13 is the SpecBAS equivalent to \n and lets you include line feed in strings.
• "FAT" :D ...... – Adam Varhegyi Feb 16 '18 at 14:27
## JavaScript (ES6), 165 bytes
f=s=>{n=s[0];e.textContent=' @ \\@/|@__@|@/ \\@|@|\\@\\/@/'.substr(n*3-3,3)+(n>1?
|
:
/|\\
)+(n<5|n>6?'/ \\':'/|\\');s=s.slice(1)+n;s[0]-n&&setTimeout(f,750,s)}
f("123245762")
<pre id=e>
# JavaScript (ES6), 210 bytes
s=>setInterval(_=>(c=console).clear(i=0)&c.log(, @
/|\\
/ \\,\\@/
|
/ \\,|@_
|
/ \\,_@|
|
/ \\,@/
|
/|\\, \\@
|
/|\\,|@|
|
/ \\,\\@\\
|
/ \\,/@/
|
/ \\.split,[s[i++%s.length]]),750)
• How does this handle the rules about excluding 0 and duplicates? – Morgan Thrapp Apr 6 '16 at 13:48
• @Morgan'Venti'Thrappuccino I was under the impression that invalid inputs don't need defined behaviour, since it says How it fails in case of invalid input is completely up to you. I just handle invalid input by displaying an animation. :P None of the other existing answers checked this either, however the answer the OP just posted does, so maybe he did want us to handle invalid input. – user81655 Apr 6 '16 at 14:01
• "I just handle invalid input by displaying an animation. :P" Lol.. That's one way of looking at it. xD I'll edit the question. – Kevin Cruijssen Apr 6 '16 at 14:43
# Mathematica, 252 bytes
i=0;Dynamic[Uncompress["1:eJxTTMoPSuNkYGAoZgESPpnFJcFCQIaCQ4yBoZF+TUwMmFaIiQELx8Q46IMEFGrgwoJA4RqHeCyi8Q41aKICQFEUA2qg5gIlHdCEIeaimyAMcQTEWWj26aO7DQDaqDEh"][[FromDigits[#~StringTake~{i=i~Mod~StringLength@#+1}]]],UpdateInterval->3/4,TrackedSymbols->{}]&
Would be nice if someone could create a GIF. Run in a notebook.
# Python3, 338 bytes
import os,time
p=" @ \n/|\\\n/ \\","\\@/\n | \n/ \\","|@_\n | \n/ \\","_@|\n | \n/ \\","@/ \n | \n/|\\"," \\@\n | \n/|\\","|@|\n | \n/ \\","\\@\\\n | \n/ \\","/@/\n | \n/ \\"
i=input()
for j in range(len(i)):(i[j]in"123456789"and i[j]!=i[(j+1)%len(i)])or exit()
while 1:[[time.sleep(0.75),os.system("clear"),print(p[int(j)-1])]for j in i]
# Java 8, 663636634631596355 354 bytes
Just for the lols I tried to make the program in Java. Admittedly I'm pretty bad at golfing and regexes, so it can most likely be golfed (a lot?) more. Nevertheless, here it is in Java 7.
Now almost two years later and I almost halved the code in Java 8. Why did I ever made those rules about validating the input and requiring a full program, though... >.> I hate my past self now..
interface M{static void main(String[]a)throws Exception{if(!a[0].matches("[1-9]{2,100}")|a[0].matches("(.).*\\1|.*(.)\\2.*"))return;for(;;)for(int c:a[0].getBytes()){c-=48;System.out.printf("%s%n%s%n%s%n",c<2?" @ ":c<3?"\\@/":c<4?"|@_":c<5?"_@|":c<6?"@/ ":c<7?" \\@":c<8?"|@|":c<9?"\\@\\":"/@/",c<2?"/|\\":" | ",c%7>4?"/|\\":"/ \\");Thread.sleep(750);}}}
Explanation:
Try it online. (After it has timed out after 60 sec.)
interface M{ // Class
static void main(String[]a) // Mandatory main-method
throws Exception{ // Required throws for the Thread.sleep
if(!a[0].matches("[1-9]{2,100}")
// Validate 2-100 integers only containing 1-9
|a[0].matches("(.).*\\1|.*(.)\\2.*")
// Validate no adjacent duplicated char (with wrap-around)
return; // If either isn't valid, stop the program
for(;;) // Loop indefinitely
for(int c:a[0].getBytes()){// Inner loop over the characters of the input
c-=48; // Convert character-code to integer
System.out.printf("%s%n%s%n%s%n",
// Print:
c<2?" @ ":c<3?"\\@/":c<4?"|@_":c<5?"_@|":c<6?"@/ ":c<7?" \\@":c<8?"|@|":c<9?"\\@\\":"/@/",
// The top part of the stick figure
c<2?"/|\\":" | " // The middle part of the stick figure
c%7>4?"/|\\":"/ \\"); // The bottom part of the stick figure
Thread.sleep(750);}}} // Sleep 750 ms
Gif:
(Note: Old gif, since it's clearly jdk1.8+ now.)
• Usually we say Java 7 to refer to this version of Java as 1.7 can confuse some to think that it is Java 1 revision 7 – GamrCorps Apr 6 '16 at 15:11
• The long if statement could be turned into something like p(new String[]{" @ ",y,"|@_", ... }[c-49]);if(c==49){p(x);l();}else if(c==53){w();p(x);}else d();. Also I think that defining l and w wastes characters, now that they're only used twice. Consider letting s instead be char[] s=a[0].toCharArray()`; all the other operations with it are much shorter, then. – Alex Meiburg Aug 1 '16 at 17:13
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2020-05-30 22:41:30
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{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.24781998991966248, "perplexity": 4801.591646611907}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347410352.47/warc/CC-MAIN-20200530200643-20200530230643-00184.warc.gz"}
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http://wiki.tekkotsu.org/index.php?title=RobotMath:_Solving_Triangles,_Part_1&oldid=2300
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RobotMath: Solving Triangles, Part 1
Contents
Significance of Triangles
Head constructed from a triangular mesh
Triangles are everywhere. They are the subject of trigonometry (Greek for measurement of triangles), which underlies the higher mathematics used in engineering, physics, and computer graphics. The human head at right is constructed from a mesh of triangles, as are all the characters, objects, and scenery in your favorite computer game.
The triangle below has sides of length a, b, and c and angles A, B, and C. You can see that angle C is formed by sides a and b, placing it opposite side c. Similarly, angle B is formed by sides a and c and is opposite side b.
Given enough information about a triangle, it is possible to calculate all the remaining values. For example, if we know the lengths a and b and the angle C, we can calculate the angles A and B and the length c. Or, if we know the length c and the angles A and B, we can calculate the lengths a and b and the angle C. Some of these calculations are easier than others, but none are terribly difficult. Many simple robotics reduce to solving a triangle, as you'll see below.
Here's a simple rule that will help you solve for the angles: The angles of a triangle always sum to 180°.
$A + B + C = 180^\circ = \pi\,\mbox{radians}$
So if you know any two angles of a triangle, you immediately know the third angle as well.
Solving A Robot Arm
The robot arm below has two joints: a shoulder and an elbow, and two links: an upper arm of length l1, and a lower arm, shown in blue, of length l2. The tip of the arm is called the end effector. It might be a gripper, or a tool such as a screwdriver or a welding torch, but for our purposes we will treat it as just a point.
We denote the shoulder angle by θ1 and the elbow angle by θ2. We adopt the convention that the shoulder joint is at the origin of our coordinate system, and a joint angle of zero means the associated link is pointing straight out, not bent. So when θ1 = 0° the upper arm is pointing straight ahead along the x axis. When θ2 = 0°, the lower arm is pointing in the same direction as the upper arm and the entire arm can be viewed as one segment of length l1+l2 pointing in direction θ1.
Problem: Suppose we rotate the shoulder joint to the right so that the upper arm is at angle θ1 relative to the x axis. Then we rotate the elbow joint to the left so that the lower arm is at angle θ2 relative to the upper arm. We will choose θ2 such that the end effector at the tip of the arm lies on the x axis, as shown. (You'll learn later how to calculate θ2 to put the end-effector where you want it.) What is the value of the angle β as a function of θ1 and θ2? Note that angles α and θ2 are supplementary, meaning they sum to 180° or π radians.
Solution:
Our knowledge about the sum of the angles of a triangle tells us that $\theta_1 + \alpha + \beta = \pi \quad$
Therefore $\beta = \pi - \theta_1 - \alpha \quad$
And our knowledge about supplementary angles tells us that $\alpha = \pi - \theta_2 \quad$
Therefore $\beta = \pi - \theta_1 - (\pi-\theta_2) = \theta_2-\theta_1 \quad$
Solving for the lengths of the sides of a triangle, such as side S above, or for the angles when only one angle and two sides are initially known, requires a little more work. The trick is to divide the triangle into two right triangles, as shown below. Right triangles, where one of the angles is known to be 90°, have nice mathematical properties that make them easy to work with. Any triangle can be decomposed into two right triangles by dropping a normal h from a vertex to the opposite side. (The term normal means the line is perpendicular to the line it intersects, i.e., the angle between them is 90°.) We can solve any triangle by decomposing it this way. You'll see how in a later lesson. First we need to learn more about right triangles.
Right Triangles
Right angles have nice mathematical properties that make them easy to work with. The side opposite the right angle is the longest side of the triangle, and is called the hypotenuse. Pick either one of the other two angles and call it θ. The other two sides are called the adjacent side and the opposite side because of their relationship to θ.
The most common way to draw a triangle in a math textbook is with the adjacent side extending from the origin in the positive direction along the x axis, and the opposite side parallel to the positive y axis, as shown above. If the lengths of the sides are x, y, and r, then the vertices are at (0,0), (x,0), and (x,y). The best-known property of right triangles is that they obey the Pythagorean Theorem: the square of the hypotenuse is equal to the sum of the squares of the other two sides.
$r^2 = x^2 + y^{2\quad}$
There are many proofs of the Pythagorean Theorem, but they are too complicated to go into here. See [Wikipedia] for details.
Another important property of right triangles is that the various ratios of their sides x, y, and r are completely determined by θ, so if we know θ plus the length of just one of the sides, we can determine the lengths of the other two sides. To do this, we make use of the sine, cosine, and tangent functions.
Solving Right Triangles with Sine, Cosine, and Tangent
The sine of the angle θ (written sin θ in mathematical shorthand) is determined by drawing a right triangle with angle θ and measuring the ratio of the "opposite" side (i.e., the side opposite θ) to the hypotenuse. The triangle can be of any size; the ratio of the sides will be the same as long as θ is the same.
$\sin\theta = \frac{\mbox{opposite}_\theta}{\mbox{hypotenuse}} = \frac{y}{r}$
Given any right triangle, ff we know r and θ, we can solve for y:
$y = r\sin\theta^{\quad}$
Or if we know y and θ, we can solve for r:
$r = \frac{y}{\sin\theta}$
Actually it's not enough to know θ to solve for y or r: we need to know sin θ. In earlier times people looked up values of trigonometric functions such as sine in lengthy printed tables, but today they are built in to calculators and computers.
The cosine and tangent functions are defined similarly:
$\cos\theta = \frac{\mbox{adjacent}_\theta}{\mbox{hypotenuse}} = \frac{x}{r}$
$\tan\theta = \frac{\mbox{opposite}_\theta}{\mbox{adjacent}_\theta} = \frac{y}{x}$
We don't actually need all three of these functions to solve a right triangle; using just sin and the Pythagorean theorem we can solve for any side given any other side. But it's convenient to have all three functions at our disposal.
Problems:
1. Given x and θ, solve for y using the tangent function.
2. Given x and θ, solve for y using cosine and the Pythagorean theorem.
3. What is the value of sin(θ) divided by cos(θ) ?
4. Harder: given r and tan θ, solve for x.
5. Prove that sin2(θ)+cos2(θ)=1. Hint: start with the Pythagorean theorem, plus x = r cos(θ) and y = r sin(θ).
6. Draw a diagram to help you solve this problem. A robot facing north detects an east-west wall 3 meters ahead. If it turns by 30 degrees, how far can it travel before it hits the wall?
7. A flying robot finds itself 1000 meters above a dry lakebed. If it cuts its engines and descends on a 7° glidepath, how far will it travel over the ground before it lands?
The Third Angle
The functions sin(θ), cos(θ), and tan(θ) are defined based on a triangle containing an angle θ and another angle equal to 90°, making it a right triangle. What about the third angle, which we'll call φ?
The hypotenuse, which is the side opposite the right angle, is still r, but notice that the other side adjacent to θ (side x) is opposite to φ, and the side opposite θ (side y) is adjacent to φ. Therefore:
$\sin\phi = \frac{\mbox{opposite}_\phi}{\mbox{hypotenuse}} = \frac{x}{r} = \cos\theta$
$\cos\phi = \frac{\mbox{adjacent}_\phi}{\mbox{hypotenuse}} = \frac{y}{r} = \sin\theta$
$\tan\phi = \frac{\mbox{opposite}_\phi}{\mbox{adjacent}_\phi} = \frac{x}{y} = \frac{1}{\tan\theta} = \cot\theta$
Here, cot is the abbreviation for the cotangent function, which is the reciprocal of the tangent. Less commonly seen are the reciprocal of cosine, called the secant, and the reciprocal of sine, called the cosecant.
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2019-05-21 15:27:24
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|
https://www.digi.com/resources/documentation/digidocs/embedded/dey/3.2/cc6plus/yocto_t_change-partition-table_emmc
|
You can customize the partition table of a Digi embedded module using a series of U-Boot commands.
The eMMC is partitioned using a GPT. The partition table (partition names, sizes, and UUIDs) is described in a U-Boot environment variable.
The syntax for describing a GPT as a string is defined at doc/README.gpt file in the U-Boot source code:
`````` Format of partitions layout:
"partitions=uuid_disk=...;name=u-boot,size=60MiB,uuid=...;
name=kernel,size=60MiB,uuid=...;"
or
"partitions=uuid_disk=${uuid_gpt_disk};name=${uboot_name},
size=${uboot_size},uuid=${uboot_uuid};"
The fields 'name' and 'size' are mandatory for every partition.
The field 'start' is optional.
The fields 'uuid' and 'uuid_disk' are optional if CONFIG_RANDOM_UUID is
enabled. A random uuid will be used if omitted or they point to an empty/
non-existent environment variable. The environment variable will be set to
the generated UUID.``````
A size of "-" (without the MiB suffix) indicates that the partition will use the remaining available space on the media, so the dash is usually used for the last partition in the table.
Once the partition table is defined as a string, you can write the partition table to the media using U-Boot’s `gpt` command.
Digi provides the following predefined variables:
• Partition tables
• `parts_linux`: partition table suitable for storing the Digi Embedded Yocto Linux system
• UUIDs
• `uuid_disk`: predefined UUID for the media
• `part1_uuid..part9_uuid`: predefined UUIDs for the partitions
• Scripts
• `partition_mmc_linux`: script for writing the partition table in variable `parts_linux` to the MMC device index in variable `mmcdev`
## Change the partition table (run-time)
• Edit the partition table of your selected operating system with `env edit` command:
``=> env edit parts_linux``
• Modify the command to fit your desired partition table, but note the following:
• Do not change the names of the system partitions. These partitions are meant to have their original names for system updates to work. For Digi Embedded Yocto these are: safe, linux, recovery, rootfs, update.
• You can resize system partitions, but make sure your system will fit in them.
• The update partition must be approximately the size of the rootfs partition to fit system updates.
• You can add partitions as necessary, but you must assign a different name and UUID.
• (Optional) Set `mmcdev` to the desired media. By default this variable should point to the MMC media index you are booting from (eMMC or microSD). You don’t need to set `mmcdev` unless you want to write the GPT to a specific MMC media other than the one you booted from.
• Save the environment with `saveenv`.
• Run the script that matches your operating system to write the new partition table to the MMC media:
``=> run partition_mmc_linux``
## Change the default partition table (build-time)
You may want to change the default partition table so that all devices ship with the same partition layout. To do that:
• Open your platform’s common header file in U-Boot source code at `include/configs/ccimx6_common.h`.
• Locate the partition table definition for your module’s eMMC capacity (constants `LINUX_4GB_PARTITION_TABLE`, `LINUX_8GB_PARTITION_TABLE` or `LINUX_16GB_PARTITION_TABLE`):
``````#define LINUX_4GB_PARTITION_TABLE \
"\"uuid_disk=${uuid_disk};" \ "start=2MiB," \ "name=linux,size=64MiB,uuid=${part1_uuid};" \
"name=recovery,size=64MiB,uuid=${part2_uuid};" \ "name=rootfs,size=1536MiB,uuid=${part3_uuid};" \
"name=update,size=1536MiB,uuid=${part4_uuid};" \ "name=safe,size=16MiB,uuid=${part5_uuid};" \
"name=safe2,size=16MiB,uuid=${part6_uuid};" \ "name=data,size=-,uuid=${part7_uuid};" \
"\""
#define LINUX_8GB_PARTITION_TABLE \
"\"uuid_disk=${uuid_disk};" \ "start=2MiB," \ "name=linux,size=64MiB,uuid=${part1_uuid};" \
"name=recovery,size=64MiB,uuid=${part2_uuid};" \ "name=rootfs,size=3GiB,uuid=${part3_uuid};" \
"name=update,size=3GiB,uuid=${part4_uuid};" \ "name=safe,size=16MiB,uuid=${part5_uuid};" \
"name=safe2,size=16MiB,uuid=${part6_uuid};" \ "name=data,size=-,uuid=${part7_uuid};" \
"\""
#define LINUX_16GB_PARTITION_TABLE \
"\"uuid_disk=${uuid_disk};" \ "start=2MiB," \ "name=linux,size=64MiB,uuid=${part1_uuid};" \
"name=recovery,size=64MiB,uuid=${part2_uuid};" \ "name=rootfs,size=7GiB,uuid=${part3_uuid};" \
"name=update,size=7GiB,uuid=${part4_uuid};" \ "name=safe,size=16MiB,uuid=${part5_uuid};" \
"name=safe2,size=16MiB,uuid=${part6_uuid};" \ "name=data,size=-,uuid=${part7_uuid};" \
"\""``````
• Edit the appropriate constant in the header file using the syntax described in Change the partition table (run-time). If using DualBoot, the constants are called `LINUX_DUALBOOT_4GB_PARTITION_TABLE`, `LINUX_DUALBOOT_8GB_PARTITION_TABLE`, and `LINUX_DUALBOOT_16GB_PARTITION_TABLE`.
• Save your changes and recompile U-Boot.
• Reset the variable to default value that corresponds to your operating system:
``=> env default parts_linux``
You can also reset the whole environment to default values:
``=> env default -a``
• Save the environment with `saveenv`.
• Reset the device (or verify that `mmcdev` points to the MMC index of the media where the partition table will be written)
• Write the new partition table to the media that corresponds to your operating system:
``=> run partition_mmc_linux``
The `mmcpart`, `mmcroot`, and `recoverycmd` U-Boot variables are used by the `dboot` and `update` commands as well as by the installation, recovery, and boot scripts generated by Digi Embedded Yocto. The values of these variables must reflect any changes made to the partition table, and vice versa. Make sure the variable syntax is correct; incorrect syntax may lead to unexpected results.
|
2022-07-01 02:28:08
|
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|
http://wiki.custommapmakers.org/index.php?title=Compiling:q3map2
|
Compiling:q3map2
Official Support Forum @Splashdamage
Q3Map2 is a BSP compiler for games based on the Quake III Arena engine. It compiles .map files, which are editable with an editor, into .bsp files, which are understandable for the game and not editable. It currently supports the following platforms:
• ioUrban Terror
• Nexuiz
• Open Arena
• Quake 3 Arena/Team Arena
• Soldier of Fortune II
• Star Trek Elite Force
• Star Wars Jedi Knight: Jedi Academy
• Star Wars Jedi Knight II: Jedi Outcast
• Tenebrae (Quake1 Engine Modification Project)
• Tremulous
• War§ow
• Wolfenstein: Enemy Territory
note: Future support for Call of Duty is a possibility.
Q3Map2 was designed to replace the Q3Map.exe that comes with QERadiant, GtkRadiant and GMAX Tempest. However, there are significant enhancements that require a little twiddling to use, such as faster lighting and enhanced surface production.
Usage
Q3Map2 is a command-line utility. In general, users make use of Q3Map2 in one of three ways:
It should be noted that the default command lines given in the GtkRadiant bsp menu are by no means a complete showcase of the available Q3Map2 options and switches. You may edit the bsp menu command lines from the GtkRadiant preferences, but for total control of your Q3Map2 compile writing a batch file or using one of the front ends is probably a better idea.
Q3Map2 command lines generally follow the format:
"C:\path\to\Q3Map2.exe" [<general option>] [<major switch> [<minor switch> <minor switch>...]] "C:\path\to\maps\mapname.map"
Switches
General options
-connect <hostname/ip address>
• Enables output of Q3Map2 logging to a remote host. This switch allows all other (non-solitary) switches.
-info
• Analyzes a compiled bsp and outputs information to the screen or log. This switch is solitary, and allows for only the path to a compiled bsp (not .map, it should be noted) afterward.
-game <quake3|wolf|et|etut|ef|jk2|ja|sof2|tenebrae|qfusion>
• Enables support for games other than Quake III: Arena. This switch allows all other (non-solitary) switches.
Game switch "q3ut4" is commonly refered in Urbanterror mapping tutorials. It actually does not exist, q3map2 silently defaults to quake3 settings.
-fs_game <mod name>
• Enables support for game mods other than the basic game defined in your -game switch. This switch allows all other (non-solitary) switches.
-fs_basepath <"C:\path\to\game\directory">
• Required in order for Q3Map2 to read assets directories of games other than Quake III: Arena. This switch allows all other (non-solitary) switches.
-convert [-format <ase|map|quake3|wolf|et|etut|ef|jk2|ja|sof2|tenebrae|qfusion>]
• Converts a compiled .bsp to another format. This switch is solitary, and allows for only the path to a compiled bsp (not .map, it should be noted) afterward.
• With no -format sub-switch, -convert converts a compiled .bsp to an .ase model. Q3Map2 -convert outputs mapname.ase to the "maps" directory.
• Other formats are available via the -format sub-switch:
• map "decompiles" a .bsp back to a .map (most entities are lost, as is texture alignment information). Q3Map2 -convert -format map outputs mapname_c.map in the "maps" directory.
• quake3 "cross-compiles" a .bsp from some other game's .bsp format (this "other" game is specified in the -game switch) to Quake III: Arena. For example, to cross-compile from Wolfenstein: Enemy Territory to Quake III: Arena, "-game et -convert -format quake3" would be used. This feature is still a work in progress, and should be treated as experimental for the time being; differences in surfaceparm bitflags can cause weirdness in cross-compiled .bsps. Q3Map2 -convert -format quake3 outputs mapname_c.bsp in the "maps" directory.
• the other game shortnames work similarly to quake3, but for their respective games instead.
-scale <N.N>
• Scales a compiled .bsp by the prescribed factor. For example, Q3Map2 -scale 0.25 will output a new .bsp that is 25% of the original .bsp's size, while Q3Map2 -scale 2.0 will output a new .bsp that is twice as large. Q3Map2 -scale outputs mapname_s.bsp in the "maps" directory. This switch is solitary, and allows for only the path to a compiled bsp (not .map, it should be noted) afterward.
-export
• Exports internal lightmaps from a compiled .bsp to external .tga images. This switch is solitary, and allows for only the path to a compiled bsp (not .map, it should be noted) afterward.
-import
• Imports external .tga lightmaps back into a compiled .bsp. Imported lightmaps will only work on the unmodified BSP they were exported from. This switch is solitary, and allows for only the path to a compiled bsp (not .map, it should be noted) afterward.
-threads <number of threads Q3Map2 should use>
• Specifies a number of threads to use during compiling. On Windows, Q3Map2 automatically detects the number of CPUs present, and sets the threadcount accordingly. The -threads switch can be used to override this behavior.
• Linux users will need to manually set -threads in order to activate SMP support in Q3Map2.
-v
• Enables verbose mode. This is generally a good idea.
-rename
• Used to fix an issue with misc_model entities in SOF2. If your misc_models show up unlit and completely black, use the -rename switch.
Major switches
-bsp
• Compiles a .map file into a binary space partition (BSP) file for use with the Quake III: Arena engine. Also writes a .prt (portal information) file and a .srf (surface) file. It is not necessary to add -bsp to your command line; it is the default switch.
-vis
• Creates visibility sets based on the portal file.
-light
• Calculates lighting data. With no -light sub-switches enabled, less than desirable output will be achieved.
Minor switches
The three major switches (-bsp, -vis, and -light) all have many options that can be accessed via their respective minor switches. There are quite a few minor switches, so in the interest of readability the lists of minor switches have been organized on separate pages of the Q3Map2 wiki.
BSP phase minor switches
Q3Map2 BSP phase minor switches have their own page at Compiling:q3map2/BSP
VIS phase minor switches
Q3Map2 VIS phase minor switches have their own page at Compiling:q3map2/VIS
Light phase minor switches
Q3Map2 Light phase minor switches have their own page at Compiling:q3map2/LIGHT
Q3Map2-specific entities
_decal
• Specifies a decal to be projected. Should contain one or more patch meshes and target an info_null entity. The targetting line drawn between the center of the _decal entity and the targetted info_null is the axis and distance of projection. It helps to think of the _decal mesh as if it were a light gel, the info_null were the target of a spotlight, and that you were "shining" this decal onto map geometry.
_skybox
• The _skybox entity is a true miracle: It specifies the origin of a skybox (a wholly contained, seperate area of the map), similar to some games' portal skies. When compiled with Q3Map2, the skybox surfaces will be visible from any place where sky is normally visible. It will cast shadows on the normal parts of the map, and can be used with cloud layers and other effects. Please see the Mapping:Sky for usage info.
Q3Map2-specific entity keys
Q3Map2-specific entity keys have their own page at Compiling:q3map2/Entity keys.
FS_q3map_Radbump_4a is a modification of the Q3Map2 compiler developed by TwentySeven. It integrates support for normal, bump, and specular maps.
FS_q3map_Radbump_4a is available from [Q3Map2 bumpmapping modifications at Urbanterror forums]
A brief usage description [FS_q3map_Radbump_4a at Urbanterror level design tutorials]
|
2018-02-21 17:03:40
|
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|
https://pychubby.readthedocs.io/en/latest/source/CLI.html
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# CLI¶
pychubby also offers a simple Command Line Interface that exposes some of the functionality of the Python package.
## Usage¶
After installation of pychubby an entry point pc becomes available.
To see the basic information write pc --help:
Usage: pc [OPTIONS] COMMAND [ARGS]...
Automated face warping tool.
Options:
--help Show this message and exit.
Commands:
list List available actions.
perform Take an action.
To perform actions one uses the perform subcommand. pc perform --help:
Usage: pc perform [OPTIONS] COMMAND [ARGS]...
Take an action.
Options:
--help Show this message and exit.
Commands:
Chubbify Make a chubby face.
LinearTransform Linear transformation.
OpenEyes Open eyes.
...
...
...
The syntax for all actions is identical. The positional arguments are
1. Input image path (required)
2. Output image path (not required)
If the output image path is not provided the resulting image is simply going to be plotted.
All the options correspond to the keyword arguments of the constructor of the respective action classes in pychubby.actions module.
To give a specific example let us use the Smile action. To get info on the parameters write pc perform Smile --help:
Usage: pc perform Smile [OPTIONS] INP_IMG [OUT_IMG]
Make a smiling face.
Options:
--scale FLOAT
--help Show this message and exit.
In particular, one can then warp an image in the following fashion
pc perform Smile --scale 0.3 img_cousin.jpg img_cousin_smiling.jpg
## Limitations¶
The features that are unavailable via the CLI are the following:
1. AbsoluteMove, Lambda and Pipeline actions
2. Different actions for different people
3. Lower level control
Specifically, if the user provides a photo with multiple faces the same action will be performed on all of them.
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2021-11-29 21:52:47
|
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https://math.stackexchange.com/questions/2558228/injective-homomorphism-example
|
# Injective Homomorphism Example
Ok, so there is this problem that quite confuses me because the example seems too easy (at least thats what I think):
Find an example of a group homomorphism $\Phi: S_3 \rightarrow S_4$ that is injective.
I think the example is just $\Phi(\sigma)=\sigma$, this is the only example I can think about but I am not sure if it is correct. Can anyone tell me if this example is valid? If not what other examples could there be?
• If $\sigma\in S_3$, then how can $\sigma$ be in $S_4$? – Lord Shark the Unknown Dec 9 '17 at 8:06
• Technically $\sigma \in S_3$ is also in $S_4$ no? – Aurora Borealis Dec 9 '17 at 8:46
## 1 Answer
Your example is (almost) right. If you concretely take $S_n$ as the permutations of $\{1,2,...,n\}$ then any $\sigma \in S_3$ can be mapped to $g(\sigma) \in S_4$ by having $g(\sigma)(4)=4,$ while for $k=1,2,3$ put $g(\sigma)(k)=\sigma(k).$
• Wait so I dont get it, is it correct to say it is $\Phi(\sigma)=\sigma$ is indeed an injective homomorphism, for all $\sigma \in S_3?$ – Aurora Borealis Dec 9 '17 at 8:47
• @AuroraBorealis Not technically, since you want $\Phi(\sigma)$ to be in $S_4.$ And if $\sigma$ is in $S_3$ that's not in $S_4,$ unless in a loose sense in which you always map $4$ to itself. – coffeemath Dec 9 '17 at 8:50
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2019-07-16 14:54:41
|
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https://en.wikibooks.org/wiki/LMIs_in_Control/pages/Insensitive_Disk_Region_Design
|
# LMIs in Control/pages/Insensitive Disk Region Design
Insensitive Disk Region Design
Similar to the insensitive strip region design problem, insensitive disk region design is another way with which robust stabilization can be achieved where closed-loop eigenvalues are placed in particular regions of the complex plane where the said regions has an inner boundary that is insensitive to perturbations of the system parameter matrices.
## The System
Suppose we consider the following linear system that needs to be controlled:
{\displaystyle {\begin{aligned}{\begin{cases}\rho x&=Ax+Bu,\\y&=Cx\\\end{cases}}\end{aligned}}}
where ${\displaystyle x\in \mathbb {R} ^{n}}$, ${\displaystyle y\in \mathbb {R} ^{m}}$, and ${\displaystyle u\in \mathbb {R} ^{r}}$ are the state, output and input vectors respectively, and ${\displaystyle \rho }$ represents the differential operator (in the continuous-time case) or one-step shift forward operator (i.e., ${\displaystyle \rho x(k)=x(k+1)}$) (in the discrete-time case). Then the steps to obtain the LMI for insensitive strip region design would be obtained as follows.
## The Data
Prior to obtaining the LMI, we need the following matrices: ${\displaystyle A}$, ${\displaystyle B}$, and ${\displaystyle C}$.
## The Optimization Problem
Consider the above linear system as well as 2 positive scalars ${\displaystyle \gamma }$ and ${\displaystyle q}$. Then the output feedback control law ${\displaystyle u=Ky}$ would be designed such that:
{\displaystyle {\begin{aligned}{\eta }&=||A+BKC+qI||<{\gamma }\\\end{aligned}}}
Recalling the definition, we have:
{\displaystyle {\begin{aligned}{\mathbb {D} _{q,\eta }}&=\{{s|s\in \mathbb {C} ,|s+q|<{\eta }}\}\\&=\{x+jy|x,y\in \mathbb {R} ,{(x+q)^{2}}+{y^{2}}<{\eta ^{2}}\}\\\end{aligned}}}
and
{\displaystyle {\begin{aligned}{\mathbb {D} _{q,r}}&=\{{s|s\in \mathbb {C} ,|s+q|<{\gamma }}\}\\&=\{x+jy|x,y\in \mathbb {R} ,{(x+q)^{2}}+{y^{2}}<{\gamma ^{2}}\}\\\end{aligned}}}
Letting ${\displaystyle K}$ being the solution to the above problem, then
{\displaystyle {\begin{aligned}{\lambda _{i}}(A+BKC)&\in {\mathbb {D} _{q,\eta }}\subset {\mathbb {D} _{q,r}},&i=1,2,...,n\\\end{aligned}}}
## The LMI: Insensitive Strip Region Design
Using the above info, we can convert the given problem into an LMI, which - after using Schur compliment Lemma - results in the following:
{\displaystyle {\begin{aligned}{\begin{cases}{\text{min }}\gamma \\{\text{s.t. }}{\begin{bmatrix}-{\gamma }I&&(A+BKC+qI)\\{(A+BKC+qI)^{T}}&&-{\gamma }I\end{bmatrix}}&<0\\\end{cases}}\end{aligned}}}
## Conclusion:
For Schur stabilization, we can choose to solve the problem with ${\displaystyle q=0}$. Schur stability is achieved when ${\displaystyle {\gamma }{\leq }1}$. Alternately, if ${\displaystyle {\gamma }}$ is greater than (but very close to) 1, then Schur stability is also achieved when ${\displaystyle {\eta }=||A+BKC+qI||_{2}{\leq }1}$.
## Implementation
• Example Code - A GitHub link that contains code (titled "InsensitiveDiskRegion.m") that demonstrates how this LMI can be implemented using MATLAB-YALMIP.
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2020-09-20 15:31:26
|
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|
https://code.tutsplus.com/tutorials/working-with-indexeddb-part-3--net-36220?ec_unit=translation-info-language
|
# Working With IndexedDB - Part 3
Welcome to the final part of my IndexedDB series. When I began this series my intent was to explain a technology that is not always the most... friendly one to work with. In fact, when I first tried working with IndexedDB, last year, my initial reaction was somewhat negative ("Somewhat negative" much like the Universe is "somewhat old."). It's been a long journey, but I finally feel somewhat comfortable working with IndexedDB and I respect what it allows. It is still a technology that can't be used everywhere (it sadly missed being added to iOS7), but I truly believe it is a technology folks can learn and make use of today.
In this final article, we're going to demonstrate some additional concepts that build upon the "full" demo we built in the last article. To be clear, you must be caught up on the series or this entry will be difficult to follow, so you may also want to check out part one.
## Counting Data
Let's start with something simple. Imagine you want to add paging to your data. How would you get a count of your data so you can properly handle that feature? I've already shown you how you can get all your data and certainly you could use that as a way to count data, but that requires fetching everything. If your local database is huge, that could be slow. Luckily the IndexedDB spec provides a much simpler way of doing it.
The count() method, run on an objectStore, will return a count of data. Like everything else we've done this will be asynchronous, but you can simplify the code down to one call. For our note database, I've written a function called doCount() that does just this:
Remember - if the code above is a bit hard to follow, you can break it up into multiple blocks. See the earlier articles where I demonstrated this. The result handler is passed a result value representing the total number of objects available in the store. I modified the UI of our demo to include an empty span in the header.
The final thing I need to do is simply add a call to doCount when the application starts up and after any add or delete operation. Here is one example from the success handler for opening the database.
You can find the full example in the zip you downloaded as fulldemo2. (As an FYI, fulldemo1 is the application as it was at the end of the previous article.)
## Filter As You Type
For our next feature, we're going to add a basic filter to the note list. In the earlier articles in this series I covered how IndexedDB does not allow for free form search. You can't (well, not easily) search content that contains a keyword. But with the power of ranges, it is easy to at least support matching at the beginning of a string.
If you remember, a range allows us to grab data from a store that either begins with a certain value, ends with a value, or lies in between. We can use this to implement a basic filter against the title of our note fields. First, we need to add an index for this property. Remember, this can only be done in the onupgradeneeded event.
Next, I added a simple form field to the UI:
Then I added a "keyup" handler to the field so I'd see immediate updates while I type.
Notice how I'm calling displayNotes. This is the same function I used before to display everything. I'm going to update it to support both a "get everything" action as well as a "get filtered" type action. Let's take a look at it.
To be clear, the only change here is at the bottom. Opening a cursor with or without a range gives us the same type of event handler result. That's handy then as it makes this update so trivial. The only complex aspect is in actually building the range. Notice what I've done here. The input, filter, is what the user typed. So imagine this is "The". We want to find notes with a title that begins with "The" and ends in any character. This can be done by simply setting the far end of the range to a high ASCII character. I can't take credit for this idea. See the StackOverflow link in the code for attribution.
You can find this demo in the fulldemo3 folder. Note that this is using a new database so if you've run the previous examples, this one will be empty when you first run it.
While this works, it has one small problem. Imagine a note titled, "Saints Rule." (Because they do. Just saying.) Most likely you will try to search for this by typing "saints". If you do this, the filter won't work because it is case sensitive. How do we get around it?
One way is to simply store a copy of our title in lowercase. This is relatively easy to do. First, I modified the index to use a new property called titlelc.
Then I modified the code that stores notes to create a copy of the field:
Finally, I modified the search to simply lowercase user input. That way if you enter "Saints" it will work just as well as entering "saints."
That's it. You can find this version as fulldemo4.
## Working With Array Properties
For our final improvement, I'm going to add a new feature to our Note application - tagging. This will
let you add any number of tags (think keywords that describe the note) so that you can later find other
notes with the same tag. Tags will be stored as an array. That by itself isn't such a big deal. I mentioned in the beginning of this series that you could easily store arrays as properties. What is a bit more complex is handling the search. Let's begin by making it so you can add tags to a note.
First, I modified my note form to have a new input field. This will allow the user to enter tags separated by a comma:
I can save this by simply updating my code that handles Note creation/updating.
Notice that I'm defaulting the value to an empty array. I only populate it if you typed something in. Saving this is as simple as appending it to the object we pass to IndexedDB:
That's it. If you write a few notes and open up Chrome's Resources tab, you can actually see the data being stored.
Now let's add tags to the view when you display a note. For my application, I decided on a simple use case for this. When a note is displayed, if there are tags I'll list them out. Each tag will be a link. If you click that link, I'll show you a list of related notes using the same tag. Let's look at that logic first.
This function (a new addition to our application) handles the note display code formally bound to the table cell click event. I needed a more abstract version of the code so this fulfills that purpose. For the most part it's the same, but note the logic to check the length of the tags property. If the array is not empty, the content is updated to include a simple list of tags. Each one is wrapped in a link with a particular class I'll use for lookup later. I've also added a div specifically to handle that search.
At this point, I've got the ability to add tags to a note as well as display them later. I've also planned to allow the user to click those tags so they can find other notes using the same tag. Now here comes the complex part.
You've seen how you can fetch content based on an index. But how does that work with array properties? Turns out - the spec has a specific flag for dealing with this: multiEntry. When creating an array-based index, you must set this value to true. Here is how my application handles it:
That handles the storage aspect well. Now let's talk about search. Here is the click handler for the tag link class:
There's quite a bit here - but honestly - it is very similar to what we've dicussed before. When you click a tag, my code begins by grabbing the text of the link for the tag value. I create my transaction, objectstore, and index objects as you've seen before. The range is new this time. Instead of creating a range from something and to something, we can use the only() api to specify that we want a range of only one value. And yes - that seemed weird to me as well. But it works great. You can see then we open the cursor and we can iterate over the results as before. There is a bit of additional code to handle cases where there may be no matches. I also take note of the original note, i.e. the one you are viewing now, so that I don't display it as well. And that's really it. I've got one last bit of code that handles click events on those related notes so you can view them easily:
You can find this demo in the folder fulldemo5.
## Conclusion
I sincerely hope that this series has been helpful to you. As I said in the beginning, IndexedDB was not a technology I enjoyed using. The more I worked with it, and the more I began to wrap my head around how it did things, the more I began to appreciate how much this technology could help us as web developers. It definitely has room to grow, and I can definitely see people preferring to use wrapper libraries to simplify things, but I think the future for this feature is great!
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2022-11-30 02:49:30
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https://www.semanticscholar.org/paper/Smoothing-estimates-for-non-dispersive-equations-Ruzhansky-Sugimoto/34eb8c6fb60a86d7542b786ed4315d9c790eef0c
|
# Smoothing estimates for non-dispersive equations
@article{Ruzhansky2015SmoothingEF,
title={Smoothing estimates for non-dispersive equations},
author={Michael Ruzhansky and Mitsuru Sugimoto},
journal={Mathematische Annalen},
year={2015},
volume={365},
pages={241-269}
}
• Published 3 August 2015
• Mathematics
• Mathematische Annalen
This paper describes an approach to global smoothing problems for non-dispersive equations based on ideas of comparison principle and canonical transformation established in authors’ previous paper (Ruzhansky and Sugimoto, Proc Lond Math Soc, 105:393–423, 2012), where dispersive equations were treated. For operators $$a(D_x)$$a(Dx) of order m satisfying the dispersiveness condition $$\nabla a(\xi )\ne 0$$∇a(ξ)≠0 for $$\xi \not =0$$ξ≠0, the global smoothing estimate \begin{aligned} {\left…
By using commutator methods, we show uniform resolvent estimates and obtain globally smooth operators for self-adjoint injective homogeneous operators $H$ on graded groups, including Rockland
• Mathematics
• 2020
Smoothing (and decay) spacetime estimates are discussed for evolution groups of self-adjoint operators in an abstract setting. The basic assumption is the existence (and weak continuity) of the
In the first part of this thesis we study the propagation of Wigner measures linked to solutions of the Schrodinger equation with potentials presenting conical singularities and show that they are
• Mathematics
Springer Proceedings in Mathematics & Statistics
• 2019
Our aim in this work is to give some quantitative insight on the dispersive effects exhibited by solutions of a semiclassical Schrodinger-type equation in $$\mathbf{R}^d$$. We describe quantitatively
• Mathematics
• 2018
Our aim in this work is to give some quantitative insight on the dispersive effects exhibited by solutions of a semiclassical Schrödinger-type equation in Rd . We describe quantitatively the
ABSTRACT By using commutator methods, we show uniform resolvent estimates and obtain globally smooth operators for self-adjoint injective homogeneous operators H on graded groups, including Rockland
## References
SHOWING 1-10 OF 42 REFERENCES
• Mathematics
• 2003
The local $L^2$-mapping property of Fourier integral operators has been established in H\"ormander \cite{H} and in Eskin \cite{E}. In this paper, we treat the global $L^2$-boundedness for a class of
• Mathematics
• 2012
This paper describes a new approach to global smoothing problems for dispersive and non‐dispersive evolution equations based on the global canonical transforms and the underlying global microlocal
• Mathematics
• 1988
Is it possible for time evolution partial differential equations which are reversible and conservative to smooth locally the initial data? For the linear wave equation, for instance, the answer is
Soit f dans l'espace de Schwartz S(R n ) et soit S t f(x)=u(x,t)=∫ Rn exp(ix•ξ)•exp(it|ξ| 2 )•f^(ξ)dξ, x∈R n , t∈R. f^ est la transformee de Fourier de f. On considere u solution de l'equation de
• Mathematics
• 1994
We study well-posedness of the initial value problem for the generalized Benjamin-Ono equation tu + ukdxU = 0, k e , in Sobolev spaces HS(R). For small data and higher nonlinearities (k > 2) new
• Mathematics
• 2005
ABSTRACT The local L 2-mapping property of Fourier integral operators has been established in Hörmander (1971) and in Eskin (1970). In this article, we treat the global L 2-boundedness for a class of
ABSTRACT We discuss smoothing effects of dispersive-type pseudodifferential equations whose principal part is not necessarily elliptic. For equations with constant coefficients, a restriction theorem
• Mathematics
• 2004
This paper deals with the critical case of the global smoothing estimates for the Schrödinger equation. Although such estimates fail for critical orders of weights and smoothing, it is shown that
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2023-02-05 04:24:45
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|
https://socratic.org/questions/how-do-you-find-the-derivative-of-the-function-g-t-4t-t-1
|
# How do you find the derivative of the function g(t)=(4t)/(t+1)?
Nov 9, 2016
$g ' \left(t\right) = \frac{4}{t + 1} ^ 2$
#### Explanation:
Use the quotient rule. The quotient rule states that the quotient of two functions, such as some function $y = \frac{u}{v}$, where $u$ and $v$ are functions of $x$, has a derivative of $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{du}}{\mathrm{dx}} \cdot v - u \cdot \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$.
So, applying this to the function at hand, we see that:
$g \left(t\right) = \frac{4 t}{t + 1}$
$g ' \left(t\right) = \frac{\left(\frac{d}{\mathrm{dt}} 4 t\right) \cdot \left(t + 1\right) - 4 t \left(\frac{d}{\mathrm{dt}} \left(t + 1\right)\right)}{t + 1} ^ 2$
$\textcolor{w h i t e}{g ' \left(t\right)} = \frac{4 \left(t + 1\right) - 4 t \left(1\right)}{t + 1} ^ 2$
$\textcolor{w h i t e}{g ' \left(t\right)} = \frac{4}{t + 1} ^ 2$
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2021-01-20 19:45:10
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https://www.seamplex.com/wasora/realbook/real-003-lag.html
|
# First-order lags
• Difficulty: 003/100
• Author: jeremy theler
• Keywords: end_time, dt, lag, HEADER,
Whenever the special variable end_time is non-zero, wasora enters into transient mode. These examples introduce transient problems by illustrating how first-order lags can be used to filter signals.
## lag.was
This example generates a signal $$r(t)$$ which is zero except for $$a < t < b$$, where it takes the value one. Then, the signal $$y(t)$$ is computed as a first-order lag of $$r(t)$$ with a characteristic time $$\tau$$. The output consists of three columns containing $$t$$, $$r(t)$$ and $$y(t)$$. By using the keyword HEADER a commented line is pre-prended to the output with a textual representation of the expressions passed to PRINT which are automatically understood by “qdp”: “https://bitbucket.org/gtheler/qdp” so a reasonable figure can be obtained with virtually no effort, as shown in terminal mimic where the output of wasora is piped to qdp.
# this is a transient problem and lasts 5 units of time
end_time = 5
# each time step is equal to 1/20th of a unit of time
dt = 1/20
# some parameters, which we define as constants
CONST a b tau
a = 1
b = 3
tau = 1.234
# signal r is equal to zero except for a < t < b
r = 0
r[a:b] = 1
# signal y is equal to signal r fitered through a lag
# of characteristic time tau
y = lag(r, tau)
PRINT t r y
# exercise: investigate how the result of the lag
# depends on the time step
$wasora lag.was | qdp -o lag$
## compact.was
Instead of writing the long input shown in lag.was, we could have obtained the same result with a couple of lines. Indeed, the terminal shows that the output of this input is the same as the one of the previous longer example.
# the preceeding example could have been written in fewer
# lines as follows (although the SPOT rule is broken)
end_time = 5
dt = 1/20
PRINT t heaviside(t-1)-heaviside(t-3) lag(heaviside(t-1)-heaviside(t-3),1.234)
$wasora lag.was > lag.dat$ wasora compact.was > compact.dat
$diff -s lag.dat compact.dat Files lag.dat and compact.dat are identical$ rm lag.dat compact.dat
$ The reported difference is due to the presence of the HEADER keyword in the first input so qdp can automatically label the bullets. ## quasi-sine.was Not only does this example illustrate the usage of a first-order lag, but also of a point-wise defined function~$$s(t)$$ (more on one-dimensional functions in section ref{007-functions}). In this case, the data is interpolated using the Akima method, and end_time is set to the variable s_b which contains the last value of the one-dimensional function s (incidentally, s_a contains the first value). FUNCTION s(t) INTERPOLATION akima DATA { 0 1 1 1+0.25 2 1-0.25 3 1+0.5 4 1-0.5 5 1+0.75 6 1-0.75 7 1+0.75 8 1-0.75 9 1+0.9 10 1-0.9 12 1 15 1 18 1 20 1 } end_time = s_b PRINT t s(t) lag(s(t),1) HEADER $ wasora quasi-sine.was | qdp -o quasi-sine
\$
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2018-12-12 18:48:20
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https://questioncove.com/updates/542ee383e4b032bb1c8221b9
|
OpenStudy (anonymous):
The balance of a loan is $2,570 in January, and the monthly payment is$125.50. The relationship between the loan balance, y, and the number of monthly payments made since January, x, can be represented by the equation y = 2,570 - 125.50x. In what months does the loan balance, y, meet the condition, $1,600 < y <$2,000?
OpenStudy (anonymous):
hey mail me na
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2017-08-21 14:01:35
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|
http://math.stackexchange.com/questions/52741/in-general-what-courses-must-you-take-in-order-to-get-a-bs-degree-in-pure-math?answertab=oldest
|
# In general, what courses must you take in order to get a BS degree in pure math? [closed]
In general, what courses must you take in order to get a BS degree in pure math?
And what is a topic of graduation dissertation you wrote?
-
The answer depends strongly on your institution, so ask your institution. I don't think there is anything sensible to be said in general. – Qiaochu Yuan Jul 20 '11 at 20:44
@Tommy L: For example, graduation dissertations are not a universal requirement, and are, in many countries, uncommon. – André Nicolas Jul 20 '11 at 21:37
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2014-03-08 08:31:08
|
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|
https://socratic.org/questions/how-do-you-find-the-derivative-of-2-sin-x-sin-2x
|
# How do you find the derivative of 2 sin x + sin 2x?
By using $\mathrm{ds} \in u = \mathrm{du} \cos u$
$f \left(x\right) = 2 \sin x + \sin 2 x$
$f ' \left(x\right) = 2 \cos x + 2 \cos 2 x$
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2022-07-06 02:03:48
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http://www.cirosantilli.com/elf-hello-world/
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# ELF Hello World Tutorial
Introductory analysis of a simple example of the Executable and Linkable File format.
Extracted from my Stack Overflow answer.
1. Introduction
2. Minimal ELF file
3. Generate the example
4. Object hd
5. Executable hd
6. Global file structure
7. Section vs segment
8. ELF header
9. Section header table
10. Sections
1. Index 0 section
2. .data section
3. .text section
4. SHT_STRTAB
5. .shstrtab
6. .symtab
1. STT_FILE
2. STT_SECTION
3. STT_NOTYPE
4. SHT_SYMTAB on the executable
7. .strtab
8. .rela.text
11. Program header table
12. Backlinks
## Introduction
ELF is the dominating file format for Linux. It competes with Mach-O for OS X and PE for Windows.
ELF supersedes .coff, which supersedes a.out.
### Standards
ELF is specified by the LSB:
The LSB basically links to other standards with minor extensions, in particular:
A handy summary can be found at:
man elf
### How to learn
Spin like mad between:
### Specified file formats
The ELF standard specifies multiple file formats:
• Object files (.o).
Intermediate step to generating executables and other formats:
Source code
|
| Compilation
|
v
Object file
|
| Linking
|
v
Executable
Object files exist to make compilation faster: with make, we only have to recompile the modified source files based on timestamps.
We have to do the linking step every time, but it is much less expensive.
• Executable files (no standard Linux extension).
This is what the Linux kernel can actually run.
• Archive files (.a).
Libraries meant to be embedded into executables during the Linking step.
• Shared object files (.so).
Libraries meant to be loaded when the executable starts running.
• Core dumps.
Such files may be generated by the Linux kernel when the program does naughty things, e.g. segfault.
They exist to help debugging the program.
In this tutorial, we consider only object and executable files.
### Implementations
• Compiler toolchains generate and read ELF files.
Sane compilers should use a separate standalone library to do the dirty work. E.g., Binutils uses BFD (in-tree and canonical source).
• Operating systems read and run ELF files.
Kernels cannot link to a library nor use the C stlib, so they are more likely to implement it themselves.
This is the case of the Linux kernel 4.2 which implements it in th file fs/binfmt_elf.c.
• Specialized libraries. Examples:
## Minimal ELF file
It is non-trivial to determine what is the smallest legal ELF file, or the smaller one that will do something trivial in Linux.
Some impressive attempts:
In this example we will consider a saner hello world example that will better capture real life cases.
## Generate the example
Let’s break down a minimal runnable Linux x86-64 example:
section .data
hello_world db "Hello world!", 10
hello_world_len equ - hello_world section .text global _start _start: mov rax, 1 mov rdi, 1 mov rsi, hello_world mov rdx, hello_world_len syscall mov rax, 60 mov rdi, 0 syscall Compiled with: nasm -w+all -f elf64 -o 'hello_world.o' 'hello_world.asm' ld -o 'hello_world.out' 'hello_world.o' TODO: use a minimal linker script with -T to be more precise and minimal. Versions: • NASM 2.10.09 • Binutils version 2.24 (contains ld) • Ubuntu 14.04 We don’t use a C program as that would complicate the analysis, that will be level 2 :-) ## Object hd hd hello_world.o Gives: 00000000 7f 45 4c 46 02 01 01 00 00 00 00 00 00 00 00 00 |.ELF............| 00000010 01 00 3e 00 01 00 00 00 00 00 00 00 00 00 00 00 |..>.............| 00000020 00 00 00 00 00 00 00 00 40 00 00 00 00 00 00 00 |........@.......| 00000030 00 00 00 00 40 00 00 00 00 00 40 00 07 00 03 00 |....@.....@.....| 00000040 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| * 00000080 01 00 00 00 01 00 00 00 03 00 00 00 00 00 00 00 |................| 00000090 00 00 00 00 00 00 00 00 00 02 00 00 00 00 00 00 |................| 000000a0 0d 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| 000000b0 04 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| 000000c0 07 00 00 00 01 00 00 00 06 00 00 00 00 00 00 00 |................| 000000d0 00 00 00 00 00 00 00 00 10 02 00 00 00 00 00 00 |................| 000000e0 27 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |'...............| 000000f0 10 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| 00000100 0d 00 00 00 03 00 00 00 00 00 00 00 00 00 00 00 |................| 00000110 00 00 00 00 00 00 00 00 40 02 00 00 00 00 00 00 |........@.......| 00000120 32 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |2...............| 00000130 01 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| 00000140 17 00 00 00 02 00 00 00 00 00 00 00 00 00 00 00 |................| 00000150 00 00 00 00 00 00 00 00 80 02 00 00 00 00 00 00 |................| 00000160 a8 00 00 00 00 00 00 00 05 00 00 00 06 00 00 00 |................| 00000170 04 00 00 00 00 00 00 00 18 00 00 00 00 00 00 00 |................| 00000180 1f 00 00 00 03 00 00 00 00 00 00 00 00 00 00 00 |................| 00000190 00 00 00 00 00 00 00 00 30 03 00 00 00 00 00 00 |........0.......| 000001a0 34 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |4...............| 000001b0 01 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| 000001c0 27 00 00 00 04 00 00 00 00 00 00 00 00 00 00 00 |'...............| 000001d0 00 00 00 00 00 00 00 00 70 03 00 00 00 00 00 00 |........p.......| 000001e0 18 00 00 00 00 00 00 00 04 00 00 00 02 00 00 00 |................| 000001f0 04 00 00 00 00 00 00 00 18 00 00 00 00 00 00 00 |................| 00000200 48 65 6c 6c 6f 20 77 6f 72 6c 64 21 0a 00 00 00 |Hello world!....| 00000210 b8 01 00 00 00 bf 01 00 00 00 48 be 00 00 00 00 |..........H.....| 00000220 00 00 00 00 ba 0d 00 00 00 0f 05 b8 3c 00 00 00 |............<...| 00000230 bf 00 00 00 00 0f 05 00 00 00 00 00 00 00 00 00 |................| 00000240 00 2e 64 61 74 61 00 2e 74 65 78 74 00 2e 73 68 |..data..text..sh| 00000250 73 74 72 74 61 62 00 2e 73 79 6d 74 61 62 00 2e |strtab..symtab..| 00000260 73 74 72 74 61 62 00 2e 72 65 6c 61 2e 74 65 78 |strtab..rela.tex| 00000270 74 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |t...............| 00000280 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| 00000290 00 00 00 00 00 00 00 00 01 00 00 00 04 00 f1 ff |................| 000002a0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| 000002b0 00 00 00 00 03 00 01 00 00 00 00 00 00 00 00 00 |................| 000002c0 00 00 00 00 00 00 00 00 00 00 00 00 03 00 02 00 |................| 000002d0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| 000002e0 11 00 00 00 00 00 01 00 00 00 00 00 00 00 00 00 |................| 000002f0 00 00 00 00 00 00 00 00 1d 00 00 00 00 00 f1 ff |................| 00000300 0d 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| 00000310 2d 00 00 00 10 00 02 00 00 00 00 00 00 00 00 00 |-...............| 00000320 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| 00000330 00 68 65 6c 6c 6f 5f 77 6f 72 6c 64 2e 61 73 6d |.hello_world.asm| 00000340 00 68 65 6c 6c 6f 5f 77 6f 72 6c 64 00 68 65 6c |.hello_world.hel| 00000350 6c 6f 5f 77 6f 72 6c 64 5f 6c 65 6e 00 5f 73 74 |lo_world_len._st| 00000360 61 72 74 00 00 00 00 00 00 00 00 00 00 00 00 00 |art.............| 00000370 0c 00 00 00 00 00 00 00 01 00 00 00 02 00 00 00 |................| 00000380 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| 00000390 ## Executable hd hd hello_world.out Gives: 00000000 7f 45 4c 46 02 01 01 00 00 00 00 00 00 00 00 00 |.ELF............| 00000010 02 00 3e 00 01 00 00 00 b0 00 40 00 00 00 00 00 |..>.......@.....| 00000020 40 00 00 00 00 00 00 00 10 01 00 00 00 00 00 00 |@...............| 00000030 00 00 00 00 40 00 38 00 02 00 40 00 06 00 03 00 |....@.8...@.....| 00000040 01 00 00 00 05 00 00 00 00 00 00 00 00 00 00 00 |................| 00000050 00 00 40 00 00 00 00 00 00 00 40 00 00 00 00 00 |..@.......@.....| 00000060 d7 00 00 00 00 00 00 00 d7 00 00 00 00 00 00 00 |................| 00000070 00 00 20 00 00 00 00 00 01 00 00 00 06 00 00 00 |.. .............| 00000080 d8 00 00 00 00 00 00 00 d8 00 60 00 00 00 00 00 |...............| 00000090 d8 00 60 00 00 00 00 00 0d 00 00 00 00 00 00 00 |...............| 000000a0 0d 00 00 00 00 00 00 00 00 00 20 00 00 00 00 00 |.......... .....| 000000b0 b8 01 00 00 00 bf 01 00 00 00 48 be d8 00 60 00 |..........H....| 000000c0 00 00 00 00 ba 0d 00 00 00 0f 05 b8 3c 00 00 00 |............<...| 000000d0 bf 00 00 00 00 0f 05 00 48 65 6c 6c 6f 20 77 6f |........Hello wo| 000000e0 72 6c 64 21 0a 00 2e 73 79 6d 74 61 62 00 2e 73 |rld!...symtab..s| 000000f0 74 72 74 61 62 00 2e 73 68 73 74 72 74 61 62 00 |trtab..shstrtab.| 00000100 2e 74 65 78 74 00 2e 64 61 74 61 00 00 00 00 00 |.text..data.....| 00000110 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| * 00000150 1b 00 00 00 01 00 00 00 06 00 00 00 00 00 00 00 |................| 00000160 b0 00 40 00 00 00 00 00 b0 00 00 00 00 00 00 00 |..@.............| 00000170 27 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |'...............| 00000180 10 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| 00000190 21 00 00 00 01 00 00 00 03 00 00 00 00 00 00 00 |!...............| 000001a0 d8 00 60 00 00 00 00 00 d8 00 00 00 00 00 00 00 |...............| 000001b0 0d 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| 000001c0 04 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| 000001d0 11 00 00 00 03 00 00 00 00 00 00 00 00 00 00 00 |................| 000001e0 00 00 00 00 00 00 00 00 e5 00 00 00 00 00 00 00 |................| 000001f0 27 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |'...............| 00000200 01 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| 00000210 01 00 00 00 02 00 00 00 00 00 00 00 00 00 00 00 |................| 00000220 00 00 00 00 00 00 00 00 90 02 00 00 00 00 00 00 |................| 00000230 08 01 00 00 00 00 00 00 05 00 00 00 07 00 00 00 |................| 00000240 08 00 00 00 00 00 00 00 18 00 00 00 00 00 00 00 |................| 00000250 09 00 00 00 03 00 00 00 00 00 00 00 00 00 00 00 |................| 00000260 00 00 00 00 00 00 00 00 98 03 00 00 00 00 00 00 |................| 00000270 4c 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |L...............| 00000280 01 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| 00000290 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| 000002a0 00 00 00 00 00 00 00 00 00 00 00 00 03 00 01 00 |................| 000002b0 b0 00 40 00 00 00 00 00 00 00 00 00 00 00 00 00 |..@.............| 000002c0 00 00 00 00 03 00 02 00 d8 00 60 00 00 00 00 00 |...............| 000002d0 00 00 00 00 00 00 00 00 01 00 00 00 04 00 f1 ff |................| 000002e0 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| 000002f0 11 00 00 00 00 00 02 00 d8 00 60 00 00 00 00 00 |...............| 00000300 00 00 00 00 00 00 00 00 1d 00 00 00 00 00 f1 ff |................| 00000310 0d 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| 00000320 00 00 00 00 04 00 f1 ff 00 00 00 00 00 00 00 00 |................| 00000330 00 00 00 00 00 00 00 00 2d 00 00 00 10 00 01 00 |........-.......| 00000340 b0 00 40 00 00 00 00 00 00 00 00 00 00 00 00 00 |..@.............| 00000350 34 00 00 00 10 00 02 00 e5 00 60 00 00 00 00 00 |4..............| 00000360 00 00 00 00 00 00 00 00 40 00 00 00 10 00 02 00 |........@.......| 00000370 e5 00 60 00 00 00 00 00 00 00 00 00 00 00 00 00 |...............| 00000380 47 00 00 00 10 00 02 00 e8 00 60 00 00 00 00 00 |G..............| 00000390 00 00 00 00 00 00 00 00 00 68 65 6c 6c 6f 5f 77 |.........hello_w| 000003a0 6f 72 6c 64 2e 61 73 6d 00 68 65 6c 6c 6f 5f 77 |orld.asm.hello_w| 000003b0 6f 72 6c 64 00 68 65 6c 6c 6f 5f 77 6f 72 6c 64 |orld.hello_world| 000003c0 5f 6c 65 6e 00 5f 73 74 61 72 74 00 5f 5f 62 73 |_len._start.__bs| 000003d0 73 5f 73 74 61 72 74 00 5f 65 64 61 74 61 00 5f |s_start._edata._| 000003e0 65 6e 64 00 |end.| 000003e4 ## Global file structure An ELF file contains the following parts: • ELF header. Points to the position of the section header table and the program header table. • Section header table (optional on executable). Each has e_shnum section headers, each pointing to the position of a section. • N sections, with N <= e_shnum (optional on executable) • Program header table (only on executable). Each has e_phnum program headers, each pointing to the position of a segment. • N segments, with N <= e_phnum (only on executable) The order of those parts is not fixed: the only fixed thing is the ELF header that must be the first thing on the file: Generic docs say: Although the figure shows the program header table immediately after the ELF header, and the section header table following the sections, actual files may differ. Moreover, sections and segments have no specified order. Only the ELF header has a fixed position in the file. In pictures: sample object file with three sections: +-------------------+ | ELF header |---+ +---------> +-------------------+ | e_shoff | | |<--+ | Section | Section header 0 | | | |---+ sh_offset | Header +-------------------+ | | | Section header 1 |---|--+ sh_offset | Table +-------------------+ | | | | Section header 2 |---|--|--+ +---------> +-------------------+ | | | | Section 0 |<--+ | | +-------------------+ | | sh_offset | Section 1 |<-----+ | +-------------------+ | | Section 2 |<--------+ +-------------------+ But nothing (except sanity) prevents the following topology: +-------------------+ | ELF header |---+ e_shoff +-------------------+ | | Section 1 |<--|--+ +---------> +-------------------+ | | | | |<--+ | sh_offset | Section | Section header 0 | | | | |------|---------+ | Header +-------------------+ | | | | Section header 1 |------+ | | Table +-------------------+ | | | Section header 2 |---+ | sh_offset +---------> +-------------------+ | sh_offset | | Section 2 |<--+ | +-------------------+ | | Section 0 |<---------------+ +-------------------+ But some newbies may prefer PNGs :-) ## Section vs segment We will get into more detail later, but it is good to have it in mind now: • section: exists before linking, in object files. One ore more sections will be put inside a single segment by the linker. Major information sections contain for the linker: • is this section • raw data to be loaded into memory, e.g. .data, .text, etc. • or metadata about other sections, that will be used by the linker, but disappear at runtime e.g. .symtab, .srttab, .rela.text • segment: exists after linking, in the executable file. Contains information about how each segment should be loaded into memory by the OS, notably location and permissions. See also: ## ELF header readelf -h hello_world.o: Magic: 7f 45 4c 46 02 01 01 00 00 00 00 00 00 00 00 00 Class: ELF64 Data: 2's complement, little endian Version: 1 (current) OS/ABI: UNIX - System V ABI Version: 0 Type: REL (Relocatable file) Machine: Advanced Micro Devices X86-64 Version: 0x1 Entry point address: 0x0 Start of program headers: 0 (bytes into file) Start of section headers: 64 (bytes into file) Flags: 0x0 Size of this header: 64 (bytes) Size of program headers: 0 (bytes) Number of program headers: 0 Size of section headers: 64 (bytes) Number of section headers: 7 Section header string table index: 3 readelf -h hello_world.out: Magic: 7f 45 4c 46 02 01 01 00 00 00 00 00 00 00 00 00 Class: ELF64 Data: 2's complement, little endian Version: 1 (current) OS/ABI: UNIX - System V ABI Version: 0 Type: EXEC (Executable file) Machine: Advanced Micro Devices X86-64 Version: 0x1 Entry point address: 0x4000b0 Start of program headers: 64 (bytes into file) Start of section headers: 272 (bytes into file) Flags: 0x0 Size of this header: 64 (bytes) Size of program headers: 56 (bytes) Number of program headers: 2 Size of section headers: 64 (bytes) Number of section headers: 6 Section header string table index: 3 Bytes in the object file: 00000000 7f 45 4c 46 02 01 01 00 00 00 00 00 00 00 00 00 |.ELF............| 00000010 01 00 3e 00 01 00 00 00 00 00 00 00 00 00 00 00 |..>.............| 00000020 00 00 00 00 00 00 00 00 40 00 00 00 00 00 00 00 |........@.......| 00000030 00 00 00 00 40 00 00 00 00 00 40 00 07 00 03 00 |....@.....@.....| Executable: 00000000 7f 45 4c 46 02 01 01 00 00 00 00 00 00 00 00 00 |.ELF............| 00000010 02 00 3e 00 01 00 00 00 b0 00 40 00 00 00 00 00 |..>.......@.....| 00000020 40 00 00 00 00 00 00 00 10 01 00 00 00 00 00 00 |@...............| 00000030 00 00 00 00 40 00 38 00 02 00 40 00 06 00 03 00 |....@.8...@.....| Structure represented: # define EI_NIDENT 16 typedef struct { unsigned char e_ident[EI_NIDENT]; Elf64_Half e_type; Elf64_Half e_machine; Elf64_Word e_version; Elf64_Addr e_entry; Elf64_Off e_phoff; Elf64_Off e_shoff; Elf64_Word e_flags; Elf64_Half e_ehsize; Elf64_Half e_phentsize; Elf64_Half e_phnum; Elf64_Half e_shentsize; Elf64_Half e_shnum; Elf64_Half e_shstrndx; } Elf64_Ehdr; Manual breakdown: • 0 0: EI_MAG = 7f 45 4c 46 = 0x7f 'E', 'L', 'F': ELF magic number • 0 4: EI_CLASS = 02 = ELFCLASS64: 64 bit elf • 0 5: EI_DATA = 01 = ELFDATA2LSB: little endian data • 0 6: EI_VERSION = 01: format version • 0 7: EI_OSABI (only in 2003 Update) = 00 = ELFOSABI_NONE: no extensions. • 0 8: EI_PAD = 8x 00: reserved bytes. Must be set to 0. • 1 0: e_type = 01 00 = 1 (big endian) = ET_REl: relocatable format On the executable it is 02 00 for ET_EXEC. • 1 2: e_machine = 3e 00 = 62 = EM_X86_64: AMD64 architecture • 1 4: e_version = 01 00 00 00: must be 1 • 1 8: e_entry = 8x 00: execution address entry point, or 0 if not applicable like for the object file since there is no entry point. On the executable, it is b0 00 40 00 00 00 00 00. The kernel puts the RIP directly on that value when executing. It can be configured by the linker script or -e. But it will segfault if you set it too low: http://stackoverflow.com/questions/2187484/why-is-the-elf-execution-entry-point-virtual-address-of-the-form-0x80xxxxx-and-n • 2 0: e_phoff = 8x 00: program header table offset, 0 if not present. 40 00 00 00 on the executable, i.e. it starts immediately after the ELF header. • 2 8: e_shoff = 40 7x 00 = 0x40: section header table file offset, 0 if not present. • 3 0: e_flags = 00 00 00 00 Arch specific. i386 docs say: The Intel386 architecture defines no flags; so this member contains zero. • 3 4: e_ehsize = 40 00 : size of this elf header. TODO why this field needed? Isn’t the size fixed? • 3 6: e_phentsize = 00 00 : size of each program header, 0 if not present. 38 00 on executable: it is 56 bytes long • 3 8: e_phnum = 00 00 : number of program header entries, 0 if not present. 02 00 on executable: there are 2 entries. • 3 A: e_shentsize and e_shnum = 40 00 07 00: section header size and number of entries • 3 E: e_shstrndx (Section Header STRing iNDeX) = 03 00: index of the .shstrtab section. ## Section header table Array of Elf64_Shdr structs. Each entry contains metadata about a given section. e_shoff of the ELF header gives the starting position, 0x40 here. e_shentsize and e_shnum from the ELF header say that we have 7 entries, each 0x40 bytes long. So the table takes bytes from 0x40 to 0x40 + 7 + 0x40 - 1 = 0x1FF. Some section names are reserved for certain section types: http://www.sco.com/developers/gabi/2003-12-17/ch4.sheader.html#special_sections e.g. .text requires a SHT_PROGBITS type and SHF_ALLOC + SHF_EXECINSTR readelf -S hello_world.o: There are 7 section headers, starting at offset 0x40: Section Headers: [Nr] Name Type Address Offset Size EntSize Flags Link Info Align [ 0] NULL 0000000000000000 00000000 0000000000000000 0000000000000000 0 0 0 [ 1] .data PROGBITS 0000000000000000 00000200 000000000000000d 0000000000000000 WA 0 0 4 [ 2] .text PROGBITS 0000000000000000 00000210 0000000000000027 0000000000000000 AX 0 0 16 [ 3] .shstrtab STRTAB 0000000000000000 00000240 0000000000000032 0000000000000000 0 0 1 [ 4] .symtab SYMTAB 0000000000000000 00000280 00000000000000a8 0000000000000018 5 6 4 [ 5] .strtab STRTAB 0000000000000000 00000330 0000000000000034 0000000000000000 0 0 1 [ 6] .rela.text RELA 0000000000000000 00000370 0000000000000018 0000000000000018 4 2 4 Key to Flags: W (write), A (alloc), X (execute), M (merge), S (strings), l (large) I (info), L (link order), G (group), T (TLS), E (exclude), x (unknown) O (extra OS processing required) o (OS specific), p (processor specific) struct represented by each entry: typedef struct { Elf64_Word sh_name; Elf64_Word sh_type; Elf64_Xword sh_flags; Elf64_Addr sh_addr; Elf64_Off sh_offset; Elf64_Xword sh_size; Elf64_Word sh_link; Elf64_Word sh_info; Elf64_Xword sh_addralign; Elf64_Xword sh_entsize; } Elf64_Shdr; ## Sections ### Index 0 section Contained in bytes 0x40 to 0x7F. The first section is always magic: http://www.sco.com/developers/gabi/2003-12-17/ch4.sheader.html says: If the number of sections is greater than or equal to SHN_LORESERVE (0xff00), e_shnum has the value SHN_UNDEF (0) and the actual number of section header table entries is contained in the sh_size field of the section header at index 0 (otherwise, the sh_size member of the initial entry contains 0). There are also other magic sections detailed in Figure 4-7: Special Section Indexes. #### SHT_NULL In index 0, SHT_NULL is mandatory. Are there any other uses for it: http://stackoverflow.com/questions/26812142/what-is-the-use-of-the-sht-null-section-in-elf ? ### .data section .data is section 1: 00000080 01 00 00 00 01 00 00 00 03 00 00 00 00 00 00 00 |................| 00000090 00 00 00 00 00 00 00 00 00 02 00 00 00 00 00 00 |................| 000000a0 0d 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| 000000b0 04 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| • 80 0: sh_name = 01 00 00 00: index 1 in the .shstrtab string table Here, 1 says the name of this section starts at the first character of that section, and ends at the first NUL character, making up the string .data. .data is one of the section names which has a predefined meaning http://www.sco.com/developers/gabi/2003-12-17/ch4.strtab.html These sections hold initialized data that contribute to the program’s memory image. • 80 4: sh_type = 01 00 00 00: SHT_PROGBITS: the section content is not specified by ELF, only by how the program interprets it. Normal since a .data section. • 80 8: sh_flags = 03 7x 00: SHF_ALLOC and SHF_EXECINSTR: http://www.sco.com/developers/gabi/2003-12-17/ch4.sheader.html#sh_flags, as required from a .data section • 90 0: sh_addr = 8x 00: TODO: standard says: If the section will appear in the memory image of a process, this member gives the address at which the section’s first byte should reside. Otherwise, the member contains 0. but I don’t understand it very well yet. • 90 8: sh_offset = 00 02 00 00 00 00 00 00 = 0x200: number of bytes from the start of the program to the first byte in this section • a0 0: sh_size = 0d 00 00 00 00 00 00 00 If we take 0xD bytes starting at sh_offset 200, we see: 00000200 48 65 6c 6c 6f 20 77 6f 72 6c 64 21 0a 00 |Hello world!.. | AHA! So our "Hello world!" string is in the data section like we told it to be on the NASM. Once we graduate from hd, we will look this up like: readelf -x .data hello_world.o which outputs: Hex dump of section '.data': 0x00000000 48656c6c 6f20776f 726c6421 0a Hello world!. NASM sets decent properties for that section because it treats .data magically: http://www.nasm.us/doc/nasmdoc7.html#section-7.9.2 Also note that this was a bad section choice: a good C compiler would put the string in .rodata instead, because it is read-only and it would allow for further OS optimizations. • a0 8: sh_link and sh_info = 8x 0: do not apply to this section type. http://www.sco.com/developers/gabi/2003-12-17/ch4.sheader.html#special_sections • b0 0: sh_addralign = 04 = TODO: why is this alignment necessary? Is it only for sh_addr, or also for symbols inside sh_addr? • b0 8: sh_entsize = 00 = the section does not contain a table. If != 0, it means that the section contains a table of fixed size entries. In this file, we see from the readelf output that this is the case for the .symtab and .rela.text sections. ### .text section Now that we’ve done one section manually, let’s graduate and use the readelf -S of the other sections. [Nr] Name Type Address Offset Size EntSize Flags Link Info Align [ 2] .text PROGBITS 0000000000000000 00000210 0000000000000027 0000000000000000 AX 0 0 16 .text is executable but not writable: if we try to write to it Linux segfaults. Let’s see if we really have some code there: objdump -d hello_world.o gives: hello_world.o: file format elf64-x86-64 Disassembly of section .text: 0000000000000000 <_start>: 0: b8 01 00 00 00 mov0x1,%eax
5: bf 01 00 00 00 mov $0x1,%edi a: 48 be 00 00 00 00 00 movabs$0x0,%rsi
11: 00 00 00
14: ba 0d 00 00 00 mov $0xd,%edx 19: 0f 05 syscall 1b: b8 3c 00 00 00 mov$0x3c,%eax
20: bf 00 00 00 00 mov $0x0,%edi 25: 0f 05 syscall If we grep b8 01 00 00 on the hd, we see that this only occurs at 00000210, which is what the section says. And the Size is 27, which matches as well. So we must be talking about the right section. This looks like the right code: a write followed by an exit. The most interesting part is line a which does: movabs$0x0,%rsi
to pass the address of the string to the system call. Currently, the 0x0 is just a placeholder. After linking happens, it will be modified to contain:
4000ba: 48 be d8 00 60 00 00 movabs $0x6000d8,%rsi This modification is possible because of the data of the .rela.text section. ### SHT_STRTAB Sections with sh_type == SHT_STRTAB are called string tables. They hold a null separated array of strings. Such sections are used by other sections when string names are to be used. The using section says: • which string table they are using • what is the index on the target string table where the string starts So for example, we could have a string table containing: Data: \0 a b c \0 d e f \0 Index: 0 1 2 3 4 5 6 7 8 The first byte must be a 0. TODO rationale? And if another section wants to use the string d e f, they have to point to index 5 of this section (letter d). Notable string table sections: • .shstrtab • .strtab ### .shstrtab Section type: sh_type == SHT_STRTAB. Common name: section header string table. The section name .shstrtab is reserved. The standard says: This section holds section names. This section gets pointed to by the e_shstrnd field of the ELF header itself. String indexes of this section are are pointed to by the sh_name field of section headers, which denote strings. This section does not have SHF_ALLOC marked, so it will not appear on the executing program. readelf -x .shstrtab hello_world.o Gives: Hex dump of section '.shstrtab': 0x00000000 002e6461 7461002e 74657874 002e7368 ..data..text..sh 0x00000010 73747274 6162002e 73796d74 6162002e strtab..symtab.. 0x00000020 73747274 6162002e 72656c61 2e746578 strtab..rela.tex 0x00000030 7400 t. The data in this section has a fixed format: http://www.sco.com/developers/gabi/2003-12-17/ch4.strtab.html If we look at the names of other sections, we see that they all contain numbers, e.g. the .text section is number 7. Then each string ends when the first NUL character is found, e.g. character 12 is \0 just after .text\0. ### .symtab Section type: sh_type == SHT_SYMTAB. Common name: symbol table. First the we note that: • sh_link = 5 • sh_info = 6 For SHT_SYMTAB sections, those numbers mean that: • strings that give symbol names are in section 5, .strtab • the relocation data is in section 6, .rela.text A good high level tool to disassemble that section is: nm hello_world.o which gives: 0000000000000000 T _start 0000000000000000 d hello_world 000000000000000d a hello_world_len This is however a high level view that omits some types of symbols and in which the symbol types . A more detailed disassembly can be obtained with: readelf -s hello_world.o which gives: Symbol table '.symtab' contains 7 entries: Num: Value Size Type Bind Vis Ndx Name 0: 0000000000000000 0 NOTYPE LOCAL DEFAULT UND 1: 0000000000000000 0 FILE LOCAL DEFAULT ABS hello_world.asm 2: 0000000000000000 0 SECTION LOCAL DEFAULT 1 3: 0000000000000000 0 SECTION LOCAL DEFAULT 2 4: 0000000000000000 0 NOTYPE LOCAL DEFAULT 1 hello_world 5: 000000000000000d 0 NOTYPE LOCAL DEFAULT ABS hello_world_len 6: 0000000000000000 0 NOTYPE GLOBAL DEFAULT 2 _start The binary format of the table is documented at http://www.sco.com/developers/gabi/2003-12-17/ch4.symtab.html The data is: readelf -x .symtab hello_world.o Which gives: Hex dump of section '.symtab': 0x00000000 00000000 00000000 00000000 00000000 ................ 0x00000010 00000000 00000000 01000000 0400f1ff ................ 0x00000020 00000000 00000000 00000000 00000000 ................ 0x00000030 00000000 03000100 00000000 00000000 ................ 0x00000040 00000000 00000000 00000000 03000200 ................ 0x00000050 00000000 00000000 00000000 00000000 ................ 0x00000060 11000000 00000100 00000000 00000000 ................ 0x00000070 00000000 00000000 1d000000 0000f1ff ................ 0x00000080 0d000000 00000000 00000000 00000000 ................ 0x00000090 2d000000 10000200 00000000 00000000 -............... 0x000000a0 00000000 00000000 ........ The entries are of type: typedef struct { Elf64_Word st_name; unsigned char st_info; unsigned char st_other; Elf64_Half st_shndx; Elf64_Addr st_value; Elf64_Xword st_size; } Elf64_Sym; Like in the section table, the first entry is magical and set to a fixed meaningless values. #### STT_FILE Entry 1 has ELF64_R_TYPE == STT_FILE. ELF64_R_TYPE is continued inside of st_info. Byte analysis: • 10 8: st_name = 01000000 = character 1 in the .strtab, which until the following \0 makes hello_world.asm This piece of information file may be used by the linker to decide on which segment sections go: e.g. in ld linker script we write: segment_name : { file(section) } to pick a section from a given file. Most of the time however, we will just dump all sections with a given name together with: segment_name : { *(section) } • 10 12: st_info = 04 Bits 0-3 = ELF64_R_TYPE = Type = 4 = STT_FILE: the main purpose of this entry is to use st_name to indicate the name of the file which generated this object file. Bits 4-7 = ELF64_ST_BIND = Binding = 0 = STB_LOCAL. Required value for STT_FILE. • 10 13: st_shndx = Symbol Table Section header Index = f1ff = SHN_ABS. Required for STT_FILE. • 20 0: st_value = 8x 00: required for value for STT_FILE • 20 8: st_size = 8x 00: no allocated size Now from the readelf, we interpret the others quickly. #### STT_SECTION There are two such entries, one pointing to .data and the other to .text (section indexes 1 and 2). Num: Value Size Type Bind Vis Ndx Name 2: 0000000000000000 0 SECTION LOCAL DEFAULT 1 3: 0000000000000000 0 SECTION LOCAL DEFAULT 2 TODO what is their purpose? #### STT_NOTYPE Then come the most important symbols: Num: Value Size Type Bind Vis Ndx Name 4: 0000000000000000 0 NOTYPE LOCAL DEFAULT 1 hello_world 5: 000000000000000d 0 NOTYPE LOCAL DEFAULT ABS hello_world_len 6: 0000000000000000 0 NOTYPE GLOBAL DEFAULT 2 _start hello_world string is in the .data section (index 1). It’s value is 0: it points to the first byte of that section. _start is marked with GLOBAL visibility since we wrote: global _start in NASM. This is necessary since it must be seen as the entry point. Unlike in C, by default NASM labels are local. ##### SHN_ABS hello_world_len points to the special st_shndx == SHN_ABS == 0xF1FF. 0xF1FF is chosen so as to not conflict with other sections. st_value == 0xD == 13 which is the value we have stored there on the assembly: the length of the string Hello World!. This means that relocation will not affect this value: it is a constant. This is small optimization that our assembler does for us and which has ELF support. If we had used the address of hello_world_len anywhere, the assembler would not have been able to mark it as SHN_ABS, and the linker would have extra relocation work on it later. #### SHT_SYMTAB on the executable By default, NASM places a .symtab on the executable as well. This is only used for debugging. Without the symbols, we are completely blind, and must reverse engineer everything. You can strip it with objcopy, and the executable will still run. Such executables are called stripped executables. ### .strtab Holds strings for the symbol table. This section has sh_type == SHT_STRTAB. It is pointed to by sh_link == 5 of the .symtab section. readelf -x .strtab hello_world.o Gives: Hex dump of section '.strtab': 0x00000000 0068656c 6c6f5f77 6f726c64 2e61736d .hello_world.asm 0x00000010 0068656c 6c6f5f77 6f726c64 0068656c .hello_world.hel 0x00000020 6c6f5f77 6f726c64 5f6c656e 005f7374 lo_world_len._st 0x00000030 61727400 art. This implies that it is an ELF level limitation that global variables cannot contain NUL characters. ### .rela.text Section type: sh_type == SHT_RELA. Common name: relocation section. .rela.text holds relocation data which says how the address should be modified when the final executable is linked. This points to bytes of the text area that must be modified when linking happens to point to the correct memory locations. Basically, it translates the object text containing the placeholder 0x0 address: a: 48 be 00 00 00 00 00 movabs$0x0,%rsi
11: 00 00 00
to the actual executable code containing the final 0x6000d8:
4000ba: 48 be d8 00 60 00 00 movabs $0x6000d8,%rsi 4000c1: 00 00 00 It was pointed to by sh_info = 6 of the .symtab section. readelf -r hello_world.o gives: Relocation section '.rela.text' at offset 0x3b0 contains 1 entries: Offset Info Type Sym. Value Sym. Name + Addend 00000000000c 000200000001 R_X86_64_64 0000000000000000 .data + 0 The section does not exist in the executable. The actual bytes are: 00000370 0c 00 00 00 00 00 00 00 01 00 00 00 02 00 00 00 |................| 00000380 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 |................| The struct represented is: typedef struct { Elf64_Addr r_offset; Elf64_Xword r_info; Elf64_Sxword r_addend; } Elf64_Rela; So: • 370 0: r_offset = 0xC: address into the .text whose address this relocation will modify • 370 8: r_info = 0x200000001. Contains 2 fields: • ELF64_R_TYPE = 0x1: meaning depends on the exact architecture. • ELF64_R_SYM = 0x2: index of the section to which the address points, so .data which is at index 2. The AMD64 ABI says that type 1 is called R_X86_64_64 and that it represents the operation S + A where: • S: the value of the symbol on the object file, here 0 because we point to the 00 00 00 00 00 00 00 00 of movabs$0x0,%rsi
• A: the addend, present in field r_added
This address is added to the section on which the relocation operates.
This relocation operation acts on a total 8 bytes.
• 380 0: r_addend = 0
So in our example we conclude that the new address will be: S + A = .data + 0, and thus the first thing in the data section.
#### .rel.text
Besides sh_type == SHT_RELA, there also exists SHT_REL, which would have section name .text.rel (not present in this object file).
Those represent the same struct, but without the addend, e.g.:
typedef struct {
Elf64_Addr r_offset;
Elf64_Xword r_info;
} Elf64_Rela;
The ELF standard says that in many cases the both can be used, and it is just a matter of convenience.
## Program header table
Only appears in the executable.
Contains information of how the executable should be put into the process virtual memory.
The executable is generated from object files by the linker. The main jobs that the linker does are:
• determine which sections of the object files will go into which segments of the executable.
In Binutils, this comes down to parsing a linker script, and dealing with a bunch of defaults.
You can get the linker script used with ld --verbose, and set a custom one with ld -T.
• do relocation according to the .rela.text section. This depends on how the multiple sections are put into memory.
readelf -l hello_world.out gives:
Elf file type is EXEC (Executable file)
Entry point 0x4000b0
There are 2 program headers, starting at offset 64
Program Headers:
Type Offset VirtAddr PhysAddr
FileSiz MemSiz Flags Align
LOAD 0x0000000000000000 0x0000000000400000 0x0000000000400000
0x00000000000000d7 0x00000000000000d7 R E 200000
LOAD 0x00000000000000d8 0x00000000006000d8 0x00000000006000d8
0x000000000000000d 0x000000000000000d RW 200000
Section to Segment mapping:
Segment Sections...
00 .text
01 .data
On the ELF header, e_phoff, e_phnum and e_phentsize told us that there are 2 program headers, which start at 0x40 and are 0x38 bytes long each, so they are:
00000040 01 00 00 00 05 00 00 00 00 00 00 00 00 00 00 00 |................|
00000050 00 00 40 00 00 00 00 00 00 00 40 00 00 00 00 00 |..@.......@.....|
00000060 d7 00 00 00 00 00 00 00 d7 00 00 00 00 00 00 00 |................|
00000070 00 00 20 00 00 00 00 00 |.. ..... |
and:
00000070 01 00 00 00 06 00 00 00 | ........|
00000080 d8 00 00 00 00 00 00 00 d8 00 60 00 00 00 00 00 |...............|
00000090 d8 00 60 00 00 00 00 00 0d 00 00 00 00 00 00 00 |...............|
000000a0 0d 00 00 00 00 00 00 00 00 00 20 00 00 00 00 00 |.......... .....|
Structure represented http://www.sco.com/developers/gabi/2003-12-17/ch5.pheader.html:
typedef struct {
Elf64_Word p_type;
Elf64_Word p_flags;
Elf64_Off p_offset;
Elf64_Addr p_vaddr;
Elf64_Addr p_paddr;
Elf64_Xword p_filesz;
Elf64_Xword p_memsz;
Elf64_Xword p_align;
} Elf64_Phdr;
Breakdown of the first one:
• 40 0: p_type = 01 00 00 00 = PT_LOAD: this is a regular segment that will get loaded in memory.
• 40 4: p_flags = 05 00 00 00 = execute and read permissions. No write: we cannot modify the text segment. A classic way to do this in C is with string literals: http://stackoverflow.com/a/30662565/895245 This allows kernels to do certain optimizations, like sharing the segment amongst processes.
• 40 8: p_offset = 8x 00 TODO: what is this? Standard says:
This member gives the offset from the beginning of the file at which the first byte of the segment resides.
But it looks like offsets from the beginning of segments, not file?
• 50 0: p_vaddr = 00 00 40 00 00 00 00 00: initial virtual memory address to load this segment to
• 50 8: p_paddr = 00 00 40 00 00 00 00 00: unspecified effect. Intended for systems in which physical addressing matters. TODO example?
• 60 0: p_filesz = d7 00 00 00 00 00 00 00: size that the segment occupies in memory. If smaller than p_memsz, the OS fills it with zeroes to fit when loading the program. This is how BSS data is implemented to save space on executable files. i368 ABI says on PT_LOAD:
The bytes from the file are mapped to the beginning of the memory segment. If the segment’s memory size (p_memsz) is larger than the file size (p_filesz), the ‘‘extra’’ bytes are defined to hold the value 0 and to follow the segment’s initialized area. The file size may not be larger than the memory size.
• 60 8: p_memsz = d7 00 00 00 00 00 00 00: size that the segment occupies in memory
• 70 0: p_align = 00 00 20 00 00 00 00 00: 0 or 1 mean no alignment required. TODO why is this required? Why not just use p_addr directly, and get that right? Docs also say:
p_vaddr should equal p_offset, modulo p_align
The second segment (.data) is analogous. TODO: why use offset 0x0000d8 and address 0x00000000006000d8? Why not just use 0 and 0x00000000006000d8?
Then the:
Section to Segment mapping:
section of the readelf tells us that:
• 0 is the .text segment. Aha, so this is why it is executable, and not writable
• 1 is the .data` segment.
TODO where does this information come from? http://stackoverflow.com/questions/23018496/section-to-segment-mapping-in-elf-files
Comments
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2017-06-24 20:47:19
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https://jp.mathworks.com/help/signal/ug/kaiser-window.html?lang=en
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## Kaiser Window
The Kaiser window is an approximation to the prolate spheroidal window, for which the ratio of the mainlobe energy to the sidelobe energy is maximized. For a Kaiser window of a particular length, the parameter β controls the relative sidelobe attenuation. For a given β, the relative sidelobe attenuation is fixed with respect to window length. The statement kaiser(n,beta) computes a length n Kaiser window with parameter beta.
As β increases, the relative sidelobe attenuation decreases and the mainlobe width increases. This screen shot shows how the relative sidelobe attenuation stays approximately the same for a fixed β parameter as the length is varied.
Examples of Kaiser windows with length 50 and β parameters of 1, 4, and 9 are shown in this example.
To create these Kaiser windows using the MATLAB® command line, type the following:
n = 50; w1 = kaiser(n,1); w2 = kaiser(n,4); w3 = kaiser(n,9); [W1,f] = freqz(w1/sum(w1),1,512,2); [W2,f] = freqz(w2/sum(w2),1,512,2); [W3,f] = freqz(w3/sum(w3),1,512,2); plot(f,20*log10(abs([W1 W2 W3]))) grid legend('\beta = 1','\beta = 4','\beta = 9')
To create these Kaiser windows using the MATLAB command line, type the following:
w1 = kaiser(50,4); w2 = kaiser(20,4); w3 = kaiser(101,4); [W1,f] = freqz(w1/sum(w1),1,512,2); [W2,f] = freqz(w2/sum(w2),1,512,2); [W3,f] = freqz(w3/sum(w3),1,512,2); plot(f,20*log10(abs([W1 W2 W3]))) grid legend('length = 50','length = 20','length = 101')
### Kaiser Windows in FIR Design
There are two design formulas that can help you design FIR filters to meet a set of filter specifications using a Kaiser window. To achieve a relative sidelobe attenuation of –α dB, the β (beta) parameter is
$\beta =\left\{\begin{array}{cc}0.1102\left(\alpha -8.7\right),& \alpha >50,\\ 0.5842{\left(\alpha -21\right)}^{0.4}+0.07886\left(\alpha -21\right),& 50\ge \alpha \ge 21,\\ 0,& \alpha <21.\end{array}$
For a transition width of $\Delta \omega$ rad/sample, use the length
$n=\frac{\alpha -8}{2.285\Delta \omega }+1$.
Filters designed using these heuristics will meet the specifications approximately, but you should verify this. To design a lowpass filter with cutoff frequency 0.5$\pi$ rad/sample, transition width 0.2$\pi$ rad/sample, and 40 dB of attenuation in the stopband, try
[n,wn,beta] = kaiserord([0.4 0.6]*pi,[1 0],[0.01 0.01],2*pi); h = fir1(n,wn,kaiser(n+1,beta),'noscale');
The kaiserord function estimates the filter order, cutoff frequency, and Kaiser window beta parameter needed to meet a given set of frequency domain specifications.
The ripple in the passband is roughly the same as the ripple in the stopband. As you can see from the frequency response, this filter nearly meets the specifications:
fvtool(h,1)
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2021-10-23 15:40:18
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http://beta.mxnet.io/api/gluon-related/_autogen/mxnet.optimizer.Adam.html
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class mxnet.optimizer.Adam(learning_rate=0.001, beta1=0.9, beta2=0.999, epsilon=1e-08, lazy_update=True, **kwargs)[source]
This class implements the optimizer described in Adam: A Method for Stochastic Optimization, available at http://arxiv.org/abs/1412.6980.
If the storage types of grad is row_sparse, and lazy_update is True, lazy updates are applied by:
for row in grad.indices:
m[row] = beta1 * m[row] + (1 - beta1) * rescaled_grad[row]
v[row] = beta2 * v[row] + (1 - beta2) * (rescaled_grad[row]**2)
w[row] = w[row] - learning_rate * m[row] / (sqrt(v[row]) + epsilon)
The lazy update only updates the mean and var for the weights whose row_sparse gradient indices appear in the current batch, rather than updating it for all indices. Compared with the original update, it can provide large improvements in model training throughput for some applications. However, it provides slightly different semantics than the original update, and may lead to different empirical results.
Otherwise, standard updates are applied by:
rescaled_grad = clip(grad * rescale_grad + wd * weight, clip_gradient)
m = beta1 * m + (1 - beta1) * rescaled_grad
v = beta2 * v + (1 - beta2) * (rescaled_grad**2)
w = w - learning_rate * m / (sqrt(v) + epsilon)
This optimizer accepts the following parameters in addition to those accepted by Optimizer.
For details of the update algorithm, see adam_update.
Parameters: beta1 (float, optional) – Exponential decay rate for the first moment estimates. beta2 (float, optional) – Exponential decay rate for the second moment estimates. epsilon (float, optional) – Small value to avoid division by 0. lazy_update (bool, optional) – Default is True. If True, lazy updates are applied if the storage types of weight and grad are both row_sparse.
__init__(learning_rate=0.001, beta1=0.9, beta2=0.999, epsilon=1e-08, lazy_update=True, **kwargs)[source]
Initialize self. See help(type(self)) for accurate signature.
Methods
__init__([learning_rate, beta1, beta2, …]) Initialize self. create_optimizer(name, **kwargs) Instantiates an optimizer with a given name and kwargs. create_state(index, weight) Creates auxiliary state for a given weight. create_state_multi_precision(index, weight) Creates auxiliary state for a given weight, including FP32 high precision copy if original weight is FP16. register(klass) Registers a new optimizer. set_learning_rate(lr) Sets a new learning rate of the optimizer. set_lr_mult(args_lr_mult) Sets an individual learning rate multiplier for each parameter. set_lr_scale(args_lrscale) [DEPRECATED] Sets lr scale. set_wd_mult(args_wd_mult) Sets an individual weight decay multiplier for each parameter. update(index, weight, grad, state) Updates the given parameter using the corresponding gradient and state. update_multi_precision(index, weight, grad, …) Updates the given parameter using the corresponding gradient and state.
Attributes
learning_rate opt_registry
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2019-03-25 14:30:56
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https://socratic.org/questions/what-is-the-oxidation-number-of-s-in-h2so3
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# What is the oxidation number of S in H2SO3?
$+ 4$
So we have 2 hydrogen which each has a charge of $+ 1$, and we have 3 oxygen which each has a charge of $- 2$. If we cancel these out we would have that sulfur should have a charge of $+ 4$ to give a total charge of $0$.
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2022-08-10 15:15:57
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http://mathhelpforum.com/pre-calculus/77808-need-help-evaluating-trig-limits-print.html
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# Need help with evaluating Trig Limits
• March 9th 2009, 04:17 PM
Kasper
Need help with evaluating Trig Limits
Hey, this is my first time posting here so I will try to keep this brief!
So I am redoing pre-calculus to upgrade my math for post-secondary, and it's been a year since I did this in grade 12 so I'm pretty rusty. =(
I am going through limits at the moment and I came across the equation,
lim sinAx = A/B
x→0 Bx
Or otherwise
lim sinx = 1
x→0 x
Now this has helped me greatly in evaluating limits, but I remember there being a cosine limit something like this that also equals 1 which I could use to isolate and simplify the problem, but I just can't remember and I can't seem to find it in my textbooks, does anyone know of anything like this?
Thanks you very much for the help! (Hi)
Kasper
• March 9th 2009, 05:28 PM
GaloisTheory1
Quote:
Originally Posted by Kasper
Hey, this is my first time posting here so I will try to keep this brief!
So I am redoing pre-calculus to upgrade my math for post-secondary, and it's been a year since I did this in grade 12 so I'm pretty rusty. =(
I am going through limits at the moment and I came across the equation,
lim sinAx = A/B
x→0 Bx
Or otherwise
lim sinx = 1
x→0 x
Now this has helped me greatly in evaluating limits, but I remember there being a cosine limit something like this that also equals 1 which I could use to isolate and simplify the problem, but I just can't remember and I can't seem to find it in my textbooks, does anyone know of anything like this?
Thanks you very much for the help! (Hi)
Kasper
i'n not sure if there is one for cosine, but it is true that:
$\lim_{x \rightarrow 0} \frac{x}{\sin{x}}=1$
• March 10th 2009, 08:02 PM
GaloisTheory1
Quote:
Originally Posted by GaloisTheory1
i'n not sure if there is one for cosine, but it is true that:
$\lim_{x \rightarrow 0} \frac{x}{\sin{x}}=1$
the reason is that in calculus, you will/have learned that
$\lim_{x \rightarrow 0} \frac{x}{\sin{x}}=1$ is in an indeterminant form so you can use l'hopital's rule to get the limit.
however,
$\lim_{x \rightarrow 0} \frac{x}{\cos{x}}=1$
is not in an indeterminant form. if this doesn't make sense now, just wait until you take calculus.
• March 11th 2009, 02:17 AM
HallsofIvy
I believe you are thinking of
$\lim_{x\rightarrow 0} \frac{1- cos(x)}{x}= 0$
• March 11th 2009, 07:39 AM
Kasper
Quote:
Originally Posted by HallsofIvy
I believe you are thinking of
$\lim_{x\rightarrow 0} \frac{1- cos(x)}{x}= 0$
That's it! Thanks alot, that's what I was looking for.(Rock)
|
2014-08-23 18:42:25
|
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|
https://www.physicsforums.com/threads/semi-direct-product.212738/
|
# Semi Direct Product
1. Feb 2, 2008
### mathusers
thnx for helping on the previous post.
heres the next one, as usual any hints on how to approach the question would be greatly appreciated. i will attempt the questions with the hints :)
(1)
describe explicitly all homomorphisms
$\varphi : C_4 \rightarrow Aut(C_5)$
(2)
For each such homomorphism $\varphi$ describe the semidirect product $C_5 \rtimes_{\varphi} C_4$ in terms of generators and relations.
(3) How many distinct isomorphism types of groups of the form $C_5 \rtimes_{\varphi} C_4$ are there?
2. Feb 3, 2008
### morphism
It will help to know what Aut(C_5) is. Hint: It's cyclic.
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2017-07-26 20:56:33
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|
http://pisarenko.net/blog/2014/04/13/cyclo-cross-bicycle-build-part-2/
|
This is the part 2 of cyclocross bicycle build series. Jump to part 1.
Disassembly preparation
With the bicycle, materials, tools and parts figured out I was ready to begin the disassembly process. Taking things apart is fun and bicycles are not exception to this rule. As a guide I reversed the bike assembly order from Parking Tool website. The process starts with the chain removal and continue along the drive train until the levers and controls are taken off.
With the tools and parts laid before me I went into the process with ecstasy.
At first I hoped to borrow a bike stand from a friend or a colleague. Nobody replied so I purchased my own. It is a lot more convenient to work on bicycles on a proper bike stand. I used to rely on a kickstand or improvisation. With a task such as a whole bicycle assembly ad-hoc solutions become unwieldy and frustrating. I purchased Park Tool PRS-1 stand.
Cube Cross Disc bicycle looks good on its own. I really like the color scheme. I have purchased a couple of “pimping” parts to improve aesthetics even further. Things like blue bolts and cable housing, blue jockey wheels and quick release axles.
Getting down to it
There is nothing inherently difficult in taking a bicycle apart when the right tools are at hand. It’s actually a precise and highly satisfying work.
Part by part the bicycle was slowly dissolving until only a frame and wheels remained. A bicycle is a really good illusion. Initially, it appears complex, intricate and intimidating with all the seemingly difficult adjustments. However, once a bicycle is stripped of parts comes a realization of its simplicity and elegance.
The disassembly went on and on until the bicycle looked like this. Note that for the following shot I’ve already put different wheels on. I’ve left the old cables inside to guide the insertion of new ones.
Shimano 105 component groupset used to power this bike. Now it’s going to be replaced with Shimano Ultegra 6800. I will have a bunch of parts that I need to sell. I don’t fancy cluttering my place with too many random spare parts and pieces. Hopefully, there are enough bike enthusiasts at work to claim most of them.
Wheels
I have never had tubeless road bicycle tires. The assembly process was a bit unusual. Putting road tires on the Stan’s Alpha 340 wheelset was relatively painless. But I had typical problems putting cyclocross Raven tires on Alpha 400 even though I had installed the recommended additional rim strip. The problem was the tires did not seal and air went out. Tires would not inflate sufficiently to seal the gaps and air would get out. The solution was to unscrew the valve and let it move inside of the tire for initial setup. Then, I had to pump really hard to reach a critical pressure point. At the same time I was throwing the wheel up slightly so that the tire would fall into place. Still at the same time I screwed the valve back in. This maddening sequence went on until I’ve noticed that the air is not escaping as fast as it did initially. An air compressor would come in handy because it would be much easier to reach critical inflation point. Some folks report success with CO2 cartridges.
My complete order from notubes.com contained the following:
• Alpha 340 cyclocross wheelset, Shimano 11 speed
• Alpha 400 cyclocross wheelset, Shimano 11 speed
• Stan’s quart of tire sealant
• Stan’s sealant injector
• 2x Hutchison 3 tubeless tire
• 2x Raven tubeless tire
• Core remover (to inject sealant into tire without removing it)
Stans Alpha 400 wheelset with Raven cyclocross tires (700x35). I’ll use this wheelset mostly for winter commuting. I also want to try riding off road courses on this bike.
Stans Alpha 340 wheelset with Hutchison 3 road tires (700x23). Beautiful.
Did I mention that Stan’s 3.30 hubs freewheel sound gorgeous?
Newly discovered alternatives
I discovered that it is possible to buy decent bicycle frames directly from Chinese manufacturers. Initially, I was skeptical but as I read people’s reports it appears that the quality is acceptable. A carbon cyclocross frame costs 500-600$- much lower than about 2000$ most manufacturers would ask for. It’s also possible to slap a custom paint job on it!
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2018-08-20 18:22:04
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|
http://www.chegg.com/homework-help/questions-and-answers/3b-derive-following-string-grammar-figure-71-draw-corresponding-syntax-tree-a1b2c3-71-lang-q4649400
|
3b. Derive the following string with the grammar of Figure 7.1 and draw the corresponding syntax tree: a1b2c3
7.1 Languages, Grammars, and Parsing
Every language has an alphabet. Formally, an alphabet is a finite, nonempty set of
characters. For example, the C++ alphabet is the nonempty set
{ a, b, c, d, e, f, g, h, i , j, k, l , m, n,
o, p, q, r, s, t, u, v, w, x, y, z, A, B,
C, D, E, F, G, H, I, J, K, L, M, N, O, P,
Q, R, S, T, U, V, W, X, Y, Z, 0, 1, 2, 3,
4, 5, 6, 7, 8, 9, +, —, *, /, =, <, >, [,
], (, ), {, }, ., ,, :, ;, &, !, %, ' , ",
_, \, #, ?, ^, | , ~}
The alphabet for Pep/8 assembly language is similar except for the punctuation
characters, as shown in the following set:
{ a, b, c, d, e, f, g, h, i , j, k, l , m, n,
o, p, q, r, s, t, u, v, w, x, y, z, A, B,
C, D, E, F, G, H, I, J, K, L, M, N, O, P,
Q, R, S, T, U, V, W, X, Y, Z, 0, 1, 2, 3,
4, 5, 6, 7, 8, 9, \, ., ,, :, ;, ' , "}
Another example of an alphabet is the alphabet for the language of real num-
bers, not in scientific notation. It is the set
{ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, +, —, .}
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2014-09-03 00:18:42
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https://socratic.org/questions/a-weather-balloon-has-a-maximum-volume-of-7-50-10-3-l-the-balloon-contains-195-l
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# A weather balloon has a maximum volume of 7.50 * 10^3 L. The balloon contains 195 L of helium gas at a pressure of 0.993 atm. What will be the pressure when the balloon is at maximum volume?
Approx. $0.03 \cdot a t m \ldots \ldots \ldots$
We use old $\text{Boyle's Law}$, ${P}_{1} {V}_{1} = {P}_{2} {V}_{2}$ at constant temperature.
And we solve for ${P}_{2} = \frac{{P}_{1} {V}_{1}}{V} _ 2 = \frac{195 \cdot L \times 0.993 \cdot a t m}{7.5 \times {10}^{3} L}$
$= 2.58 \times {10}^{-} 2 \cdot a t m .$
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2022-01-28 18:59:44
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https://cinc.rud.is/web/packages/cdccovidview/reference/laboratory_confirmed_hospitalizations.html
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This function grabs all data for all networks, catchments, seasons, and ages. In the future there will be ways of selecting just the desired target areas.
laboratory_confirmed_hospitalizations()
data frame
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2021-12-05 05:33:28
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|
https://ccssmathanswers.com/eureka-math-grade-8-module-3-lesson-8/
|
# Eureka Math Grade 8 Module 3 Lesson 8 Answer Key
## Engage NY Eureka Math 8th Grade Module 3 Lesson 8 Answer Key
### Eureka Math Grade 8 Module 3 Lesson 8 Example Answer Key
Example 1.
In the picture below, we have triangle ABC that has been dilated from center O by a scale factor of r=$$\frac{1}{2}$$. It is noted by A’B’C’. We also have triangle A”B”C”, which is congruent to triangle A’B’C’ (i.e., △A’B’C’≅△A”B”C”).
Describe the sequence that would map triangle A”B”C” onto triangle ABC.
→ Based on the definition of similarity, how could we show that triangle A”B”C” is similar to triangle ABC?
→ To show that △A” B” C”~△ABC, we need to describe a dilation followed by a congruence.
→ We want to describe a sequence that would map triangle A”B”C” onto triangle ABC. There is no clear way to do this, so let’s begin with something simpler: How can we map triangle A’B’C’ onto triangle ABC? That is, what is the precise dilation that would make triangle A’B’C’ the same size as triangle ABC?
→ A dilation from center O with scale factor r=2
→ Remember, our goal was to describe how to map triangle A”B”C” onto triangle ABC. What precise dilation would make triangle A”B”C” the same size as triangle ABC?
→ A dilation from center O with scale factor r=2 would make triangle A”B”C” the same size as triangle ABC.
→ (Show the picture below with the dilated triangle A”B”C” noted by A”’B”’C”’.) Now that we know how to make triangle A”B”C” the same size as triangle ABC, what rigid motion(s) should we use to actually map triangle A”B”C” onto triangle ABC? Have we done anything like this before?
→ Problem 2 of the Problem Set from Lesson 2 was like this. That is, we had two figures dilated by the same scale factor in different locations on the plane. To get one to map to the other, we just translated along a vector.
→ Now that we have an idea of what needs to be done, let’s describe the translation in terms of coordinates. How many units and in which direction do we need to translate so that triangle A”’B”’C”’ maps to triangle ABC?
→ We need to translate triangle A”’B”’C”’ 20 units to the left and 2 units down.
→ Let’s use precise language to describe how to map triangle A”B”C” onto triangle ABC. We need information about the dilation and the translation.
→ The sequence that would map triangle A”B”C” onto triangle ABC is as follows: Dilate triangle A”B”C” from center O by scale factor r=2. Then, translate the dilated triangle 20 units to the left and 2 units down.
→ Since we were able to map triangle A”B”C” onto triangle ABC with a dilation followed by a congruence, we can write that triangle A”B”C” is similar to triangle ABC, in notation, △A” B” C”~△ABC.
Example 2.
In the picture below, we have triangle DEF that has been dilated from center O, by scale factor r=3. It is noted by D’ E’ F’. We also have triangle D” E” F”, which is congruent to triangle D’ E’ F’ (i.e., △D’ E’ F’≅
△D”E”F”).
→ We want to describe a sequence that would map triangle D”E”F” onto triangle DEF. This is similar to what we did in the last example. Can someone summarize the work we did in the last example?
→ First, we figured out what scale factor r would make the triangles the same size. Then, we used a sequence of translations to map the magnified figure onto the original triangle.
→ What is the difference between this problem and the last?
→ This time, the scale factor is greater than one, so we need to shrink triangle D”E”F” to the size of triangle DEF. Also, it appears as if a translation alone does not map one triangle onto another.
→ Now, since we want to dilate triangle D”E”F” to the size of triangle DEF, we need to know what scale factor r to use. Since triangle D”E”F” is congruent to D’E’F’, then we can use those triangles to determine the scale factor needed. We need a scale factor so that |OF|=r|OF’|. What scale factor do you think we should use, and why?
→ We need a scale factor r=$$\frac{1}{3}$$ because we want |OF|=r|OF’|.
→ What precise dilation would make triangle D”E”F” the same size as triangle DEF?
→ A dilation from center O with scale factor r=$$\frac{1}{3}$$ would make triangle D”E”F” the same size as triangle DEF.
→ (Show the picture below with the dilated triangle D”E”F” noted by D”’E”’F”’.) Now we should use what we know about rigid motions to map the dilated version of triangle D”E”F” onto triangle DEF. What should we do first?
→ We should translate triangle D”’E”’F”’ 2 units to the right.
→ (Show the picture below, the translated triangle noted in red.) What should we do next (refer to the translated triangle as the red triangle)?
→ Next, we should reflect the red triangle across the x-axis to map the red triangle onto triangle DEF.
→ Use precise language to describe how to map triangle D”E”F” onto triangle DEF.
→ The sequence that would map triangle D”E”F” onto triangle DEF is as follows: Dilate triangle D”E”F” from center O by scale factor r=$$\frac{1}{3}$$ . Then, translate the dilated image of triangle D”E”F”, noted by D”’ E”’ F”’, two units to the right. Finally, reflect across the x-axis to map the red triangle onto triangle DEF.
→ Since we were able to map triangle D”E”F” onto triangle DEF with a dilation followed by a congruence, we can write that triangle D”E”F” is similar to triangle DEF. (In notation: △D”E”F”~△DEF)
Example 3.
In the diagram below, △ABC ~△A’B’C’. Describe a sequence of a dilation followed by a congruence that would prove these figures to be similar.
→ Let’s begin with the scale factor. We know that r|AB|=|A’B’|. What scale factor r makes △ABC the same size as △A’B’C’?
→ We know that r⋅2=1; therefore, r=$$\frac{1}{2}$$ makes △ABC the same size as △A’ B’ C’.
→ If we apply a dilation from the origin of scale factor r=$$\frac{1}{2}$$, then the triangles are the same size (as shown and noted by triangle A”B”C”). What sequence of rigid motions would map the dilated image of △ABC onto △A’B’C’?
→ We could translate the dilated image of △ABC, △A”B”C”, 3 units to the right and 4 units down and then reflect the triangle across line A’B’.
→ The sequence that would map △ABC onto △A’B’C’ to prove the figures similar is a dilation from the origin by scale factor r=$$\frac{1}{2}$$, followed by the translation of the dilated version of △ABC 3 units to the right and 4 units down, followed by the reflection across line A’B’.
Example 4.
In the diagram below, we have two similar figures. Using the notation, we have △ABC ~△DEF. We want to describe a sequence of the dilation followed by a congruence that would prove these figures to be similar.
→ First, we need to describe the dilation that would make the triangles the same size. What information do we have to help us describe the dilation?
→ Since we know the length of side $$\overline{A C}$$ and side $$\overline{D F}$$, we can determine the scale factor.
→ Can we use any two sides to calculate the scale factor? Assume, for instance, that we know that side $$\overline{A C}$$ is
18 units in length and side $$\overline{E F}$$ is 2 units in length. Could we find the scale factor using those two sides, $$\overline{A C}$$ and $$\overline{E F}$$? Why or why not?
→ No, we need more information about corresponding sides. Sides $$\overline{A C}$$ and $$\overline{D F}$$ are the longest sides of each triangle (they are also opposite the obtuse angle in the triangle). Side $$\overline{A C}$$ does not correspond to side $$\overline{E F}$$. If we knew the length of side $$\overline{B C}$$, we could use sides $$\overline{B C}$$ and $$\overline{E F}$$.
→ Now that we know that we can find the scale factor if we have information about corresponding sides, how would we calculate the scale factor if we were mapping △ABC onto △DEF?
→ |DF|=r|AC|, so 6=r⋅18, and r=$$\frac{1}{3}$$.
→ If we were mapping △DEF onto △ABC, what would the scale factor be?
→ |AC|=r|DF|, so 18=r∙6, and r=3.
→ What is the precise dilation that would map △ABC onto △DEF?
→ Dilate △ABC from center O, by scale factor r=$$\frac{1}{3}$$.
→ (Show the picture below with the dilated triangle noted as △A’B’C’.) Now we have to describe the congruence. Work with a partner to determine the sequence of rigid motions that would map △ABC onto △DEF.
→ Translate the dilated version of △ABC 7 units to the right and 2 units down. Then, rotate d degrees around point E so that segment B’C’ maps onto segment EF. Finally, reflect across line EF.
Note that “d degrees” refers to a rotation by an appropriate number of degrees to exhibit similarity. Students may choose to describe this number of degrees in other ways.
→ The sequence of a dilation followed by a congruence that proves △ABC ~△DEF is as follows: Dilate △ABC from center O by scale factor r=$$\frac{1}{3}$$. Translate the dilated version of △ABC 7 units to the right and 2 units down. Next, rotate around point E by d degrees so that segment B’C’ maps onto segment EF, and then reflect the triangle across line EF.
Example 5.
→ Knowing that a sequence of a dilation followed by a congruence defines similarity also helps determine if two figures are in fact similar. For example, would a dilation map triangle ABC onto triangle DEF (i.e., is △ABC ~△DEF)?
→ No. By FTS, we expect the corresponding side lengths to be in proportion and equal to the scale factor. When we compare side $$\overline{A C}$$ to side $$\overline{D F}$$ and $$\overline{B C}$$ to $$\overline{E F}$$, we get $$\frac{18}{6}$$≠$$\frac{15}{4}$$.
→ Therefore, the triangles are not similar because a dilation does not map one to the other.
Example 6.
→ Again, knowing that a dilation followed by a congruence defines similarity also helps determine if two figures are in fact similar. For example, would a dilation map Figure A onto Figure A’ (i.e., is Figure A ~ Figure A’)?
→ No. Even though we could say that the corresponding sides are in proportion, there exists no single rigid motion or sequence of rigid motions that would map a four-sided figure to a three-sided figure. Therefore, the figures do not fulfill the congruence part of the definition for similarity, and Figure A is not similar to Figure A’.
### Eureka Math Grade 8 Module 3 Lesson 8 Exercise Answer Key
Exercises
Allow students to work in pairs to describe sequences that map one figure onto another.
Exercise 1.
Triangle ABC was dilated from center O by scale factor r=$$\frac{1}{2}$$. The dilated triangle is noted by A’B’C’. Another triangle A”B”C” is congruent to triangle A’B’C’ (i.e., △A”B”C”≅△A’B’C’). Describe a dilation followed by the basic rigid motion that would map triangle A”B”C” onto triangle ABC.
Triangle A”B”C” needs to be dilated from center O, by scale factor r=2 to bring it to the same size as triangle ABC. This produces a triangle noted by A”’B”’C”’. Next, triangle A”’B”’C”’ needs to be translated 4 units up and 12 units left. The dilation followed by the translation maps triangle A”B”C” onto triangle ABC.
Exercise 2.
Describe a sequence that would show △ABC ~△A’ B’ C’.
Since r|AB|=|A’ B’ |, then r⋅2=6, and r=3. A dilation from the origin by scale factor r=3 makes △ABC the same size as △A’B’C’. Then, a translation of the dilated image of △ABC ten units right and five units down, followed by a rotation of 90 degrees around point C’, maps △ABC onto △A’ B’ C’, proving the triangles to be similar.
Exercise 3.
Are the two triangles shown below similar? If so, describe a sequence that would prove △ABC ~△A’B’C’. If not, state how you know they are not similar.
Yes, △ABC ~△A’B’C’. The corresponding sides are in proportion and equal to the scale factor:
$$\frac{10}{15}$$=$$\frac{4}{6}$$=$$\frac{12}{18}$$=$$\frac{2}{3}$$=r
To map triangle ABC onto triangle A’B’C’, dilate triangle ABC from center O, by scale factor r=$$\frac{2}{3}$$. Then, translate triangle ABC along vector $$\overrightarrow{A A^{\prime}}$$. Next, rotate triangle ABC d degrees around point A.
Exercise 4.
Are the two triangles shown below similar? If so, describe the sequence that would prove △ABC ~△A’B’C’. If not, state how you know they are not similar.
Yes, triangle △ABC ~△A’B’C’. The corresponding sides are in proportion and equal to the scale factor:
$$\frac{4}{3}$$=$$\frac{8}{6}$$=$$\frac{4}{3}$$=$$1.3 \overline{3}$$; $$\frac{10.67}{8}$$=1.33375; therefore, r=1.33 which is approximately equal to $$\frac{4}{3}$$
To map triangle ABC onto triangle A’B’C’, dilate triangle ABC from center O, by scale factor r=$$\frac{4}{3}$$. Then, translate triangle ABC along vector $$\overrightarrow{A A^{\prime}}$$. Next, rotate triangle ABC 180 degrees around point A’.
### Eureka Math Grade 8 Module 3 Lesson 8 Problem Set Answer Key
Students practice dilating a curved figure and describing a sequence of a dilation followed by a congruence that maps one figure onto another.
Question 1.
In the picture below, we have triangle DEF that has been dilated from center O by scale factor r=4. It is noted by D’E’F’. We also have triangle D”E”F”, which is congruent to triangle D’E’F’ (i.e., △D’E’F’≅△D”E”F”). Describe the sequence of a dilation, followed by a congruence (of one or more rigid motions), that would map triangle D”E”F” onto triangle DEF.
First, we must dilate triangle D”E”F” by scale factor r=$$\frac{1}{4}$$ to shrink it to the size of triangle DEF. Next, we must translate the dilated triangle, noted by D”’E”’F”’, one unit up and two units to the right. This sequence of the dilation followed by the translation would map triangle D”E”F” onto triangle DEF.
Question 2.
Triangle ABC was dilated from center O by scale factor r=$$\frac{1}{2}$$. The dilated triangle is noted by A’B’C’. Another triangle A”B”C” is congruent to triangle A’B’C’ (i.e., △A”B”C”≅△A’B’C’). Describe the dilation followed by the basic rigid motions that would map triangle A”B”C” onto triangle ABC.
Triangle A”B”C” needs to be dilated from center O by scale factor r=2 to bring it to the same size as triangle ABC. This produces a triangle noted by A”’B”’C”’. Next, triangle A”’B”’C”’ needs to be translated 18 units to the right and two units down, producing the triangle shown in red. Next, rotate the red triangle d degrees around point B so that one of the segments of the red triangle coincides completely with segment BC. Then, reflect the red triangle across line BC. The dilation, followed by the congruence described, maps triangle A”B”C” onto triangle ABC.
Question 3.
Are the two figures shown below similar? If so, describe a sequence that would prove the similarity. If not, state how you know they are not similar.
No, these figures are not similar. There is no single rigid motion, or sequence of rigid motions, that would map Figure A onto Figure B.
Question 4.
Triangle ABC is similar to triangle A’B’C’ (i.e., △ABC ~△A’B’C’). Prove the similarity by describing a sequence that would map triangle A’B’C’ onto triangle ABC.
The scale factor that would magnify triangle A’B’C’ to the size of triangle ABC is r=3. The sequence that would prove the similarity of the triangles is a dilation from center O by a scale factor of r=3, followed by a translation along vector $$\overrightarrow{A^{\prime} A}$$, and finally, a reflection across line AC.
Question 5.
Are the two figures shown below similar? If so, describe a sequence that would prove △ABC ~△A’B’C’. If not, state how you know they are not similar.
Yes, the triangles are similar. The scale factor that triangle ABC has been dilated is r=$$\frac{1}{5}$$. The sequence that proves the triangles are similar is as follows: dilate triangle A’B’C’ from center O by scale factor r=5; then, translate triangle A’B’C’ along vector $$\overrightarrow{C^{\prime} C}$$; next, rotate triangle A’B’C’ d degrees around point C; and finally, reflect triangle A’B’C’ across line AC.
Question 6.
Describe a sequence that would show △ABC ~△A’ B’ C’.
Since r|AB|=|A’ B’|, then r∙3=1 and r = $$\frac{1}{3}$$. A dilation from the origin by scale factor r$$\frac{1}{3}$$ makes △ABC the same size as △A’B’C’. Then, a translation of the dilated image of △ABC four units down and one unit to the right, followed by a reflection across line A’ B’, maps △ABC onto △A’ B’ C’, proving the triangles to be similar.
In the picture below, we have triangle DEF that has been dilated from center O by scale factor r=$$\frac{1}{2}$$. The dilated triangle is noted by D’E’F’. We also have a triangle D”EF, which is congruent to triangle DEF (i.e., △DEF≅△D”EF). Describe the sequence of a dilation, followed by a congruence (of one or more rigid motions), that would map triangle D’E’F’ onto triangle D”EF.
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2022-07-04 02:33:47
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|
https://heart.bmj.com/highwire/markup/97842/expansion?width=1000&height=500&iframe=true&postprocessors=highwire_tables%2Chighwire_reclass%2Chighwire_figures%2Chighwire_math%2Chighwire_inline_linked_media%2Chighwire_embed
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Table 2
Scores for the SF-36 questionnaire of nine patients (A to I) compared with a reference population matched for age and sex
• Normal score was defined as within 1 SD of the mean of the reference population.
• Patient H suffers from aortic regurgitation and patient I from asthma.
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2020-09-25 08:05:57
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http://mikehigginbottom.com/2016/01/
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## Why Should I Not Use Pointers?
So if pointers are so great, why not only use pointers? For two reasons, both of which are consequences of the fact that pointers are a method of indirection. Indirection is, at heart, adding a step to the set of steps needed to get at the information you need. This makes the process of getting that information more complicated.
With that complexity comes, firstly, increased risk – you have to manage memory yourself and this is fraught with peril and requires both experience and discipline. The object has to exist somewhere for a pointer to have something to point to, so you now have three things to manage – the pointer, the pointee and the relationship between them.
Secondly, the added complexity of indirection brings inefficiency. This can be inefficiency in terms of CPU cycles or in terms of memory usage. Pointers are often an unnecessary overhead e.g. for(i=0; i<10; i++) would be overly verbose if written using pointer notation. If you’re just dealing locally with simple built-in types that don’t take up much space then there’s often little to be gained by using pointers. It’s only when you start trying to pass things around, particularly large things, that the overhead of indirection becomes smaller than the cycles or space wasted by not using it.
## Avoid Passing Large Chunks Of Data
Because of C’s pass by value semantics, passing data into functions using parameters requires pushing and popping that data onto and off the stack, byte by byte. Using return values to pass data out of a function back to the callee also requires copying the data onto the stack.
If you’re just passing or returning an int, then this is no big deal, but if you’re passing big data structures then all that copying can soon mount up.
The solution is to pass pointers into the function instead:
Using a pointer means passing only a few bytes rather than the whole structure. For a large array for example, instead of saying “here’s the contents of the array”, pointers allow you to say, “here’s where you can find the array already in memory”.
## Why Does My Function Not Update My Variable?
So, as a result of some aberrant brain misfiring you’ve decided to start learning to code in C. You pretty quickly come across the idea of functions and your mind is blown.
Then you try to write a function and all is not well. Not well at all. In fact things are looking a bit peaky. You’re not getting the output you expect. Again. Gah. Why was it that you chose C again?
After a stupidly long time spent debugging you finally narrow down the problem to what is obviously nonsensical behaviour. C is clearly broken and you’ve just discovered a fatal flaw that has gone undetected in it for the past fifty years.
This classic and frequent problem encountered at some stage by pretty much all newbies (unless of course they cheated and read the manual) is pretty much universally heralded by “WTF is my function not updating my variable?”
C passes parameters by value – it makes local copy variables of parameters passed into functions. Passing in a pointer means you get a copy of the pointer but that copy is still pointing to the original memory. Pointers are a way to bypass C’s ‘pass by value’ mechanism. Sometimes I want changes I make to be reflected in the source of the data so I need to use reference semantics through pointers. Sometimes I want to manipulate data independently of the original data so I use copy or value semantics. The difference may appear subtle but it is important – a classic and frequent problem encountered by newbies is “Why does my function not update my variable?”
## Pointers As Function Parameters
Suppose you’re writing a function to double a given number. You might naively assume that this would be a reasonable solution:
If you run it you’ll find number remains unchanged. This is, you might be surprised to discover, by design. It’s down to C’s pass by value semantics.
Instead, use referential semantics as follows to allow the doubleIt function to have access to the number variable.
If you run this you’ll find this function works as you would hope and does indeed double the number passed to it.
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2017-09-24 19:15:24
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https://meta.stackexchange.com/questions/353893/bug-which-mistakenly-accuses-the-writer-for-shouting-and-issues-a-warning
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Bug which mistakenly accuses the writer for shouting and issues a warning
Today, while typing an answer on Math.SE, I needed to use the factorial expression. While typing in my answer, I received the following warning:
DON'T SHOUT! All-caps posts are considered rude and will quickly be deleted.
Surprisingly, there were no letters in my answer, yet it warned me about typing in capitals.
Discuss (if possible): the mechanism of the CAPS detection function.
Bug to be fixed: training the algorithm to read for real alphabets instead of the number of exclamation marks.
Further research (on Chemistry, Physics and primary Meta): it was found that none of them really respond to caps or the above expression. For that reason, I checked my bug again on Math.SE, and it worked again.
Exact code:
$100! - 101! + ... - 109! + 110!$
Reference:
The third reference linked is stuck at evaluation and the bug has not been fixed. As per a comment :
This is caused by a blacklist entry for pattern ^(?-i:\P{Ll}+){15,}$, which matches 15 or more consecutive non-lowercase characters. Too bad you can't just use ^(?-i:\p{Lu}+){15,}$, since spaces and punctuation should not be considered to break the consecutive count. – Chris Jester-Young
The above report is definitely not valid now, look below:
looking at the above, please note there is a small a. Now look at the below one:
Now since it looks like it must be networkwide, here is an example of meta-stackexchange:
Update: After a bit of research on the S.E. network, I could not find any documentation to the warning triggered by the system. So before flagging a [bug] I would like to discuss how is the feature implemented and why is it site specific, i.e. only implemented on MSE and not on the sites namely physics, chemistry, meta stackexchange. The purpose of asking why is it site specific is that as far as I know, the backend code which can trigger such features makes it apply to the whole network.
• Yeah...and it doesn't show the error if the post does include caps, exclamation points, rude words etc. etc. etc. – Ollie Sep 5 at 19:54
• You have to click out of the text field to see any warnings. I tried entering the same text as you (a string of uppercase A's) and it was detected by the system. – Sonic the Masked Werehog Sep 7 at 5:07
• @SonictheMaskedWerehog now see the update, I typed a small "a" ago at the end (seen at the screenshot) – Queenie Goldstein Sep 7 at 5:14
• @SonictheMaskedWerehog Now I saw your screenshot, I think the "Test" word possibly failed the bug....and...that mouse pointer is BIG – Queenie Goldstein Sep 7 at 5:21
• Cagliostro, I understood that you replied to the wrong person; from the context of your comment. I've tested this on meta.math.se and a few other sites which support MathJax, as far as I can tell it ONLY :) occurs on the main math.se site and does not occur on MSE (here). --- Please demonstrate to yourself and give us a link to at least one more site where this occurs. An error that occurs on only one site should be posted on their meta, and not the main meta.; it would be off-topic here without any additional sites. At least you're up > 250! – Rob Sep 7 at 6:24
• @Rob testing is in progress on other sites ... I asked this question here assuming it is a networkwide implementation but I am surprised that a site can implement it's own features!(or the other sites rejected the global implementation). I am aware that there are certain differences between real meta and math meta for the reputations but this capitalization looks like a feature that should be on all sites. Seeing the test results, i think probably it was a hurry to put it on primary meta but I'll also wait to see the developer team comment on site specification of this feature – Queenie Goldstein Sep 7 at 6:31
• @Rob I have temporarily remove the bug tag so that we can discuss on the implementation of features which are site specific and features which are network wide – Queenie Goldstein Sep 7 at 8:27
• Maybe the page is reading your mind and anticipating that you are about to start SHOUTING! – GhostCat Sep 7 at 10:52
• @Rob the additional tags were supplied by another user. Second...studying your comment, I think I should re-include the bug tag. I undid the bug tag considering this line of yours -->more exceptions and special circumstances, the more cost and time intensive _I'll reinclude if you suggest this again because in no way am happy with the implementation which gets falsly triggered and can be avoided using just 1 letter not in caps. – Queenie Goldstein Sep 7 at 12:17
• It is possible, and not unheard of, to have a [discussion] [bug] [feature-request] tagged question - but I think the "discussion" has been had, it should work properly (not fouled by a small 'a'), and it's a "bug" that the feature doesn't work correctly on Math.SE; you'd like to "feature-request" that this be implemented on English language sites. Let's leave it for a bit, and not have Edititous (too much editing). :) --- BTW: I've deleted some of my comments, they are still available to Devs & Mods. – Rob Sep 7 at 13:19
• @Ollie Yes, I tried to test on math meta but did not know about the sandbox....so when I flagged my question for closure it was deleted and then i was told about the sandbox...so I'll remove that comm. – Queenie Goldstein Sep 9 at 4:06
• @Cagliostro copying and pasting your exact code doesn't provide you with an error here on Meta, funnily. – Ollie Sep 17 at 17:06
• @Ollie yes that has been seen(the second reference and the last image and even some comments), the capital warning is very centric and probably coded specifically into one site and forgotten by the developer to remove/upgrade – Queenie Goldstein Sep 17 at 17:50
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2020-09-30 09:05:06
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http://zbmath.org/?q=an:0676.65072
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# zbMATH — the first resource for mathematics
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##### Operators
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Diagonally implicit Runge-Kutta-Nyström methods for oscillatory problems. (English) Zbl 0676.65072
A number of second order two and three stage diagonally implicit Runge- Kutta-Nyström (DIRKN) methods with low phase error are developed for the initial value problem $\left(i\right)\phantom{\rule{1.em}{0ex}}{y}^{\text{'}\text{'}}=f\left(t,y\right),$ $\left(ii\right)\phantom{\rule{1.em}{0ex}}y\left(0\right)={y}_{0},$ ${y}^{\text{'}}\left(0\right)={y}_{0}^{\text{'}}·$ The definition of phase error and amplitude error for the Runge-Kutta-Nyström algorithm associated with (i), (ii) has been developed by the authors in an earlier paper [ibid. 24, 595-617 (1987; Zbl 0624.65058)]. The phase error $\phi$ ($\omega$ h) and amplification error $\alpha$ ($\omega$ h) are defined as extensions of the explicitly known formulas for deviation of the phase and amplitude of the RKN solution of (i), (ii) from the true solution when $f=-{\omega }^{2}y$. An RKN method is said to be dispersive and dissipative of order q and r if $\left(iii\right)\phantom{\rule{1.em}{0ex}}\phi \left(\omega h\right)=O\left({h}^{q+1}\right),$ $\left(iv\right)\phantom{\rule{1.em}{0ex}}\alpha \left(\omega h\right)=O\left({h}^{r+1}\right)·$ The authors construct two-stage second order DIRKN algorithms in which $q=4,6$, one of which is unconditionally strongly stable. Also three-stage algorithms are constructed with $q=6,8,10$, one of which is P-stable. Results of computation of actual dispersion error are presented using the new algorithms on five test problems. The new algorithms are better than standard DIRKN methods in case the dispersion error grows larger than the truncation error.
Reviewer: J.B.Butler jun.
##### MSC:
65L05 Initial value problems for ODE (numerical methods) 34A34 Nonlinear ODE and systems, general 34C10 Qualitative theory of oscillations of ODE: zeros, disconjugacy and comparison theory
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2014-03-11 17:10:53
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https://space.stackexchange.com/questions/14317/how-well-can-voyager-1-separate-earth-signals-from-solar-noise-these-days
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# How well can Voyager 1 separate Earth signals from Solar noise these days?
This is a follow-up question to Can Voyager 1 receive signals from Earth? and this and this answer. UPDATED: see additional information and discussion below.
As it continues to move farther from the Sun, the angular separation between the Earth and Sun continues to decrease, can its antenna actually resolve the two and limit the noise from the sun? (seems to be about 0.4 degrees at opposition now) For that matter, how strong IS the noise from the sun relative to earth transmissions, considering the passband of Voyager's electronics - is is it a serious issue to begin with? As earth oscillates in its orbit - is there a seasonal effect?
Besides the mind-boggling large distances an weak signal, the problem I'm talking about here is that - as seen from Voyager 1 (and 2), the earth is only a small fraction of 1° away from the sun, which is a powerful and noisy radio source.
Voyager receives at around 2 GHz1, so its 3.7m diameter dish can not separate the two. Even a quiet sun is almost $$\text{10}^6$$ Jy. The electronics is circa 1970, if it has an input bandwidth presented to the front-end of 10MHz, the sun will be a million times stronger than the 20kW Deep Space Network DSN signal.
above: data for the Sun, planets, Pluto, Voyager 1 and Voyager 2, from January 1, 1969 (a good year to start things) until now. Dots are now. Data is from NASA JPL Horizons.
above: Angular separation between the earth and the sun in degrees, as seen from Voyager 1 (heavy, blue) and Voyager 2 (light, green). The dips to near-zero stop happening once the spacecraft left the plane of the ecliptic.
above: example of the type of settings I used to get the data in ecliptic coordinates.
above: Partial screen shots from DSN Now: https://eyes.nasa.gov/dsn/dsn.html, just for fun because this is downlink at 8 GHz rather than uplink at 2 GHz.
1 from All About Circuits' Communicating Over Billions of Miles: Long Distance Communications in the Voyager Spacecraft (a fun read!):
The uplink carrier frequency of Voyager 1 is 2114.676697 MHz and 2113.312500 MHz for Voyager 2. The uplink carrier can be modulated with command and/or ranging data. Commands are 16-bps, Manchester-encoded, biphase-modulated onto a 512 HZ square wave subcarrier.
from DESCANSO Design and Performance Summary Series Article 4: Voyager Telecommunications as discussed in this answer.
• This has been on the back burner in my mind since you posted. I't starting to bug me so I'll do some research and post an answer when i get some free time. Mar 7, 2016 at 18:48
• @AndrewW. I wouldn't want you to leave Voyager 1 on the back burner too long! If you can just add a short answer with some kind of a link. Am I right - about 0.4 degrees max at opposition? Yikes! Is the sun actually very noisy?
– uhoh
Jun 4, 2016 at 14:59
• I inquired about this at JPL's Space Flight Operations Facility, and they said that they can, and do, still talk with the Voyager probes, usually on a daily basis, and that they still have clear communication with both. Jul 19, 2016 at 19:59
• @uhoh I'm not sure about every day, but I see one or both of them on there fairly often, labeled as VGR1 and VGR2. Voyager 1 is talking right now. Jul 20, 2016 at 1:21
• Woops, I forgot path loss in the DSN signal the received signal is P.G.A.4.pi/d^2. Here's a link showing an example of link budget from the DSN: propagation.gatech.edu/ECE6390/project/Fall2010/Projects/group7/… Jul 21, 2016 at 19:05
According to this source, the quiescent Sun produces about 10-20 Watts per square meter per Hertz at 2 GHz at Earth's orbit. At a distance of 152 AU, it will be a factor of 1522 weaker or about 4 x 10-25 W/m2/Hz. Collected with a 12 foot dish (~10 square meters) gives 4 x 10-24 W/Hz for solar noise.
Thermal noise in the front end is kT W/Hz. Assuming a noise temperatue of, say, 250K, this is 1.4 x 10-23 x 250 = 3.5 x 10-21 W/Hz
So the solar noise is much less than the thermal noise from Voyager's front-end amplifier. The fact that the antenna provides no discrimination doesn't matter.
The radio noise from the Sun can be over 1000 times larger during a solar storm. It would then be comparable to the front-end noise and could be a problem.
• This sounds pretty conclusive, and surprisingly low, thanks! Related: What exactly is the interaction that blocked Juno's data downlink near solar conjunction? and pointing out that Stars are not particularly "loud" in radio are answers to: How far have individual stars been seen by radio telescopes?
– uhoh
Apr 9, 2021 at 7:47
• @uhoh Yes, the situation is very different the other way round. The downlink antenna has more than 100x the area of Voyager and the cryogenic receiver noise is probabably 10x less. Plus the sun is now at 1 AU, so there's that reduction by152^2 also missing. I have no idea what the plasma around the sun does to the transmission, but it can't be good. The only positive thing is that the directivity of the big down-link antenna is much better, especially at X-band. Apr 9, 2021 at 19:21
• @uhoh It is remarkable to think that, within that narrow beamwidth and that narrow bandwidth, the Goldstone DSN antenna is much more luminous than the Sun. Apr 9, 2021 at 19:25
• Yes indeed, the world looks very different in radio! Luckily, while Earth always appears close to the Sun from the Voyagers, they don't usually appear close to the Sun from Earth.
– uhoh
Apr 9, 2021 at 19:28
I was fortunate to have worked on Voyager, and other projects at JPL from 1970 to 1975. I was also fortunate to have had Solomon Golomb, PhD, as my advisor and mentor in Electrical Engineering graduate school at USC in the late 1960's. I wanted to study communications theory, and Dr. Golomb was the only professor at USC who was involved in that area. According to information from USC published at the time of his death, Dr. Golomb's research when I was his research assistant is the reason NASA can separate faint radio signals sent from spacecraft from much stronger background noise. (Incidentally, this research is given the credit for us having CD's, DVD's, and cell phones.) I really did not and do not really understand the research that I helped this reknowned mathematician who was a full professor of Electrical Engineering perform. I do know that without him, we would not be able to decipher Voyager's signals.
I believe that the Voyager receiver uses a PLL, phase locked loop to acquire the signal and filter out the solar noise. Also, the solar radiation is different to the transmitted signal which Voyager can use to differentiate between the noise with a band pass filter and some other electronics.
• Can you tell how you know this, and what it implies? Aug 13, 2016 at 13:53
The separation is still not too hard, for two reasons. One, the sun emits incoherent noise, while the DSN signal is carefully designed for coherent detection and processing by the spacecraft receivers. Two, the sun is a broadband source, which divides its power in the band roughly equally across the whole thing, while the DSN signal occupies only a tiny fraction of the band, so its power density is comparable. However, as the Voyagers continue to get farther away, the data rates they can comfortably support continue to decrease.
The main reference text for what follows is the DSN Telecommunications Link Design Handbook (Module 202), to which the DESCANSO document linked above directs the reader who wants even more detail than that contained. The DESCANSO text is also useful, answering parts of this question starting about five pages after the beam width graphic posted above, but I'll get to that a bit later.
Phase Coherence
The uplink signal carrier phase is tracked with the classic Costas variant of the digital phase locked loop. Costas, John P., "Synchronous communications", Proceedings of the IRE 44 (12) 1713–1718, 1956, doi:10.1109/jrproc.1956.275063.
The frequency which the PLL tracks measures the Doppler shift caused by the spacecraft motion (for deep space, always receding very rapidly). The beauty of this scheme is that the whole point of a phase-locked loop is to search the input spectrum for anything that might be coherent, and latch onto it while discarding everything else. There are comments below about what happens when the sun gets very close to the line of sight back to Earth, but the problem is not the total power output of the sun; it is the scintillation $$-$$ the variation of power with time $$-$$ that forces the PLL to run with so large a tracking bandwidth that it can't lock up properly.
Since the frequency of a carrier equals the rate-of-change of carrier phase, the Downlink Channel supports Doppler measurement by extracting the phase of the downlink carrier (Reference 1).
The Receiver and Ranging Processor (RRP) accepts the signal from the IDC and extracts carrier phase with a digital phase-locked loop (Reference 2). The loop is configured to track the phase of a phase-shift keyed signal with residual carrier, a suppressed carrier, or a QPSK signal.
There is an additional loss to the carrier loop signal-to-noise ratio when tracking a residual carrier with non-return-to-zero symbols in the absence of a subcarrier. This loss is due to the presence of data sidebands overlaying the residual carrier in the frequency domain and therefore increasing the effective noise level for carrier synchronization. In this case, $$\rho_L$$ must be calculated as (Reference 3)
$$S_L$$ = squaring loss of the Costas loop (Reference 4),
The one-sided, noise-equivalent, carrier loop bandwidth is denoted $$B_L$$. The user may choose to change $$B_L$$ during a tracking pass, and this can be implemented without losing phase-lock, assuming the change is not too large. There are limits on the carrier loop bandwidth. $$B_L$$ can be no larger than 200 Hz. The lower limit on $$B_L$$ is determined by the phase noise on the downlink. In general, the value selected for $$B_L$$ should be small in order to maximize the carrier loop signal-to-noise ratio. On the other hand, $$B_L$$ must be large enough that neither of the following variables becomes too large: the static phase error due to Doppler dynamics and the contribution to carrier loop phase error variance due to phase noise on the downlink. The best $$B_L$$ to select will depend on circumstances. Often, it will be possible to select a $$B_L$$ of less than 1 Hz. A larger value for $$B_L$$ is necessary when there is significant uncertainty in the downlink Doppler dynamics, when the downlink is one-way (or two-way non-coherent) and originates with a less stable oscillator (such as an Auxiliary Oscillator), or when the Sun-Earth-probe angle is small (so that solar phase scintillations are present on the downlink).
The user may select either a type 2 or type 3 carrier loop. Both loop types are perfect, meaning that the loop filter implements a true accumulation. In the presence of a persistent Doppler acceleration, a type 2 loop will periodically slip cycles.
They don't have a reference for explaining type 2 versus type 3 filters. A recent one I found interesting was P. Kanjiya, V. Khadkikar and M. S. E. Moursi, "Obtaining Performance of Type-3 Phase-Locked Loop Without Compromising the Benefits of Type-2 Control System," IEEE Transactions on Power Electronics 33(2) 1788-1796, 2018, doi: 10.1109/TPEL.2017.2686440.
References they did put in the above are:
1. P. W. Kinman, "Doppler Tracking of Planetary Spacecraft", IEEE Transactions on Microwave Theory and Techniques 40(6) 1199-1204, 1992.
2. J. B. Berner and K. M. Ware, "An Extremely Sensitive Digital Receiver for Deep Space Satellite Communications", Eleventh Annual International Phoenix Conference on Computers and Communications, pp. 577-584, Scottsdale, Arizona, April 1-3, 1992.
3. J. Lesh, "Tracking Loop and Modulation Format Considerations for High Rate Telemetry", DSN Progress Report 42-44, Jet Propulsion Laboratory, Pasadena, CA, pp. 117-124, April 15, 1978.
4. M. K. Simon and W. C. Lindsey, "Optimum Performance of Suppressed Carrier Receivers with Costas Loop Tracking", IEEE Transactions on Communications 25(2) 215-227, 1977.
Power Density
10 MHz is the full range of the RF spectrum allocated for the use of deep space research satellite communication, from 2010 MHz to 2020 MHz. However, the DSN does not fill that uniformly, and the spacecraft do not pay close attention to all channels at once. This is like your radio, which picks up the entire 20 MHz FM band (88.7 to 108.7 MHz), but only listens to one 200 kHz channel at a time. The difference is that, in order for this to work with the Voyagers, the channel instantaneous bandwidth needs to be very small.
The central result of communication theory is the channel capacity formula (Shannon 1948), which relates the theoretical maximum bit rate, $$C$$, to the occupied signal bandwidth, $$B$$ (though giving a rigorous theoretical definition of bandwidth is tricky), and the relative power of signal, $$S$$, and noise, $$N$$, as
$$C=B\log_2\left(1+\frac{S}{N}\right)$$
Note this uses the ratio $$S/N$$ not expressed in decibels, so 20 dB SNR means plug in $$S/N$$ = 100 to obtain $$C=B\log_2(101)\approx 6.66 B$$. If $$S = N$$, so $$S/N = 1$$, then $$C = B\log_2(2)=B$$. The quantity $$1 + S/N$$ equals $$(S+N)/N$$, which is what the display on a spectrum analyzer actually shows when you look at the amplitude of the Fourier transform as a function of offset frequency. If the "hill" you see on the screen is 3 dB high, that means $$(S+N)/N=2$$, so $$S=N$$. This statistic is sometimes called the Lawson-Uhlenbeck deflection ratio, in honor of the classic textbook Lawson, J. L., & Uhlenbeck, G. E., Threshold Signals, MIT Radiation Lab Series, Volume 24, New York, McGraw Hill (1950).
The data rates actually used in DSN for Voyager seem ridiculously tiny by near-Earth standards, but channel capacity and power spectral density tell us why it needs to be like that. Voyager Telecommunications on page 60 says
the command processor assembly (CPA) and the command modulator assembly (CMA) clock out the command bit stream, modulate the command subcarrier, and provide the modulated subcarrier to the station’s exciter for modulation of the RF uplink carrier. The command bit rates, the command subcarrier frequency, and the command modulation index (suppression of the uplink carrier) are controlled through standards and limits tables.
...directs the station to turn command modulation on and selects the 16-bps command rate and a calibrated “buffer” in the station’s CMA. The CMA produces the command subcarrier, which produces a 512-Hz square wave to match the subcarrier tracking-loop best-lock frequency in the Voyager CDU.
Exact numbers from here on depend on how exactly you prefer to define "bandwidth", and just how much of the DSN signal's power fits inside each one. The basic idea is the sun spreads its power nearly uniformly across the whole 10 MHz, as a blackbody in a narrow band is probably the closest nature gets to the theoretically-beloved "additive white Gaussian noise" (AWGN). When you view that on a power spectral density display, you get the result of dividing by bandwidth.
That is, since each bin in your histogram shows, for example, 1 Hz of bandwidth, then only one ten-millionth of the sun's power falls in each bin. If the whole DSN signal falls entirely within one such bin $$-$$ which is entirely possible if it is operating as an unmodulated carrier $$-$$ its SNR in that one bin is ten million times what its average SNR is over the whole 10 MHz.
Even when somewhat wider than that, there is still a considerable advantage; for example, if we use 512 Hz as the nominal bandwidth, then in those 512 of your 10 million bins, the DSN SNR will be 10,000,000/512 = 19,500 times higher than its average over all 10 million bins. The "actual" bandwidth of a square wave depends strongly on the author's preference, since it is a sum of all the odd harmonics, which gives the classic $$\sin(x)/x$$ shape in the frequency domain, with sidelobes that fall rather slowly. However, if we take 512 Hz at face value for $$B$$, then using 16 bps for $$C$$ means we only need $$1/32 = C/B = \log_2(1+S/N)$$, so SNR could theoretically be as low as -16.5 dB and DSN could still get the message through, even before considering things like error-correction coding gain.
Different tasks performed with different modulation schemes require different levels of SNR for equivalently acceptable performance, but explaining what they all are is quite complicated. Instead, I refer you to pages 64 and 65 of Voyager Telecommunications, which shows plots of how the ratio of telemetry power to background noise varies by year and day of year (the curves) versus SNR needed for certain bit rates to provide desired performance (the horizontal dashed and dotted lines). In those plots (which are actually for X band, not S band, but the idea is the same and they're the only ones I could find), variation based on time of year is as large as the difference between 2020 and 2005, but less than the difference between 2005 and 1995 (earlier years have bigger gaps than later ones).
Since you "modified" your question: The purpose of an antenna is to receive a signal not to filter noise. The Power voyager receives, could be sun + earth, earth - sun, sun - earth or only sun. The Modulation used is still unknown, so i cant calculate the needed SNR for it to work.
But the normal radiation level shouldn't be a problem for the voyager probe. Think of the sun as a constant emitter. Only flares create spikes. The spectrum of these radio waves will have a Gaussian bell curve (Watts/ Hz). The Engineers of Nasa will probably have picked a Frequency spectrum that is on one of the lower ends of this bell curve. The radio power of the sun decreases faster per Meter away from its origin then that of earth. Since earth uses directional communication and the sun is just a big ball of energy. So the true problem will be, to always point at voyager correctly.
• I'm pretty sure that it depends on many things. If you search for words like Voyager 1, high gain antenna, uplink (means from DSN to Voyager) and start reading, there's a lot there. The sun is pretty noisy at ~2GHz and the high gain antenna can not separate the earth from the sun. It transmits at ~8GHz but (if I read correctly) uplink is always at the lower frequency. There is a lot of work the 1970's space-hardened electronics has to do before demodulation. It's really incredible stuff!
– uhoh
Jul 20, 2016 at 13:38
• You cant calculate the Signal to Noise ratio without the energy put into the signal. Yes they state the Hertz and the kb/s, but these numbers dont help calculating the actual Watts the probe gets to use to decide what a signal means. The work you are referring is probably the signal transformation to a bit stream. (Demodulating the signal -> deciding what bit / byte that is -> checking against check sum)Again, without the used technique it is for me impossible to say what the signal to noise ratio is or, what the actual needed ratio is.
– Git
Jul 20, 2016 at 13:53
• That's a downlink signal, just to illustrate the "...mind-boggling large distances an weak signal...". So far I haven't seen an uplink. I'll add that to the label, Any receiver system has some frequency limitation before the first active stage. I read somewhere that the feed horn has a 20MHz bandpass for example, but there may also be a tuned circuit. Since the sun is broadband, the wider the frequency range exposed to the front-end (first active amplification state) the more chance of overloading it, especially if the sun is active. It's a real-world problem, not just a demodulation problem
– uhoh
Jul 22, 2016 at 0:15
• The frequency of the sun does not pose a problem, since it always stays the same. So you just have to use a Filter for these Frequencies, their Amplitudes. So you tune to your specific frequency, add a low high pass filter to get just the amplitudes you are interested, and then transform your signal back to lose the carrier. Then all the nice digital stuff takes place. The signal does not lose so much strength over distance since it originated from a dish not just a wire. But the sun does.
– Git
Jul 22, 2016 at 8:19
• @Git 1) the sun is a broadband source including some energy at the uplink frequency. 2) The inverse quare law applies just as much to the transmit antenna as it does to the sun. . Apr 9, 2021 at 6:30
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2023-03-25 13:12:07
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https://www.hackmath.net/en/math-problem/15013?tag_id=102,9
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The prison ball
Calculate the density of the material that the prison ball is made from if you know its diameter is 15cm and its weight is approximately 2.3kg. With the help of mathematical-physicochemical tables estimate what material the ball is made from.
Result
h = 1301.534 kg/m3
Solution:
$D=15 \ cm \rightarrow m=15 / 100 \ m=0.15 \ m \ \\ m=2.3 \ \text{kg} \ \\ \ \\ r=D/2=0.15/2=\dfrac{ 3 }{ 40 }=0.075 \ \text{m} \ \\ \ \\ V=\dfrac{ 4 }{ 3 } \cdot \ \pi \cdot \ r ^3=\dfrac{ 4 }{ 3 } \cdot \ 3.1416 \cdot \ 0.075 ^3 \doteq 0.0018 \ \text{m}^3 \ \\ \ \\ h=m/V=2.3/0.0018 \doteq 1301.5338 \doteq 1301.534 \ \text{kg/m}^3$
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2020-05-30 20:22:08
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https://chemistry.stackexchange.com/questions/55757/why-is-difluoride-a-stronger-oxidizing-agent-than-dichloride
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# Why is difluoride a stronger oxidizing agent than dichloride? [closed]
I am preparing for a chemistry olympiad and I am stuck with the question that why $$\ce{F_{2}}$$ is stronger oxidizing agent than $$\ce{Cl_{2}}$$.
## closed as off-topic by orthocresol♦, Jon Custer, Todd Minehardt, Wildcat, A.K.Jul 30 '16 at 23:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
If this question can be reworded to fit the rules in the help center, please edit the question.
• Fluorine has a more positive redox potential (probably the most positive) at +2.87 V thus it spontaneously oxidises most everything else as they have a lower potential. Cl has a potential 1.35 V. Note also that Oxidation is loss of electrons, Reduction gain, thus F is reduced in oxidising Cl. – porphyrin Jul 30 '16 at 9:39
If you think about it, strong oxidizing agents must have a high "desire" or propensity to be reduced themselves (as the nature of redox reactions is such). We'll build up on the explanation from the ground up.
Comparing both fluorine and chlorine gas, both are halogens and both only need two electrons each to be reduced to fluoride and chloride ions respectively:
$$\ce{X2 + 2e- <=> 2 X-}$$
Since it appears that the two are similar in most respects, we should look at the clearest differences between Fluorine and Chlorine instead: the number of protons and the number of electrons.
Fluorine is a row 2 element with 9 protons an electronic configuration 1s2 2s2 2p5, whereas chlorine is a row 3 element with 17 protons and electronic configuration 1s2 2s2 2p6 3s2 3p5. Although chlorine has more protons to attract electrons (and therefore be reduced / act as an oxidizing agent), it also has two inner shells worth of electrons that create a shielding effect to reduce the attraction. Fluorine, however, only has one inner shell to do this.
From a simplistic physics point of view, electronic attraction is governed by the charge and the inverse square of the distance. You can approximate that the shielding effect will make the effective charge of the atomic nucleus the same in both halogens, so distance becomes the key deciding factor. Since chlorine is larger, and therefore the attracted electron further than it would be if it was instead bound to fluorine, the attraction (and therefore oxidizing power) is greater in fluorine than in chlorine.
However, looking at the electron affinity (EA) values you'd find that fluorine doesn't follow the group trend of decreasing EA and there's a very good discussion on that over at ChemLibre. Long story short, it's because fluorine is so small that you have to account for the electronic repulsion as well which reduces the bond strength of fluorine vs chlorine (36.6 vs 58 kcal/mol)
There is another measurement which seems to be consistent with this whole fluorine being more oxidizing than chlorine though - and that is reduction potential. The reason that there is a difference between the values for reduction potential and electron affinity even though they are both supposed to be measuring the same thing (that is the energy change related to accepting an electron), the key is the environment in which they are measured. Electron affinities are measured in the gas phase where it is purely the energy related to the electron transfer whereas reduction potentials are measured in aqueous solution and so factors such as hydration energy start to play a role.
As your textbook correctly points out, the high hydration energy is due to fluorine's higher charge density compared to that of chlorine.
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2019-08-24 02:44:29
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https://en-academic.com/dic.nsf/enwiki/2505013
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# Mass distribution
Mass distribution
Mass distribution is a term used in physics and mechanics and describes the spatial distribution of mass within a solid body. In principle, it is relevant also for gases or liquids, but on earth their mass distribution is almost homogeneous.
## Astronomy
In astronomy mass distribution has decisive influence on the development e.g. of nebulae, stars and planets. The mass distribution of a solid defines its center of gravity and influences its dynamical behaviour - e.g. the oscillations and eventual rotation.
## Mathematical modelling
A mass distribution can be modeled as a measure. This allows point masses, line masses, surface masses, as well as masses given by a volume density function. Alternatively the latter can be generalized to a distribution. For example, a point mass is represented by a delta function defined in 3-dimensional space. A surface mass on a surface given by the equation f(x,y,z) = 0 may be represented by a density distribution g(x,y,z) δ (f(x,y,z)), where $g/\mid \nabla f \mid$ is the mass per unit area.
The mathematical modelling can be done by potential theory, by numerical methods (e.g. a great number of mass points), or by theoretical equilibrium figures.
## Geology
In geology the aspects of rock density are involved.
## Rotating solids
Rotating solids are affected considerably by the mass distribution, either if they are homogeneous or inhomogeneous - see Torque, moment of inertia, wobble, imbalance and stability.[disambiguation needed ]
## Related topics
Wikimedia Foundation. 2010.
### Look at other dictionaries:
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• distribution massique — masių pasiskirstymas statusas T sritis fizika atitikmenys: angl. mass distribution vok. Massenverteilung, f rus. распределение масс, n pranc. distribution de masses, f; distribution massique, f … Fizikos terminų žodynas
• Mass drug administration — Distribution of antimalarials in Italy in the 1930s The administration of drugs to whole populations irrespective of disease status is referred to as mass drug administration (MDA). This article describes the administration of antimalarial drugs… … Wikipedia
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2021-06-20 20:07:36
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https://mathoverflow.net/questions/315572/can-mk-be-interpreted-in-a-class-theory-about-an-abstract-hierarchy-principle
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# Can MK be interpreted in a class theory about an abstract hierarchy principle + an accessibility principle?
The following is a first order MONOSORTED class theory, that is primarily motivated by an abstract hierarchy principle. It extends first order logic with equality, its language has only two extra-logical primitives those of equality $$="$$ and class membership $$\in"$$. A class function $$F$$ is said to be hierarchical if and only if it sends ordinals to sets in such a manner that $$F(A) \subsetneq F(B) \leftrightarrow A < B$$. Now this theory stipulates that every hierarchical function if the class of all limit ordinals that are members of elements of its domain is equal in cardinality to some set, then the range of that function is a set. So this theory combines an accessibility principle with an abstract hierarchy principle. I tend to think that the result is equi-interpretable with $$MK$$ , however I'm not sure. Here is the formal exposition of it:
Define: $$Set(X) \iff \exists Y (X \in Y)$$
Axioms:
Extensionality: $$\forall A,B [\forall X(X \in A \leftrightarrow X \in B) \to A=B]$$
Foundation: $$\exists Y (Y \in X) \to \exists Y \in X (\not \exists C \in Y (C \in X))$$
Comprehension: if $$\varphi(Y,W_1,..,W_n)$$ is a formula in which only symbols $$Y,W_1,..,W_n"$$ occur free (and only free), and in which the symbol $$X"$$ doesn't occur, then : $$\forall W_1,..,W_n \exists X \forall Y [Y \in X \leftrightarrow Set(Y) \wedge \varphi(Y,W_1,..,W_n)]$$ is an axiom
Define: $$X= \{Y|\varphi(Y,W_1,..,W_n)\} \iff \forall Y [Y \in X \leftrightarrow Set(Y) \wedge \varphi(Y,W_1,..,W_n)]$$
Set existence: $$\exists X (Set(X))$$
Pairing: $$\forall A,B,X [Set(A) \wedge Set(B) \wedge X=\{A,B\} \to Set(X)]$$
Union: $$\forall A,X [Set(A) \wedge X=\{Y|\exists Z \in A (Y \in Z)\} \to Set(X)]$$
Power:$$\forall A,X [Set(A) \wedge X=\{Y|\forall Z \in Y (Z \in A)\} \to Set(X)]$$
Subsets: $$\forall X,Y [ Set(X) \wedge Y \subset X \to Set(Y)]$$
Hierarchy: $$\forall F \big{[} \forall M \in F \ \exists A,B (A \in ORD \wedge M= \langle A,B \rangle) \wedge \\ \forall A,B,X,Y ( \langle A,X \rangle \in F \wedge \langle B, Y \rangle \in F \to [A < B \leftrightarrow X \subsetneq Y] \wedge [A=B \leftrightarrow X=Y]) \wedge \\ \exists S (Set(S) \wedge |S|= | \{L | \exists A,B (\langle A,B \rangle \in F \wedge L \in A) \wedge \not \exists K (L=K \cup \{K\}) \}|) \\ \to \\ Set(range(F)) \big{]}$$
Where $$ORD$$ is the class of all Von Neumann ordinals that are sets.
Question: Is this theory equi-interpretable with $$MK$$?
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2019-02-19 16:03:59
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https://www.transtutors.com/questions/swings-incorporated-makes-one-product-a-wooden-swing-made-entirely-by-hand-the-compa-2573087.htm
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# Swings Incorporated makes one product, a wooden swing made entirely by hand. The company planned ...
Swings Incorporated makes one product, a wooden swing made entirely by hand. The company planned to make 4,000 swings during the month, but because of unexpected demand was able to complete 4,200 swings. In order to be able to meet the excess demand, the company hired less experience workers. The standard cost of each swing is as follows: Standard materials 120 feet of lumber Standard cost $4 per foot Standard labor hours 25 hours Standard labor rate$15 per hour Because the company was not able to get its usual high grade of lumber, more lumber was needed. The company purchased 530,000 feet of lumber and used 515,000 feet of lumber at $3.50 per foot. Material price variances are the responsibility of the purchasing manager, while material quantity variances are the responsibility of the production manager. The company paid total labor charges of$1,365,000 because 26 hours were needed to complete each swing using the lower-quality lumber. Calculate each of the following and state whether the variance is favorable (F) or unfavorable (U): a. labor efficiency variance b. labor rate variance c. materials quantity variance d. materials price variance
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2018-06-18 05:53:49
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https://www.physicsforums.com/threads/area-of-figure-resulting-from-unit-square-transformation.816464/
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# Area of figure, resulting from unit square transformation
1. May 29, 2015
### Myr73
1- Let the linear operator on R^2 have the following matrix:
A = 1 0
-1 3
What is the area of the figure that results from applying this transformation to the unit square?
2- I am abit confused here, I thought that the matrix for the unit square would be,
0 0
1 0
0 1
1 1....
But apparently that is not it.
And after finding the unit square matrix, I blv I would find the images of it by computing A^Tx
I am not sure though
2. May 29, 2015
### DEvens
A square would be four points. You seem to have collected four points that make up one particular unit square, and arranged them as a 2x4 array. That is kind of odd.
What does it mean to be a linear operator on R^2? What does the operator A operate on? Hint: R^2 means there are two real numbers involved. But there seem to be 8 real numbers involved in your square. So how is that going to work?
3. May 29, 2015
### Myr73
oh. ok, so are they asking to find the Ax? where x would be the unit square.. so like
[10. ] [ 1 0]
[ 1 3 ] [ 01 ] ...
Last edited: May 29, 2015
4. May 30, 2015
anyone?
5. Jun 1, 2015
### HallsofIvy
That matrix is NOT the unit square. The unit square is a geometric figure, not a matrix, that has vertices at (0, 0), (1, 0), (1, 1), and (0, 1).
The bottom side, from (0, 0) to (1, 0), the vector $\begin{bmatrix}1 \\ 0 \end{bmatrix}$, is mapped into
$$\begin{bmatrix}1 & 0 \\ -1 & 3\end{bmatrix}\begin{bmatrix}1 \\ 0 \end{bmatrix}= \begin{bmatrix}1 \\ -1\end{bmatrix}$$
The left side, from (0, 0) to (0, 1), the vector $\begin{bmatrix}0 \\ 1 \end{bmatrix}$, is mapped into
$$\begin{bmatrix}1 & 0 \\ -1 & 3 \end{bmatrix}\begin{bmatrix}0 \\ 1 \end{bmatrix}= \begin{bmatrix}0 \\ 3\end{bmatrix}$$
The other two sides get mapped into equivalent vectors so this is a parallelogram with vertices at (0, 0), (1, -1), (1, 2), and (0, 3).
It should be easy to find the area of that parallelogram.
(And, it is worth noting that this matrix has determinant 3.)
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2018-05-22 02:44:38
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https://www.physicsforums.com/threads/lagrangian-of-simple-pendulum.664335/
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# Lagrangian of simple pendulum
1. Jan 14, 2013
### cpsinkule
1. The problem statement, all variables and given/known data
2 masses, m$_{1}$ and m$_{2}$ are fixed at the endpoints of a rigid rod of length l. mass m$_{1}$ is attached to a horizontal bar so that it may move in the x direction freely, but not in the y direction. let θ be the angle the rod makes with the vertical, what is the corresponding Lagrangian of the system if it is assumed to be in the uniform gravitational field g?
2. Relevant equations
Euler Lagrange eqns
3. The attempt at a solution
The only issue I am running into with this problem is in the kinetic energy terms. My kinetic terms are : $\frac{1}{2}$(m$_{1}$+m$_{2}$)$\dot{x}$$^{2}$+$\frac{1}{2}$m$_{2}$l$^{2}$$\dot{θ}$$^{2}$ but the book proposes a solution the same as mine except with the added term $\frac{1}{2}$m$_{2}$(2l$\dot{x}$$\dot{θ}$cosθ). I am not understanding where this term comes from, I thought I took care of the x velocity dependence in the first term.
2. Jan 14, 2013
### TSny
Find expressions for the x and y coordinates of m2 in terms of the x coordinate of m1 and θ.
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2018-03-24 12:26:27
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https://mathematica.stackexchange.com/questions/236029/matrix-multiplication-of-non-commuting-objects
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# Matrix multiplication of non-commuting objects
I have two matrices, $$A$$ and $$B$$. The elements of these matrices are some abstract non-commuting objects, which I'm just representing as variables. For now, I don't care about having Mathematica know about the specifics of these commutation relations - all I care about is that it doesn't assume that they commute, so that, for example, when I calculate something like $$\text{Tr}(A B)$$, the expression that it returns keeps the elements of $$A$$ in front of the elements of $$B$$. Of course for something as simple as $$\text{Tr}(AB)$$ I can just rearrange the elements by hand, but I would like to have the ability to do this for more complex expressions, with potentially more than two matrices. It seems Mathematica has some built-in functionality for non-commutative multiplication, but this is only for scalars - is there something that generalizes this to matrices of non-commuting objects?
As Dot is a special case of Inner with Dot[##] == Inner[Times,##, Plus], we can just tell Mathematica to use NonCommutativeMultiply in the first slot to take the place of regular multiplication.
ncdot[ten1_, ten2_] := Inner[NonCommutativeMultiply, ten1, ten2, Plus]
ncdot[{x, y}, {u, v}]
ncdot[{{x, y}, {z, w}}, {{a, b}, {c, d}}]
Tr@%
x ** u + y ** v
{{x ** a + y ** c, x ** b + y ** d}, {w ** c + z ** a, w ** d + z ** b}}
w ** d + x ** a + y ** c + z ** b
This contrasts the automatic canonical sorting MMA does under the hood of the individual symbols as in
{x, y}.{u, v}
(*and*)
{{x, y}, {z, w}}.{{a, b}, {c, d}} // Tr
u x + v y
d w + a x + c y + b z
To use this for several items (e.g. matrices $$A,B,C$$), we incorporate/redefine Fold into the definition as
ncdot[tens__] := Fold[Inner[NonCommutativeMultiply, ##] &, {tens}]
(*Without the inclusion of the fourth argument of Inner, the default is Plus*)
Which we can use as
ncdot @@ Table[i[j, k], {i, {a, b, c}}, {j, 2}, {k, 2}];
% /. (i : (a | b | c))[j_, k_] :> Subscript[i, j\[InvisibleComma]k] // MatrixForm
$$\left( \begin{array}{cc} \left(a_{1,1}\text{**}b_{1,1}+a_{1,2}\text{**}b_{2,1}\right)\text{**}c_{1,1}+\left(a_{1,1}\text{**}b_{ 1,2}+a_{1,2}\text{**}b_{2,2}\right)\text{**}c_{2,1} & \left(a_{1,1}\text{**}b_{1,1}+a_{1,2}\text{**}b_{2,1}\right)\text{**}c_{1,2}+\left(a_{1,1}\text{**}b_{ 1,2}+a_{1,2}\text{**}b_{2,2}\right)\text{**}c_{2,2} \\ \left(a_{2,1}\text{**}b_{1,1}+a_{2,2}\text{**}b_{2,1}\right)\text{**}c_{1,1}+\left(a_{2,1}\text{**}b_{ 1,2}+a_{2,2}\text{**}b_{2,2}\right)\text{**}c_{2,1} & \left(a_{2,1}\text{**}b_{1,1}+a_{2,2}\text{**}b_{2,1}\right)\text{**}c_{1,2}+\left(a_{2,1}\text{**}b_{ 1,2}+a_{2,2}\text{**}b_{2,2}\right)\text{**}c_{2,2} \\ \end{array} \right)$$
Maybe this?
Tr[{{a, b}, {c, d}}.{{x, y}, {z, w}}] /. Times -> NonCommutativeMultiply
Tr[{{a, b}, {c, d}}.{{x, y}, {z, w}} /. Times -> NonCommutativeMultiply]
a ** x + b ** z + c ** y + d ** w
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2021-07-23 16:19:31
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http://mathhelpforum.com/number-theory/111733-how-close-fermats-theorem-print.html
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# How close to Fermat's theorem?
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• November 1st 2009, 11:41 AM
wonderboy1953
How close to Fermat's theorem?
MS = Magic Squares(s)
Multigrade example (through to the fifth power, a pentagrade):
1+15+22+50+57+71 = 2+11+27+45+61+70 = 5+6+35+37+66+67 = 216
12+152+222+502+572+712 = 22+112+272+452+612+702
= 52+62+352+372+662+672 = 11,500
13+153+223+503+573+713 = 23+113+273+453+613+703
= 53+63+353+373+663+673 = 682,128
14+154+224+504+574+714 = 24+114+274+454+614+704
= 54+64+354+374+664+674 = 42,502,564
15+155+225+505+575+715 = 25+115+275+455+615+705
= 55+65+355+375+665+675 = 2,724,334,416.
Multigrade equalities still hold by adding any integer n to every base term in both sides. Hence, the base equality always has number 1 appearing in one side (you have to imagine the exponent in this example).
I like recreational math. I was exploring multigrades in connection with MS. I was thinking outside of the box and I thought about doing a dot product with some multigrades.
There's this MS:
16 2 3 13
5 11 10 8
9 7 6 12
4 14 15 1
The first and last rows make a bigrade. I did a dot product with the numbers 1,2,3,4 on the first and last rows and there was equality on the first power, but not the second. The numbers 2,9,15 and 8 make a trigrade with 3,5,14 and 12 so I tried a 1,2,3,4 dot product with this trigrade. There's equality for the first and second powers (I like to refer to this as the first and second levels), but no equality for the third power. Strange. I tried a dot product (1-8) with the trigrade 1,6,11,16,4,7,10,13 and the remaining numbers in the MS, but still no equality beyond the second level (please note that the members of the multigrade must be arranged from low to high).
More recently I tried higher multigrades (tetragrades, pentagrades, etc.), but no equality beyond the second level (btw some multigrades may have no equality when applying the dot product). This situation reminds me of Fermat's theory which was only proven in 1994. If this is true for all cases, then the following questions arise: first, how close is this to Fermat's theorem where you can translate what I call Gauss's dot multiplier (GDM) on the multigrades into Fermat's equation? second, how easy is it to prove that GDM holds for all cases with the multigrades? (Andrew Wiles proof on Fermat's theorem ran 130 pages), third, if true, how much would this add to our understanding of math?
So that's really the story. I don't know what I stumbled onto and maybe someone out there can help out with this.
• November 2nd 2009, 08:13 AM
Media_Man
Diophantine Dizziness
This is very interesting stuff. However, I hate to tell you that it has little or nothing to do with the famed Fermat's Last Theorem. That states that no integer solutions $(x,y,z)$ exist that solve $x^n+y^n=z^n$ when $n>2$. The formulation of this theorem is extremely specific, as by simply adding a fourth term, $x^n+y^n=z^n+w^n$, there are plenty of solutions for many higher values of $n$.
This is much closer to Hilbert's Tenth problem, which asks whether there is a general method of determining if a Diophantine equation has solutions. Unfortunately, the answer is no. That means that by changing or altering a single thing in a previously known Diophantine problem, you create an entirely new problem whose solution has nothing to do with the seemingly similar original problem.
So no, your question probably cannot somehow be tied to Fermat. As for your conjecture, that no multigrade can be dotted by any vector and made into a new multigrade higher than two, you may do better to recruit the help of a computer program to accrue some evidence for such a claim, before attempting to prove it. Counterexamples tend to be easier to find than proofs.
• November 2nd 2009, 10:04 AM
wonderboy1953
Media_ an
"Counterexamples tend to be easier to find than proofs." I hope that someone does find a counterexample, otherwise this conjecture would need a proof. (btw I don't have a home computer and my time is limited at the library)
• November 3rd 2009, 09:06 AM
wonderboy1953
Media_Man - further commentary
I'm a mathematician by inclination, not a scientist meaning I'm not looking for evidence, just proof or disproof. Euler had his prime generating formula which worked for the first 41 numbers. This doesn't mean it would work for the next 41 numbers, nor even the next (which it doesn't).
Currently I don't have access to a private computer which would help speed things up, but even then at rock bottom, no computer can come up with conceiving a proof which only an intelligent creature or being can do, just check cases.
The proposition I put forth so far corresponds to Fermat's theorem. I'll keep trying to disprove my proposition and I welcome others to do so. It's either that or somehow prove it. If proven and if it translates into all Fermat cases, it may simplify the proof for Fermat's theorem and help deepen our understanding of math plus broaden our math knowledge base.
• November 4th 2009, 10:47 AM
Media_Man
Counterexample
The trigrade (2,9,15,8),(3,5,14,12) forms another trigrade when dotted with (16,4,4,1).
Sorry if I sounded dismissive. It's not that this is an easy problem, because it's not. If you were to find an important connection between multigrades and magic squares, that would be a significant discovery.
In this particular case though, I imagine that you based your conjecture on a few dozen hand calculations, when in fact, only one in ten thousand or so might actually end up working. Multigrades are extremely rare so it should come as no surprise that "dotting" one with a random vector should produce a new multigrade. If you dig through the first 20 straws in a hay stack and don't find a needle, that doesn't mean its not there. But if a computer confirms there is no needle in 90% of the haystack, that lends evidence that there may be none to be found. That is what I meant by "evidence."
I do wish you luck on your continued search for connections.
• November 4th 2009, 11:02 AM
wonderboy1953
Responding to Media_Man
"Multigrades are extremely rare..." Not true. In fact there are several methods for making them including a particular one that's extremely easy to use on any monograde to make a multigrade(s). I still stand by what I posted before.
• November 4th 2009, 03:45 PM
wonderboy1953
Clarification
As far as their number goes, it turns out that the number of multigrades is infinite (including what I call multiple multigrades involving more than one equal sign).
Again evidence isn't what's needed, proof is.
• November 4th 2009, 05:47 PM
alunw
I had trouble even understanding what a multigrade is. Possibly something got lost in translation over the web but not putting the ^ to denote taking powers makes your first post hard to understand. And you don't even have a clear conjecture, just some interesting numerical identities. Why should there even be a connection between your multigrades and Fermat's last theorem? I'm impressed that you have discovered examples of multigrades that work to ^5, but you haven't established the slightest shred of a connection between them and Fermat's last theorem. It's only Media Man's posts that allows me to know what you are conjecturing at all. It's good that you want proof or disproof of your conjecture, but for your conjecture to interest mathematicians you really need more evidence than this, and some argument to show why your conjecture might be important. To get mathematicians more interested in what you are talking about it might be a good idea to back up your claim that you have a method for producing multigrades from a monograde.
• November 5th 2009, 01:33 AM
HallsofIvy
Wonderboy1953, it would help if you would define "multigrade". I googled "multigrade" and got a lot of hits about "multigrade oils" but nothing connected with mathematics.
• November 5th 2009, 04:55 AM
Media_Man
HallsofIvy: Mathworld has a very accessible definition about multigrades, including several parametric solutions: http://mathworld.wolfram.com/MultigradeEquation.html
I haven't read any literature connecting multigrades to magic squares or FLT, but such a connection is what Wonderboy1953 is attempting to find.
• November 6th 2009, 09:17 AM
wonderboy1953
Responding
First Media-Man:
"If you were to find an important connection between multigrades and magic squares, that would be a significant discovery." The connection is that neither are valid beyond the second power for any positive number (although at this stage, my observation is still a conjecture).
"If you dig through the first 20 straws in a hay stack and don't find a needle, that doesn't mean its not there. But if a computer confirms there is no needle in 90% of the haystack, that lends evidence that there may be none to be found. That is what I meant by "evidence." The problem here is that there are an infinity of multigrades, not just 10, 100, 1,000 or even a million. So I don't know where you and alunw draw the line in the sand (I'd like to ask how you derived the 16,4,4,1 combo that creates another trigrade?).
" I haven't read any literature connecting multigrades to magic squares or FLT." You should read Alfred Posamentier's "Math Charmers" that shows a connection between MS and multigrades (which goes to infinity).
Now alunw:
"...but not putting the ^ to denote taking powers makes your first post hard to understand." sorry about that as I tried to. I'll be more careful in the future.
"...you really need more evidence than this, and some argument to show why your conjecture might be important." Its possible connection to FLT is what makes it important (I would read Fermat's Enigma by Simon Singh which may enlighten you).
"To get mathematicians more interested in what you are talking about it might be a good idea to back up your claim that you have a method for producing multigrades from a monograde." Your wish is my command so I'll offer you an example:
1 + 4 = 2 + 3, add 1 to each member:
2 + 5 = 3 + 4, now switch sides:
3 + 4 = 2 + 5, now combine the first and third equations:
1 + 3 + 4 + 4 = 2 + 2 + 3 + 5, cancel the 3's:
1^n + 4^n + 4^n = 2^n + 2^n + 5^n, valid for n = 1,2
We have just made a bigrade. You can change the numbers to whatever you like and you can repeat the method to make a trigrade and keep repeating the method to however how high you like to go with the multigrades.
Now HallsofIvy:
"Wonderboy1953, it would help if you would define "multigrade". I googled "multigrade" and got a lot of hits about "multigrade oils" but nothing connected with mathematics." Only the oils? What about the classrooms? Try to focus on math multigrades and search through 20 pages with Google.
A multigrade is an equality relationship between two groups of numbers where the sum of their powers is equal for two or more distinct, positive powers.
• November 6th 2009, 10:22 AM
Media_Man
Clarity
Wonderboy1953: We all appreciate your patience, but I think I speak for the rest of us posters when I say your original question is extremely difficult to follow. I understand what multigrades are; I understand what magic squares are. But I have yet to understand what exactly it is that you are conjecturing. I thought you were claiming that no multigrade of level>2 can be "dotted" with another vector, and create another multigrade of level>2. If that is the conjecture you were intending, then the counterexample I found makes it definitively false, since I took a trigrade, dotted it with (16,4,4,1) and made another trigrade out of it. (Sorry, by the way, I cannot enlighten you with my derivation of this result. I took my own advice and let a computer sift through thousands of candidates and that is one that popped up. You can find the code at the end of this post if you are interested.) Since you are continuing to defend your observation, I can only assume that I have misinterpreted your original question.
I know you are relying on hand calculations in making these observations. I would be more than happy to use a computer to search for a counterexample, if it exists. Obviously, this has no relevance for proving a conjecture, but is the easiest way to disprove it.
Again, what exactly is it that you are conjecturing?
int max=50;
for (int c1=-max; c1<max; c1++){
for (int c2=-max; c2<max; c2++){
for (int c3=-max; c3<max; c3++){
for (int c4=-max; c4<max; c4++){
if ((-c1+4*c2+c3-4*c4==0)
&&(-5*c1*c1+56*c2*c2+29*c3*c3-80*c4*c4==0)
&&(-19*c1*c1*c1+604*c2*c2*c2+631*c3*c3*c3-1216*c4*c4*c4==0)
&&!(c1==c2 && c2==c3 && c3==c4))
System.out.println("c1="+c1+", c2="+c2+", c3="+c3+", c4="+c4);
}}}}
• November 6th 2009, 02:27 PM
tonio
Quote:
Originally Posted by Media_Man
Wonderboy1953: We all appreciate your patience, but I think I speak for the rest of us posters when I say your original question is extremely difficult to follow. I understand what multigrades are; I understand what magic squares are. But I have yet to understand what exactly it is that you are conjecturing. I thought you were claiming that no multigrade of level>2 can be "dotted" with another vector, and create another multigrade of level>2. If that is the conjecture you were intending, then the counterexample I found makes it definitively false, since I took a trigrade, dotted it with (16,4,4,1) and made another trigrade out of it. (Sorry, by the way, I cannot enlighten you with my derivation of this result. I took my own advice and let a computer sift through thousands of candidates and that is one that popped up. You can find the code at the end of this post if you are interested.) Since you are continuing to defend your observation, I can only assume that I have misinterpreted your original question.
I know you are relying on hand calculations in making these observations. I would be more than happy to use a computer to search for a counterexample, if it exists. Obviously, this has no relevance for proving a conjecture, but is the easiest way to disprove it.
Again, what exactly is it that you are conjecturing?
int max=50;
for (int c1=-max; c1<max; c1++){
for (int c2=-max; c2<max; c2++){
for (int c3=-max; c3<max; c3++){
for (int c4=-max; c4<max; c4++){
if ((-c1+4*c2+c3-4*c4==0)
&&(-5*c1*c1+56*c2*c2+29*c3*c3-80*c4*c4==0)
&&(-19*c1*c1*c1+604*c2*c2*c2+631*c3*c3*c3-1216*c4*c4*c4==0)
&&!(c1==c2 && c2==c3 && c3==c4))
System.out.println("c1="+c1+", c2="+c2+", c3="+c3+", c4="+c4);
}}}}
Well said. I'd add also that anyone really trying to forward in this site some ideas about something he/she thinks may be important related to mathematics will better learn some basic LaTex in this site (just as I did in the last 3 weeks or so) and will make an effort to write down clearly and extensively about his/her ideas, with definitions, examples and etc.
I also can't tell so far what exactly is wonderboy trying to tell us, let alone what his conjectures about anything are.
Tonio
• November 7th 2009, 07:47 AM
wonderboy1953
Media_Man (and Tonio)
I thought you were claiming that no multigrade of level>2 can be "dotted" with another vector, and create another multigrade of level>2. First I never brought up vectors in any of my posts on this thread. I'm talking about dotting with consecutive numbers 1,2,3,4... (if it's a 2-termed multigrade on both sides of the equal sign, a^n + b^n = c^n + d^n, then I would dot with 1 and 2 with the multigrade arranged from lowest to highest numbers on both sides of the equal sign; if it's a 3-termed multigrade on both sides of the equal sign, a^n + b^n + c^n = d^n + e^n + f^n, then I would dot with 1,2,3 with the multigrade going from lowest to highest numbers on both sides of the equal sign, the number of consecutive numbers I dot with depends on the number of terms on either side of the multigrade).
I should point out that I would add a 0 to a multigrade to make it symmetric because the chances of my observation about equality to the second power improves when you do so. E.g. if you have a multigrade: a^n + b^n = c^n + d^n + e^n, then add a 0 to the left side to make it 0^n + a^n + b^n = c^n + d^n + e^n and dot multiply with 1,2,3 because you've made a three-termed multigrade on both sides of the equal sign.
A demonstration: when you have this bigrade, 4^n + 9^n + 2^n = 8^n + 1^n + 6^n for n = 1,2, rearrange from lowest to highest: 2^n + 4^n + 9^n = 1^n + 6^n + 8^n and dot multiply with 1,2,3 to get:
1 x 2^n + 2 x 4^n + 3 x 9^n = 1 x 1^n + 2 x 6^n + 3 x 8^n, you will see that you get equality for only n = 1.
Another demo: I mentioned earlier about the trigrade 2,9,15,8 and 3,5,14,12. Rearrange the numbers to 2,8,9,15 and 3,5,12,14 and dot multiply with 1,2,3,4 to get: 1 x 2^n + 2 x 8^n + 3 x 9^n + 4 x 15^n =
1 x 3^n + 2 x 5^n + 3 x 12^n + 4 x 14^n which is true for n = 1,2.
I've checked a number of multigrades with dot multiplication on the consecutive numbers and n never goes above 2 to have equality just like Fermat's equation also never has equality when n goes above 2. (btw even if you do have symmetric multigrades with its terms arranged from lowest to highest doesn't guarantee equality for n = 1 or n = 2 which also corresponds to FLT).
Would like to mention to Media_Man that I find 16,4,4,1 interesting and I thank him for his input for using his computer.
I would recommend reading Simon Singh's "Fermat's Enigma" particularly where it was proven that elliptic equations and modular forms are the same
(and I feel that my observation relates to this).
• November 7th 2009, 08:31 AM
wonderboy1953
To Media_Man
"I cannot enlighten you with my derivation of this result. I took my own advice and let a computer sift through thousands of candidates and that is one that popped up." You have already enlightened me. I hope I have enlightened you.
BTW I have came up with another method for making a multigrade two days ago starting with another multigrade which I'll demonstrate with the base numbers:
2,8,15,9 = 3,5,14,12 (a trigrade), go from right to left and add numbers that are next to each other - the end numbers add the numbers on the far left:
10^n + 23^n+ 24^n + 11^n = 8^n+19^n + 26^n + 15^n which is valid for n = 1,2,3. Simple algebra will show why this method works.
I want to thank you for helping me out with your computer (incidentally I know about half a dozen methods for making multigrades. I refrain from going into this because I don't know if the purpose of this website is for recreational math or more serious math such as helping out with homework problems - if you read the book I mentioned, you'll see why I posted my findings here as it seems to be of major importance to mathematics)
It can be a year before I get a home computer. In the meantime I look forward to your computer help, but the best thing would be to prove or disprove my conjecture. (thank you too Tonio)
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
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2015-08-05 03:03:04
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https://2021.help.altair.com/2021/feko/topics/feko/user_guide/solver_solution_methods/mom_testing_integral_equation_feko_c.htm
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# Testing and Solving the Integral Equation
The testing of the integral equation applies the integral equation over each triangle edge to obtain N equations with N unknowns which can readily be solved on a computer.
For arbitrarily shaped bodies the integral operation is much more complicated than that for a straight wire. It involves several mathematically complex derivations and pitfalls to navigate around of which some are as follows:
• When the integral equations are tested, the so-called self-terms are problematic. The testing of the integral equation at or very near the same position as the unknown leads to a (near) singularity in the matrix equation. In Feko, a computationally efficient methodology is adopted to deal with this problem.
• The testing of the integral equation applies the boundary condition (zero tangential electric field all over the surface of the conductor) at discrete points. Between these points the boundary conditions are not satisfied and this deviation is denoted the “residual”. Naturally, this residual introduces deviations from the exact physical solution. One way to minimize the residual is to minimize the average residual all over the structure. For this purpose, a set of vector weighting functions are defined. Different weighting functions were proposed and the implementation in Feko is beyond the scope of this document.
Note: Minimizing the deviation from the boundary conditions is denoted the “method of weighted residuals” or more commonly, the method of moments.
The testing of Equation 2 results in a square matrix very similar to Equation 7.
This equation can be solved for the current coefficients by using LU decomposition routines.
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2022-12-03 05:00:27
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https://web2.0calc.com/questions/fractions_23623
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+0
# Fractions
0
42
1
On Parents Day John gave his mother 2 blue flowers, 1 pink flower, and 5 white flowers. What fraction of the flowers is blue? Pink? White? Order the fractions from the greatest to the least.
Feb 18, 2022
#1
+1382
+1
Blue: $$2 \over 8$$
Pink: $$1 \over 8$$
White: $$5 \over 8$$
You can reorder them from here...
Feb 18, 2022
#1
+1382
+1
Blue: $$2 \over 8$$
Pink: $$1 \over 8$$
White: $$5 \over 8$$
You can reorder them from here...
BuilderBoi Feb 18, 2022
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2022-05-25 02:44:39
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https://electronics.stackexchange.com/questions/288709/what-is-the-best-way-to-evaluate-gps-antenna-performance-by-receiver-data-only
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# What is the best way to evaluate GPS antenna performance by receiver data only?
I'd like to test GPS Antenna performance, but I don't have access to any expensive RF equipment. What is the best way to evaluate GPS antenna performance by using only standard NMEA sentence type data? The Sirf and U-blox based receivers I am using allows me to enable almost any standard NMEA sentence.
One crude way of testing an antenna is seeing if it actually gets a lock -- however, that's the most basic test. I'd like to evaluate further.
Some ideas that I have:
1. Measuring number of satellites it can lock to
2. Looking at all the signal strengths of all satellites
3. Adding up all signal strengths of all satellites that it can lock to.
4. Look at the HDOP and VDOP values -- however, perhaps this is not helpful or meaningful as this only depends on the geometry of the locked satellites.
• What are your performance metrics - what would qualify as Good and Better for you? Where will it be mounted? What is the GPS information used for? Are you comparing good and great, or average and poor antennas - are you looking to weed out complete disasters or to get the absolute best from a selection of available antennas? – tomnexus Feb 25 '17 at 20:21
• @tomnexus, all antennas are active, magnetic mount antennas placed on top of a roof of a vehicle. All are typically using a 25mm patch, with a builtin LNA, and 3-5m cable length hooked up to a receiver inside a vehicle. I am comparing good and great Antennas and I am trying to find the absolute best from a selection of available antennas. Antennas range from $5 / each to$100 / each. – Adam B Feb 26 '17 at 2:14
• Adding reference I just found: digikey.ca/en/pdf/t/tallysman-wireless/gnss-antenna-performance – Adam B Feb 26 '17 at 5:18
• – Adam B Feb 26 '17 at 5:20
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2019-10-23 00:28:02
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https://www.bbc.co.uk/editorialguidelines/guidelines/war-terror-emergencies/demonstrations
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# Editorial Guidelines
## Demonstrations, Disturbances and Riots
11.4.8
Comprehensive coverage of demonstrations, disturbances and riots is an important part of our news reporting. However, in addition to the specific guidelines concerning accuracy and impartiality (see 11.4.2 above), it is important that:
• we assess the risk that, by previewing likely prospects of disturbances, we might encourage them
• we withdraw immediately if we suspect we are inflaming the situation
• when reporting live, we must either install a delay, or cut away and record material for use in an edited report, if the level of violence or disorder becomes too graphic.
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2019-06-24 12:30:03
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|
https://www.scala-lang.org/files/archive/spec/2.11/06-expressions.html
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Expressions
Expr ::= (Bindings | id | _') =>' Expr
| Expr1
Expr1 ::= if' (' Expr )' {nl} Expr [[semi] else' Expr]
| while' (' Expr )' {nl} Expr
| try' ({' Block }' | Expr) [catch' {' CaseClauses }'] [finally' Expr]
| do' Expr [semi] while' (' Expr ')'
| for' ((' Enumerators )' | {' Enumerators }') {nl} [yield'] Expr
| throw' Expr
| return' [Expr]
| [SimpleExpr .'] id =' Expr
| SimpleExpr1 ArgumentExprs =' Expr
| PostfixExpr
| PostfixExpr Ascription
| PostfixExpr match' {' CaseClauses }'
PostfixExpr ::= InfixExpr [id [nl]]
InfixExpr ::= PrefixExpr
| InfixExpr id [nl] InfixExpr
PrefixExpr ::= [-' | +' | ~' | !'] SimpleExpr
SimpleExpr ::= new' (ClassTemplate | TemplateBody)
| BlockExpr
| SimpleExpr1 [_']
SimpleExpr1 ::= Literal
| Path
| _'
| (' [Exprs] )'
| SimpleExpr .' id s
| SimpleExpr TypeArgs
| SimpleExpr1 ArgumentExprs
| XmlExpr
Exprs ::= Expr {,' Expr}
BlockExpr ::= ‘{’ CaseClauses ‘}’
| ‘{’ Block ‘}’
Block ::= BlockStat {semi BlockStat} [ResultExpr]
ResultExpr ::= Expr1
| (Bindings | ([implicit'] id | _') :' CompoundType) =>' Block
Ascription ::= :' InfixType
| :' Annotation {Annotation}
| :' _' *'
Expressions are composed of operators and operands. Expression forms are discussed subsequently in decreasing order of precedence.
Expression Typing
The typing of expressions is often relative to some expected type (which might be undefined). When we write "expression $e$ is expected to conform to type $T$", we mean:
1. the expected type of $e$ is $T$, and
2. the type of expression $e$ must conform to $T$.
The following skolemization rule is applied universally for every expression: If the type of an expression would be an existential type $T$, then the type of the expression is assumed instead to be a skolemization of $T$.
Skolemization is reversed by type packing. Assume an expression $e$ of type $T$ and let $t_1[\mathit{tps}_1] >: L_1 <: U_1 , \ldots , t_n[\mathit{tps}_n] >: L_n <: U_n$ be all the type variables created by skolemization of some part of $e$ which are free in $T$. Then the packed type of $e$ is
$T$ forSome { type $t_1[\mathit{tps}\_1] >: L_1 <: U_1$; $\ldots$; type $t_n[\mathit{tps}\_n] >: L_n <: U_n$ }.
Literals
SimpleExpr ::= Literal
Typing of literals is as described here; their evaluation is immediate.
The Null Value
The null value is of type scala.Null, and is thus compatible with every reference type. It denotes a reference value which refers to a special “null” object. This object implements methods in class scala.AnyRef as follows:
• eq($x\,$) and ==($x\,$) return true iff the argument $x$ is also the "null" object.
• ne($x\,$) and !=($x\,$) return true iff the argument x is not also the "null" object.
• isInstanceOf[$T\,$] always returns false.
• asInstanceOf[$T\,$] returns the default value of type $T$.
• ## returns 0.
A reference to any other member of the "null" object causes a NullPointerException to be thrown.
Designators
SimpleExpr ::= Path
| SimpleExpr .' id
A designator refers to a named term. It can be a simple name or a selection.
A simple name $x$ refers to a value as specified here. If $x$ is bound by a definition or declaration in an enclosing class or object $C$, it is taken to be equivalent to the selection $C$.this.$x$ where $C$ is taken to refer to the class containing $x$ even if the type name $C$ is shadowed at the occurrence of $x$.
If $r$ is a stable identifier of type $T$, the selection $r.x$ refers statically to a term member $m$ of $r$ that is identified in $T$ by the name $x$.
For other expressions $e$, $e.x$ is typed as if it was { val $y$ = $e$; $y$.$x$ }, for some fresh name $y$.
The expected type of a designator's prefix is always undefined. The type of a designator is the type $T$ of the entity it refers to, with the following exception: The type of a path $p$ which occurs in a context where a stable type is required is the singleton type $p$.type.
The contexts where a stable type is required are those that satisfy one of the following conditions:
1. The path $p$ occurs as the prefix of a selection and it does not designate a constant, or
2. The expected type $\mathit{pt}$ is a stable type, or
3. The expected type $\mathit{pt}$ is an abstract type with a stable type as lower bound, and the type $T$ of the entity referred to by $p$ does not conform to $\mathit{pt}$, or
4. The path $p$ designates a module.
The selection $e.x$ is evaluated by first evaluating the qualifier expression $e$, which yields an object $r$, say. The selection's result is then the member of $r$ that is either defined by $m$ or defined by a definition overriding $m$.
This and Super
SimpleExpr ::= [id .'] this'
| [id '.'] super' [ClassQualifier] .' id
The expression this can appear in the statement part of a template or compound type. It stands for the object being defined by the innermost template or compound type enclosing the reference. If this is a compound type, the type of this is that compound type. If it is a template of a class or object definition with simple name $C$, the type of this is the same as the type of $C$.this.
The expression $C$.this is legal in the statement part of an enclosing class or object definition with simple name $C$. It stands for the object being defined by the innermost such definition. If the expression's expected type is a stable type, or $C$.this occurs as the prefix of a selection, its type is $C$.this.type, otherwise it is the self type of class $C$.
A reference super.$m$ refers statically to a method or type $m$ in the least proper supertype of the innermost template containing the reference. It evaluates to the member $m'$ in the actual supertype of that template which is equal to $m$ or which overrides $m$. The statically referenced member $m$ must be a type or a method. <!-- explanation: so that we need not create several fields for overriding vals -->
If it is a method, it must be concrete, or the template containing the reference must have a member $m'$ which overrides $m$ and which is labeled abstract override.
A reference $C$.super.$m$ refers statically to a method or type $m$ in the least proper supertype of the innermost enclosing class or object definition named $C$ which encloses the reference. It evaluates to the member $m'$ in the actual supertype of that class or object which is equal to $m$ or which overrides $m$. The statically referenced member $m$ must be a type or a method. If the statically referenced member $m$ is a method, it must be concrete, or the innermost enclosing class or object definition named $C$ must have a member $m'$ which overrides $m$ and which is labeled abstract override.
The super prefix may be followed by a trait qualifier [$T\,$], as in $C$.super[$T\,$].$x$. This is called a static super reference. In this case, the reference is to the type or method of $x$ in the parent trait of $C$ whose simple name is $T$. That member must be uniquely defined. If it is a method, it must be concrete.
Example
Consider the following class definitions
class Root { def x = "Root" }
class A extends Root { override def x = "A" ; def superA = super.x }
trait B extends Root { override def x = "B" ; def superB = super.x }
class C extends Root with B {
override def x = "C" ; def superC = super.x
}
class D extends A with B {
override def x = "D" ; def superD = super.x
}
The linearization of class C is {C, B, Root} and the linearization of class D is {D, B, A, Root}. Then we have:
(new A).superA == "Root",
(new C).superB = "Root", (new C).superC = "B",
(new D).superA == "Root", (new D).superB = "A", (new D).superD = "B",
Note that the superB function returns different results depending on whether B is mixed in with class Root or A.
Function Applications
SimpleExpr ::= SimpleExpr1 ArgumentExprs
ArgumentExprs ::= (' [Exprs] )'
| (' [Exprs ,'] PostfixExpr :' _' *' ')'
| [nl] BlockExpr
Exprs ::= Expr {,' Expr}
An application $f$($e_1 , \ldots , e_m$) applies the function $f$ to the argument expressions $e_1 , \ldots , e_m$. If $f$ has a method type ($p_1$:$T_1 , \ldots , p_n$:$T_n$)$U$, the type of each argument expression $e_i$ is typed with the corresponding parameter type $T_i$ as expected type. Let $S_i$ be type type of argument $e_i$ $(i = 1 , \ldots , m)$. If $f$ is a polymorphic method, local type inference is used to determine type arguments for $f$. If $f$ has some value type, the application is taken to be equivalent to $f$.apply($e_1 , \ldots , e_m$), i.e. the application of an apply method defined by $f$.
The function $f$ must be applicable to its arguments $e_1 , \ldots , e_n$ of types $S_1 , \ldots , S_n$.
If $f$ has a method type $(p_1:T_1 , \ldots , p_n:T_n)U$ we say that an argument expression $e_i$ is a named argument if it has the form $x_i=e'_i$ and $x_i$ is one of the parameter names $p_1 , \ldots , p_n$. The function $f$ is applicable if all of the following conditions hold:
• For every named argument $x_i=e_i'$ the type $S_i$ is compatible with the parameter type $T_j$ whose name $p_j$ matches $x_i$.
• For every positional argument $e_i$ the type $S_i$ is compatible with $T_i$.
• If the expected type is defined, the result type $U$ is compatible to it.
If $f$ is a polymorphic method it is applicable if local type inference can determine type arguments so that the instantiated method is applicable. If $f$ has some value type it is applicable if it has a method member named apply which is applicable.
Evaluation of $f$($e_1 , \ldots , e_n$) usually entails evaluation of $f$ and $e_1 , \ldots , e_n$ in that order. Each argument expression is converted to the type of its corresponding formal parameter. After that, the application is rewritten to the function's right hand side, with actual arguments substituted for formal parameters. The result of evaluating the rewritten right-hand side is finally converted to the function's declared result type, if one is given.
The case of a formal parameter with a parameterless method type =>$T$ is treated specially. In this case, the corresponding actual argument expression $e$ is not evaluated before the application. Instead, every use of the formal parameter on the right-hand side of the rewrite rule entails a re-evaluation of $e$. In other words, the evaluation order for =>-parameters is call-by-name whereas the evaluation order for normal parameters is call-by-value. Furthermore, it is required that $e$'s packed type conforms to the parameter type $T$. The behavior of by-name parameters is preserved if the application is transformed into a block due to named or default arguments. In this case, the local value for that parameter has the form val $y_i$ = () => $e$ and the argument passed to the function is $y_i$().
The last argument in an application may be marked as a sequence argument, e.g. $e$: _*. Such an argument must correspond to a repeated parameter of type $S$* and it must be the only argument matching this parameter (i.e. the number of formal parameters and actual arguments must be the same). Furthermore, the type of $e$ must conform to scala.Seq[$T$], for some type $T$ which conforms to $S$. In this case, the argument list is transformed by replacing the sequence $e$ with its elements. When the application uses named arguments, the vararg parameter has to be specified exactly once.
A function application usually allocates a new frame on the program's run-time stack. However, if a local function or a final method calls itself as its last action, the call is executed using the stack-frame of the caller.
Example
Assume the following function which computes the sum of a variable number of arguments:
def sum(xs: Int*) = (0 /: xs) ((x, y) => x + y)
Then
sum(1, 2, 3, 4)
sum(List(1, 2, 3, 4): _*)
both yield 10 as result. On the other hand,
sum(List(1, 2, 3, 4))
would not typecheck.
Named and Default Arguments
If an application might uses named arguments $p = e$ or default arguments, the following conditions must hold.
• For every named argument $p_i = e_i$ which appears left of a positional argument in the argument list $e_1 \ldots e_m$, the argument position $i$ coincides with the position of parameter $p_i$ in the parameter list of the applied function.
• The names $x_i$ of all named arguments are pairwise distinct and no named argument defines a parameter which is already specified by a positional argument.
• Every formal parameter $p_j:T_j$ which is not specified by either a positional or a named argument has a default argument.
If the application uses named or default arguments the following transformation is applied to convert it into an application without named or default arguments.
If the function $f$ has the form $p.m$[$\mathit{targs}$] it is transformed into the block
{ val q = $p$
q.$m$[$\mathit{targs}$]
}
If the function $f$ is itself an application expression the transformation is applied recursively on $f$. The result of transforming $f$ is a block of the form
{ val q = $p$
val $x_1$ = expr$_1$
$\ldots$
val $x_k$ = expr$_k$
q.$m$[$\mathit{targs}$]($\mathit{args}_1$)$, \ldots ,$($\mathit{args}_l$)
}
where every argument in $(\mathit{args}_1) , \ldots , (\mathit{args}_l)$ is a reference to one of the values $x_1 , \ldots , x_k$. To integrate the current application into the block, first a value definition using a fresh name $y_i$ is created for every argument in $e_1 , \ldots , e_m$, which is initialised to $e_i$ for positional arguments and to $e'_i$ for named arguments of the form $x_i=e'_i$. Then, for every parameter which is not specified by the argument list, a value definition using a fresh name $z_i$ is created, which is initialized using the method computing the default argument of this parameter.
Let $\mathit{args}$ be a permutation of the generated names $y_i$ and $z_i$ such such that the position of each name matches the position of its corresponding parameter in the method type ($p_1:T_1 , \ldots , p_n:T_n$)$U$. The final result of the transformation is a block of the form
{ val q = $p$
val $x_1$ = expr$_1$
$\ldots$
val $x_l$ = expr$_k$
val $y_1$ = $e_1$
$\ldots$
val $y_m$ = $e_m$
val $z_1$ = $q.m\$default\$i[\mathit{targs}](\mathit{args}_1), \ldots ,(\mathit{args}_l)$
$\ldots$
val $z_d$ = $q.m\$default\$j[\mathit{targs}](\mathit{args}_1), \ldots ,(\mathit{args}_l)$
q.$m$[$\mathit{targs}$]($\mathit{args}_1$)$, \ldots ,$($\mathit{args}_l$)($\mathit{args}$)
}
Signature Polymorphic Methods
For invocations of signature polymorphic methods of the target platform $f$($e_1 , \ldots , e_m$), the invoked function has a different method type ($p_1$:$T_1 , \ldots , p_n$:$T_n$)$U$ at each call site. The parameter types $T_ , \ldots , T_n$ are the types of the argument expressions $e_1 , \ldots , e_m$ and $U$ is the expected type at the call site. If the expected type is undefined then $U$ is scala.AnyRef. The parameter names $p_1 , \ldots , p_n$ are fresh.
Note
On the Java platform version 7 and later, the methods invoke and invokeExact in class java.lang.invoke.MethodHandle are signature polymorphic.
Method Values
SimpleExpr ::= SimpleExpr1 _'
The expression $e$ _ is well-formed if $e$ is of method type or if $e$ is a call-by-name parameter. If $e$ is a method with parameters, $e$ _ represents $e$ converted to a function type by eta expansion. If $e$ is a parameterless method or call-by-name parameter of type =>$T$, $e$ _ represents the function of type () => $T$, which evaluates $e$ when it is applied to the empty parameterlist ().
Example
The method values in the left column are each equivalent to the eta-expanded expressions on the right.
placeholder syntax eta-expansion
math.sin _ x => math.sin(x)
math.pow _ (x1, x2) => math.pow(x1, x2)
val vs = 1 to 9; vs.fold _ (z) => (op) => vs.fold(z)(op)
(1 to 9).fold(z)_ { val eta1 = z; val eta2 = 1 to 9; op => eta2.fold(eta1)(op) }
Some(1).fold(??? : Int)_ { val eta1 = () => ???; val eta2 = Some(1); op => eta2.fold(eta1())(op) }
Note that a space is necessary between a method name and the trailing underscore because otherwise the underscore would be considered part of the name.
Type Applications
SimpleExpr ::= SimpleExpr TypeArgs
A type application $e$[$T_1 , \ldots , T_n$] instantiates a polymorphic value $e$ of type [$a_1$ >: $L_1$ <: $U_1, \ldots , a_n$ >: $L_n$ <: $U_n$]$S$ with argument types $T_1 , \ldots , T_n$. Every argument type $T_i$ must obey the corresponding bounds $L_i$ and $U_i$. That is, for each $i = 1 , \ldots , n$, we must have $\sigma L_i <: T_i <: \sigma U_i$, where $\sigma$ is the substitution $[a_1 := T_1 , \ldots , a_n := T_n]$. The type of the application is $\sigma S$.
If the function part $e$ is of some value type, the type application is taken to be equivalent to $e$.apply[$T_1 , \ldots ,$ T$_n$], i.e. the application of an apply method defined by $e$.
Type applications can be omitted if local type inference can infer best type parameters for a polymorphic functions from the types of the actual function arguments and the expected result type.
Tuples
SimpleExpr ::= (' [Exprs] )'
A tuple expression ($e_1 , \ldots , e_n$) is an alias for the class instance creation scala.Tuple$n$($e_1 , \ldots , e_n$), where $n \geq 2$. The empty tuple () is the unique value of type scala.Unit.
Instance Creation Expressions
SimpleExpr ::= new' (ClassTemplate | TemplateBody)
A simple instance creation expression is of the form new $c$ where $c$ is a constructor invocation. Let $T$ be the type of $c$. Then $T$ must denote a (a type instance of) a non-abstract subclass of scala.AnyRef. Furthermore, the concrete self type of the expression must conform to the self type of the class denoted by $T$. The concrete self type is normally $T$, except if the expression new $c$ appears as the right hand side of a value definition
val $x$: $S$ = new $c$
(where the type annotation : $S$ may be missing). In the latter case, the concrete self type of the expression is the compound type $T$ with $x$.type.
The expression is evaluated by creating a fresh object of type $T$ which is initialized by evaluating $c$. The type of the expression is $T$.
A general instance creation expression is of the form new $t$ for some class template $t$. Such an expression is equivalent to the block
{ class $a$ extends $t$; new $a$ }
where $a$ is a fresh name of an anonymous class which is inaccessible to user programs.
There is also a shorthand form for creating values of structural types: If {$D$} is a class body, then new {$D$} is equivalent to the general instance creation expression new AnyRef{$D$}.
Example
Consider the following structural instance creation expression:
new { def getName() = "aaron" }
This is a shorthand for the general instance creation expression
new AnyRef{ def getName() = "aaron" }
The latter is in turn a shorthand for the block
{ class anon\$X extends AnyRef{ def getName() = "aaron" }; new anon\$X }
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2019-08-19 11:46:09
|
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https://stats.stackexchange.com/questions/400781/explosive-arma-processes-are-stationary
|
# Explosive AR(MA) processes are stationary?
According to Theorem 8.8 in Time Series A.W. van der Vaart an ARMA process $$\phi (L)X_t=\theta(L)\epsilon_t$$ has a unique stationary solution $$X_t=\psi(L)\epsilon_t$$ with $$\psi=\theta/\phi$$ if $$\phi$$ has no roots on the complex unit circle. This would imply that the explosive process, with $$\rho>1$$, is a stationary process $$X_t=\rho X_{t-1}+\epsilon_t$$ with stationary solution $$X_t=\sum_{i=1}^\infty \rho^{-i}\epsilon_{t+i}$$.
Now indeed $$\sum_{i=1}^{\infty} \rho^{-i} < \infty$$ so that weak stationarity can be proved by using this representation.
However, here on stackexchange I see a lot of question/answers that suggest that the process above is not stationary (see for example Are explosive ARMA(1, 1) processes stationary?, Non-Stationary: Larger-than-unit root). In particular, the accepted answer of the latter question claims that the process is non-stationary by simulating a series and showing it displays explosive trending behaviour.
I think the only way to reconcile the theorem I mention above and the plots in the accepted answer of (Non-Stationary: Larger-than-unit root) is the following: the explosive process is indeed stationary but non-ergodic, that is, we cannot find the statistical properties of $$X_t$$ such as $$\mathbb{E}(X_t)=\mu$$ by observing a single infinitely long sample path of the explosive process, mathematically: $$\lim_{t \to \infty}\frac{1}{t}\sum_{t=1}X_t \neq\mathbb{E}X_t$$
• @corey979 I am baffled by the notion that an explosive process could be considered stationary, and would undoubtedly experience wonder and delight if I were shown that it is so. That said: the variances of explosive processes are functions of time, and the means of explosive processes are functions of time, and perturbations to explosive processes give stronger effects as more time passes since they occurred, so I am not understanding how an explosive process could be stationary in any sense. Apr 2, 2019 at 15:47
• @Alexis, "variances of explosive processes are functions of time" - sure about this? Apr 2, 2019 at 17:21
• Excellent question on a subtle topic! Apr 2, 2019 at 17:22
• @Alexis, you have to turn time back Apr 11, 2019 at 19:23
• @Aksakal I spent some time starring at the solution, and it went click: you not only need a time machine for some point in the future, but for all futures. So you are quite correct when you say "I won't like it!" :) Apr 12, 2019 at 16:32
Yes, there is a stationary solution for $$\rho>1$$ in AR(1) process: $$X_t=\rho X_{t-1}+\varepsilon_t$$ I'm not sure you'll like it though: $$X_t=-\sum_{k=1}^\infty\frac 1 {\rho^k}\varepsilon_{t+k}$$ Notice the index: $$t+k$$, you'd need DeLorean to use this in practice.
When $$\rho>1$$ the process is not invertible.
• Argh! I am so busy right now. I wanna go play with this in simulation to appreciate why that is stationary, but that is going to have to wait until at least tonight. Thank you, as always, for pushing my understanding on time series. Apr 2, 2019 at 17:38
• @Alexis, look at 4.5.3 example to simulate here maths.qmul.ac.uk/~bb/TimeSeries/TS_Chapter4_5.pdf basically, it's stationary but not causal, i.e. your today depends on tomorrow. Apr 2, 2019 at 17:41
• Stangely appropriate to what I am currently busy with: time-varying confounding and g-estimation of causal effects in Hernán & Robins. :) Apr 2, 2019 at 17:42
• Am I right about the non-ergodicity of the process? Apr 2, 2019 at 17:57
• @Joogs, I don't think so. Apr 2, 2019 at 18:02
First we can write the model in reverse AR(1) form as:
$$X_{t} = \frac{1}{\rho} X_{t+1} - \frac{\epsilon_{t+1}}{\rho}.$$
Suppose you now define the observable values using the filter:
$$X_t = - \sum_{k=1}^\infty \frac{\epsilon_{t+k}}{\rho^k}.$$
You can confirm by substitution that both the original AR(1) form and the reversed form hold in this case. As pointed out in an excellent answer to a related question by Michael, this means that the model is not identified unless we exclude this solution by definition.
|
2022-06-30 17:56:34
|
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|
https://socratic.org/questions/how-do-you-complete-the-square-to-find-the-vertex-for-y-x-2-3x-2
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# How do you complete the square to find the vertex for y=x^2-3x+2?
To complete the square half the coefficient of x, that is $- \frac{3}{2}$, then square it, that is $\frac{9}{4}$. Now add and subtract it to the expression as follows:
y=${x}^{2} - 3 x + \frac{9}{4} - \frac{9}{4} + 2$
=${\left(x - \frac{3}{2}\right)}^{2} - \frac{1}{4}$. That is how 'complete the square' is done.
Vertex is $\left(\frac{3}{2} , - \frac{1}{4}\right)$
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2020-06-02 10:25:07
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http://tatome.de/zettelkasten/zettelkasten.php?tag=neural-fields
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# Show Tag: neural-fields
Select Other Tags
The in-vitro study of the rat intermediate SC by Lee and Hall did not find evidence for the long-range inhibitory/short-range excitatory connection pattern theorized by proponents of the neural-field theory of SC fixation.
Cuijpers and Erlhagen use neural fields to implement Bayes' rule for combining the activities of neural populations spatially encoding probability distributions.
Beck et al. argue that simply adding time point-to-time point responses of a population code will integrate the information optimally if the noise in the input is what they call "Poisson-like".
That is somewhat expected as in a Poisson distribution with mean $\lambda$ the variance is $\lambda$ and the standard deviation is $\sqrt{\lambda}$ and adding population responses is equivalent to counting spikes over a longer period of time, thus increasing the mean of the distribution.
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2018-10-17 00:50:53
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https://www.physicsforums.com/threads/trigonometry-sinus-inequality.231852/
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# Trigonometry sinus inequality
1. Apr 28, 2008
### Physicsissuef
1. The problem statement, all variables and given/known data
Find the solutions for x:
$$sinx<\frac{\sqrt{3}}{2}$$
2. Relevant equations
3. The attempt at a solution
Will the solution for x be
$$x\in (\frac{-\pi}{2} , \frac{\pi}{3})$$ ??
or
$$x\in (\frac{5\pi}{6} , \frac{-\pi}{2})$$ or
$$x\in (\frac{5\pi}{6} , \frac{\pi}{3})$$
2. Apr 28, 2008
### Tom Mattson
Staff Emeritus
1.) Take out a piece of graph paper.
2.) Graph the function $f(x)=\sin(x)$ over a few periods.
3.) On the same grid, graph the line $y=\sqrt{3}/2$.
The solution should be obvious after that.
3. Apr 28, 2008
### Physicsissuef
I see $$x\in (\frac{-\pi}{2} , \frac{\pi}{3})$$, but I am not sure, because all of this solutions are acceptable.
4. Apr 28, 2008
### Tom Mattson
Staff Emeritus
Yes, clearly there are infinitely many intervals on which the inequality is satisfied. The solution is the union of all of them.
5. Apr 28, 2008
### Physicsissuef
But if I write
$$x\in (\frac{-\pi}{2} +2k\pi , \frac{\pi}{3}+2k\pi)$$
It will be useless to make union of all of them, right?
6. Apr 28, 2008
### Tom Mattson
Staff Emeritus
I don't agree that those are the intervals that satisfy the inequality. Look at the graph again. You should see that one of the intervals for which $\sin(x)<\sqrt{3}/2$ is:
$$\left(\frac{2\pi}{3},\frac{7\pi}{3}\right)$$
That interval isn't captured by any value of $k$ in your solution.
What makes you say that? The problem statement was not restricted to some subset of the real line. If you want to correctly answer the question then you must express it as the union of all of the intervals on which the inequality is satisfied.
7. Apr 28, 2008
### Physicsissuef
They don't? I think they do. I draw trigonometric circular and they fit very well.
They are same with
$$x\in (\frac{5\pi}{6} , \frac{-\pi}{2})$$
8. Apr 28, 2008
### Tom Mattson
Staff Emeritus
If you draw the picture that I advised you to draw, then it is perfectly obvious that your solutions do not work.
Here is a simple counterexample. The inequality is satisfied for $x=\pi$. Now look closely at your solutions.
For $k=0$, the interval is:
$$\left(\frac{-\pi}{2},\frac{\pi}{3}\right )$$.
For $k=1$, the interval is:
$$\left(\frac{3\pi}{2},\frac{7\pi}{3}\right )$$.
Your solution misses $x=\pi$, and it is clearly wrong. Again: Draw the picture I advised you to draw, and it really ought to be clear.
9. Apr 29, 2008
### Physicsissuef
Yes, sorry. I realized that is $$\left(\frac{2\pi}{3},\frac{7\pi}{3}\right)$$
10. Apr 29, 2008
### BrendanH
As I am usually finicky, I just want to point at that, if you intend to include the constant 'k' in your solution, you should identify that it is an element of the natural numbers only.
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2017-02-27 16:44:35
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https://worldsworstsportsblog.com/tag/year-of-the-pitcher/
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## Did Run Production Change in 2010?June 2, 2011
Posted by tomflesher in Baseball, Economics.
Tags: , ,
Part of the narrative of last year’s season was the compelling “Year of the Pitcher” storyline prompted by an unusual number of no-hitters and perfect games. Though it’s too early in the season to say the same thing is happening this year, a few bloggers have suggested that run production is down in 2011 and we might see the same sort of story starting again.
As a quick and dirty check of this, I’d like to compare production in the 2000-2009 sample I used in a previous post to production in 2010. This will introduce a few problems, notably that using one year’s worth of data for run production will lead to possibly spurious results for the 2010 data and that the success of the pitchers may be a result of the strategy used to generate runs. That is, if pitchers get better, and strategy doesn’t change, then we see pitchers taking advantage of inefficiencies in strategy. If batting strategy stays the same and pitchers take advantage of bad batting, then we should see a change in the structure of run production since the areas worked over by hitters – for example, walks and strikeouts – will see shifts in their relative importance in scoring runs.
Hypothesis: A regression model of runs against hits, doubles, triples, home runs, stolen bases, times caught stealing, walks, times hit by pitch, sacrifice bunts, and sacrifice flies using two datasets, one with team-level season-long data for each year from 2000 to 2009 and the other from 2010 only, will yield statistically similar beta coefficients.
Method: Chow test.
Result: There is a difference, significant at the 90% but not 95% level. That might be a result of a change in strategy or of pitchers exploiting strategic inefficiencies.
R code behind the cut.
## Spitballing: Blanton in the Phillies’ RotationFebruary 25, 2011
Posted by tomflesher in Baseball.
Tags: , , , , , , , , , , , , ,
The Phillies have one of the best rotations, on paper, in baseball today. Although some people are measured in their optimism, including Jayson Stark, I think the important thing to remember is that we’re arguing over whether they’re “the best ever,” not if they’re going to be competitive. Rotations that bring this kind of excitement at the beginning of the year are few and far between. The Mets, for example, aren’t drawing this kind of expectation – guys like R.A. Dickey and Mike Pelfrey are solid, but they don’t have the deserved reputations of Roy Halladay, Roy Oswalt, Cliff Lee, Cole Hamels, and Joe Blanton.
I’m hardly the first to say it, but Joe Blanton seems to be the odd man out. He’ll be making about \$8.5 million next year. Blanton faced 765 batters last year, fourth behind Halladay, Hamels, and Kyle Kendrick. Immediately behind Blanton was Jamie Moyer with 460 batters faced. For the record, the fifth-most-active pitcher faced 362 batters in 2009 (Chan Ho Park) and 478 in 2008 (Adam Eaton). Let’s take that number and adjust it to about 550 batters faced, since Blanton will get more starts than most fifth starters and he’ll stay in longer since he’s a proven quantity. In a normal year, the Phils face about 6200 batters, so that means Blanton’s 550 will be about 9% of the team’s total. (That figure is robust even in last year’s Year of the Pitcher with depressed numbers of batters faced.)
According to J.C. Bradbury’s Hot Stove Economics, this yields an average marginal revenue product of 3.15 million. This figure is based on the average rate that pitchers prevent runs and the average revenue of an MLB team. Obviously, Blanton is a better than the average pitcher (ignoring his negative Wins Above Replacement last year) and the Phillies make more money than most teams, but this is a pretty damning figure.
The other thing to take into account is that Blanton’s marginal wins aren’t worth as much to the Phillies now that they have a four-ace rotation. He won’t get every start and he won’t be a 20-game winner. Even if he were, he’ll be providing insurance wins – he might have an extra ten wins over a AAA-level replacement, but chances are that those wins won’t make the difference between making the playoffs and missing them when you figure in the Phillies’ solid bullpen and run production.
Instead, let’s say Blanton goes to the White Sox, just to pick a team. Jake Peavy and Edwin Jackson combined for 765 batters faced, so plug Blanton in for Freddy Garcia with 671 batters faced – a worst-case scenario. That would be 10.85 % of the batters faced, bringing him up to about 3.8 million. In this case, though, you have a team who finished 6 games back and missed the playoffs. If you replace Garcia with Blanton, you stand a very good chance to make the playoffs. That’s another way of saying that the Phillies’ 6-game lead over Atlanta (the NL wild card team) was worth less than the Twins’ 6-game lead over the White Sox (when neither team had as many wins as the AL wild card).
Economists would refer to this as a diminishing marginal returns situation – when you have fewer wins, around the middle of the pack, each additional win is worth a little less. This captures the idea that taking a 110-win team and giving them 111 wins would cost a lot of money and not yield much extra benefit, but a 90-win team making 91 wins might let them overtake another team.
The upshot of all of this? Trade Blanton for prospects. Rely on the bullpen and develop a future starter. Roy Halladay won’t be competitive forever.
## Diagnosing the ALDecember 22, 2010
Posted by tomflesher in Baseball, Economics.
Tags: , , , , , ,
In the previous post, I crunched some numbers on a previous forecast I’d made and figured out that it was a pretty crappy forecast. (That’s the fun of forecasting, of course – sometimes you’re right and sometimes you’re wrong.) The funny part of it, though, is that the predicted home runs per game for the American League was so far off – 3.4 standard errors below the predicted value – that it’s highly unlikely that the regression model I used controls for all relevant variables. That’s not surprising, since it was only a time trend with a dummy variable for the designated hitter.
There are a couple of things to check for immediately. The first is the most common explanation thrown around when home runs drop – steroids. It seems to me that if the drop in home runs were due to better control of performance-enhancing drugs, then it should mostly be home runs that are affected. For example, intentional walks should probably be below expectation, since intentional walks are used to protect against a home run hitter. Unintentional walks should probably be about as expected, since walks are a function of plate discipline and pitcher control, not of strength. On-base percentage should probably drop at a lower magnitude than home runs, since some hits that would have been home runs will stay in the park as singles, doubles, or triples rather than all being fly-outs. There will be a drop but it won’t be as big. Finally, slugging average should drop because a loss in power without a corresponding increase in speed will lower total bases.
I’ll analyze these with pretty new R code behind the cut.
## What Happened to Home Runs This Year?December 22, 2010
Posted by tomflesher in Baseball, Economics.
Tags: , , , , , , , ,
1 comment so far
I was talking to Jim, the writer behind Apparently, I’m An Angels Fan, who’s gamely trying to learn baseball because he wants to be just like me. Jim wondered aloud how much the vaunted “Year of the Pitcher” has affected home run production. Sure enough, on checking the AL Batting Encyclopedia at Baseball-Reference.com, production dropped by about .15 home runs per game (from 1.13 to .97). Is that normal statistical variation or does it show that this year was really different?
In two previous posts, I looked at the trend of home runs per game to examine Stuff Keith Hernandez Says and then examined Japanese baseball’s data for evidence of structural break. I used the Batting Encyclopedia to run a time-series regression for a quadratic trend and added a dummy variable for the Designated Hitter. I found that the time trend and DH control account for approximately 56% of the variation in home runs per year, and that the functional form is
$\hat{HR} = .957 - .0188 \times t + .0004 \times t^2 + .0911 \times DH$
with t=1 in 1955, t=2 in 1956, and so on. That means t=56 in 2010. Consequently, we’d expect home run production per game in 2010 in the American League to be approximately
$\hat{HR} = .957 - .0188 \times 56 + .0004 \times 3136 + .0911 \approx 1.25$
That means we expected production to increase this year and it dropped precipitously, for a residual of -.28. The residual standard error on the original regression was .1092, so on 106 degrees of freedom, so the t-value using Texas A&M’s table is 1.984 (approximating using 100 df). That means we can be 95% confident that the actual number of home runs should fall within .1092*1.984, or about .2041, of the expected value. The lower bound would be about 1.05, meaning we’re still significantly below what we’d expect. In fact, the observed number is about 3.4 standard errors below the expected number. In other words, we’d expect that to happen by chance less than .1% (that is, less than one tenth of one percent) of the time.
Clearly, something else is in play.
## Matt Garza, Fifth No-Hitter of 2010July 26, 2010
Posted by tomflesher in Baseball.
Tags: , , , , , ,
1 comment so far
Tonight, Matt Garza pitched the fifth no-hitter of 2010. He joins Edwin Jackson, Roy Halladay, Dallas Braden, and Ubaldo Jimenez in the Year of the Pitcher club.
As I pointed out when Jackson hit his no-hitter, no-hit games are probably Poisson distributed. Let’s update the chart.
The Poisson distribution has probability density function
$f(n; \lambda)=\frac{\lambda^n e^{-\lambda}}{n!}$
Maintaining our prior rate of 2.45 no-hitters per season, that means $\lambda = 2.45$. Our function is then
$f(n; \lambda = 2.5)=\frac{2.45^n (.0864)}{n!}$
The probabilities remain the same:
n p cumulative 0 0.0863 0.0863 1 0.2114 0.2977 2 0.2590 0.5567 3 0.2115 0.7683 4 0.1296 0.8978 5 0.0635 0.9613 6 0.0259 0.9872 7 0.0091 0.9963 8 0.0028 0.9991 9 0.0008 0.9998 10 0.0002 1.0000
And though the expectation (E(49)) and cumulative expectation (C(49)) remain the same, the observed values shift slightly:
E(49) Observed C(49) Total 4.23 5 4.23 5 10.36 11 14.59 16 12.69 8 27.28 24 10.36 17 37.65 41 6.35 1 43.99 42 3.11 5 47.10 47 1.27 1 48.37 48 0.44 0 48.82 48 0.14 1 48.95 49 0.04 0 48.99 49 0.01 0 49.00 49
The tailing observations (say, for 4+ no-hitters) don’t quite match the expected frequencies, but the cumulative values match quite nicely. There might be some unobserved variables that explain the weirdness in the upper tail. Still, cumulatively, we have 47 seasons with 5 or fewer no-hitters, which is almost exactly what’s expected. This is unusual, but not outside the realm of statistical expectation.
## Back when it was hard to hit 55…July 8, 2010
Posted by tomflesher in Baseball, Economics.
Tags: , , , , , , , , ,
Last night was one of those classic Keith Hernandez moments where he started talking and then stopped abruptly, which I always like to assume is because the guys in the truck are telling him to shut the hell up. He was talking about Willie Mays for some reason, and said that Mays hit 55 home runs “back when it was hard to hit 55.” Keith coyly said that, while it was easy for a while, it was “getting hard again,” at which point he abruptly stopped talking.
Keith’s unusual candor about drug use and Mays’ career best of 52 home runs aside, this pinged my “Stuff Keith Hernandez Says” meter. After accounting for any time trend and other factors that might explain home run hitting, is there an upward trend? If so, is there a pattern to the remaining home runs?
The first step is to examine the data to see if there appears to be any trend. Just looking at it, there appears to be a messy U shape with a minimum around t=20, which indicates a quadratic trend. That means I want to include a term for time and a term for time squared.
Using the per-game averages for home runs from 1955 to 2009, I detrended the data using t=1 in 1955. I also had to correct for the effect of the designated hitter. That gives us an equation of the form
$\hat{HR} = \hat{\beta_{0}} + \hat{\beta_{1}}t + \hat{\beta_{2}} t^{2} + \hat{\beta_{3}} DH$
The results:
Estimate Std. Error t-value p-value Signif B0 0.957 0.0328 29.189 0.0001 0.9999 t -0.0188 0.0028 -6.738 0.0001 0.9999 tsq 0.0004 0.00005 8.599 0.0001 0.9999 DH 0.0911 0.0246 3.706 0.0003 0.9997
We can see that there’s an upward quadratic trend in predicted home runs that together with the DH rule account for about 56% of the variation in the number of home runs per game in a season ($R^2 = .5618$). The Breusch-Pagan test has a p-value of .1610, indicating a possibility of mild homoskedasticity but nothing we should get concerned about.
Then, I needed to look at the difference between the predicted number of home runs per game and the actual number of home runs per game, which is accessible by subtracting
$Residual = HR - \hat{HR}$
This represents the “abnormal” number of home runs per year. The question then becomes, “Is there a pattern to the number of abnormal home runs?” There are two ways to answer this. The first way is to look at the abnormal home runs. Up until about t=40 (the mid-1990s), the abnormal home runs are pretty much scattershot above and below 0. However, at t=40, the residual jumps up for both leagues and then begins a downward trend. It’s not clear what the cause of this is, but the knee-jerk reaction is that there might be a drug use effect. On the other hand, there are a couple of other explanations.
The most obvious is a boring old expansion effect. In 1993, the National League added two teams (the Marlins and the Rockies), and in 1998 each league added a team (the AL’s Rays and the NL’s Diamondbacks). Talent pool dilution has shown up in our discussion of hit batsmen, and I believe that it can be a real effect. It would be mitigated over time, however, by the establishment and development of farm systems, in particular strong systems like the one that’s producing good, cheap talent for the Rays.
## Tough LossesJuly 8, 2010
Posted by tomflesher in Baseball.
Tags: , , , , , , , , ,
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2021-01-21 00:43:16
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https://www.ibpsadda.com/online-test/IBPS-PO-Reasoning-Ability-Test-34-880
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# IBPS PO Reasoning Ability Questions with Answers Practice online test 34
Description: free IBPS PO Reasoning Ability Questions with Answers Practice test 34 for IBPS PO Preliminary and Main online test Prepare bank PO banking mock exams adda
1 . Read the following information carefully and answer the questions given below.
A word arrangement machine, when given an input, line of words rearranges it following a particular rule. The following is the illustration of the input and the steps of arrangement.
Input : pull the cover and then push into
Step I : pull the then and cover push into
Step II : then the pull into push cover and
Step III :into pull the then and cover push
Step IV :into pull and then the cover push and so on
Input:Try your best until you get goal. Which of the following steps would be 'get goal try until you your best'?
Step II Step III Step V Step VI None of these
2 . Read the following information carefully and answer the questions given below.
A word arrangement machine, when given an input, line of words rearranges it following a particular rule. The following is the illustration of the input and the steps of arrangement.
Input : pull the cover and then push into
Step I : pull the then and cover push into
Step II : then the pull into push cover and
Step III :into pull the then and cover push
Step IV :into pull and then the cover push and so on
If Step VI of an input is 'deep gutter ball into the has fallen' Which of the following would definitely be the input?
has the ball fallen into deep gutter ball has fallen into the deep gutter deep gutter has fallen into the ball gutter has deep ball fallen into the None of these
3 . Read the following information carefully and answer the questions given below.
A word arrangement machine, when given an input, line of words rearranges it following a particular rule. The following is the illustration of the input and the steps of arrangement.
Input : pull the cover and then push into
Step I : pull the then and cover push into
Step II : then the pull into push cover and
Step III :into pull the then and cover push
Step IV :into pull and then the cover push and so on
If Step IV of an input is we can't measure the depth without scale what would be step VII?
scale we the measure can't depth without the we scal e wi thout depth can't measure without we scale the can't measure depth the we depth wi thout scal e can't measure None of these
4 . Read the following information carefully and answer the questions given below.
A word arrangement machine, when given an input, line of words rearranges it following a particular rule. The following is the illustration of the input and the steps of arrangement.
Input : pull the cover and then push into
Step I : pull the then and cover push into
Step II : then the pull into push cover and
Step III :into pull the then and cover push
Step IV :into pull and then the cover push and so on
Input:standing hard always is impossible for all. Which of the following will be step VIII for this input?
hard all standing is impossible for always hard all impossible is standing for always impossible all hard always for standing is impossible all for always hard standing is None of these
5 . Read the following information carefully and answer the questions given below.
A word arrangement machine, when given an input, line of words rearranges it following a particular rule. The following is the illustration of the input and the steps of arrangement.
Input : pull the cover and then push into
Step I : pull the then and cover push into
Step II : then the pull into push cover and
Step III :into pull the then and cover push
Step IV :into pull and then the cover push and so on
If step I of an input is 'play and jump until you tired fully', what would be step VI of the input given above?
jump fully tired you and play unti tired fully jump until play and you tired fully play until jump and you play fully tired you and jump until None of these
6 . In these questions, relationship between different elements is shown in the statements. These statements are followed by two conclusions. Mark your answer
(1) If only conclusion I follows.
(2) If only conclusion II follows.
(3) If either conclusion I or II follows.
(4) If neither conclusion I nor II follows.
(5) If both conclusions I and II follow.
$Statements :$
A < B, B < C, C = D, D > E
$Conclusions :$
I. B = D
II. B < D
if only Conclusion I is follows. if only Conclusion II is follows if either I or II is follows. if neither I nor II is follows. if both I and II are follows.
7 . In these questions, relationship between different elements is shown in the statements. These statements are followed by two conclusions. Mark your answer
(1) If only conclusion I follows.
(2) If only conclusion II follows.
(3) If either conclusion I or II follows.
(4) If neither conclusion I nor II follows.
(5) If both conclusions I and II follow.
$Statements :$
M = N $\geq$ O < P = Q $\leq$ R
$Conclusions :$
I. N $\geq$ P II. R > N
if only Conclusion I is follows. if only Conclusion II is follows if either I or II is follows. if neither I nor II is follows. if both I and II are follows.
8 . In these questions, relationship between different elements is shown in the statements. These statements are followed by two conclusions. Mark your answer
(1) If only conclusion I follows.
(2) If only conclusion II follows.
(3) If either conclusion I or II follows.
(4) If neither conclusion I nor II follows.
(5) If both conclusions I and II follow.
$Statements :$
S < T, T< U = W, W < X
$Conclusions :$
I. S $\geq$ W II. W $\geq$ T
if only Conclusion I is follows. if only Conclusion II is follows if either I or II is follows. if neither I nor II is follows. if both I and II are follows.
9 . In these questions, relationship between different elements is shown in the statements. These statements are followed by two conclusions. Mark your answer
(1) If only conclusion I follows.
(2) If only conclusion II follows.
(3) If either conclusion I or II follows.
(4) If neither conclusion I nor II follows.
(5) If both conclusions I and II follow.
$Statements :$
G < H, I < G H < J, J $\leq$ K
$Conclusions :$
I. H < K
II. H > I
if only Conclusion I is follows. if only Conclusion II is follows if either I or II is follows. if neither I nor II is follows. if both I and II are follows.
10 . In these questions, relationship between different elements is shown in the statements. These statements are followed by two conclusions. Mark your answer
(1) If only conclusion I follows.
(2) If only conclusion II follows.
(3) If either conclusion I or II follows.
(4) If neither conclusion I nor II follows.
(5) If both conclusions I and II follow.
$Statements :$
C < B, K $\geq$ G, G = M, B $\leq$ K, M $\leq$ B
$Conclusions :$
I. M $\leq$ K II. C = G
if only Conclusion I is follows. if only Conclusion II is follows if either I or II is follows. if neither I nor II is follows. if both I and II are follows.
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2018-12-10 06:34:51
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https://math.stackexchange.com/questions/3150808/exercise-involving-the-regularity-axiom
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Exercise involving the Regularity Axiom
I am not sure how to proceed with the following question and would appreciate some help:
Let $$A,B$$ and $$C$$ be sets.Further suppose that $$A\in B$$ and $$B \in C$$.Using the Regularity Axiom show that $$C \notin A$$.
My attempt:
Since $$A \in B$$ and $$B \in C$$ therefore $$C:=$${{$$A$$}} or equivalently {$$A$$}$$\in C$$. By the regularity axiom, since $$C \neq \emptyset$$, then there exists an element $$x$$ such that $$x \in C$$ and $$x \cap C= \emptyset$$. With {$$A$$} being the only element in $$C$$,it follows that {$$A$$}$$\cap C= \emptyset$$. To prove that $$C \notin A$$, suppose instead that $$C \in A$$ and derive a contradiction.
With $$C \in A$$ and {$$A$$} $$\in C$$ then by the pairing axiom the set {$$A,C$$} exists and hence $$A \cap$${$$A,C$$}$$=C$$ and $$C \cap$${$$A,C$$}$$=\emptyset$$....i do not really know how to proceed...looking at other sets and pairing them up does not yield any contradiction for me...
Help!
There are a few issues to address.
Since $$A \in B$$ and $$B \in C$$ therefore $$C:=$${{$$A$$}} or equivalently {$$A$$}$$\in C$$.
You seem to be assuming that $$A$$ is the only element of $$B$$ and that $$B$$ is the only element of $$C.$$ However, this has not been given.
By the regularity axiom, since $$C \neq \emptyset$$, then there exists an element $$x$$ such that $$x \in C$$ and $$x \cap C= \emptyset$$.
Very true.
With {$$A$$} being the only element in $$C$$,it follows that {$$A$$}$$\cap C= \emptyset$$.
Well, again, we don't actually know for sure that $$\{A\}$$ is an element of $$C,$$ though it is a subset of an element of $$C.$$
To prove that $$C \notin A$$, suppose instead that $$C \in A$$ and derive a contradiction.
Ah! Now that's a good idea!
With $$C \in A$$ and {$$A$$} $$\in C$$ then by the pairing axiom the set {$$A,C$$} exists and hence $$A \cap$${$$A,C$$}$$=C$$ and $$C \cap$${$$A,C$$}$$=\emptyset$$....i do not really know how to proceed...looking at other sets and pairing them up does not yield any contradiction for me...
Try showing that $$\{A,B,C\}$$ is a set.
Once that's done, from here, we can use Regularity to get our contradiction. By Regularity, one of $$A,B,C$$ must have empty intersection with $$\{A,B,C\},$$ but $$C\in A,$$ $$A\in B,$$ and $$B\in C,$$ so this is not possible.
We could proceed even more easily with a direct proof. We first prove that $$\{A,B,C\}$$ is a set. Next, since $$A\in B$$ and $$B\in C,$$ we must have that $$A\cap\{A,B,C\}=\emptyset$$ by Regularity, so in particular, $$C\notin A.$$
• Thank you for your nice response ! Would the following be correct : Assuming that $C \in A$ and with {$A,B,C$}$\neq \emptyset$ then $A \cap${A,B,C}$=C$.But by the regularity axiom, there exists an $x \in${$A,B,C$} with $x \cap${$A,B,C$}$= \emptyset$...is this right so far? if it is ..what allows me to choose x=A and thereby conclude a contradiction? Thank you! – HalfAFoot Mar 16 at 21:04
• You're close. Rather, we can say that $C\in A\cap\{A,B,C\}.$ However, we're not done, yet! We still need to justify that $B\cap\{A,B,C\}$ and $C\cap\{A,B,C\}$ are non-empty, and that $\{A,B,C\}$ is even a set! – Cameron Buie Mar 16 at 21:09
• Thank you very much! Applying the pairing axiom on (i) $A$ with itself and (ii) $B$ and $C$ yield the following sets {$A,A$} and {$B,C$}.Collecting these two sets into a family F and applying the union axiom suggests the existence of a set that consists of all elements belonging to at least one set in F hence {$A,B,C$} is assured.Now the following hold : (i)$C \in A \cap$ {$A,B,C$} ; (ii) $A \in B \cap$ {$A,B,C$} ; (iii) $B \in C \cap$ {$A,B,C$}...By the regularity axiom one of A,B,C must be disjoint from the set {$A,B,C$}...but (i)-(iii) seem to be non empty! – HalfAFoot Mar 16 at 21:34
• Nicely done! I've edited my answer to lead you in that direction, and to pose an alternative to proof by contradiction. – Cameron Buie Mar 16 at 21:35
• We can but try! – Cameron Buie Mar 16 at 21:42
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2019-09-21 15:16:09
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http://mathhelpforum.com/algebra/46044-solved-check-algebric-operation-print.html
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# [SOLVED] Check this algebric operation
• August 15th 2008, 05:29 PM
Neenoon
[SOLVED] Check this algebric operation
$(a^\frac {1}{3}-b^\frac {1}{3})$ $(a^\frac {2}{3}+a^\frac {1}{3} b^\frac {1}{3}+b^\frac {2}{3})$
$(a^{\frac {1}{3}+\frac {2}{3}})+(a^{\frac {1}{3}+\frac {1}{3}}b^{\frac {1}{3}})-(b^{\frac {1}{3}}a^{\frac {2}{3}})-(b^{\frac {1}{3}+\frac {1}{3}}a^{\frac {1}{3}})-(b^{\frac {1}{3}+\frac {2}{3}})
$
$
(a^{\frac {3}{3}})+(a^{\frac {2}{3}}b^\frac {1}{3})-(a^{\frac {2}{3}}b^\frac {1}{3})-(a^{\frac {1}{3}}b^\frac {2}{3})-(b^{\frac {3}{3}})
$
Please tell me this is right? (Doh)
$
a-(a^{\frac {1}{3}}b^\frac {2}{3})-b
$
• August 15th 2008, 06:21 PM
Soroban
Hello, Neenoon!
You dropped a term . . . There are six multiplications.
$(a^{\frac{1}{3}} -b^{\frac{1}{3}})\,(a^{\frac{2}{3}}+a^{\frac{1}{3}} b^{\frac{1}{3}} + b^{\frac{2}{3}})$
. . $= \;(a^{\frac{1}{3}})(a^{\frac{2}{3}}) + (a^{\frac{1}{3}})(a^{\frac{1}{3}}b^{\frac{1}{3}}) + (a^{\frac{1}{3}})(b^{\frac{2}{3}})$ . $- (b^{\frac{1}{3}})(a^{\frac{2}{3}}) - (b^{\frac{1}{3}})(a^{\frac{1}{3}}b^{\frac{1}{3}}) - (b^{\frac{1}{3}})(b^{\frac{2}{3}})$
. . $= \;a + \underbrace{a^{\frac{2}{3}}b^{\frac{1}{3}} + a^{\frac{1}{3}}b^{\frac{2}{3}} - a^{\frac{2}{3}}b^{\frac{1}{3}} - a^{\frac{1}{3}}b^{\frac{2}{3}}}_{\text{These cancel out}} - b$
. . $= \;a - b$
• August 15th 2008, 06:21 PM
Prove It
Quote:
Originally Posted by Neenoon
$(a^\frac {1}{3}-b^\frac {1}{3})$ $(a^\frac {2}{3}+a^\frac {1}{3} b^\frac {1}{3}+b^\frac {2}{3})$
$(a^{\frac {1}{3}+\frac {2}{3}})+(a^{\frac {1}{3}+\frac {1}{3}}b^{\frac {1}{3}})-(b^{\frac {1}{3}}a^{\frac {2}{3}})-(b^{\frac {1}{3}+\frac {1}{3}}a^{\frac {1}{3}})-(b^{\frac {1}{3}+\frac {2}{3}})
$
$
(a^{\frac {3}{3}})+(a^{\frac {2}{3}}b^\frac {1}{3})-(a^{\frac {2}{3}}b^\frac {1}{3})-(a^{\frac {1}{3}}b^\frac {2}{3})-(b^{\frac {3}{3}})
$
Please tell me this is right? (Doh)
$
a-(a^{\frac {1}{3}}b^\frac {2}{3})-b
$
Let's see...
$(a^\frac{1}{3}-b^\frac{1}{3})(a^\frac{2}{3}+a^\frac{1}{3}b^\frac{ 1}{3}+b^\frac{2}{3})
=a^\frac{1}{3}(a^\frac{2}{3}+a^\frac{1}{3}b^\frac{ 1}{3}+b^\frac{2}{3})-b^\frac{1}{3}(a^\frac{2}{3}+a^\frac{1}{3}b^\frac{1 }{3}+b^\frac{2}{3})$
$=a+a^\frac{2}{3}b^\frac{1}{3}+a^\frac{1}{3}b^\frac {2}{3}-a^\frac{2}{3}b^\frac{1}{3}-a^\frac{1}{3}b^\frac{2}{3}-b
=a-b$
This is actually a case of the difference of two cubes rule...
$a^3-b^3=(a-b)(a^2+ab+b^2)$
• August 16th 2008, 05:57 AM
Neenoon
I'll get back to math on monday, and check this out!
I totally messed this one up... and I thought I was doing better. (Doh) (Angry) (Crying)
Thank you so much! It totally sounds right.
... but the actual answer is a-b, right?
• August 16th 2008, 06:08 AM
vishalgarg
hi neoon
it is a simple case of x^3 - y^3
where x = a^1/3
n y = b^1/3
so ans will be a-b
• August 19th 2008, 04:08 PM
Neenoon
yes vishalgarg, totally simple ... (Surprised) (Headbang) (Sweating) (Sweating) (Sweating) (Sweating)
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2015-05-29 15:18:06
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https://ftp.aimsciences.org/article/doi/10.3934/dcds.2009.25.1181
|
# American Institute of Mathematical Sciences
December 2009, 25(4): 1181-1193. doi: 10.3934/dcds.2009.25.1181
## Notes on the asymptotically self-similar singularities in the Euler and the Navier-Stokes equations
1 Department of mathematics, Sungkyunkwan University, 300 Cheoncheon-dong, Jangan-gu, Suwon, 440-746, South Korea, South Korea
Received November 2008 Revised May 2009 Published September 2009
We exclude a type of asymptotically self-similar singularities which are the limiting cases of the results in [5] for the Euler and Navier-Stokes equations in dimension three.
Citation: Dongho Chae, Kyungkeun Kang, Jihoon Lee. Notes on the asymptotically self-similar singularities in the Euler and the Navier-Stokes equations. Discrete and Continuous Dynamical Systems, 2009, 25 (4) : 1181-1193. doi: 10.3934/dcds.2009.25.1181
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2020 Impact Factor: 1.392
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2022-05-23 11:31:08
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