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Is it correct? $\tan x + \cot x \ge 2$ proof The question is to prove $\tan x + \cot x \ge 2$ when $x$ is an acute angel.
This is what I did
$$\begin{align}
\tan x + \cot x &\ge 2\\
\frac{1}{\sin x \cos x} &\ge 2\\
\left(\frac{1}{\sin x \cos x}\right) - 2 &\ge 0\\
\left(\frac{1 - 2\sin x \cos x}{\sin x \cos x}\right) &\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &\ge 0\\
\left(\frac{(\sin x - \cos x)^2}{\sin x \cos x}\right) &\ge 0\\
\end{align}$$
Both nominator and denominator will never be negative because nominator is powered to two and cosx & sinx are positive when angel is acute.
Is it correct? Is there another way to solve?
|
Even if your implications are correct, your logic is flawed.
Logically, what you have done is similar to
$$\begin{align}
+1&=-1
\\ \Rightarrow (+1)^2&=(-1)^2
\\ \Rightarrow 1&=1.
\end{align}$$
Now $1=1$ so $+1=-1$ must be true also.
Deriving a true statement says nothing about the truth of the hypothesis.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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"answer_id": 6
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|
Proving formula for sum of squares with binomial coefficient $$\sum_{k=0}^{n-1}(k^2)= \binom{n}{3} + \binom{n+1}{3}$$
How should I prove that it is the correct formula for sum of squares?
Should I use induction to prove the basis? Any help is appreciated.
|
Using repeated differences we get
$$
\begin{array}{llll}
0 & 1 & 5 & 14 & \\
1 & 4 & 9 & \\
3 & 5 & \\
2& \\
\end{array}
$$
The last difference repeats forever, because $k^2$ has degree $2$.
Newton's interpolation formula then gives us this formula for the sequence of sums:
$$
0 \binom{n-1}{0} + 1 \binom{n-1}{1} + 3 \binom{n-1}{2} + 2 \binom{n-1}{3}
$$
Using Pascal's rule repeatedly we get
$$
\binom{n-1}{1} + \binom{n-1}{2} + \binom{n-1}{2} + \binom{n-1}{2} + \binom{n-1}{3} + \binom{n-1}{3}
$$
$$
=\binom{n}{2} + \binom{n-1}{2} + \binom{n}{3} + \binom{n-1}{3}
$$
$$
=\binom{n}{2} + \binom{n}{3} + \binom{n-1}{2} + \binom{n-1}{3}
$$
$$
=\binom{n+1}{3} + \binom{n}{3}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Closed-form of $\int_0^{\pi/2}\frac{\sin^2x\arctan\left(\cos^2x\right)}{\sin^4x+\cos^4x}\,dx$ I have just seen two active posts about integrals of inverse trigonometric function, $\arctan(x)$, here on MSE. So I decide to post this question. This integral comes from a friend of mine (it's not a homework problem) and we have tried to evaluate it but no success so far. I have discussed it in chatroom with @Chris'ssis but she gave me a horrible closed-form without proof. You may have a look here and here. My friend doesn't know the closed-form either. Here is the problem:
$$\int_0^{\pi/2}\frac{\sin^2x\arctan\left(\cos^2x\right)}{\sin^4x+\cos^4x}\,dx$$
Any idea? Any help would be appreciated. Thanks in advance.
|
$\newcommand{\al}[1]{\begin{align}#1\end{align}} \renewcommand{\Im}{\operatorname{Im}}$Result:
$$I = \frac{\pi \left[2 \pi +\log \left(-4 \sqrt{17+13 \sqrt{2}}+8 \sqrt{2}+13\right)-4 \tan ^{-1}\left(\sqrt{\frac{1}{\sqrt{2}}-\frac{1}{2}}+\frac{1}{\sqrt{2}}+1\right)\right]}{8 \sqrt{2}} \\\approx 0.299397.$$
The evaluation of this integral by hand is not as tedious as I thought at first. It turns out I had already considered this integral last week, in a different form.
First substitute $\tan x = t$, as suggested by FDP in the comments. This gives
$$
I = \int_0^\infty dt \frac{t^2 \arctan \left( {\frac{1}{1+t^2}}\right)}{1+t^4}.
$$
Now observe that
$$
\al{
\arctan \left( {\frac{1}{1+t^2}} \right) &= \Im \log(i + 1 + t^2) = \Im \left[\log(i+1) + \log\left(1 + \frac{1-i}{2} t^2\right) \right] \\&= \frac{\pi}{4} + \Im \log\left(1 + \frac{1-i}{2} t^2\right).
}
$$
The integral splits up into a trivial part and a less trivial part. We will now find the latter.
Consider the integral
$$
J(a) = \int_0^\infty dx \frac{x^2}{1+x^4} \log(1+a x^2).
$$
We have $J(0) = 0$, and
$$
\al{
J'(a) &= \int_0^\infty dx \frac{x^4}{(1+x^4)(1+a x^2)}
\\&= \frac{1}{1+a^2} \int_0^\infty dx \left[ \frac{1}{1+ a x^2} + \frac{a x^2 - 1}{1+x^4} \right]
\\&= \frac{\pi /2}{1+a^2}\left[\frac{1}{\sqrt a} + \frac{a-1}{\sqrt2}\right].
}
$$
This is straightforward to integrate with respect to $a$. (For the first term, just substitute $\sqrt a = u$ and use partial fractions.)
The result is
$$
J(a) = \frac{\pi}{2 \sqrt 2} \left[ \log \left(a+\sqrt{2} \sqrt{a}+1\right)+2 \arctan\left(\sqrt{2} \sqrt{a}+1\right) \right]
$$
After plugging in the limit and tedious simplification (see Appendix to this answer), we obtain
$$
\Im J\left(\frac{1-i}{2} \right) = -\frac{\pi}{2 \sqrt{2}}\left\{\operatorname{arccoth}\left[\sqrt{7+5 \sqrt{2}}+\sqrt{2}+1\right] + \arctan\left[\frac{1}{41} \left(2 \sqrt{89 \sqrt{2}-119}+2 \sqrt{2}+7\right)\right] \right\}
$$
Putting everything together,
$$
\al{
I &= \frac \pi 4 \int_0^\infty dt \frac{t^2}{1+t^4} + \Im \int_0^\infty dt \frac{t^2 \log\left(1 + \frac{1-i}{2} t^2\right)}{1+t^4}
\\&= \frac{\pi^2}{8 \sqrt 2} + \Im J\left(\frac{1-i}{2} \right)
\\&= \frac{\pi}{2 \sqrt{2}}\left\{ \frac \pi 4-\operatorname{arccoth}\left[\sqrt{7+5 \sqrt{2}}+\sqrt{2}+1\right] - \arctan\left[\frac{1}{41} \left(2 \sqrt{89 \sqrt{2}-119}+2 \sqrt{2}+7\right)\right] \right\}.
}
$$
This is numerically equal to the claimed result, which I obtained via a completely different route.
Throughout I have not worried explicitly about choosing the correct branch of $\log$. If this led to any mistakes, please point it out to me.
Appendix
Here I sketch how $\Im J\left(\frac{1-i}{2} \right)$ can be calculated. For example, consider the term
$$
\Im \log\left(1 + \frac{1-i}{2} + \sqrt 2 \sqrt{\frac{1-i}{2}} \right) = \Im \log\left(\frac 3 2 - \frac i 2 + 2^{1/4}\left(\cos \frac \pi 8 - i \sin \frac \pi 8 \right) \right).
$$
Using the half-angle formulas for sine and cosine, we can write it as
$$
\al{
-\arctan \left(\frac{1+ \sqrt[4]{2} \sqrt{2-\sqrt{2}}}{3 + \sqrt[4]{2} \sqrt{2+\sqrt{2}}}\right) &= - \arctan\left(\frac{2 \sqrt{5 \sqrt{2}-7}+1}{7-2 \sqrt{2}}\right)
\\&= - \arctan\left[\frac{1}{41} \left(2 \sqrt{89 \sqrt{2}-119}+2 \sqrt{2}+7\right)\right].
}
$$
To obtain the first equality, multiply numerator and denominator by a certain factor to get rid of the fourth roots. For the second, multiply by another factor to get rid of all the roots. Of course this simplification is just for aesthetics.
Moral: introduce a parameter in such a way that the integral with respect to the parameter becomes simple, and use Mathematica to simplify the end result.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Simplifying $\frac{x^6-1}{x-1}$ I have this:
$$\frac{x^6-1}{x-1}$$
I know it can be simplified to $1 + x + x^2 + x^3 + x^4 + x^5$
Edit : I was wondering how to do this if I didn't know that it was the same as that.
|
By indeterminate coefficients:
$$(x-1)(ax^5+bx^4+cx^3+dx^2+ex+f)\\=ax^6+(b-a)x^5+(c-b)x^4+(d-c)x^3+(e-d)x^2+(f-e)x-f.$$
After identification, $$a=b=c=d=e=f=1.$$
|
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|
Integral of Rational Functions $$\int \frac{dx}{ax^2 + bx + c} \quad \text{for} \quad 4ac-b^2 >0$$
then
$$\begin{align}
ax^2 + bx + c
&= a\biggl(x+\frac{b}{2a}\biggr)^2 + \frac{4ac-b^2}{4a} \\
&= \Biggl(\sqrt{a}\biggl(x+\frac{b}{2a}\biggr)\Biggr)^2 + \Biggl(\sqrt{\frac{4ac-b^2}{4a}}\Biggr)^2
\end{align}$$
implies
$$ \int \frac{dx}{ax^2 + bx + c} = \int \frac{dx}{\Bigl(\sqrt{a}\bigl(x+\frac{b}{2a}\bigr)\Bigr)^2 + \Bigr(\sqrt{\frac{4ac-b^2}{4a}}\Bigr)^2} = \frac{2\sqrt{a}}{\sqrt{4ac-b^2}}\arctan\frac{2ax+b}{\sqrt{4ac-b^2}}
$$
The correct answer does not have the $\sqrt{a}$ in the numerator. I've checked over my work, but have clearly made some dumb mistake. Can anyone please let me know where such an error was made? Thank you for your help!
|
Hard to debug, since some detail is missing. To make things less messy, I would rewrite the integral as
$$ \int\frac{4a\,dx}{(2ax+b)^2+4ac-b^2}.$$
Let $4ac-b^2=K^2$. Then let $2ax+b=Ku$. We have $2a \,dx=K\,du$. So our integral is
$$\int \frac{2K\,du}{K^2u^2+K^2},$$
which is
$$\frac{2}{K}\arctan u+C.$$
Remark: I am not fond of fractions, so for completing the square I prefer to write $$ax^2+bx+c=\frac{1}{4a}\left(4a^2x^2+4abx+4c\right)=\frac{1}{4a}\left((2ax+b)^2-(b^2-4ac) \right).$$
|
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|
finding the jordan canonical with one eigenvalue Let $$A= \left[
\begin{array}{ c c }
-2 & -3 & 1 \\
0 & -2 & 0 \\
0 & 0 & -2
\end{array} \right]
$$
Determine the Jordan canonical form of A.
The only eigenvalue I found was $\lambda=-2$
So finding the eigenvector of this is
$$ \left[
\begin{array}{ c c }
0 & -3 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array} \right]. \left[
\begin{array}{ c c }
x \\
y \\
z
\end{array} \right]=0\implies -3x+z=0\iff z=3x
$$
So let $x$ and $y$ be free variables so $x=a$ and $y=b$ with $a,b\in\mathbb{F}$. So $$\left[
\begin{array}{ c c }
a \\
b \\
3b
\end{array} \right]=a\left[
\begin{array}{ c c }
1 \\
0 \\
0
\end{array} \right]+b\left[
\begin{array}{ c c }
0 \\
1 \\
3
\end{array} \right]$$
So we have two eigenvectors that are LI: $$v_1=\left[
\begin{array}{ c c }
1 \\
0 \\
0
\end{array} \right], v_2=\left[
\begin{array}{ c c }
0 \\
1 \\
3
\end{array} \right]$$
But what would $v_3$ be? Since correct me if I am wrong but our Jordan canonical matrix $J=P^{-1}AP$ where $P$ is the matrix made up of $v_1,v_2,v_3$.
|
Hint: You can find a third generalized and linearly independent eigenvector by solving (use Gaussian Elimination):
$$[A-\lambda I]v_3 = [A + 2I]v_3 = v_1$$
You should get:
$$v_3 = \left(0, -\dfrac 13, 0 \right)$$
You can then use the eigenvectors to find (this uses your current eigenvectors) and $v_3$ as:
$$J = P^{-1} A P = \begin{bmatrix}-2 &0&0 \\0 & -2 & 1 \\ 0 & 0 & -2\end{bmatrix}$$
Where $P$ is made up of your column eigenvectors.
|
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|
Series representation
I have started this problem by expanding it so that i can get some cancellation term, but couldn't reach on the correct result.I got the result like -ln4 -ln5-ln6..... Please have a look on this.
|
Consider the sum:
$$S=\sum_{n=1}^m \left(1-\frac{2n+5}{2}\ln\left(\frac{n+3}{n+2}\right)\right)=m-\frac{1}{2}\sum_{n=1}^m \ln\left(\frac{n+3}{n+2}\right)^{2n+5}$$
Since,
$$\sum_{n=1}^m \ln\left(\frac{n+3}{n+2}\right)^{2n+5}=\ln\left(\frac{4^7}{3^7}\cdot \frac{5^9}{4^9}\cdot \frac{6^{11}}{5^{11}}\cdots\frac{(m+3)^{2m+5}}{(m+2)^{2m+5}}\right)=\ln\left(\frac{4(m+3)^{2m+5}}{3^5((m+2)!)^2}\right)$$
Hence,
$$S=\ln(e^m)-\ln\left(\frac{2(m+3)^{\frac{2m+5}{2}}}{3^{5/2}(m+2)!}\right)=\ln\left(\frac{3^{5/2}}{2}\cdot \frac{e^m(m+2)!}{(m+3)^{\frac{2m+5}{2}}}\right)$$
Calculate the limit of the final expression as $m\rightarrow \infty$ using Stirling's approximation to obtain:
$$S=\ln\left(\frac{3^{5/2}}{2}\cdot \frac{\sqrt{2\pi}}{e^3}\right)=\boxed{\dfrac{1}{2}\ln(6\pi)+\ln\left(\dfrac{9}{2}\right)-3}$$
|
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|
evaluation of the integral of a certain logarithm I come across the following integral in my work
$$\int_a^\infty \log\left(\frac{x^2-1}{x^2+1}\right)\textrm{d}x,$$
with $a>1$.
Does this integral converge ? what is its value depending on $a$ ?
|
Using
$$
-i\log\left(\frac{x+i}{x-i}\right)=\pi-2\tan^{-1}(x)
$$
we get
$$
\begin{align}
\int_a^\infty\log\left(\frac{x^2-1}{x^2+1}\right)\mathrm{d}x
&=\int_a^\infty\left[\vphantom{\sum}\log(x+1)+\log(x-1)-\log(x+i)-\log(x-i)\right]\mathrm{d}x\\
&{=}+(x+1)\log(x+1)-(x+1)\\
&\hphantom{=}+(x-1)\log(x-1)-(x-1)\\
&\hphantom{=}-(x+i)\log(x+i)+(x+i)\\
&\hphantom{=}-(x-i)\log(x-i)+(x-i)\\
&=\left[\log\left(\frac{x+1}{x-1}\right)+x\log\left(\frac{x^2-1}{x^2+1}\right)-2\tan^{-1}(x)\right]_a^\infty\\
&=\log\left(\frac{a-1}{a+1}\right)+a\log\left(\frac{a^2+1}{a^2-1}\right)+2\tan^{-1}(a)-\pi
\end{align}
$$
|
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|
Prove that $\sqrt{n^2 + 2}$ is irrational
Suppose $n$ is a natural number. Prove that $\sqrt{n^2 + 2}$ is irrational.
From looking at the expression, it seems quite obvious to me that $\sqrt{n^2 + 2}$ will be irrational, since $n^2$ will be a natural number, and after adding $2$ to it, $n^2 + 2$ will no longer be a perfect square.
From that logic, I tried letting $n^2 + 2$ = ${a^2\over b^2}$ to find a contradiction, but I wasn't able to formulate a proof that worked. Any suggestions as to how I should approach this?
|
Without loss of generality we can assume $x,y\in \mathbb{N}\ (?)$.
$$n^2+2=\dfrac{x^2}{y^2} \implies x^2=y^2\left(n^2+2\right)$$
For odd or even $n$, $x,y$ can be both odd, or exactly one of them should be even.
Case 1 ($n$ odd)
Case 1.1 (both odd)
$\left(x^2\equiv1\pmod4, y^2\equiv1\pmod4\right)\land \left(x^2=y^2\left(n^2+2\right)\right) \implies 1\equiv3\pmod 4$
Case 1.2 ($x$ odd, $y$ even)
$\left(x^2\equiv1\pmod4, y^2\equiv0\pmod4\right)\land \left(x^2=y^2\left(n^2+2\right)\right) \implies 1\equiv0\pmod 4$
Case 1.3 ($x$ even, $y$ odd)
$\left(x^2\equiv0\pmod4, y^2\equiv1\pmod4\right)\land \left(x^2=y^2\left(n^2+2\right)\right) \implies 0\equiv3\pmod 4$
Case 2 ($n$ even)
Case 2.1 (both odd)
$\left(x^2\equiv1\pmod4, y^2\equiv1\pmod4\right)\land \left(x^2=y^2\left(n^2+2\right)\right) \implies 1\equiv2\pmod 4$
Case 2.2 ($x$ odd, $y$ even)
$\left(x^2\equiv1\pmod4, y^2\equiv0\pmod4\right)\land \left(x^2=y^2\left(n^2+2\right)\right) \implies 1\equiv0\pmod 4$
Case 2.3 ($x$ even, $y$ odd)
$\left(x^2\equiv0\pmod4, y^2\equiv1\pmod4\right)\land \left(x^2=y^2\left(n^2+2\right)\right) \implies 0\equiv2\pmod 4$
|
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|
Help with a tricky limit $\lim_{n\to\infty} \sum\limits_{i=1}^n (i/n)(\sqrt[2]{(i+1)/n}-\sqrt[2]{i/n})$ I have been attempting to follow the answer to a question previously asked on the site (Lebesgue integral basics), but am lost on how one might go about evaluating $$\lim_{n\to\infty} \sum\limits_{i=1}^n \frac in\left(\sqrt{(i+1)/n}-\sqrt{i/n}\right)$$ I have been unable to locate a summation formula for the sum of square roots. Any help is greatly appreciated.
|
You can manipulate the sums to get this working:
\begin{align}
\sum\limits_{i=1}^n\frac{i}{n}(\sqrt{\frac{i+1}{n}}-\sqrt{\frac{i}{n}})&=\sum\limits_{i=1}^n\frac{i}{n}\sqrt{\frac{i+1}{n}}-\sum\limits_{i=1}^n\frac{i}{n}\sqrt{\frac{i}{n}}\\&=\sum\limits_{i=1}^n\frac{i+1}{n}\sqrt{\frac{i+1}{n}}-\sum\limits_{i=1}^n\frac{1}{n}\sqrt{\frac{i+1}{n}}-\sum\limits_{i=1}^n\frac{i}{n}\sqrt{\frac{i}{n}}\\&=\sum\limits_{i=2}^{n+1}\frac{i}{n}\sqrt{\frac{i}{n}}-\sum\limits_{i=1}^n\frac{1}{n}\sqrt{\frac{i+1}{n}}-\sum\limits_{i=1}^n\frac{i}{n}\sqrt{\frac{i}{n}}\\&=(\frac{n+1}{n})^{\frac{3}{2}}-\sum\limits_{i=1}^n\frac{1}{n}\sqrt{\frac{i+1}{n}}\\&=(\frac{n+1}{n})^{\frac{3}{2}}-(\frac{1}{n})^{3/2}\sum\limits_{i=2}^{n+1}\sqrt{i}
\end{align}
Now this is a lot better. I'm going to put bounds on the sum using $\int\sqrt{x}$, because $\sqrt{x}$ is monotone increasing.
$\int\limits_1^{n+1}\sqrt{x}dx\leq\sum\limits_{i=2}^{n+1}\sqrt{i}\leq\int\limits_2^{n+2}\sqrt{x}dx$
$\frac{2}{3}(n+1)^{3/2}-\frac{2}{3}\leq\sum\limits_{i=2}^{n+1}\sqrt{i}\leq\frac{2}{3}(n+2)^{3/2}-\frac{4}{3}\sqrt{2}$
Multiplying through by $(\frac{1}{n})^{3/2}$ gives us:
$\frac{2}{3}(\frac{n+1}{n})^{\frac{3}{2}}-\frac{2}{3n^{\frac{3}{2}}}\leq(\frac{1}{n})^{3/2}\sum\limits_{i=2}^{n+1}\sqrt{i}\leq\frac{2}{3}(\frac{n+2}{n})^{\frac{3}{2}}-\frac{4\sqrt{2}}{3n^{\frac{3}{2}}}$
This directly transfers into a limit inequality:
$\frac{2}{3}\leq\lim\limits_{n\to\infty}(\frac{1}{n})^{\frac{3}{2}}\sum\limits_{i=1}^{n+1}\sqrt{i}\leq\frac{2}{3}$
We see
$\lim\limits_{n\to\infty}(\frac{1}{n})^{\frac{3}{2}}\sum\limits_{i=1}^{n+1}\sqrt{i}=\frac{2}{3}$
Our final limit then becomes
$\lim\limits_{n\to\infty}\sum\limits_{i=1}^n\frac{i}{n}(\sqrt{\frac{i+1}{n}}-\sqrt{\frac{i}{n}})=1-\frac{2}{3}=\frac{1}{3}$
|
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|
Extremely difficult log integral, real methods only $$\int_{0}^{1}\frac{x^2 + x\log(1-x)- \log(1-x) - x}{(1-x)x^2} dx$$
I tried this:
$$M_1 = \int_{0}^{1} \frac{1}{1-x} \cdot \left(\frac{x^2 + x\log(1-x) - \log(1-x) - x)}{x^2}\right) dx$$
$$M_1 = \int_{0}^{1} \frac{x^2 + x\log(1-x) - \log(1-x) - x}{(1-x)x^2} dx$$
$$M_1 = \int_{0}^{1} \frac{x\left(x + \log(1-x) - 1\right) - \log(1-x)}{(1-x)x^2} dx$$
$$M_1 = \int_{0}^{1} \frac{x + \log(1-x) - 1}{x(1-x)}dx - \int_{0}^{1} \frac{\log(1-x)}{(1-x)x^2} dx$$
but we cannot seperate as the integrals become divergent.
|
Using Paul_I's answer and the classic expansion
$$\log\left(1-x\right)=-\sum_{n=1}^\infty \frac{x^n}{n}, \quad -1<x<1,$$
we get
$$-\int_0^1\frac{\log(1-x)+x}{x^2}\mathrm dx =\sum_{n=2}^\infty\int_0^1 \frac{x^{n-2}}{n}\mathrm dx = \sum_{n=1}^\infty\frac{1}{n(n+1)}=1$$
in accordance with Mathematica.
|
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|
How to evaluate this series? I need to evaluate this series.Without using derivative.
$$A=\frac{2} {2^2} + \frac{4} {2^5}+ \frac{6} {2^8} + \cdots $$
Where the $i$ th member is calculated with the the formula below:
$$A_i=\frac{i} {2^{3i-2}}$$
Feel free to edit the tags please.Thanks for you help.
|
$$a=\frac{2}{2^2}+\frac{4}{2^5}+\frac{6}{2^8}+\frac{8}{2^11}...+ $$multiply a by $$\frac{1}{2^3} $$ then find $$a -\frac{1}{2^3}a $$
$$a -\frac{1}{2^3}a=\\\frac{2}{2^2}+\frac{4}{2^5}+\frac{6}{2^8}+\frac{8}{2^{11}}...\\-(\frac{2}{2^5}+\frac{4}{2^8}+\frac{6}{2^11}+\frac{8}{2^{14}}...)\\a -\frac{1}{2^3}a=\frac{2}{2^2}+\frac{4-2}{2^5}+\frac{6-4}{2^8}+\frac{8-6}{2^{11}}...$$now you have
$$\frac{7}{8}a=2(\frac{1}{2^2}+\frac{1}{2^5}+\frac{1}{2^8}+\frac{1}{2^{11}}...)$$
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|
Properties of prime mod $3$ We know that if $p$ is a prime congruent to $3 \mod 4$, we cannot represent it as sum of two squares. Is there a positive property of such $p$? That is, do we have any statements that say "$p$ is a prime congruent to $3 \mod 4$ iff $\underline{\mbox{a positive statement}}$ is TRUE in an unique way". For instance, we have "$p$ is a prime congruent to $1 \mod 4$ iff $p=a^2+b^2$ and $|ab|>1$ in an unique way.
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For example, we can say, a prime is $1,3 \pmod 8$ if and only if there is just one expression $p = x^2 + 2 y^2.$
You get a little flexibility by throwing in indefinite forms: a prime is $1,7 \pmod 8$ if and only if there are just two infinite sequences of expressions $p = x^2 - 2 y^2,$ under the action (and its inverse)
$$
\left(
\begin{array}{r}
x \\
y
\end{array}
\right) \mapsto
\left(
\begin{array}{rr}
3 & 4 \\
2 & 3
\end{array}
\right)
\left(
\begin{array}{r}
x \\
y
\end{array}
\right)
$$
The two orbits for the prime $7$ have base points
$$
\left(
\begin{array}{r}
3 \\
-1
\end{array}
\right)
$$
and
$$
\left(
\begin{array}{r}
3 \\
1
\end{array}
\right)
$$
|
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|
Suppose that two polar curves are given by: $R_1 = \cos(2\theta)$ and $R_2 = \sin(3\theta)$. Find the smallest positive solution exactly. Suppose that two polar curves are given by: $R_1 = \cos(2\theta)$ and $R_2 = \sin(3\theta)$. Find the smallest positive solution exactly.
I know that we are looking for the smallest positive value where the two functions intersect. But I graphed it on a polar coordinate grapher, and I don't know how you would determine the correct solution.
Any help is appreciated!
|
Using formulas:
$$\sin{(3\theta)}=3\sin{(\theta)}-4\sin^3{(\theta)}$$
$$\cos{(2\theta)}=1-2\sin^2{(\theta)}$$
We can find a cubic equation for $\sin{(\theta)}$.
We have;
\begin{equation}
\begin{aligned}
\cos{(2\theta)}&=\sin{(3\theta)} \\ \cos{(2\theta)}-\sin{(3\theta)} &= 0 \\
1-2\sin^2{(\theta)} - 3\sin{(\theta)}+4\sin^3{(\theta)} &=0
\end{aligned}
\end{equation}
We can treat this as a cubic equation just with a variable $x$ for simplicity, so we need to solve,
$$ 4x^3-2x^2-3x+1=0 $$
This is found to have 3 distinct roots
$$x_1=1, \qquad x_2=\frac{-(1+\sqrt{5})}{4}, \qquad x_3=\frac{\sqrt{5}-1}{4}. $$
Now we want $\theta$ to be the smallest positive value so we ignore $x_2$ and we can also see $0 < x_3 < x_1 \le1$. So we have $\sin(\theta)=\frac{\sqrt{5}-1}{4}$ and therefore $\theta = \frac{\pi}{10}$.
|
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|
Solve cubic equation $x^3-9 x^2-15x-6 =0$ without going Cardano
Solve the cubic equation for $x\in\mathbb{R}$ $$x^3-9 x^2-15x-6 =0$$
Note that the only real solution is $x=3+2\sqrt[3]{7}+\sqrt[3]{7^2}$. Given the regularity of this solution, can we solve for it constructively, without going full Cardano?.
Also, can we prove that there is only one real solution without using the discriminant?
|
As an alternative to the Cardano’s method, substituting $x=3+2\sqrt{14}\cosh t$ into $x^3-9 x^2-15x-6=0$ to get
$$4\cosh^3t-3\cosh t = \cosh 3t = \frac{15}{4\sqrt{14}}$$
which leads to $t=\frac13\cosh^{-1}\frac{15}{4\sqrt{14}}$ and, then, the solution
$$x=3+2\sqrt{14}\cosh\left(\frac13\cosh^{-1}\frac{15}{4\sqrt{14}}\right)$$
which is numerically equal to $3+2\sqrt[3]{7}+\sqrt[3]{49}$.
|
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|
A problem with proving using definition that $\lim_{n\to\infty}\frac {n^2-1}{n^2+1}=1$
Prove using the definition that: $$\displaystyle\lim_{n\to\infty}\frac {n^2-1}{n^2+1}=1 $$
What I did:
Let $\epsilon >0$, finding $N$: $\mid\frac {n^2-1}{n^2+1}-1\mid=\mid\frac {-2}{n^2+1}\mid\le\mid\frac {-2}{n^2}\mid=\frac {2}{n^2}\le\frac {2}{n}$ So $N=\frac 2 {\epsilon}$.
So for all $n>N$ we'll want to show that: $|a_n-l|<\epsilon \Rightarrow \mid\frac {n^2-1}{n^2+1}-1\mid=\mid\frac {-2}{n^2+1}\mid<\mid\frac {-2}{N^2} \mid=\mid\frac {-2}{\frac 1 {\epsilon^2}} \mid=\mid -2\epsilon^2 \mid=2\epsilon^2$
But that isn't smaller than $\epsilon$ so I'm doing something wrong here and I don't know what...
|
$$\left|\frac{n^2-1}{n^2+1}-1\right|=2\frac1{n^2+1}<\epsilon\iff n^2+1>\frac2\epsilon\iff n>\sqrt{\frac2\epsilon-1}$$
|
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|
Can this interesting property be proven? $$2^2+3^2+5^2+7^2+9^2+11^2=(17)^2$$
$$22^2+33^2+55^2+77^2+99^2+11^2=(143)^2$$
Also:
$$22^2+33^2+55^2+77^2+99^2+121^2=(187)^2$$
$$222^2+333^2+555^2+777^2+999^2+1221^2=(1887)^2$$
$$2222^2+3333^2+5555^2+7777^2+9999^2+12221^2=(18887)^2$$
$$22222^2+33333^2+55555^2+77777^2+99999^2+122221^2=(188887)^2$$
How can I prove this?
Are there other interesting properties like these?
|
Hint: What happens if you multiply your first identity with $111\cdots1^2$?
|
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|
Give the following linear transformation find values of parameter Find values of parameter t for which transformation is epimorphic:
$\psi([x_1,x_2,x_3,x_4])=x_1+x_2+x_3+2x_4,x_1+tx_2+x_3+3x_4,2x_1+x_2+tx_3+3x_4 $
When this transformation is epimorphic i.e. what should i look for in the reduced form of matrix of this linear transformation. My reduced matrix is:
$\begin{pmatrix}
1 & 1 & 1 & 2 \\
0 & t-1 & 0 & 1 \\
0 & -1 & t-2 & -1
\end{pmatrix} $
|
It seems the following.
The transformation $\psi$ is ephimorphic iff rank of its matrix is maximal, that is $3$.
Buy elementary transormations, which does not change the rank, we can transform the matrix as follows.
$\begin{pmatrix}
1 & 1 & 1 & 2 \\
0 & t-1 & 0 & 1 \\
0 & -1 & t-2 & -1
\end{pmatrix} $
$\begin{pmatrix}
1 & {\mathbf 0} & {\mathbf 0} & {\mathbf 0}\\
0 & t-1 & 0 & 1 \\
0 & -1 & t-2 & -1
\end{pmatrix} $
$\begin{pmatrix}
1 & 0 & 0 & 0\\
0 & t-1 & 0 & 1 \\
0 & {\mathbf t-2} & t-2 & {\mathbf 0}
\end{pmatrix} $
$\begin{pmatrix}
1 & 0 & 0 & 0\\
0 & {\mathbf 0} & 0 & 1 \\
0 & t-2 & t-2 & 0
\end{pmatrix} $
$\begin{pmatrix}
1 & 0 & 0 & 0\\
0 & 0 & 0 & 1 \\
0 & {\mathbf 0} & t-2 & 0
\end{pmatrix} $
That is the rank equals $3$ iff $t\ne 2$ .
Answer. The transformation $\psi$ is ephimorphic iff $t\ne 2$.
|
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|
What is the solid angle of the intersection loop between a cone and an off-axis sphere?
An upright (green) cone with opening angle $2a < \pi/10$ has its vertex at point O with cartesian xyz coordinates $(0,0,0)$. The cone axis (dotted line) lies in the plane $y=0$ and is parallel to the z-axis.
A (blue) sphere of radius $r$ is centred at point $P(P_x,0,P_z)$ which may lie inside or outside the cone.
The line segment connecting points $O,P$ has length $p < r$. Thus point $O$ always lies inside the sphere.
The cone surface and the sphere surface intersect in a non-planar curved line loop which includes points $L1,L2$.
From here I have found equations for the loop (where $c$ is the "cone opening parameter" defined by $c^2=(x^2+y^2)/z^2$) :-
$$
(x-P_x)^2+(y-P_y)^2+\frac{x^2+y^2}{c^2}-\frac{2.P_z}{c}\sqrt{x^2+y^2}+P_z^2=r^2
\qquad [1] $$
$$
x^2\left(1+\frac{1}{c^2}\right)-2P_x.x+y^2\left(1+\frac{1}{c^2} \right)
-2P_y.y + (P_x^2+P_y^2+P_z^2-r^2)-\frac{2P_z}{c}\sqrt{x^2+y^2}=0.
\qquad [2] $$
In this case the value of $P_y$ is zero which simplifies the above equations a little bit.
QUESTION
What (in steradians) is the solid angle $w$ subtended by the "loop" at point $P$ in terms of $a,r,P_x,P_z$?
|
Let's move everything so that the center of the sphere is in the origin. Then the sphere is simply
$$x^2+y^2+z^2=r^2$$
and the cone becomes
$$\left(x+P_x\right)^2+y^2=c^2\left(z+P_z\right)^2$$
and since you don't want a double cone, you also want
$$z+P_z>0\quad.$$
The $y=0$ plane intersects the cone in two lines, namely
$$x=\pm c(z+P_z)-P_x$$
and these intersect the sphere at
$$x_{1,2}={\frac{\pm c\left({P_z} + \sqrt{{\left(c^{2} + 1\right)} r^{2} - c^2{P_z}^{2} \pm 2 c {P_x} {P_z} - {P_x}^{2}}\right) - {P_x}}{c^{2} + 1}}\quad.$$
For a given $x$ coordinate in that plane, the points on the loop are both at
$$ z(x) = -\frac{{P_z} c^{2} - \sqrt{{\left({P_x}^{2} - {P_z}^{2}\right)} c^{2} + {\left(c^{2} + 1\right)} r^{2} + {P_x}^{2} + 2 \, {\left({P_x} c^{2} + {P_x}\right)} x}}{c^{2} + 1} $$
so the portion of the sphere in that $x$ plane and between the points on the loop will form a circular arc with a length of
$$l(x) = 2\sqrt{r^2-x^2}\arcsin\frac{z(x)}{\sqrt{r^2-x^2}}\quad.$$
To integrate these arcs into an area, you have to multiply them by the arc length element in the $y=0$ plane. This is defined by
$$\mathrm ds^2=\mathrm dx^2+\mathrm dz^2 = \mathrm dx^2\left(1+\left(\frac{\mathrm dz}{\mathrm dx}\right)^2\right)$$
which means we need the derivative
$$\frac{\mathrm dz}{\mathrm dx}=\frac{{P_x}}{\sqrt{{\left({P_x}^{2} - {P_z}^{2}\right)} c^{2} + {\left(c^{2} + 1\right)} r^{2} + {P_x}^{2} + 2 \, {\left({P_x} c^{2} + {P_x}\right)} x}}$$
to obtain
$$\mathrm ds = {\sqrt{\frac{{P_x}^{2}}{{\left({P_x}^{2} - {P_z}^{2}\right)} c^{2} + {\left(c^{2} + 1\right)} r^{2} + {P_x}^{2} + 2 \, {\left({P_x} c^{2} + {P_x}\right)} x} + 1}}\;\mathrm dx\quad.$$
So the portion of the sphere enclosed by the loop has area
$$\int_{x_1}^{x_2}l(x)\;{\sqrt{\frac{{P_x}^{2}}{{\left({P_x}^{2} - {P_z}^{2}\right)} c^{2} + {\left(c^{2} + 1\right)} r^{2} + {P_x}^{2} + 2 \, {\left({P_x} c^{2} + {P_x}\right)} x} + 1}}\;\mathrm dx\quad.$$
At least numerically this should be reasonably simple to integrate. I haven't tried yet to find a closed form for this. It might be simpler if you have explicit numbers instead of all these variables.
In the end, you can divide that area by $r^2$ to obtain the corresponding solid angle.
|
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|
Find $ \int \frac {\mathrm{d}x}{(4x^2-1)^{3/2}}$ I have trouble using trig sub. After I get that x = 2x+1, should I substitute back into the original problem's $4x^2$ with $(4(2x+1)^2)$?
|
$y=2x-1 \implies \dfrac{1}{2}dy=dx$
$\therefore\displaystyle\int\dfrac{dx}{\left(4x^2-1\right)^{\frac{3}{2}}}=\dfrac{1}{2}\displaystyle\int\dfrac{dy}{y^{\frac{3}{2}}\left(y+2\right)^{\frac{3}{2}}}$
$y=2\tan^2\theta \implies dy=4\sec^2\theta\tan\theta\ d\theta$
$\therefore\dfrac{1}{2}\displaystyle\int\dfrac{dy}{y^{\frac{3}{2}}\left(y+2\right)^{\frac{3}{2}}}=\displaystyle\int\dfrac{2\sec^2\theta\tan\theta\ d\theta}{2^3\tan^3\theta\sec^3\theta}=\dfrac{1}{4}\displaystyle\int\csc \theta \cot \theta \ d\theta$
|
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|
Partial Fraction Decomposition of Exponential Generating Functions I want to see if it is possible to write
$$ \left(\frac{x}{e^x-1}\right) \left(\frac{x^2/2! }{e^x-1-x}\right) \left(\frac{x^3/3!}{e^x-1-x-x^2/2}\right)$$
as a linear combination of the factors
$$p(x)\frac{x}{e^x-1}+q(x)\frac{x^2/2}{e^x-1-x}+r(x)\frac{x^3/3!}{e^x-1-x-x^2/2}$$
I'm specifically looking for the functions $p(x), q(x)$ and $r(x)$ so I thought I would try a partial fraction decomposition method and this would imply that
$$\frac{A}{e^x-1}+\frac{B}{e^x-1-x}+\frac{C}{e^x-1-x-x^2/2}$$
where $A=xp(x), B=\frac{1}{2}x^2q(x), C=\frac{1}{6}x^3r(x)$. So this means that
$$A(e^x-1-x)(e^x-1-x-x^2/2)+B(e^x-1)(e^x-1-x-x^2/2)+C(e^x-1)(e^x-1-x)=\frac{x^6}{12}$$
If we multiply through
$$A\left(e^{2x}-2e^x-2xe^x-\frac{x^2}{2}e^x+1+2x+\frac{3}{2}x^2+\frac{x^3}{2}\right)$$
$$+B\left(e^{2x}-2e^x-xe^x-\frac{x^2}{2}e^x+1+x+\frac{x^2}{2}\right)$$
$$+C\left(e^{2x}-2e^x-xe^x+1+x\right)$$
Thus we get
$$(A+B+C)(e^{2x}-2e^x)-(2A+B+C)xe^x-(A+B)\frac{x^2}{2}e^x+(A+B+C)+(2A+B+C)x+\left(\frac{3}{2}A+\frac{1}{2}B\right)x^2+\frac{1}{2}Ax^3=\frac{x^6}{12}$$
To me, now it seems that there is no solution since both $A+B+C=2A+B+C=0$ which means $A=0$ which then implies $B=C=0$. Is this method incorrect for such a problem and if so is there another way to approach this problem? Or did I make an error?
|
I think that your idea to decompose in simple fractions is good. You can put $K=\mathbb{Q}(x)$, replace $\exp(x)$ by $y$, and decompose the following fraction in $K(y)$:
$$\left(\frac{x}{y-1}\right) \left(\frac{x^2/2! }{y-1-x}\right) \left(\frac{x^3/3!}{y-1-x-x^2/2}\right)=\frac{A}{y-1}+\frac{B}{y-1-x}+\frac{C}{y-1-x-x^2/2}$$
To get $A$, you multiply by $y-1$ both side, simplify and put $y=1$, to get $B$, you multiply by $y-1-x$ both side, simplify and put $y=1+x$, etc .
When you are done, simply replace $y$ by $\exp(x)$.
|
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|
Find value of f(2013)? Given a function $f(x)$ such that: $f(1) + f(2) + f(3)+\cdots+f(n) = n^2f(n)$
Find the value of $f(2013)$. It is given that $f(1) = 2014$.
I tried attempting the question as a bottom-up DP, but soon realized the numbers are too irregular and large to deal with in a mechanical fashion.
|
Calculating a few of the first $n$ by hand reveals a pattern:
$$f(n) = \frac{f(1)}{T_n},$$
where $T_n$ is the $n$th triangular number.
Proof by induction:
The relationship holds for $n=1$ since $T_1 = 1.$
We assume that
$$f(x) = \sum_{k=1}^n f(k) = \frac{2}{n(n+1)} f(1).$$
This implies that
$$\frac{2}{n(n+1)} f(1) = n^2 f(n).$$
Then,
$$f(x+1) = \sum_{k=1}^{n+1} f(k) = n^2 f(n) + f(n+1) = (n+1)^2 f(n+1),$$
and
$$f(n+1) = \frac{n}{n+2}f(n) = \frac{2f(1)}{n(n+1)}\frac{n}{n+2} = \frac{2f(1)}{(n+1)(n+2)} = \frac{f(1)}{T_{n+1}}.$$
To answer the question, then,
$$f(2013) = \frac{2 \cdot 2014}{2013 \cdot 2014} = \frac{2}{2013}.$$
|
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|
How to prove $ \lim_{n\to\infty}\sum_{k=1}^{n}\frac{n+k}{n^2+k}=\frac{3}{2}$? How to prove that
$\displaystyle \lim_{n\longrightarrow\infty}\sum_{k=1}^{n}\frac{n+k}{n^2+k}=\frac{3}{2}$?
I suppose some bounds are nedded, but the ones I have found are not sharp enough (changing $k$ for $1$ or $n$ leads to the limit being between 1 and 2).
Any suggestion is welcomed.
|
Squeeze (without integral calculus).
I. $\displaystyle\sum_{k=1}^{n}\dfrac{n+k}{n^2+k} \leq \sum_{k=1}^{n}\dfrac{n+k}{n^2+1}=\frac{1}{n^2+1}\sum_{k=1}^{n}(n+k) = \frac{1}{n^2+1}(n^2+\frac{n(n+1)}{2})\rightarrow\frac{3}{2} $.
II.$\displaystyle\sum_{k=1}^{n}\dfrac{n+k}{n^2+k} \geq\sum_{k=1}^{n}\dfrac{n+k}{n^2+n}=\frac{1}{n^2+n}\sum_{k=1}^{n}(n+k) = \frac{1}{n^2+n}(n^2+\frac{n(n+1)}{2})\rightarrow\frac{3}{2} $
|
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|
Prove without induction $2^n \mid (b+\sqrt{b^2-4c})^n + (b-\sqrt{b^2-4c})^n $
Prove $2^n \mid (b+\sqrt{b^2-4c})^n + (b-\sqrt{b^2-4c})^n $ for all $n\ge 1$ and $b,c$ are integers.
Is it possible to prove this without induction?
|
We want to show$$2^n\vert\left(\left(b+\sqrt{b^2-4c}\right)^n+\left(b-\sqrt{b^2-4c}\right)^n\right).$$
Using $x-y=\frac{x^2-y^2}{x+y}$ indeed we have
$$ 2^n\vert\left(\left(b+\sqrt{b^2-4c}\right)^n+\left(\frac{4c}{b+\sqrt{b^2-4c}}\right)^n\right) \\ 2^n\vert\frac{\left(b+\sqrt{b^2-4c}\right)^{2n}+4^nc^n}{\left(b+\sqrt{b^2-4c}\right)^n} \\ 2^n\vert\frac{\left(2b^2-4c+2b\sqrt{b^2-4c}\right)^n+4^nc^n}{\left(b+\sqrt{b^2-4c}\right)^n}\\2^n\vert\frac{2^n\left(\left(b^2-2c+b\sqrt{b^2-4c}\right)^n+2^nc^n\right)}{\left(b+\sqrt{b^2-4c}\right)^n}.$$
|
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|
How to express the equations as the Square root Like? $$2\sin \frac{\pi}{16}= \sqrt{2-\sqrt{2+\sqrt{2}}}$$
What law is need to be applied here? Do I have to make the $\frac{\pi}{16}$ in a form that will be give us $\sqrt{2}$ like sin 45 degree?
|
Call $x = \sin \frac{\pi}{16} \to 1-2x^2 = \cos \frac{\pi}{8} \to \left(1-2x^2\right)^2 = \cos^2 \frac{\pi}{8} = \dfrac{1+\cos \frac{\pi}{4}}{2} = \dfrac{2+\sqrt{2}}{4} \to 1-2x^2 = \dfrac{\sqrt{2+\sqrt{2}}}{2} \to x^2 = \dfrac{2-\sqrt{2+\sqrt{2}}}{4} \to 2x = \sqrt{2-\sqrt{2+\sqrt{2}}}$.
|
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|
Maximum $\int_{0}^{y}\sqrt{x^4+\left(y-y^2\right)^2}dx$ where $y\in \left[0,1\right]$? How find maximum this integral $$\int_{0}^{y}\sqrt{x^4+\left(y-y^2\right)^2}dx$$ where $y\in \left[0,1\right]$?
|
By differentiation under the integral sign we get
\begin{align*}\left(\int_0^y\sqrt{x^4+(y-y^2)^2}\,dx\right)'&=\int_0^y\frac{2(y-y^2)(1-2y)}{2\sqrt{x^4+(y-y^2)^2}}dx+\sqrt{y^4+(y-y^2)^2}\\&=\sqrt{y^4+(y-y^2)^2}+\int_0^y\frac{y(1-y)(1-2y)}{\sqrt{x^4+(y-y^2)^2}}dx,\end{align*}
which is positive for $y\in(0,\frac12)$. Else suppose $y\in(\frac12,1)$, then it's positive if
$$\int_0^y\frac{(2y-1)(y-y^2)}{\sqrt{x^4+(y-y^2)^2}}dx=-\int_0^y\frac{y(1-y)(1-2y)}{\sqrt{x^4+(y-y^2)^2}}dx<\sqrt{y^4+(y-y^2)^2}=y\sqrt{y^2+(1-y)^2}.$$
Clearly,
$$\int_0^y\frac{(2y-1)(y-y^2)}{\sqrt{x^4+(y-y^2)^2}}dx<y\cdot\frac{(2y-1)(y-y^2)}{\sqrt{(y-y^2)^2}}=y(2y-1),$$
so it remains to check that
$$2y-1\le\sqrt{y^2+(1-y)^2}\iff 0\le y^2+(1-y)^2-(2y-1)^2=-2y^2+2y=2y(1-y)$$
So we can conclude the maximum is attained for $y=1$, where it's
$$\int_0^1\sqrt{x^4}\,dx=\int_0^1 x^2\, dx=\frac13.$$
|
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|
How to show $\lim_{n \to \infty}a_n=\frac{5^{3\cdot n}}{2^{\left(n+1\right)^2}}$? $$\lim_{n \to \infty}a_n=\dfrac{5^{3\cdot n}}{2^{\left(n+1\right)^2}}$$
I am trying to solve it using the squeeze theorem.
I have opened the expression to $$a_n=\dfrac{5^3\cdot 5^n}{2^{n^2}\cdot2^{2n}\cdot2)}$$
I think that the LHS should be $$a_n=\dfrac{2^3\cdot 2^n}{2^{n^2}\cdot2^{2n}\cdot2)}$$
But as for the RHS I do not find a bigger expression, any ideas?
|
$$ \frac{a_{n+1}}{a_{n}}=\\\frac{\frac{5^{3(n+1)}}{2^{(n+2)^2}}}{\frac{5^{3(n)}}{2^{(n+1)^2}}}=\\\frac{5^{3n+3}}{5^{3n}}\frac{2^{(n+1)^2}}{2^{(n+2)^2}}=\\125\frac{2^{n^2+2n+1}}{2^{n^2+4n+4}}=\\125\frac{1}{2^{2n+3}}=\\\frac{125}{8}\frac{1}{4^n}\\$$when n become large $$n\rightarrow \infty\\\frac{a_{n+1}}{a_{n}}=\frac{125}{8}\frac{1}{4^n} \rightarrow 0$$
|
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Find prime factorization of $2^{22} + 1$ Find the prime factorization of $2^{22} + 1$.
How can I approach this with a subtle way?
I know that $a^n -b^n = (a-b) (a^{n-1}+a^{n-2}b + \cdots + ab^{n-2}+b^{n-1})$ and $a^{2n+1} + b^{2n+1} = (a+b) (a^{2n}-a^{2n-1}b + \cdots - ab^{2n-1}+b^{2n})$. So far I have tried to apply the latter one on $4^{11} + 1$.
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$2^{22}+1=2^2\cdot2^{20}+1=4\cdot\Big(2^5\Big)^4+1^4.\quad$ But $4x^4+1=\big(2x^2-2x+1\big)\big(2x^2+2x+1\big)$.
|
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How to calculate : $\frac{1+i}{1-i}$ and $(\frac{1+\sqrt{3i}}{1-\sqrt{3i}})^{10}$ I have these problems :
How to calculate : $\frac{1+i}{1-i}$ and $(\frac{1+\sqrt{3i}}{1-\sqrt{3i}})^{10}$
For some reason this is incorrect I'll be glad to understand why, This is what I done :
I used this formula : $(\alpha,\beta)*(\gamma,\delta)=(\alpha\gamma-\beta\delta,\alpha\delta+\beta\gamma)$
And also : $(\alpha,\beta)*(\gamma,\delta)^{-1}=(\alpha,\beta)*(\frac{\gamma}{\gamma^2+\delta^2}-\frac{\delta}{\gamma^2+\delta^2})$
*
*$$\frac{1+i}{1-i}=\\(1+i)(1-i)^{-1}=\\(1+i)(\frac{1}{2}+\frac{1}{2}i)=\\(\frac{1}{2}-\frac{i^2}{2}+\frac{1}{2}i+\frac{1}{2}i)=\\(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}i+\frac{1}{2}i)=\\1+i\\ \\$$
*$$(\frac{1+\sqrt{3i}}{1-\sqrt{3i}})^{10}=\\
((1+\sqrt{3i})(1-\sqrt{3i})^{-1})^{10}=\\
((1+\sqrt{3i})(\frac{1}{1+3i}+\frac{\sqrt{3i}}{1+3i}i)^{10}=\\
(\frac{1}{1+3i}-\frac{\sqrt{3i}\sqrt{3i}}{1+3i}i+\frac{\sqrt{3i}}{1+3i}i+\frac{\sqrt{3i}}{1+3i})^{10}=\\
(\frac{4}{1+3i}+\frac{\sqrt{3i}+\sqrt{3i}}{1+3i})^{10}$$
Now I want to use De-Moivre :
$$tan(args)= \frac{\frac{\sqrt{3i}+\sqrt{3i}}{1+3i})}{\frac{4}{1+3i}}=\frac{(1+3i)(\sqrt{3i}+\sqrt{3i})}{4+12i}=\frac{\sqrt{3i}-3i\sqrt{3i}+\sqrt{3i}+3i\sqrt{3i}}{4+12i}=\frac{\sqrt{3i}+\sqrt{3i}}{4+12i}$$
But I reach to math error, when trying to calculate the args.
Any help will be appreciated.
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What you calculated in the first part is not the same quantity as what you asked in the original question: is the quantity to be computed $$\frac{1-i}{1+i},$$ or is it $$\frac{1+i}{1-i}?$$ Even if it is the latter, you have made a sign error in the fourth line: it should be $$(1+i)(\tfrac{1}{2} + \tfrac{i}{2}) = \frac{(1+i)^2}{2} = \frac{1+i+i+i^2}{2} = \frac{2i}{2} = i.$$
For your second question, you need to be absolutely sure that what you mean is $\sqrt{3i}$, rather than $i \sqrt{3}$. They are not the same. I suspect the actual question should use the latter, not the former.
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How prove this diophantine equation $(x^2-y)(y^2-x)=(x+y)^2$ have only three integer solution? HAPPY NEW YEAR To Everyone! (Now Beijing time 00:00 (2015))
Let $x,y$ are integer numbers,and such $xy\neq 0$,
Find this diophantine equation all solution
$$(x^2-y)(y^2-x)=(x+y)^2$$
I use Wolf found this equation only have two nonzero integer solution $(x,y)=(-1,1),(-1,-1)$,see wolf
$$\Longleftrightarrow x^3+y^3+x^2+y^2=xy(xy-1)$$
But How prove it?
and I found sometimes,and my problem almost similar with 2012 IMO shortlist:2012 IMO shortlist
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On the LHS the sum of the two factors is nonnegative, so $x^2-y$ and $y^2-x$ cannot be negative at the same time. Hence, if $x+y\ne0$ then both $x^2-y$ and $y^2-x$ must be positive.
If $y^2-x\ge8$ and $x^2-y\ge8$ then
$$
(x+y)^2 = (x^2-y)(y^2-x) = \\
= \frac{x^2-y}2(y^2-x) +\frac{y^2-x}2(x^2-y) +0 \ge \\
\ge
4(y^2-x) +4(x^2-y) -2(x-y)^2 = \\
= 2(x+y)^2 - 4(x+y); \\
0\le x+y \le 4.
$$
Therefore, all solutions satisfy $0\le x+y\le4$, $1\le x^2-y\le 7$ or $1\le y^2-x\le 7$.
It is a few simple cases and they all lead to finding the integer roots of certain polynomials with integer coefficients.
|
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|
Integral of an exponential I have the following: $$ I(a,b) \equiv\int_{-\infty}^\infty e^{\frac{-1}{2}\left(ax^2+\frac{b}{x^2}\right)}dx$$ where $a,b>0$. And I have the following substitution as a hint: $$y=\frac{1}{2}\left(\sqrt{a}x-\frac{\sqrt{b}}{x}\right)$$ And the integral must be evaluated. But I seem to be having trouble going through it. Many thanks in advance.
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I too got stuck on your hint. Here's how I did your integral.
First letting $x = \frac{t}{\sqrt{a}}, \; dx = \frac{dt}{\sqrt{a}}$, we get
$$ \int_{-\infty}^\infty e^{-\frac{1}{2}\left(a x^2 + \frac{b}{x^2}\right)} \, dx = \frac{1}{\sqrt{a}} I(a b), $$
where
$$ I(\omega) = \int_{-\infty}^\infty e^{-\frac{1}{2}\left(t^2 + \frac{\omega}{t^2}\right)} = 2\int_{0}^\infty e^{-\frac{1}{2}\left(t^2 + \frac{\omega}{t^2}\right)} \, dt. $$
To find $I(\omega)$, let $t = 1/z, \; dt = -dz/z^2$ to get
$$ I(\omega) = 2\int_{0}^\infty e^{-\frac{1}{2}\left(\frac{1}{z^2} + \omega z^2\right)} \, \frac{dz}{z^2}. $$
Differentiating both sides with respect to $\omega$ gives
$$ I'(\omega) = -\int_{0}^\infty e^{-\frac{1}{2}\left(\frac{1}{z^2} + \omega z^2\right)} \, dz, $$
and letting $z = \frac{y}{\sqrt{\omega}}, \; dz = \frac{dy}{\sqrt{\omega}}$,
$$ I'(\omega) = -\frac{1}{\sqrt{\omega}}\int_{0}^\infty e^{-\frac{1}{2}\left(\frac{\omega}{y^2} + y^2\right)} \, dy = -\frac{1}{2\sqrt{\omega}}I(\omega). $$
Using $I(0) = \sqrt{2\pi}$, solving the separable ODE gives
$$ I(\omega) = \sqrt{2\pi} e^{-\sqrt{\omega}}. $$
Therefore the original integral is
$$ \frac{1}{\sqrt{a}} I(a b) = \sqrt{\frac{2\pi}{a}} e^{-\sqrt{a b}}. $$
|
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Is there a lower bound for the following rational expression, in terms of $b^2$? In the following, $b$ and $r$ are both positive integers.
Is there a lower bound for the following rational expression, in terms of $b^2$?
$$L = \frac{b^2\left(2^{r+1} b^2(2^r-1) + 3\cdot 2^r - 2\right)}{\left(2^{r+1} - 1\right)(2^r b^2 + 1)}$$
What I do know is that (a crude upper bound is) $L < b^2$. My question: Is it possible to get a lower bound for $L$ in terms of $b^2$ alone?
Edit: Here is an attempt to derive a (crude) lower bound for $L$, using no more than $r \geq 1$ and the AM-GM Inequality.
Since $r \geq 1$, the numerator of $L$ is
$$b^2\left(2^{r+1} b^2(2^r-1) + 3\cdot 2^r - 2\right) \geq 4b^4 + 4b^2 = 4{b^2}(b^2 + 1).$$
Now, for the denominator of $L$, we use the AM-GM Inequality to get
$$\left(2^{r+1} - 1\right)(2^r b^2 + 1) \leq \frac{\left(2^{r+1} + 2^r b^2\right)^2}{4}.$$
Simplifying, we obtain
$$\left(2^{r+1} - 1\right)(2^r b^2 + 1) \leq \left(2^r + 2^{r-1} b^2\right)^2 < 2^{2r}(b^2 + 1)^2.$$
Consequently, I finally have:
$$L > \frac{2^{2-2r}\cdot{b^2}}{b^2 + 1}$$
which is still expressed in terms of the extra variable $r$.
I was wondering if anybody out there has any bright ideas on how to eliminate $r$ from a (hopefully) nontrivial lower bound for $L$, in terms of $b^2$.
Thanks!
|
Taking off from Matt's answer, I get:
$$\frac{1}{2}\left(1 + \frac{2^r}{2^r - 1}\right)(b^2 + 2^{-r}).$$
We need an upper bound for $\frac{1}{2}\left(1 + \frac{2^r}{2^r - 1}\right)$.
We have the upper bound:
$$\frac{1}{2}\left(1 + \frac{2^r}{2^r - 1}\right) = \frac{1}{2}\left(2 + \frac{1}{2^r - 1}\right) \leq \frac{3}{2}$$
since $r$ is a positive integer (and hence we can take $r \geq 1$).
Consequently, we now have the (adjusted) lower bound:
$$\frac{b^2\left(2^{r+1} b^2(2^r-1) + 3\cdot 2^r - 2\right)}{\left(2^{r+1} - 1\right)(2^r b^2 + 1)}=\frac{2^{r+1}(2^r-1)b^4 + (3\cdot 2^r - 2)b^2}{\left(2^{r+1} - 1\right)2^r(b^2 + 2^{-r})}=\frac{b^4 + \frac{3\cdot 2^r - 2}{2^{r+1}(2^r-1)}b^2}{\left(1 +\frac{2^r}{2^r-1}\right)(\frac{1}{2}b^2 + 2^{-r-1})}\geq \left(\frac{2}{3}{b^2}\right)\frac{b^2+3\cdot 2^{-r-1}- \frac{1}{2^{r}(2^r-1)}}{b^2+2^{-r}}\geq \left(\frac{2}{3}{b^2}\right)\frac{b^2+2^{-r}+2^{-r-1}-2^{-2r+1}}{b^2+2^{-r}}\geq \frac{2b^2}{3}.$$
I hope what I have written is correct! =)
|
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Finding an integer $n$ such that $\sin(n)$ is close to 1 Given some $\epsilon>0$, is there an efficient way to find an integer $n$ such that
$$1-\sin(n)<\epsilon$$
We all know there is always one (and many), and so I can test all $n$ from $0$ until I find a good candidate, but I ask for some efficient algorithm that given some $\epsilon$, computes quickly such an $n$.
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Consider the Taylor series of $\sin(x)$ around $\left(2k + \frac{1}{2}\right) \pi$,
\begin{align*}
\sin\left(\left(2k + \frac{1}{2}\right)\pi + x\right)
&= 1 - \frac{x^2}{2} + \frac{x^4}{24} - \ldots \\
&\le 1 - \frac{x^2}{2}
\end{align*}
as it is reasonably clear that $\frac{x^4}{24} - \frac{x^6}{120} + \ldots > 0$ for a small $x$.
So if we want $1 - \sin(n) < \epsilon$ then consider $n = \left(2k + \frac{1}{2}\right)\pi + x$, we thus obviously require
$$
\left| n - \left(2k + \frac{1}{2}\right)\pi \right| < \sqrt{2 \epsilon}.
$$
As pointed out above, this is equivalent to finding a rational approximation $\frac{2n}{4k + 1}$, which if furthermore is the convergent of a continued fraction approximation, we get
$$
\left| n - \left(2k + \frac{1}{2}\right)\pi \right| < \frac{1}{8k+2}
$$
for free.
Now, there are continued fraction formulae that do not require the knowledge of $\pi$, for example Stern's formula
$$
\frac{\pi}{2} = 1 - \frac{1}{3 - \frac{2\cdot 3}{1 - \frac{1\cdot 2}{3 - \frac{4\cdot5}{1 - \frac{3\cdot 4}{3 - \frac{6\cdot7}{1 - \cdots } } } }}},
$$
which can give you a sequence of rational approximations of $\frac{\pi}{2}$. The sequences of convergents and their numerators OEIS A001901 and denominators OEIS A046126 for this can be found, and effectively all you have to do now, as we are considering $\pi/2$ i.e. $2n$ doesn't matter any more, is find a denominator of the form $4k+1$ such that $\frac{1}{8k+2} < \sqrt{2\epsilon}$, and the corresponding $2n$ in the numerator gives our answer.
|
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|
Factor the polynomial $z^5 + 32$ in real factors The question that I have trouble solving is the following:
Factor the polynomial $z^5 + 32$ in real factors. The answer should not use trigonometric functions. (Hint: you are allowed to use the fact that: $cos(\pi/5) = \frac{1+\sqrt{5}}{4}$ and $cos(3\pi/5) = \frac{1-\sqrt{5}}{4}$.
In the previous question you were supposed to find the complex roots to the equation $z^5 = -32$ in polar form, resulting in the roots:
$$z_1 = 2e^{i\frac{\pi}{5}}$$$$z_2 = 2e^{i\frac{3\pi}{5}}$$ $$z_3 = 2e^{i\pi} = -2$$ $$z_4 = 2e^{i\frac{7\pi}{5}}$$$$ z_5 = 2e^{i\frac{9\pi}{5}} $$
My attempt at a solution:
First we know that there were only one real root to the equation $z^5 = -32$, namely $-2$.
So if we write the polynomial as the equation: $z^5 + 32 = 0$ we know that $(z+2)$ must be a factor and we have four potential factors left to find. Since we know that the polynomial $z^5 +32$ only has real coefficients, the non real roots must come in pairs.
We should then find two roots that are composed of conjugates of the roots with imaginary components to produce the other two real roots... But I am stuck here.
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Set $t=2z$, for the moment. Then
\begin{align}
z^5+32=32(t^5+1)
&=32(t+1)(t^4-t^3+t^2-t+1)\\
&=32(t+1)t^2\left(t^2+\frac{1}{t^2}-\left(t+\frac{1}{t}\right)+1\right)\\
&=32(t+1)t^2\left(\left(t+\frac{1}{t}\right)^{\!2}-\left(t+\frac{1}{t}\right)-1\right)
\end{align}
Consider $u^2-u-1=(u-\alpha)(u-\beta)$, where
$$
\alpha=\frac{1+\sqrt{5}}{2},\qquad\beta=\frac{1-\sqrt{5}}{2}
$$
and so
$$
z^5+32=32(t+1)t^2\left(t+\frac{1}{t}-\alpha\right)\left(t+\frac{1}{t}-\beta\right)
$$
which is to say
$$
z^5+32=32(t+1)(t^2-\alpha t+1)(t^2-\beta t+1)=
(z+2)(z^2-2\alpha z+4)(z^2-2\beta z+4)
$$
|
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|
sketch points given by condition on complezx number z the condition is
$|z+i| \le 3$ so its a circle with radius three centered at -i
treating $z=x+iy$
$-\sqrt{9} \le \sqrt{(x^2 )+(y+1)^2} \le \sqrt{9}$
squaring everything $-9 \le x^2 + (y+1)^2 \le 9$
$-9 \le x^2 + y^2 + 2y +1 \le 9$
I feel like I am complicating more than I need to.
|
You're really close.
Well $|z+i|\le 3$ with $z=x+iy$ we have then simply $0\le\sqrt{x^2+(y+1)^2}\le3$ ($\forall Z\in\mathbb{C},\,|Z|\ge 0$).
Thus $x^2+(y+1)^2\le3^2$ which represents the disk of radius $3$ and center $(0,-1)$.
|
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|
Showing $\binom{2n}{n} = (-4)^n \binom{-1/2}{n}$ Is there a proof for the following identity that only uses the definition of the (generalized) binomial coefficient and basic transformations?
Let $n$ be a non-negative integer.
$$\binom{2n}{n} = (-4)^n \binom{-\frac{1}{2}}{n}$$
|
$$\begin{align}
\binom{-1/2}{ n} &= (-1/2)(-3/2)(-5/2)\ldots(-(2n-1)/2)
/n! \\
&= (-1)^n 1\cdot 3\cdot 5\cdot\ldots\cdot(2n-1)/ (2^n n!) \\
&= (-1)^n (2n)!/ (2^n \cdot n! \cdot 2 \cdot 4 \cdot 6\cdot \ldots \cdot (2n)) \\
&= (-1)^n (2n)!/ (2^n n!)^2 \\
&= (-1/4)^n \binom{2n}{n}.
\end{align}$$
Hence:
$$ (1-4X)^{-1/2} $$
generates (link)
$$ \binom{2n}{n}.$$
|
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|
Find the residue at $z=-2$ for $g(z) = \frac{\psi(-z)}{(z+1)(z+2)^3}$ Find the residue at $z=-2$ for
$$g(z) = \frac{\psi(-z)}{(z+1)(z+2)^3}$$
I know that:
$$\psi(z+1) = -\gamma - \sum_{k=1}^{\infty} (-1)^k\zeta(k+1)z^k$$
Let $z \to -1 - z$ to get:
$$\psi(-z) = -\gamma - \sum_{k=1}^{\infty} \zeta(k+1)(z+1)^k$$
therefore we divide by the other part to get:
$$\frac{\psi(-z)}{(z+1)(z+2)^3} = -\frac{\gamma}{(z+1)(z+2)^3} - \sum_{k=1}^{\infty} \frac{\zeta(k+1)(z+1)^{k-1}}{(z+2)^3}$$
I have to somehow get the coefficient of $\frac{1}{z+2}$ because I want to evaluate the residue of
$g(z)$ at $z=-2$
The problem is I cant ever get a factor of $\frac{1}{z+2}$ what should I do?
|
$\psi(z)$ is regular over $\Re(z)>0$, hence $z=-2$ is a triple pole for $g(z)$. This gives:
$$\operatorname{Res}\left(g(z),z=-2\right)= \frac{1}{2}\left.\frac{d^2}{dz^2} (z+2)^3 g(z)\right|_{z=-2}=\frac{1}{2}\left.\frac{d^2}{dz^2} \frac{\psi(-z)}{z+1}\right|_{z=-2}\tag{1}$$
so:
$$\operatorname{Res}\left(g(z),z=-2\right)=\left.\frac{\psi(-z)}{(1+z)^3}+\frac{ \psi'(-z)}{(1+z)^2}+\frac{\psi''(-z)}{2(1+z)}\right|_{z=-2} = -\psi(2)+\psi'(2)-\frac{\psi''(2)}{2}$$
leading to:
$$\begin{eqnarray*}\operatorname{Res}\left(g(z),z=-2\right)&=& -(1-\gamma)+\left(1-\zeta(2)\right)-\frac{\psi''(2)}{2}\\&=&\gamma-\zeta(2)+\sum_{n\geq 0}\frac{1}{(n+2)^3}\\&=&\color{red}{\gamma-\zeta(2)+\zeta(3)-1}.\tag{2}\end{eqnarray*}$$
|
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Describe geometry of complex plane. Let $a \in \Bbb R$ and $c>0$ to be fixed. Describe the set of points $z$ such that $|z-a|-|z+a|=2c$ for every possible choice of $a$ and $c$.
Then let $a$ be a complex number using the rotation of the plane describe the locus of points satisfying the above equation.
Here is how I understand:
The set of point $z$ is the intersect of $2$ disks, $1$ disk contain all points $z$ that has center $a$ and radius $|z-a|$ and another disk contain all point $z$ that has center $-a$ and radius $|z+a|$.
Since $c>0$, $|z-a|>|z+a|$ that mean the set point $z$ we want to find is closer to $-a$ than $a$. I'm not sure if this is how they want me to describe the set of point $z$.
For the second part, is it just be the same, the only difference is on part $2$, $a$ is on complex plane not a real line.
|
with $z=x+iy$ it is equivalent to
$$\sqrt{(x-a)^2+y^2}-\sqrt{(x+a)^2+y^2}=2c$$
Does this help you?
squaring the equation $$\sqrt{(x-a)^2+y^2}=2c+\sqrt{(x+a)^2+y^2}$$
we get $$-xa-c^2=c\sqrt{(x+a)^2+y^2}$$
squaring again we get $$x^2a^2+c^4+2xac^2=c^2((x+a)^2+y^2)$$
this is equivalent to
$$x^2(a^2-c^2)+c^2(c^2-a^2)=y^2c^2$$ or
$$\frac{x^2}{c^2}-\frac{y^2}{a^2-c^2}=1$$
if $c\ne 0$ and $a^2-c^2\ne 0$
|
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|
extracting the middle term of $ (z \cos \theta + w\sin \theta )^m(- z\sin \theta + w\cos \theta )^m $ Is there a systematic way to extract the middle term of the following expression?
$$ (z \cos \theta + w\sin \theta )^m(- z\sin \theta + w\cos \theta )^m $$
This is homogeneous polynomial of degree $2m$, so I am looking for the $(zw)^m$ term.
Example $m = 2$:
$$ (z \cos \theta + w\sin \theta )^2(- z\sin \theta + w\sin \theta )^2
= \dots + (zw)^2(\cos^4 \theta - 4 \cos^2 \theta \sin^2 \theta + \sin^4 \theta) + \dots $$
The middle terms can be simplified using the cosine double angle formulas.
\begin{eqnarray}\cos^4 \theta - 4 \cos^2 \theta \sin^2 \theta + \sin^4 \theta
&=& (\cos^2 \theta - \sin^2 \theta)^2 - 2 \cos^2 \theta \sin^2 \theta \\
&=& \cos^2 2\theta - \tfrac{1}{2}\sin^2 2\theta \\
&=& \boxed{\tfrac{3}{2}\cos^2 2\theta - \tfrac{1}{2}}
\end{eqnarray}
Is it true this middle term can always be expanded as a polynomial in $x=\cos 2\theta$?
I learned of this trick while studying the Wigner D-matrices. It says the Legendre polynomials are certain matrix elements in the representation theory of $SU(2)$. Checking this fact was surprisingly hard.
|
We use the binomial theorem, to see that
$$
(z\cos\theta+w\sin\theta)^m = \sum_{k=0}^m {m\choose k}z^{m-k}\cos^{m-k}\theta \,w^k\sin^k\theta
$$
and
$$
(-z\sin\theta+w\cos\theta)^m=\sum_{k=0}^m {m\choose k}(-z)^k\sin^k\theta\,w^{m-k}\cos^{m-k}\theta.
$$
Written this way, I think it is clear that when we multiply these factors together, the coefficient in front of $z^mw^m$ becomes
$$
\sum_{k=0}^m (-1)^k{m\choose k}^2 (\cos^{2}\theta)^{m-k}(\sin^2\theta)^k.
$$
Next, since $\cos^2\theta=\frac{1}{2}(1+\cos2\theta)$ and $\sin^2\theta=\frac{1}{2}(1-\cos2\theta)$ this simplifies into
$$
\sum_{k=0}^m (-1)^k{m\choose k}^2 \bigl[\tfrac{1}{2}(1+\cos2\theta)\bigr]^{m-k}\bigl[\tfrac{1}{2}(1-\cos2\theta)\bigr]^k,
$$
a polynomial in $\cos2\theta$.
A check of the case $m=2$
In this case, the formula gives
$$
\begin{align}
&(-1)^0{2\choose 0}^2\bigl[\tfrac{1}{2}(1+\cos2\theta)\bigr]^2\\
&+(-1)^1{2\choose 1}^2\tfrac{1}{2}(1+\cos2\theta)\tfrac{1}{2}(1-\cos2\theta)\\
&+(-1)^2{2\choose 2}^2\bigl[\tfrac{1}{2}(1-\cos2\theta)\bigr]^2\\
&=\frac{3}{2}\cos^22\theta-\frac{1}{2}.
\end{align}
$$
|
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|
Hypergeometric 2F1 with negative c I've got this hypergeometric series
$_2F_1 \left[ \begin{array}{ll}
a &-n \\
-a-n+1 &
\end{array} ; 1\right]$
where $a,n>0$ and $a,n\in \mathbb{N}$
The problem is that $-a-n+1$ is negative in this case. So when I try to use Gauss's identity
$_2F_1 \left[ \begin{array}{ll}
a & b \\
c &
\end{array} ; 1\right] = \dfrac{\Gamma(c-a-b)\Gamma(c)}{\Gamma(c-a)\Gamma(c-b)}$
I give negative parameters to the $\Gamma$ function.
What other identity can I use?
I'm trying to find a closed form to this: $\sum _{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i}$
Wolfram Mathematica answered this as a closed form: $\frac{2^{-2 a} \Gamma \left(\frac{1}{2} (1-2 a)\right) \binom{a+n-1}{n} \Gamma (-a-n+1)}{\sqrt{\pi } \Gamma (-2 a-n+1)}$
But I would like to have a manual solution with proof.
This is how I got that hypergeometric series:
$\sum _{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i}$
$\dfrac{t_{i+1}}{t_{i}} = \frac{\binom{a+i+1-1}{i+1} \binom{a-i+n-2}{-i+n-1}}{\binom{a+i-1}{i} \binom{a-i+n-1}{n-i}} = \frac{(a+i) (n-i)}{(i+1) (a-i+n-1)} = \frac{(a+i) (i-n)}{ (i-a-n+1)(i+1)}$
$\sum _{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i} = _2F_1 \left[ \begin{array}{ll}
a &-n \\
-a-n+1 &
\end{array} ; 1\right]$
UPDATE
Thanks to David H, I got closer to the solution.
$\sum _{i=0}^n \binom{a+i-1}{i} \binom{a-i+n-1}{n-i} = _2F_1 \left[ \begin{array}{ll}
a &-n \\
-a-n+1 &
\end{array} ; 1\right]$
$\lim\limits_{\epsilon \to0} \frac{\Gamma (-2 a-2 \epsilon +1) \Gamma (-a-n-\epsilon +1)}{\Gamma (-a-\epsilon +1) \Gamma (-2 a-n-2 \epsilon +1)} = \frac{4^{-a} \Gamma \left(\frac{1}{2}-a\right) \Gamma (-a-n+1)}{\sqrt{\pi } \Gamma (-2 a-n+1)}$
As you can see this result is close to the expected $\frac{2^{-2 a} \Gamma \left(\frac{1}{2} (1-2 a)\right) \binom{a+n-1}{n} \Gamma (-a-n+1)}{\sqrt{\pi } \Gamma (-2 a-n+1)}$ formula. But the $\binom{a+n-1}{n}$ factor is still missing and I don't really understand why.
|
We may try to keep it simple.
Suppose we seek to evaluate
$$\sum_{k=0}^n {a-1+k\choose k} {a-1+n-k\choose n-k}.$$
It is immediately apparent that this is a convolution of two ordinary generating functions.
To see what they are re-write the sum as
$$\sum_{k=0}^n {a-1+k\choose a-1} {a-1+n-k\choose a-1}.$$
By the Newton binomial this is a convolution of
$$\frac{1}{(1-z)^a}$$
with itself, giving
$$\frac{1}{(1-z)^{2a}}.$$
Extracting coefficients we obtain
$${n+2a-1\choose 2a-1} = {n+2a-1\choose n}.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Showing ${n + a - 1 \choose a - 1} = \sum_{k = 0}^{\left\lfloor n/2 \right\rfloor} {a \choose n-2k}{k+a-1 \choose a-1}$ Prove that for integers $n \geq 0$ and $a \geq 1$, $${n + a - 1 \choose a - 1} = \sum_{k = 0}^{\left\lfloor n/2 \right\rfloor} {a \choose n-2k}{k+a-1 \choose a-1}.$$
I figured I'd post this question, which was on an assignment I did, since I thought the solution was so nice.
|
This one can also be done using complex variables.
Suppose we seek to evaluate
$$\sum_{k=0}^{\lfloor n/2\rfloor}
{a\choose n-2k} {k+a-1\choose a-1}.$$
Introduce the integral representation
$${a\choose n-2k}
=\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{a}}{z^{n-2k+1}} \; dz.$$
Note that this integral is zero when $k>\lfloor n/2\rfloor$
so we may extend the sum to infinity.
This gives for the sum the integral
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{a}}{z^{n+1}}
\sum_{k\ge 0} {k+a-1\choose a-1} z^{2k}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{a}}{z^{n+1}}
\frac{1}{(1-z^2)^a} \; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\frac{1}{(1-z)^a} \; dz.$$
This last one evaluates to
$${n+a-1\choose a-1}$$
by inspection (Newton binomial).
|
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|
If $x^2+x+1=0 $, then find the value of $(x^3+1/x^3)^3$?
If $x^2+x+1=0 $, then find the value of $(x^3+1/x^3)^3$
This thing doesn't make sense how should I use first identity to find the second one.
|
$x^2+x+1=0 \to x+1/x=-1 \to (x+1/x)^3 = -1\to x^3 + 3(x+1/x) + 1/x^3 = -1\to x^3+1/x^3 = 2 \to (x^3+1/x^3)^3 = 2^3 = 8$
|
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|
Solutions of the functional equation $f(2x) = \frac{f(x)+x}{2}$ How can I solve the following functional equation?
$$f(2x) = \frac{f(x)+x}{2},$$
for $x \in \mathbb{R}$ with $f$ being a continuous function.
|
Here's an alternative proof. Notice that by substituting in $\frac{1}{2^{n+1}}x$ for $x$, we can obtain
$$f\left(\frac{1}{2^n}x\right) = \frac{1}{2}f\left(\frac{1}{2^{n+1}}x\right) + \frac{x}{2^{n+2}}$$
Using this expression, we find that
\begin{align*}f(x) &= \frac{1}{2}f\left(\frac{1}{2}x\right) + \frac{x}{4}\\
&= \frac{1}{2}\left(\frac{1}{2}f\left(\frac{1}{4}x\right) + \frac{x}{8}\right) + \frac{x}{4}\\
&= \frac{1}{4}f\left(\frac{1}{4}x\right) + \frac{x}{4}+\frac{x}{16}\\
\\&\vdots\\
&= \frac{1}{2^n}f\left(\frac{1}{2^n}x\right) + \sum_{k=1}^{n}\frac{x}{4^k}
\end{align*}
Let's look at the behavior of this last term as $n\to \infty$. We have that $f\left(\frac{1}{2^n}x\right) \to f(0)$, which is finite, so $\frac{1}{2^n}f\left(\frac{1}{2^n}x\right) \to 0$. It follows that
$$f(x) = \sum_{k=1}^{\infty}\frac{x}{4^k} = \frac{x}{3}$$
|
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|
Is this a correct way to prove this? I've just looked at this question and sketched a way to do it my head. When I looked at the answer it looked slightly more complicated than the way I did it so I just wanted to check whether this is a correct way:
The question states:
Let $$f(x)=ax-\frac{x^3}{1+x^2}$$ where $a$ is a constant, show that
if $a\geq9/8$ then $f'(x)\geq0$ for all $x$.
My way of doing this is to first differentiate $f(x)$ so that:
$$f'(x)=a-\frac{x^2 (x^2+3)}{(x^2+1)^2}$$
thus if you assume that $a\geq9/8$ and since we know from above that $$a=f'(x)+\frac{x^2 (x^2+3)}{(x^2+1)^2}$$
then
$$ f'(x)+\frac{x^2 (x^2+3)}{(x^2+1)^2}\geq \frac{9}{8}\;\; \rightarrow\;\;f'(x)\geq\frac{9}{8}-\frac{x^2 (x^2+3)}{(x^2+1)^2} $$
then all we need to show is that:
$$ \frac{9}{8}-\frac{x^2 (x^2+3)}{(x^2+1)^2}\geq0 $$
which we can show since we can write this as:
$$\frac{1}{8}\left( \frac{x^2-3}{x^2+1} \right)^2\geq 0$$
which is obviously true since $x^2$ is greater than or equal to zero for all real $x$.
Is this a correct way of showing this or have I inadvertently assumed something (I'm not entirely sure it's correct when I 'assume $a\geq9/8$') or made done something incorrectly? Thanks
|
Not very clear. If I were you I will show in this order:
first,
$$f'(x)=a-\frac{x^2 (x^2+3)}{(x^2+1)^2}$$
Secondly,
$$ \forall x, \quad \frac{9}{8}-\frac{x^2 (x^2+3)}{(x^2+1)^2} = \frac{1}{8}\left( \frac{x^2-3}{x^2+1} \right)^2 \geq 0 $$
Thus if $a \geq \frac98$, then $\forall x$ we have
$$f'(x)=a-\frac{x^2 (x^2+3)}{(x^2+1)^2} \geq \frac{9}{8}-\frac{x^2 (x^2+3)}{(x^2+1)^2} \geq 0$$
|
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|
Integral $\int^{1}_{-1} \frac{\ln(ax^2+2bx+a)}{x^2+1}dx$ if $a>b>0$ I am trying to evaluate the following integral:
$$\int^{1}_{-1} \frac{\ln(ax^2+2bx+a)}{x^2+1}dx,$$ where $a>b>0$.
I can't really think of a way to find it. So, please give me a hint.
|
With $\sin\theta =\frac ba$\begin{align}
&\int^{1}_{-1} \frac{\ln(ax^2+2bx+a)}{x^2+1}dx\\
=& \int^{1}_{-1} \frac{\ln a}{x^2+1}dx+
\int^{1}_{-1} \frac{\ln(x^2+2x \sin \theta+1)}{x^2+1}\overset{x\to1/x}{dx}\\
=& \ \frac\pi2\ln a+\frac12 \int^{\infty}_{-\infty} \frac{\ln(x^2+2x \sin \theta+1)}{x^2+1}dx-2\int_1^\infty \frac{\ln x}{x^2+1}dx\\
= &\ \frac\pi2\ln a+\frac\pi2 \ln\left(2\cos\frac\theta2\right)-2G\\
=& \ \frac\pi2\ln2 +\frac\pi2 \ln\left( \sqrt{a^2-b^2}+a\right)-2G
\end{align}
|
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|
Limit $\lim_\limits{x\to0} \frac{\ln\left(x+\sqrt{1+x^2}\right)-x}{\tan^3(x)}$ Evaluate the given limit:
$$\lim_{x\to0} \frac{\ln\left(x+\sqrt{1+x^2}\right)-x}{\tan^3(x)} .$$
I've tried to evaluate it but I always get stuck... Obviously I need L'Hôpital's Rule here, but still get confused on the way. May someone show me what is the trick here?
Thanks.
|
This solution is based on no l'Hospital rule nor Taylor expansion. The following standard limits only are used:
\begin{eqnarray*}
\lim_{x\rightarrow 0}e^{x} &=&1. \\
\lim_{x\rightarrow 0}\frac{\tan x}{x} &=&1. \\
\lim_{x\rightarrow 0}\frac{1+x+\frac{1}{2}x^{2}-e^{x}}{x^{3}} &=&-\frac{1}{6}%
. \\
\lim_{x\rightarrow 0}\frac{\sqrt{1+x^{2}}-1-\frac{1}{2}x^{2}}{x^{3}} &=&0.
\end{eqnarray*}
We transform the original expression $f(x)=\dfrac{\ln (x+\sqrt{1+x^{2}})-x}{\tan
^{3}x}$ as follows
\begin{eqnarray*}
f(x)&=&\frac{x^{3}}{\tan ^{3}x}\cdot \dfrac{\ln (x+\sqrt{1+x^{2}})-\ln e^{x}}{%
x^{3}} \\
&=&\frac{x^{3}}{\tan ^{3}x}\cdot \dfrac{\ln (\dfrac{x+\sqrt{1+x^{2}}}{e^{x}})}{%
x^{3}} \\
&=&\frac{x^{3}}{\tan ^{3}x}\cdot \dfrac{\ln (1+\color{red}{u(x)})}{\color{red}{u(x)}}\cdot \dfrac{\color{red}{\dfrac{%
x+\sqrt{1+x^{2}}-e^{x}}{e^{x}}}}{x^{3}},\ \ \ with\ \color{red}{u(x)=\frac{x+\sqrt{1+x^{2}}-e^{x}}{e^{x}}} \\
&=&\left( \frac{x}{\tan x}\right) ^{3}\cdot \dfrac{\ln (1+u(x))}{u(x)}\cdot
\dfrac{1}{e^{x}}\cdot \left( \left( \frac{\sqrt{1+x^{2}}\color{green}{-1-\frac{1}{2}x^{2}}}{%
x^{3}}\right) +\left( \frac{\color{blue}{1}+x\color{blue}{+\frac{1}{2}x^{2}}-e^{x}}{x^{3}}\right)
\right)
\end{eqnarray*}
Since $\lim\limits_{x\rightarrow 0}u(x)=0$, and the function $t\rightarrow t^3$ is continuous, then
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\ln (x+\sqrt{1+x^{2}})-x}{\tan ^{3}x} &=&\left(
\lim_{x\rightarrow 0}\frac{x}{\tan x}\right) ^{3}\cdot \lim_{u\rightarrow 0}%
\frac{\ln (1+ u)}{u}\cdot \lim_{x\rightarrow 0}\frac{1}{e^{x}} \\
&&\cdot \left( \left( \lim_{x\rightarrow 0}\frac{\sqrt{1+x^{2}}-1-\frac{1}{2}%
x^{2}}{x^{3}}\right) +\left( \lim_{x\rightarrow 0}\frac{1+x+\frac{1}{2}%
x^{2}-e^{x}}{x^{3}}\right) \right) \\
&=&\left( 1\right) ^{3}\cdot 1\cdot \frac{1}{1}\cdot \left( \left( 0\right)
+\left( -\frac{1}{6}\right) \right) \\
&=&\color{red}{-\frac{1}{6}}.
\end{eqnarray*}
|
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|
Inverse of $3$ by $3$ matrix with non-constant entries. I'm solving a question in nonhomogenous ordinary differential equation system $x'=Px+q$, and to solve my question I need to compute the inverse of the matrix
$A=\begin{pmatrix}e^{-2t} & e^{-t} & 0 \\ -\frac{5}{4}e^{-2t} & -\frac{4}{3}e^{-t} & e^{2t} \\ -\frac{7}{4}e^{-2t} & -\frac{2}{3}e^{-t} & -e^{2t} \end{pmatrix}$
The reason I need to find the inverse is because the columns of this matrix are the independent solutions to the corresponding homogenous system $x'=Px$ and I'm using the variation of parameters method suggested here http://tutorial.math.lamar.edu/Classes/DE/NonhomogeneousSystems.aspx
How would I find the inverse of this matrix? There is no simple formula like in $2$ by $2$ matrices, and since the entries are not constant, I'm unsure how to do this easily.
|
We can factor out the variable dependence in this particular case:
$$ A=\begin{pmatrix}e^{-2t} & e^{-t} & 0 \\ -\frac{5}{4}e^{-2t} & -\frac{4}{3}e^{-t} & e^{2t} \\ -\frac{7}{4}e^{-2t} & -\frac{2}{3}e^{-t} & -e^{2t} \end{pmatrix} $$
$$ = \begin{pmatrix} 1 & 1 & 0 \\ -\frac{5}{4} & -\frac{4}{3} & 1 \\ -\frac{7}{4} & -\frac{2}{3} & -1 \end{pmatrix} \begin{pmatrix} e^{-2t} & 0 & 0 \\ 0 & e^{-t} & 0 \\ 0 & 0 & e^{2t} \end{pmatrix} = C D $$
The inverse of the diagonal matrix $D$ is found by inspection, namely:
$$ D^{-1} = \begin{pmatrix} e^{2t} & 0 & 0 \\ 0 & e^{t} & 0 \\ 0 & 0 & e^{-2t} \end{pmatrix} $$
Now $A^{-1} = D^{-1} C^{-1}$, and $C^{-1}$ is simply the inverse of a $3\times 3$ matrix with rational (constant) entries.
|
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|
Determining the best possible substitution for an integrand What substitution is best used to calculate $$\int \frac{1}{1 + \sqrt{x^2 -1}}dx$$
|
@Pp. suggests an Euler substitution of the form
\begin{align*}
x &= \frac{t^2+1}{2t} = \frac{t}{2} + \frac{1}{2t} \\
dx &= \left(\frac{1}{2} - \frac{1}{2t^2}\right)\,dt = \frac{t^2-1}{2t^2}\,dt
\end{align*}
By this method you gain being able to solve all integrals of this type: Additions, multiplications and divisions of the square root of a quadratic polynomial and $x$.
Then
\begin{align}
x^2 - 1 &= \frac{t^4+2t^2+1}{4t^2} - \frac{4t^2}{4t^2} \\
&= \frac{t^4-2t^2+1}{4t^2} = \left(\frac{t^2-1}{2t}\right)^2
\end{align}
So the integral becomes
\begin{align*}
\int\frac{1}{1+\sqrt{x^2-1}}\,dx
&= \int\frac{1}{1+\dfrac{t^2-1}{2t}} \left(\frac{t^2-1}{2t^2}\right)\,dt\\
&= \int\frac{t^2-1}{t(t^2+2t-1)}\,dt
\end{align*}
The result of the Euler substitution is the integral of a rational function, which can always be solved by using their partial fraction decomposition.
In this case the denominator splits into linear factors.
$$
\frac{t^2-1}{t(t^2+2t-1)} = \frac{A}{t} + \frac{B}{t+1-\sqrt{2}} + \frac{C}{t+1+\sqrt{2}}
$$
|
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|
Infinitely many primes of the form $6n - 1$ Prove there are infinitely many primes of the form $6n - 1$ with the following: (i) Prove that the product of two numbers of the form $6n + 1$ is also of that form. That is, show that $(6j + 1)(6k + 1) = 6m + 1$, for some choice of $m$. (ii) Show that every prime $p$ greater than $3$ is necessarily of the form $6n + 1$ or $6n − 1$.
|
(i) $(6m+1)\times(6n+1) = 36mn + 6m + 6n + 1 = 6(6mn + m + n) + 1$
(ii) To prove all primes are of the form $6n + 1$ or $6n - 1$, consider:
*
*$6x+0 = 6x$
*$6x+2 = 2(3x+1)$
*$6x+3 = 3(2x+1)$
*$6x+4 = 2(3x+2)$
*$6x+5 = 6(x+1) - 1$
(iii) To prove that there are infinitely many primes of the form $6n-1$ consider:
$(6m+1)\times(6n-1) = 36mn - 6m + 6n - 1 = 6(6mn -m +n) - 1$
$(6m-1)\times(6n-1) = 36mn - 6m - 6n + 1 = 6(6mn - m - n) + 1$
So, if we only consider numbers that are a product of primes $p > 3$, their product will always be $6x+1$ or $6x-1$
Assume $y$ is the greatest prime of the form $6x-1$
Let $v = $ product of all primes $p$ where $3 < p \le y$ so that $v$ is of the form $6x-1$ or $6x+1$
if $v$ is of the form $6x-1$, then $v+6$ is of the form $6(x+1)-1$ and is not divisible by any prime $\le y$. If it is not prime, then it must be divisible by at least one prime of the form $6x-1$. Otherwise, its form would be $6x+1$
if $v$ is of the form $6x+1$, then $v-2$ is of the form $6x+1-2 = 6x-1$ and is not divisible by any prime $\le y$. If it is not prime, then it must be divisible by at least one prime of the form $6x-1$ for the same reason as above.
Therefore, it follows that $y$ cannot be the highest prime of this form and we have proven that there are infinitely many.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that if $({x+\sqrt{x^2+1}})({y+\sqrt{y^2+1}})=1$ then $x+y=0$ Let
$$\left({x+\sqrt{x^2+1}}\right)\left({y+\sqrt{y^2+1}}\right)=1$$
Prove that $x+y=0$.
This is my solution:
Let
$$a=x+\sqrt{x^2+1}$$
and
$$b=y+\sqrt{y^2+1}$$
Then $x=\dfrac{a^2-1}{2a}$ and $y=\dfrac{b^2-1}{2b}$. Now $ab=1\implies b=\dfrac1a$. Then I replaced $x$ and $y$:
$$x+y=\dfrac{a^2-1}{2a}+\dfrac{b^2-1}{2b}=\dfrac{a^2-1}{2a}+\dfrac{\dfrac{1}{a^2}-1}{\dfrac{2}{a}}=0$$
This solution is absolutely different from solution in my book. Is my solution mathematically correct? Did I assumed something that may not be true?
|
Add other solution
since
$$\ln{(x+\sqrt{x^2+1})}+\ln{(y+\sqrt{y^2+1})}=0$$
note $f(x)=\ln{(x+\sqrt{x^2+1})}$ ia odd function and strictly increasing on real numbers,
if $$f(x)+f(y)=0\Longleftrightarrow x+y=0$$
|
{
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"url": "https://math.stackexchange.com/questions/1118742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Recurrence equation solution? Can you help me with the solution of this recurrence equation?
$$
f(n+2) = -2f(n) +3f(n+1) +n \quad\mid\quad f(1)=4 \quad\mid\quad f(2)=5
$$
Thank you.
|
suppose we define $u(n) = f(n+1) - f(n).$ then we can rewrite
$f(n+2) = -2f(n) + 3f(n+1) + n$ as $$u(n+1) = 2u(n) + n, u(1) = 1$$
we will loo for a particular solution in the form of $u(n) =an+ b $ we need
$an + a + b = 2an+ 2b + n$ is satisfied if $a = 2a+1, a+b = 2b$ which is satified if $a = b = -1$ and the homogeneous solution is $u = c2^n$ so the general solution is $$u(n) = 3*2^{n-1} - n - 1$$ which also satisfies the initial condition $u(1) = 1.$
now we need to solve $$f(n+1) - f(n) = 3*2^{n-1} - n - 1, f(1) = 4.$$ verify that $$f(n) = 3*2^{n-1} - \frac{1}{2}n^2 - n + \frac{5}{2}$$ is the solution.
|
{
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|
Convergence of $\sum_{n=1}^{\infty} \log~ ( n ~\sin \frac {1 }{ n })$ Convergence of $$\sum_{n=1}^{\infty} \log~ ( n ~\sin \dfrac {1 }{ n })$$
Attempt:
Initial Check : $\lim_{n \rightarrow \infty } \log~ ( n ~\sin \dfrac {1 }{ n }) = 0$
$\log~ ( n ~\sin \dfrac {1 }{ n }) < n ~\sin \dfrac {1 }{ n }$
$\implies \sum_{n=1}^{\infty} \log~ ( n ~\sin \dfrac {1 }{ n }) < \sum_{n=1}^{\infty} n ~\sin \dfrac {1 }{ n }$
But, $\sum_{n=1}^{\infty} n ~\sin \dfrac {1 }{ n }$ is itself a divergent sequence. Hence, I don't think the above step is of any particular use.
Could anyone give me a direction on how to move ahead.
EDIT: I did think of trying to use the power series of $\sin$. Here's what I attempted :
$\sin \dfrac {1}{n} = \dfrac {1}{n}-\dfrac {1}{n^3.3!}+\dfrac {1}{n^5.5!}-\cdots$
$\implies n \sin \dfrac {1}{n} = 1- \dfrac {1}{n^2.3!}+\dfrac {1}{n^4.5!}-\cdots$
I couldn't proceed further due to the $\log$.
Thank you for your help.
|
A slightly different way than the two current answers, dealing mostly with intuition:
$$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \ldots$$
or alternately,
$$\sin\left(\frac 1x \right) = \frac{1}{x} - \frac{1}{3!x^3} + \frac{1}{5!x^5} + \ldots$$
Multiplying by $x$ gives that
$$x\sin\left( \frac 1x \right) = 1 - \frac{1}{3!x^2} + \ldots \approx 1 - \frac{1}{6x^2},$$
and where the approximation is extremely good for high $x$ (error $\ll \frac{1}{x^4}$).
You are now interested in how $\log \left( 1 - \frac{1}{6x^2}\right)$ behaves. Fortunately, we know that
$$ \log (1 - x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \ldots$$
and thus
$$ \log \left( 1 - \frac{1}{6x^2} \right) = \frac{1}{6x^2} + \ldots \approx \frac{1}{6x^2},$$
where the error is moderately good for large $x$ (no worse than $\frac{1}{x^2}$ for sufficiently large $x$).
So at long last, you are wondering about the sum
$$ \sum_{x \geq 1} \frac{1}{6x^2} = \frac{\pi^2}{36},$$
though the particular value doesn't matter, but only the fact that it converges.
|
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|
Finding an order of a coset in $A/B$ where $A$ is a free abelian group and $B$ is a subgroup. Let $A$ be a free abelian group with basis $x_1,x_2,x_3$ and let $B$ be a subgroup of A generated by $x_1+x_2+4x_3, 2x_1-x_1+2x_3$. In the group $A/B$ find the order of the coset $(x_1+2x_3)+B$.
How can I find this order? Just so you can reply more directly, I shall find a basis $f_1,f_2,f_3$ such that $d_1\cdot f_1,d_2\cdot f_2,d_3\cdot f_3$ is a basis of $B$ where $d_i|d_{i+1}$.
$Attempt$: so by $Smith$ algorithm, $\begin{pmatrix}1 & 2 \\1 & -1 \\4 & 2 \\\end{pmatrix}$ $\underrightarrow{R_2\to R_2-R_1} \begin{pmatrix}1 & 2 \\0 & -3 \\4 & 2 \\\end{pmatrix}$ $\underrightarrow{R_3\to R_3-4R_1} \begin{pmatrix}1 & 2 \\0 & -3 \\0 & -6 \\\end{pmatrix}$ $\underrightarrow{R_3\to R_3-2R_1} \begin{pmatrix}1 & 2 \\0 & -3 \\0 & 0 \\\end{pmatrix}$ $\underrightarrow{C_2\to C_2-2C_1} \begin{pmatrix}1 & 0 \\0 & -3 \\0 & 0 \\\end{pmatrix}$$\underrightarrow{C_2\to -C_2} \begin{pmatrix}1 & 0 \\0 & 3 \\0 & 0 \\\end{pmatrix}$ and therefore, $d_1=1, \space d_2=3, \space d_3=0.$
Therefore, while $M= \begin{pmatrix}1 & 2 \\1 & -1 \\4 & 2 \\\end{pmatrix}$, there exist $Q,P\in GL_n(\Bbb{Z})$ such that $QMP^{-1}= \begin{pmatrix}1 & 0& 0 \\0 & 3 & 0\\0 & 0 & 0\\\end{pmatrix}$ and $P^{-1}$ columns are a basis of $A$ that fulfills the aforementioned requirements. One way to find $P^{-1}$ is to classify the identity element where the classification is the inverse of the row operations I used above in the opposite direction: from the end to the start: $\begin{pmatrix}1 & 0& 0 \\0 & 1 & 0\\0 & 0 & 1\\\end{pmatrix}$ $\underrightarrow{R_3\to R_3 + 2R_1}\begin{pmatrix}1 & 0& 0 \\0 & 1 & 0\\0 & 2 & 1\\\end{pmatrix}$ $\underrightarrow{R_3\to R_3+4R_1}\begin{pmatrix}1 & 0& 0 \\0 & 1 & 0\\4 & 2 & 1\\\end{pmatrix}$ $\underrightarrow{R_2\to R_2+R_1}\begin{pmatrix}1 & 0& 0 \\1 & 1 & 0\\4 & 2 & 1\\\end{pmatrix}$ $\Rightarrow \{x_1+x_2+4x_3, x_2+2x_3 , x_3\}$ is the desired basis, where $\Rightarrow \{x_1+x_2+4x_3, 3\cdot (x_2+2x_3
)\}$ is a basis of $B$. Furthermore: $A/B\cong \Bbb{Z}_{f_1}\times \Bbb{Z}_{f_2}\times \Bbb{Z}_{f_3}/1\cdot \Bbb{Z}_{f_1}\times 3\cdot \Bbb{Z}_{f_2}\times 0\cdot \Bbb{Z}_{f_3} \cong \Bbb{Z}/\Bbb{Z}\times \Bbb{Z}/3\Bbb{Z}\cong \Bbb{Z}/3\Bbb{Z}$. How shall I continue?
|
If you stick to the "$QAP^{-1}$ convention" and do things horizontally, then $(Q^{-1})^{\top}$ is what gives you the basis and $P^{-1})^{\top}$ is what gives you the generators. To avoid transposing, take $\begin{pmatrix}1&1&4\\2&-1&-2\end{pmatrix}$.
Using elementary row and column operations and recording them in matrix form, I get that $$QAP^{-1}=\begin{pmatrix}1&0&0\\0&3&0\end{pmatrix}$$
where $Q=\begin{pmatrix}1&0\\-2&1\end{pmatrix}$ and $P=\begin{pmatrix}1&1&4\\0&-1&-2\\0&0&1\end{pmatrix}$, since $P^{-1}=\begin{pmatrix}1&1&-2\\0&-1&2\\0&0&1\end{pmatrix}$.
As it is known, our good generators are now given by $(f_1',f_2')=Q(f_1,f_2)^{\top}$ where $f_1=e_1+e_2+4e_3$ and $f_2=2e_1-e_2+2e_3$, so that $f_1'=f_1$ and $f_2'= -3e_2-6e_3=-3(e_2+2e_3)$.
And our good basis is given by $(e_1',e_2',e_3')=P(e_1,e_2,e_3)^{\top}$, so that $e_1'=f_1'=f_1$ and $e_2'=-e_2-2e_3$ and $f_2'=3e_2'$.
Now $x_1+2x_3=(1,0,2)= (1,1,4)+(0,-1,-2)=e_1'+e_2'$. I say this has order $3$ in the quotient.
|
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|
Limit of non-linear multi-variable function I'm trying to prove the limit of the following function is $0$:
$\lim_{(x,y) \to (1,-1)} {x^3} - {2xy^2} + 1$
I know that I'm trying to find a $\delta$ s.t $ 0 < \sqrt{(x - 1)^2 + (y + 1)^2} < \delta $ which implies $|{x^3} - {2xy^2} + 1| < \epsilon$
I tried factoring $x$ so that $|x({x^2} - {2y^2}) + 1|$ and then adding 1 and - 1 using the triangle inequality to try to simplify into the expression I wanted, but I don't seem to be getting anywhere.
I think I should try to get $|\sqrt{y^2}|$ so I could say it's less than $|\sqrt{ (x - 1)^2 + (y + 1)^2}|$, and subsequently define what $\delta$ is, but I'm just having a hard time how I could factorize this with the $+1$ in the problem. Any suggestions?
|
We have $$\begin{align}|x^3-2xy^2+1|&=|(x-1+1)^3-2(x-1+1)(y+1-1)^2+1|\\&=\left|(x-1)^3+3(x-1)^2+3(x-1)-2(x-1)\left[(y+1)^2+2(x+1)\right]-2\left[(y+1)^2-2(y+1)\right]\right|\end{align}$$
Idea: We have those $x-1$ and $y+1$ in the condition $\sqrt{(x-1)^2+(y+1)^2}<\delta$, we make them appear in the expression $x^3-2xy^2+1$ by writing $x$ as $x-1+1$ and $y$ as $y+1-1$.
If we choose $\delta=\epsilon/19$ we get
$$|x-1|<\epsilon/19$$
and
$$|y+1|<\epsilon/19$$
from where
$$\begin{align}&\left|(x-1)^3+3(x-1)^2+3(x-1)-2(x-1)\left[(y+1)^2+2(x+1)\right]-2\left[(y+1)^2-2(y+1)\right]\right|\\&\leq|x-1|^3+3|x-1|^2+3|x-1|+2|x-1|\left[|y+1|^2+2|x+1|\right]+2\left[|y+1|^2+2|y+1|\right]\\&\leq 19\epsilon/19=\epsilon\end{align}$$
|
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|
Show that if $m^2 + n^2 $ is divisible by $4$, then $mn$ is also divisible by $4$. Show that if $m$ and $n$ are integers such that $m^2 + n^2 $ is divisible by $4$, then $mn$ is also divisible by $4$.
I am not sure where to begin.
|
$(m+n)^2=m^2+n^2+2mn$, let $m^2+n^2=4k$ then $(m+n)^2=4k+2mn$ , now as RHS is divisible by $2$ LHS will also be divisible by $2$ but LHS is square of some quantity thus if $2$ divides it $4$ will also divide now consider $(m+n)^2-4k=2mn$. Now LHS is divisible by $4$ so $2mn$ is divisible by $4$ thus $mn$ is divisible by $2\implies$ atleast one of $m$ or $n$ is even but as $m^2+n^2$ is even , both of them must be even and therefore $mn$ is divisible by $4$.
|
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|
Inequality $x^4+y^4+(x^2+1)(y^2+1)\ge x^3(1+y) +y^3(1+x)+x+y$ for $x,y \in\mathbb{R}$ Prove for $x,y \in\mathbb{R}$ that such inequality exists ;
$x^4+y^4+(x^2+1)(y^2+1)\ge x^3(1+y) +y^3(1+x)+x+y$
And here is what I realised ;
because $(x^2+1)(y^2+1) >=1$
and $x^4+y^4 \ge 0$
$1\ge x^3(1+y) +y^3(1+x)+x+y$
I'd prefer a hint
|
Hint: First expand the inequality. Then expand $(x-y)^4$ and use it to get rid of the $x^3y+xy^3$ terms. Beat the rest with repeated usages of AM-GM in the form $a^2+b^2\ge 2ab$ with various choices of $a,b$.
A short solution: It's equivalent to
$$\frac14(x-y)^4+\frac14(x^2-y^2)^2+\frac12(x^2+1)(x-1)^2+\frac12(y^2+1)(y-1)^2,$$
which is obviously greater or equal to zero.
More detail:
After expanding the terms and multiplying by 2, your inequality becomes
$$2x^4+2y^4+2x^2y^2+2x^2+2y^2+2\ge 2x^3y+2xy^3+2x^3+2y^3+2x+2y$$
Note that $(x-y)^4+(x^2-y^2)^2\ge 0\implies x^4+y^4+2x^2y^2\ge 2x^3y+2xy^3$, so it suffices to show
$$x^4+y^4+2x^2+2y^2+2\ge 2x^3+2y^3+2x+2y$$
We have $(x-1)^2+(y-1)^2\ge 0\implies x^2+y^2+2\ge 2x+2y$, so it reduces to
$$x^4+y^4+x^2+y^2\ge 2x^3+2y^3\iff x^2(x-1)^2+y^2(y-1)^2\ge 0$$
|
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|
Inequality $(a+b)^2 + (a+b+4c)^2\ge \frac{kabc}{a+b+c}$ for $a,b,c \in\mathbb{R}$ Find biggest constans k such that $(a+b)^2 + (a+b+4c)^2\ge \frac{kabc}{a+b+c}$ is true for any $a,b,c \in\mathbb{R}$
Could you check up my solution? I'm not sure it's ok -
$(a+b)^2 + (a+b+4c)^2 \ge 0$
and
$0\ge \frac{kabc}{a+b+c}$
so
$0 \ge k$
|
We need to worry only about $a+b+c> 0^\dagger$ and the inequality is homogeneous, so we may set $a+b+c=1$ and $c \in (0, 1]$
Then we have $ab \le \frac14(1-c)^2$ and we need to find largest $k$ satisfying
$$(1-c)^2+(1+3c)^2 \ge \frac{k}4 c(1-c)^2 \implies \frac{k}4 \le \frac1c+\frac1c\left(\frac{1+3c}{1-c}\right)^2$$
Minimising that gets you $k \le 100$ and when $a=b=\frac25, c=\frac15$, we have equality, so $k=100$ is indeed the largest possible.
$^\dagger$ if $a, b, c \in \mathbb R$, cases with $a+b+c=0$ or $\dfrac{abc}{a+b+c} \to \infty$ occur and no $k$ will satisfy the inequality.
|
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|
Given that a,b,c are distinct positive real numbers, prove that (a + b +c)( 1/a + 1/b + 1/c)>9
Given that $a,b,c$ are distinct positive real numbers, prove that $(a + b +c)\big( \frac1{a}+ \frac1{b} + \frac1{c}\big)>9$
This is how I tried doing it:
Let $p= a + b + c,$ and $q=\frac1{a}+ \frac1{b} + \frac1{c}$.
Using AM>GM for $p, q$, I get:
$$\frac{p+q}{2} > {(pq)}^{1/2}$$
$$\sqrt{(a + b +c)\bigg(\frac1{a}+ \frac1{b} + \frac1{c}\bigg)} < \frac{\big(a+\frac1{a} + b+\frac1{b} + c+\frac1{c}\big)}2$$
And for any $x\in \mathbb{R}, \space \space x+\frac1{x}≥2.$
Thus,
$$(a + b +c)\bigg(\frac1{a}+ \frac1{b} + \frac1{c}\bigg)<9, $$
which is the opposite of what had to be proven. What did I do wrong?
|
By the AM-GM inequality,
$$\frac{a}{b}+\frac{b}{a}\geq 2 $$
and so on, so:
$$ (a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) = 3+\sum_{cyc}\left(\frac{b}{a}+\frac{a}{b}\right) \geq 3+6 = \color{red}{9}$$
as wanted.
|
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|
Volume of figure between $x^2+y^2+z^2=16$ and $ x^2+y^2=6z$ if $z\geq 0$ I have a problem where I have to find volume of figure formed, when $x^2+y^2+z^2=16$ and $ x^2+y^2=6z$ intersects if $z\geq 0$.
Here is a graphic for clarity:
So far I have transformed the problem to cylindrical coordinates (given the fact that we are dealing with sphere):
$x=r\cos{\varphi}; \quad y=r\sin{\varphi}; \quad z=z \quad \Rightarrow \begin{cases} x^2+y^2+z^2=16 \quad \Rightarrow & r^2+z^2=4^2 \\ x^2+y^2=6z \quad \Rightarrow& r^2=6z\end{cases}$
Then I can say that $-z^2+16=6z \quad \Rightarrow -z^2-6z+16=0 \quad \Rightarrow z_1=2$ and $z_2=-8$ (we don't need this one)
Then $r^2=6z \quad \Rightarrow r=\sqrt{12}=2\sqrt{3}$
So the volume would now be:
$$\int _0^{2 \pi }\int _0^{2 \sqrt{3}}\int _{\frac{r^2}{6}}^{\sqrt{-r^2-16}}dzdrd\varphi=\int _0^{2 \pi }\int _0^{2 \sqrt{3}}\left(\sqrt{-r^2-16}-\frac{r^2}{6}\right)drd\varphi$$
At this moment I realize that I do not know how to integrate this integral.
I was hoping to see Step-by-step solution in WolframAlpha, but no success there (The definite integral of inner integrand is complex)
How would I have to proceed with my problem?
EDIT:
People pointed out that I have forgotten to transform $dxdydz$ to $rd\varphi dr dz$
Also I should have $\sqrt{16-r^2}$ instead of $\sqrt{-16-r^2}$.
So now the volume would be calculated by:
$$\int _0^{2 \pi }\int _0^{2 \sqrt{3}}\int _{\frac{r^2}{6}}^{\sqrt{16-r^2}}rdzdrd\varphi=\int _0^{2 \pi }\int _0^{2 \sqrt{3}}\left (r \cdot \left (\sqrt{16-r^2}-\frac{r^2}{6} \right )\right)drd\varphi$$
|
Thank to @Lucian I realized, that, the problem could be solved easily, by calculating volume when rotating $x=\sqrt{16-z^2}$ and $x=\sqrt{6z}$ about the $Oz$ axis. Therefore my volume would be:
$$\int_{2 \sqrt{3}}^4 \pi \left(16-z^2\right) \, dz+\int_0^{2 \sqrt{3}} \pi 6 z \, dz=\frac{236 \pi }{3}-24 \sqrt{3} \pi$$
(If anyone sees something wrong, please indicate my mistakes)
|
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|
Where am I going wrong in my linear Diophantine solution? Let $-2x + -7y = 9$. We find integer solutions $x, y$.
These solutions exist iff $\gcd(x, y) \mid 9$. So,
$-7 = -2(4) + 1$
then
$-2 = 1(-2)$
so the gcd is 1, and $1\mid9$. OK.
In other words,
$(-7)(1) + (-2)(-4) = 1$
Multiply by 9 on each side to get
$(-7)(9) + (-2)(-36) = 9$
$\begin{align} a = -7\\
x_0 = 9\\
b = -2\\
y_0 = -36\\
\end{align}$
Then we end up with
$x' = 9 - 7t, y' = -36 + 2t$
But when $t = 2$, we get $x'(2) = 9 - 7(2) = -5$ and $y'(2) = -36 + 2(2) = -32$ but clearly that's not a solution....
Where am I going wrong?
|
$x_0$ should be $-36$ and $y_0$ should be $9$. Somehow they got switched.
|
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|
Why does solving the spherical Bessel equation using Frobenius series produce two quadratic equations for the exponents at the singularity? The spherical Bessel equation is:
$$x^2y'' + 2xy' + (x^2 - \frac{5}{16})y = 0$$
If I seek a Frobenius series solution, I will have:
\begin{align*}
&\quad y = \sum_{n = 0}^{\infty} a_nx^{n + r} \\
&\implies y' = \sum_{n = 0}^{\infty} (n + r)a_nx^{n + r - 1} \\
&\implies y'' = \sum_{n = 0}^{\infty} (n + r)(n + r - 1)a_nx^{n + r - 2}
\end{align*}
Substituting into the ODE of interest:
\begin{align*}
&\quad x^2y'' + 2xy' + (x^2 - \frac{5}{16})y = 0 \\
&\equiv \sum_{n = 0}^{\infty} (n + r)(n + r - 1)a_nx^{n + r} + \sum_{n = 0}^{\infty} 2(n + r)a_nx^{n + r} + \sum_{n = 0}^{\infty} a_nx^{n + r + 2} + \sum_{n = 0}^{\infty} \frac{-5}{16}a_nx^{n + r} = 0 \\
&\equiv \sum_{n = 0}^{\infty} [(n + r)(n + r - 1) + 2(n + r) - \frac{5}{16}]a_nx^{n + r} + \sum_{n = 2}^{\infty} a_{n - 2}x^{n + r} = 0 \\
&\equiv [r(r-1) + 2r - (5/16)]a_0 + [(r+1)(r) + 2(r+1) - (5/16)]a_1 + \\
&\quad \sum_{n = 2}^{\infty} ([(n + r)(n + r - 1) + 2(n + r) - \frac{5}{16}]a_n + a_{n-2})x^{n + r} = 0 \\
&\implies [r(r-1) + 2r - (5/16)] = 0 \wedge \\
&\quad [(r+1)(r) + 2(r+1) - (5/16)] = 0 \wedge \\
&\quad [(n + r)(n + r - 1) + 2(n + r) - \frac{5}{16}]a_n + a_{n-2} = 0
\end{align*}
The first conjunct is the standard indicial equation specifying $r$. The second conjunct is yet another quadratic that $r$ has to satisfy. Is there a mistake? Or should $a_1 = 0$?
|
The main indicial is the the first equation. You get two roots from the first equation so you should set $a_1=0$. If there is a common root between the first and second equations, then you can consider that root and set both $a_1$ and $a_0$ nonzero.
|
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|
Prove that the equation $a^2+b^2=c^2+3$ has infinitely many integer solutions $(a,b,c)$. Prove that the equation $a^2+b^2=c^2+3$ has infinitely many integer solutions $(a,b,c)$.
My attempt:
$(a+1)(a-1)+(b+1)(b-1)=c^2+1$
This form didn't help so I thought of $\mod 3$, but that didn't help either. Please help. Thank you.
|
Note that the requirement is simply to demonstrate infinitely many solutions - we are not required to find all solutions.
Setting $c=b+1$, we see that $a^2 = c^2-b^2+3 = 2b+4$
Therefore for any even $a>2$, we can choose $b=\frac{a^2-4}{2}$ and $c=b+1$. This gives infinitely many solutions as required.
|
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|
Relationship between trigonometric and hyperbolic sine Why is the following identity true?
$$ \sin(ix) = i\sinh(x)$$
When I do the calculation, I get this:$$\sin(ix) = \frac{{e^{i(ix)}}-e^{-i(ix)}}{2i}=\frac{e^{-x}-e^x}{2i}=-\frac{e^x-e^{-x}}{2i}=-\left(\frac{\sinh(x)}{i}\right)$$
|
Another derivation:
$$\displaystyle \sin(ix) = ix - \frac {(ix)^3}{3!} + \frac {(ix)^5}{5!} - \frac {(ix)^7}{7!}+ \cdots$$
$$ = ix + \frac {ix^3} {3!} + \frac {ix^5}{5!} + \frac {ix^7}{7!} + \cdots $$
$$ = i \left({x + \frac {x^3}{3!} + \frac {x^5}{5!} + \frac {x^7}{7!} + \cdots }\right)= i \sinh x$$
|
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|
find the length of rectangle based on area of frame A rectangular picture is $3$ cm longer than its width, $(x+3)$. A frame $1$ cm wide is placed around the picture. The area of the frame alone is $42 \text{cm}^2$. Find the length of the picture.
I have tries:
$(x+6)(x+3) = 42 \\
x^2+3x+6x+12 = 42 \\
x^2+9x+12x-42 = 0 \\
\\
x^2 + 21 -42$
|
The area of the frame is the difference between the area of the rectangle bordered by the outside of the frame and the area of the picture. Since the frame adds $1~\text{cm}$ on each side of the picture, the outer rectangle has length $x + 8~\text{cm}$ and width $x + 5~\text{cm}$. Thus, the area of the frame is
\begin{align*}
(x + 8~\text{cm})(x + 5~\text{cm}) - (x + 6~\text{cm})(x + 3~\text{cm}) & = 42~\text{cm}^2\\
(x^2 + 13x~\text{cm} + 40~\text{cm}^2) - (x^2 + 9x~\text{cm} + 18~\text{cm}^2 & = 42~\text{cm}^2\\
4x~\text{cm} + 22~\text{cm}^2 & = 42~\text{cm}^2\\
4x~\text{cm} & = 20~\text{cm}^2\\
x~\text{cm} & = 5~\text{cm}
\end{align*}
Hence the length of the picture is $x + 6~\text{cm} = 5~\text{cm} + 6~\text{cm} = 11~\text{cm}$.
|
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|
The inequality. Regional olympiad 2015 Let $a, b, c$ - the positive real numbers, and $ab+bc+ca=1$
Prove that $\sqrt{a+\frac{1}{a}}+\sqrt{b+\frac{1}{b}}+\sqrt{c+\frac{1}{c}} \geqslant 2(\sqrt{a}+\sqrt{b}+\sqrt{c})$
Probably, we should use these facts:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{abc}$
$(a+b+c)^2 = a^2 + b^2 + c^2 + 2$
But I don't know how to use them. Please, help.
|
By Cauchy-Schwarz we have
$$
\sqrt{a + \frac{1}{a}} = \sqrt{a + b + c + \frac{bc}{a}} \geq \frac{1}{2}\left(\sqrt{a} + \sqrt{b} + \sqrt{b} + \sqrt{\frac{bc}{a}}\right)
$$
and similarly
$$
\sqrt{b + \frac{1}{b}} \geq \frac{1}{2}\left(\sqrt{a} + \sqrt{b} + \sqrt{b} + \sqrt{\frac{ca}{b}}\right)
$$
and
$$
\sqrt{c + \frac{1}{c}} \geq \frac{1}{2}\left(\sqrt{a} + \sqrt{b} + \sqrt{b} + \sqrt{\frac{ab}{c}}\right).
$$
Thus
$$
\sqrt{a + \frac{1}{a}} + \sqrt{b + \frac{1}{b}} + \sqrt{c + \frac{1}{c}}
\geq \frac{1}{2}\left(3\left(\sqrt{a} + \sqrt{b} + \sqrt{c}\right) + \frac{\sqrt{a}\sqrt{b}}{\sqrt{c}} + \frac{\sqrt{b}\sqrt{c}}{\sqrt{a}} + \frac{\sqrt{c}\sqrt{a}}{\sqrt{b}} \right) \geq 2\left(\sqrt{a} + \sqrt{b} + \sqrt{c}\right).
$$
|
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How to prove $p(z-y)(zy)(z^2-yz+y^2) \mid x^p-(z-y)^p \Rightarrow x=z-y$? Assume $p>2$ prime and $1<x<y<z$ coprime. How to prove the following: $$p(z-y)(zy)(z^2-yz+y^2) \mid x^p-(z-y)^p \Rightarrow x=z-y$$
I remember it as an extra exercise which I couldn't solve when I was a student. It still bothers me...
I can only proof a partial statement: the case that $p \nmid \phi \big( p(z-y)(zy)(z^2-yz+y^2) \big)$ whereby $\phi(m)=m\displaystyle\prod_{\substack{q \mid m \\ q \ prime}}{(1-\frac{1}{q})}$ is Euler's totient function.
Let $n=p(z-y)(zy)(z^2-yz+y^2)$ and $\text{rad}(n)=\displaystyle\prod_{\substack{p \mid n \\ p \ prime}}{p}$ the radical of $n$.
We know that for $q>2$ prime, $s^k \equiv 1 \pmod{q}$ has $\gcd(k,q-1)$ solutions (See link). Thus for all prime divisors $q>2$ of $n$ with $p \nmid q-1$ the relation $s^p \equiv t^p \pmod{q}$ implies $q \mid s-t$. Stated differently, the map $s \mapsto s^p \pmod{q}$ is injective on $[1 \dots q]$. Also for the case $q=2$.
We conclude that $n \mid x^p-(z-y)^p$ and $p \nmid \phi(n)$ implies $\text{rad}(n) \mid x-(z-y)$. For the last step I must assume that $y<\text{rad}(p(z-y)(zy)(z^2-yz+y^2))$ to ensure that $x,z-y \in [1 \dots \text{rad}(n)]$. This assumption seems obvious but I've never seen any proof of this statement (See question).
Who can help me with the missing parts?
|
The following reasoning is derived from exercise 1.19.13 in the excellent book about polynomials with Barbeau. This exercise is based on problem A2 from the 37th Putnam Competition in 1976. All lower case symbols used are representing non-negative integers unless specified differently.
Define $F_n=(x+y)^n-x^n-y^n$, $G_n=(x+y)^n+x^n+y^n$, $P=xy(x+y)$, and $Q=x^2+xy+y^2$. We have: $F_0=-1$, $G_0=3$, $F_1=0$, $G_2=2Q$, $F_3=3P$, $G_4=2Q^2$, $F_5=5PQ$ and $G_6=2Q^3+3P^2$. The polynomials $F_2, F_4,F_6,G_1,G_3$ and $G_5$ can not be written in terms of $P$ and $Q$. By multiplication one can easily verify that
$$F_n=QF_{n-2}+PG_{n-3}$$
$$G_n=QG_{n-2}+PF_{n-3}$$
In the first line we can replace $G_{n-3}$ and $F_{n-2}$
$$F_n=Q(QF_{n-4}+PG_{n-5})+P(QG_{n-5}+PF_{n-6})=Q^2F_{n-4}+2PQG_{n-5}+P^2F_{n-6}$$
Thus $F_n \equiv Q^2F_{n-4}+P^2F_{n-6} \pmod{PQ}$. Suppose $n=6k+j$ and $0 \le j < 6$. This replacement can be repeated $k$ times until we end with the term $Q^{2k}F_{n-4k}+P^{2k}F_j$ modulo $PQ$.
We conclude that if and only if $j=1,5$ the last term $P^{2k}F_j$ vanishes modulo $PQ$.
Let $n-4k=4m+i$. Note that assuming $j=1,5$ implies $i=1,3$. After $m$ steps we end with the term $Q^{2(k+m)}F_{i}$ modulo $PQ$, and as $F_i \equiv 0 \pmod{P}$ we conclude that if $xy(x+y) \not =0$
$$xy(x+y)(x^2+xy+y^2) \mid (x+y)^n-x^n-y^n \Leftrightarrow n\equiv{1,5} \pmod{6}$$
The next step is to look at non-negative coprime solutions of
$$xy(x+y)(x^2+xy+y^2) \mid (x+y)^n-y^n$$
If we assume that $n\equiv{1,5} \pmod{6}$, rewriting yields $(x+y)^n-y^n=F_n+x^n$ and we can use $F_n \equiv 0 \pmod{PQ}$. Thus the problem reduces to solving
$$x^n = k*xy(x+y)(x^2+xy+y^2)$$
A solution is $x=0$. Suppose now $x>0$. This implies $y>0$ because $y=0$ is impossible. As $y \mid x^n$, we conclude that prime divisors of $y$ are also prime divisors of $x$. The condition that $x,y$ are coprime implies that the only option is $y=1$. The relation becomes $x^n = k*x(x+1)(x^2+x+1)$ which has no positive solution as both $x,x+1$ are coprime.
In total we find that if $n\equiv{1,5} \pmod{6}$, $x=0$ is the only solution.
Now we proof the following statement if $n=1,2,3$ or $n=1,5 \pmod{6}$, and $0<x<y<z$ coprime
$$(z-y)(zy)(z^2-yz+y^2) \mid x^n-(z-y)^n \Rightarrow x=z-y$$
A trivial solution is $x=z-y$. Assume $x \not = z-y$ and thus $k \not = 0$ if $x^n-(z-y)^n=k*(z-y)(zy)(z^2-yz+y^2)$. Note that $k$ can be negative.
Case $n=1,2,3$. Suppose $x>z-y$, thus $k>0$. We have $x^n-(z-y)^n=k*(z-y)(zy)(z^2-yz+y^2)> k*(1)(y^2)(y^2-y^2+y^2)= k*y^4$ as $z>y$. Thus $x^n> k*y^4+(z-y)^n$ which implies $y<x \Rightarrow \Leftarrow x<y$.
Now suppose $x<z-y$, thus $k<0$. Note that $y^2-yz+y^2=(z-y)^2+yz$. We have $(z-y)^n-x^n=(-k)*(z-y)(zy)(z^2-yz+y^2)$ $= (-k)*zy(z-y)^3+(-k)*(z-y)(zy)^2$. Thus $(z-y)^n > (-k)*zy(z-y)^3+x^n \Rightarrow \Leftarrow$. In total we conclude that there is only one solution.
Case $n=1,5 \pmod{6}$. Define $t=z-y$ and note that $(z-y)(zy)(z^2-yz+y^2)$ $=ty(y+t)(t^2+ty+y^2)$ which divides $x^n-(z-y)^n=\big((x-(z-y)\big)+t)^n-t^n$. Now we can apply our earlier finding to conclude that $x-(z-y)=0$. QED
A final remark. The question posed has as extra condition that $p$ is a prime larger than 2. It is well-know that for primes $p>3$ we have $p \equiv 1,5 \pmod{6}$, and by Fermat's Little Theorem $p \mid (x+y)^p-x^p-y^p$. Unfortunately, this condition put me on the wrong track for a long time :(
|
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Sequence $a_{n+1}=\sqrt{1+\frac{1}{2}a_n^2}$ I am trying but cant figure out anything.
$a_{n+1}=\sqrt{1+\frac{1}{2}a_n^2}$
I am trying to proove that $a_n^2-2<0$.
Getting $$a_{n+1} -a_n=\dots=\frac{2-a_n^2}{2\left(\sqrt{1+\frac{1}{2}a_n^2} +a_n\right)}$$
Then I have no clue how to proove it since I am not given $a_1$.Induction doesnt seem to work nor any contradiction.
|
Hints:
1) $\dfrac{1}{2}a_{n}^2-a_{n+1}^2=\dfrac{1}{2}a_{n+1}^2-a_{n+2}^2=\ldots=\dfrac{1}{2}a_{n+i}^2-a_{n+i+1}^2=-1$
Therefore...
2) Proving that $a_n^2-2<0$ is the same as proving that $a_{n+1}^2-2<0$, and
3) Proving that $a_n^2-2<0$ is the same as proving that $a_{n-1}^2-2<0$
|
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Limit of $(\cos{xe^x} - \ln(1-x) -x)^{\frac{1}{x^3}}$ So I had the task to evaluate this limit
$$ \lim_{x \to 0} (\cos{(xe^x)} - \ln(1-x) -x)^{\frac{1}{x^3}}$$
I tried transforming it to:
$$ e^{\lim_{x \to 0} \frac{ \ln{(\cos{xe^x} - \ln(1-x) -x)}}{x^3}}$$
So I could use L'hospital's rule, but this would just be impossible to evaluate without a mistake. Also, I just noticed this expression is not of form $\frac{0}{0}$.
Any solution is good ( I would like to avoid Taylor series but if that's the only way then that's okay).
I had this task on a test today and I failed to do it.
|
First notice that $$\cos (xe^x) = \sum_{n=0}^{\infty}(-1)^n \frac{(xe^x)^{2n}}{2n!}$$
and $$\ln (1 - x) = -\sum_{n=0}^{\infty} \frac{x^n}{n}$$
Thus $$\cos (xe^x) - \ln (1 - x) = \sum_{n=0}^{\infty}(-1)^n \frac{(xe^x)^{2n}}{2n!} +\sum_{n=0}^{\infty}\frac{x^n}{n} = 1 + x - \frac{2x^3}{3} - O(x^4) $$
Therefore we have
$$\ln (1 - \frac{2x^3}{3}) = - \frac{2x^3}{3} - \frac{2x^6}{9} - O(x^9)$$
Finally
$$\begin{align}\lim_{x \to 0} \frac{\ln (\cos xe^x - \ln (1 - x) - x)}{x^3} &= \lim_{x \to 0} -\frac{2}{3} - \frac{2x^3}{9} - O(x^6) =\color{red}{ -\frac{2}{3}}\end{align}$$
Now you may find your limit.
|
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finding the minimum value of $\frac{x^4+x^2+1}{x^2+x+1}$ given $f(x)=\frac{x^4+x^2+1}{x^2+x+1}$.
Need to find the min value of $f(x)$.
I know it can be easily done by polynomial division but my question is if there's another way
(more elegant maybe) to find the min?
About my way: $f(x)=\frac{x^4+x^2+1}{x^2+x+1}=x^2-x+1$. (long division)
$x_{min}=\frac{-b}{2a}=\frac{1}{2}$. (when $ax^2+bx+c=0$)
So $f(0.5)=0.5^2-0.5+1=\frac{3}{4}$
Thanks.
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Multiply the fraction by $\frac{x^2-1}{x^2-1}$:$$\frac{(x^2-1)(x^4+x^2+1)}{(x+1)(x-1)(x^2+x+1)}=\frac{x^6-1}{(x+1)(x^3-1)}=\frac{x^3+1}{x+1}=x^2-x+1$$
|
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|
Question about Lagrange multiplier and maximum point Find the maximum of $\log{x}+\log{y}+3\log{z}$ on portion of the sphere $x^2 + y^2 + z^2 =5r^2$
where $x,y,z>o $
I found that maximum is $5\log{r} + 3\log{\sqrt{3}}$ at $(r,r,3\sqrt{3})$
And Use this result to prove that for real positive $a, b, c$,
$abc^3$ is less than or equal to $27(a+b+\frac{c}{5})^5$
Can anyone help me prove this inequality given that we have a maximum?
|
There are two typos:
The maximum happens at $(r,r,\sqrt{3}r)$, and the right hand side of the inequality should be $27(\frac{a+b+c}{5})^5$.
Suppose $a=x^2,b=y^2,c=z^2$. Use the fact that $\log{x}+\log{y}+3\log{z}\leq 5\log{r} + 3\log{\sqrt{3}}$ when $x,y,z$ satisfies the sphere equation.
So
$$\log{xyz^3}\leq\log{3\sqrt{3}r^5}\\
\Rightarrow \log{x^2y^2z^6}\leq \log{27r^{10}}\\
\Rightarrow \log{abc^3}\leq \log{27(\frac{a+b+c}{5})^5}\\
\Rightarrow abc^3\leq 27(\frac{a+b+c}{5})^5$$
I used the fact that $r^2=\frac{x^2+y^2+z^2}{5}$
Some log properties I used:
$$\log{a+b}=\log{a}+\log{b}\\
\log{a^r}=r\log{a}$$
|
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By viewing the polynomials as a difference of two squares, factorise the following polynomials. By viewing the polynomials as a difference of two squares, factorise the following polynomial:
$$x^4+x^2+1.$$
I searched but couldn't find a way to solve this
Edit: By using Hans Lundmark hint, I get:
$$(x^2+1)^2-x^2$$
Is it fully factorized?
|
We don't factorization:
$$\dfrac2{x^4+ax^2+1}=\dfrac{x^2+1}{x^4+ax^2+1}-\dfrac{x^2-1}{x^4+ax^2+1}=\dfrac{1+\dfrac1{x^2}}{x^2+a+\dfrac1{x^2}}+\dfrac{1-\dfrac1{x^2}}{x^2+a+\dfrac1{x^2}}$$
Now as $\displaystyle\int\left(1\pm\dfrac1{x^2}\right)dx=x\mp\dfrac1x,$
for the first integral, choose $x-\dfrac1x=u$ and $x+\dfrac1x=v$ for the second
Use $x^2+\dfrac1{x^2}=\left(x+\dfrac1x\right)^2-2=\left(x-\dfrac1x\right)^2+2$
|
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Implicit differentiation of $e^{x^2+y^2} = xy$ I just want to reconfirm the steps needed to answer this question. Thank you
Find $\dfrac{dy}{dx}$ in the followng:
$$e^{\large x^2 + y^2}= xy$$
I got this so far.
$\newcommand{\dd}{\mathrm{d}}\frac{\dd x}{\dd y} e^{x^2}\cdot e^{y^2} = \frac{\dd x}{\dd y} xy$
$u=e^{x^2}$: $\frac{\dd u}{\dd x}=2x(e^{x^2})$
$v=e^{y^2}$: $\frac{\dd v}{\dd x}=2y(e^{y^2})\frac{\dd y}{\dd x}$
$u=x$: $\frac{\dd u}{\dd x}=1$
$v=y$: $\frac{\dd v}{\dd x}=\frac{\dd y}{\dd x}$
$$\begin{align}
e^{x^2}\cdot2y(e^{y^2})\frac{\dd y}{\dd x} + e^{y^2}\cdot 2x(e^{x^2}) &= x \frac{\dd y}{\dd x} + y\\
e^{x^2}\cdot2y(e^{y^2})\frac{\dd y}{\dd x} - x \frac{\dd y}{\dd x} &= y - e^{y^2}\cdot 2x(e^{x^2})\\
\frac{\dd y}{\dd x} \left[e^{x^2}\cdot2y(e^{y^2})-x\right] &= y - e^{y^2}\cdot 2x(e^{x^2})\\
\frac{\dd y}{\dd x} &= \frac{y - e^{y^2}\cdot 2x(e^{x^2})}{e^{x^2}\cdot2y(e^{y^2})-x}
\end{align}$$
|
$$ e^{x^2+y^2} =xy
\\ e^{x^2+y^2}\; [x^2+y^2]' =[x]'\, y+x\,[y]'
\\ e^{x^2+y^2}\; [2x+2y\,\color{blue}{y'}] =[1]\, y+x\,[\color{blue}{y'}]
\\ 2xe^{x^2+y^2}+2ye^{x^2+y^2}\,\color{blue}{y'} =y+x\,\color{blue}{y'}
\\ 2ye^{x^2+y^2}\,\color{blue}{y'}-x\,\color{blue}{y'} =y-2xe^{x^2+y^2}
\\ (2ye^{x^2+y^2}-x)\,\color{blue}{y'} =y-2xe^{x^2+y^2}
\\ \color{blue}{y' = \dfrac{y-2xe^{x^2+y^2}}{2ye^{x^2+y^2}-x}} $$
|
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Summation inductional proof: $\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$ Having the following inequality
$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}<2$$
To prove it for all natural numbers is it enough to show that:
$\frac{1}{(n+1)^2}-\frac{1}{n^2} <2$ or $\frac{1}{(n+1)^2}<2-\frac{1}{n^2} $
|
$$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots+\frac{1}{n^2}+...\uparrow\frac{\pi^2}{6}=1,6449340668482...\lt2.$$
See: http://mathworld.wolfram.com/PiFormulas.html
|
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|
For odd $m\ge3$, does it follow: $\frac{x^m + y^m}{x+y} + (xy)\frac{x^{m-2} + y^{m-2}}{x+y} = x^{m-1} + y^{m-1}$ Unless I am making a mistake, I am calculating that:
$$\frac{x^m + y^m}{x+y} + (xy)\frac{x^{m-2} + y^{m-2}}{x+y} = x^{m-1} + y^{m-1}$$
Here's my reasoning:
*
*$\dfrac{x^m + y^m}{x+y} = x^{m-1} - x^{m-2}y - xy^{m-2} + x^{m-3}y^2 + x^2y^{m-3} + \dots + x^{\frac{m-1}{2}}y^{\frac{m-1}{2}} + y^{m-1}$
*$\dfrac{x^{m-2} + y^{m-2}}{x+y} = x^{m-3} - x^{m-4}y - xy^{m-4} + x^{m-5}y^2 + x^2y^{m-5} + \dots + x^{\frac{m-3}{2}}y^{\frac{m-3}{2}} + y^{m-3}$
*$(xy)\dfrac{x^{m-2} + y^{m-2}}{x+y} = x^{m-2}y - x^{m-3}y^2 - x^2y^{m-3} + x^{m-4}y^3 + x^3y^{m-4} + \dots + x^{\frac{m-1}{2}}y^{\frac{m-1}{2}} + xy^{m-2}$
*So that, $\dfrac{x^m + y^m}{x+y} + (xy)\dfrac{x^{m-2} + y^{m-2}}{x+y} = x^{m-1} + y^{m-1}$
*I am figuring that $\dfrac{x^{m-2} + y^{m-2}}{x+y}$ has exactly 2 terms less than $\dfrac{x^m + y^m}{x+y}$
Is this reasoning correct? For each value that I test, it seems correct.
Thanks,
-Larry
|
Well, I would do it that way :
$$ x^m = x\ x^{m-1} $$
So :
$$\begin{align}\frac{x^m + y^m}{x+y} + (xy)\frac{x^{m-2} + y^{m-2}}{x+y} &= \frac{x^m + y^m+ (xy)\ x^{m-2} + (xy)\ y^{m-2}}{x+y} \\
&=\frac{x^m + y^m+ y\ x^{m-1} + x\ y^{m-1}}{x+y} \\
&=\frac{ x\ x^{m-1} + y\ y^{m-1}+ y\ x^{m-1} + x\ y^{m-1}}{x+y}\\
&=\frac{ (x+y)\ x^{m-1} + (x+y)\ y^{m-1}}{x+y}\\
&= x^{m-1} + y^{m-1}\end{align}$$
|
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|
Evaluate $\int{\frac{\sqrt[3]{x+8}}{x}}dx$ So I've tried solving the equation below by using $u=x+8$, and I get $\int{\frac{\sqrt[3]{u}}{u-8}}du$ which doesn't seem to lead anywhere, I've also tried taking the $ln$ top and bottom, but I don't know how to proceed. Any hints?
$$\int{\frac{\sqrt[3]{x+8}}{x}}dx$$
Update: Using partial fraction decomposition, $A=2, B=-2, C= -8$ (thank you to all the helpful posts) and after some tedious calculations, the answer I obtained was: $3{\sqrt[3]{x+8}}+24(2ln|\sqrt[3]{x+8}-2|-2ln|\sqrt[3]{x+8}+2|+{\frac{4}{\sqrt[3]{x+8}+2}})+C$
|
Let $x=u^3-8$, then $\mathrm dx =3u^2\mathrm du$ and we get the integral
$$\int\frac{u}{u^3-8}3u^2\mathrm du$$
Last integral can be solved by partial fractions since
\begin{align*}
\frac{3u^3}{u^3-8}&=3+\frac{24}{u^3-8}\\
&=3+\frac{A}{u-2}+\frac{Bu+C}{u^2+2u+4}
\end{align*}
where $A$, $B$ and $C$ are constants we must find.
|
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|
Let $a$, $b$ and $c$ be the three sides of a triangle. Show that $\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3$.
Let $a$, $b$ and $c$ be the three sides of a triangle.
Show that $$\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3\,.$$
A full expanding results in:
$$\sum_{cyc}a(a+b-c)(a+c-b)\geq3\prod_{cyc}(a+b-c),$$ or
$$\sum_{cyc}(a^3-ab^2-ac^2+2abc)\geq\sum_{cyc}(-3a^3+3a^2b+3a^2c-2abc),$$ but it becomes very ugly.
|
By C-S $\sum\limits_{cyc}\frac{a}{b+c-a}=\sum\limits_{cyc}\frac{a^2}{ab+ac-a^2}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(2ab-a^2)}\geq3$, where the last inequality it's just
$$\sum\limits_{cyc}(a-b)^2\geq0.$$
Done!
|
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|
Why is $\lim_{n\to\infty} n(e - (1+\frac{1}{n})^n) = \frac{e}{2}$ I'm having trouble understanding why
$$\lim_{n\to\infty} n(e - (1+\tfrac{1}{n})^n) = \frac{e}{2}$$
Can someone offer me a proof for this?
|
Using: $\ln(1 + \frac{1}{n}) \sim \frac{1}{n} - \frac{1}{2n^2}$
And: $e^{-\frac{1}{2n}} \sim 1 - \frac{1}{2n} $
$$n(e - (1 + \frac{1}{n})^n) = n(e - e^{n\ln(1 + \frac{1}{n})}) = ne(1 - e^{n\ln(1 + \frac{1}{n}) - 1}) \sim ne(1 - e^{1 - \frac{1}{2n} - 1}) = ne(1 - e^{-\frac{1}{2n}}) = ne(1 - (1 - \frac{1}{2n}) = ne(\frac{1}{2n}) = \frac{e}{2}$$
$\therefore$ etc.
|
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|
Proof of a limit of a sequence I want to prove that $$\lim_{n\to\infty} \frac{2n^2+1}{n^2+3n} = 2.$$
Is the following proof valid?
Proof
$\left|\frac{2n^2+1}{n^2+3n} - 2\right|=\left|\frac{1-6n}{n^2+3n}\right| =\frac{6n-1}{n(n+3)} $ (because $n \in \mathbb N^+)$. $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$
We have $n \ge 1 \implies 6n -1 > n + 3 \implies \frac{n+3}{n(n+3)} < \frac{6n-1}{n^2+3n}.$
Let $\epsilon > 0$ be given.
Note that $ \frac{6n-1}{n^2+3n}< \epsilon \iff \frac{n+3}{n(n+3)} < \epsilon \iff n > \frac{1}{\epsilon}.$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (**)$
By the Archimedean Property of $\mathbb R$, $\exists N \in \mathbb N^+$ such that $N > \frac{1}{\epsilon}.$
If $n \ge N$, then $n > \frac{1}{\epsilon}$, and from $(*)$ and $(**)$ it follows that $\left|\frac{2n^2+1}{n^2+3n} - 2\right| < \epsilon$.
Therefore $\lim_{n\to\infty} \frac{2n^2+1}{n^2+3n} = 2.$
|
Your proof is incorrect: you want to show that
$$\frac{6n-1}{n^2-3n}<\varepsilon$$
so what you need is
$$\frac{6n-1}{n^2-3n}<{\rm something}$$ or $6n-1<$something, that is an estimate of $6n-1$ from above, not from below.
|
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|
Example of contour integration Could someone help me evaluate the following integral with contour integration ?
$$\int_{0}^{2\pi}\frac{d\theta}{(a+b\cos\theta)^2}.$$
Constraints are: $a>b>0$.
|
Using the parametrization $z = e^{i\theta}$, $0 \le \theta \le 2\pi$ for the unit circle, we can write the integral as the contour integral
$$\oint_{|z| = 1} \dfrac{1}{\left(a + b\frac{z + z^{-1}}{2}\right)^2} \frac{dz}{iz},$$
which can be rewritten
$$\oint_{|z| = 1} \dfrac{4z}{i(bz^2 + 2az + b)^2}\, dz,$$
or
$$\frac{4}{ib^2}\oint \frac{z}{(z^2 + \frac{2a}{b}z + 1)^2}\, dz.$$
The roots of $z^2 + 2(a/b)z + 1$ are
$$z_0 = \frac{-a-\sqrt{a^2 - b^2}}{b}\quad \text{and} \quad z_1 = \frac{-a + \sqrt{a^2 - b^2}}{b}.$$
The point $z_0$ lies outside the circle since
$$|z_0| = \frac{a + \sqrt{a^2 - b^2}}{b} > \frac{a}{b} > 1.$$
So $z_1$ lies inside the circle (shown by computation or by use of the fact $z_0z_1 = 1$ and $|z_0| > 1$). Hence, the function $z/(z - z_0)^2$ is analytic inside and on $|z| = 1$. Since $z^2 + 2(b/a)z + 1 = (z - z_0)(z - z_1)$, it follows from Cauchy's differentiation formula that
\begin{align}\frac{4}{ib^2}\oint_{|z| = 1} \frac{z}{(z + \frac{2b}{a}z + 1)^2}\, dz&= \frac{4}{ib^2} \cdot 2\pi i \frac{d}{dz}\bigg|_{z = z_1} \frac{z}{(z - z_0)^2}\\
&= \frac{8\pi}{b^2} \cdot \frac{-(z_1 + z_0)}{(z_1 - z_0)^3}\\
&= \frac{8\pi}{b^2} \cdot \dfrac{\frac{2a}{b}}{\frac{8(a^2 - b^2)\sqrt{a^2 - b^2}}{b^3}}\\
&= \frac{2\pi a}{(a^2 - b^2)\sqrt{a^2 - b^2}}.
\end{align}
|
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|
Find the sum $\sum_{n=1}^{50}\frac{1}{n^4+n^2+1}$ Find the sum $\sum_{n=1}^{50}\frac{1}{n^4+n^2+1}$
$$\begin{align}\frac{1}{n^4+n^2+1}&
=\frac{1}{n^4+2n^2+1-n^2}\\
&=\frac{1}{(n^2+1)^2-n^2}\\
&=\frac{1}{(n^2+n+1)(n^2-n+1)}\\
&=\frac{1-n}{2(n^2-n+1)}+\frac{1+n}{2(n^2+n+1)}\\
\end{align}$$
For $n={1,2,3}$ it is not giving telescooping series.
$=\frac{1}{3}+\frac{3}{14}+\frac{2}{13}+0-\frac{1}{6}-\frac{1}{7}$
|
Partial answer.
Rewrite the terms as:
$$\frac{1}{2}\left(\frac{n+1}{n^2+n+1} - \frac{n-1}{1+(n-1)+(n-1)^2}\right) = \\\frac{1}{2}\left(\frac{1}{n^2+n+1}+f(n)- f(n-1)\right)$$ where $f(n)=\frac{n}{1+n+n^2}$. In particular $\sum_{n=1}^\infty \left(f(n)-f(n-1)\right)=-f(0)=0$.
So we've reduced it to:
$$\frac12\sum \frac{1}{n^2+n+1}$$
Not sure how to compute that sum.
|
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|
geometric proof of $2\cos{A}\cos{B}=\cos{(A+B)}+\cos{(A-B)}$ I have seen geometric proof of identities
$$\cos{(A+B)}=\cos{A}\cos{B}-\sin{A}\sin{B}$$
and
$$\cos{(A-B)}=\cos{A}\cos{B}+\sin{A}\sin{B}$$
By adding two equation, $$2\cos{A}\cos{B}=\cos{(A+B)}+\cos{(A-B)}$$.
But how to prove this by geometry?
Thank you.
|
$$\begin{align}
2 \cos A \cos B &= \cos(A-B)+\cos(A+B) \\[6pt]
2 \sin A \,\sin B &= \cos(A-B)-\cos(A+B)
\end{align}$$
Note. Although not labeled (yet), these identities are also evident:
$$\begin{align}
2 \,\sin A \cos B &= \sin(A+B)+\sin(A-B) \\[6pt]
2 \cos A \,\sin B &= \sin(A+B)-\sin(A-B)
\end{align}$$
|
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|
Simplification of $\sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$ I was trying to simplify $\sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$. Numerical evaluation suggested that the answer is $\sqrt{2}$ and it checked out when I substituted $\sqrt{2}$ in the equation $x= \sqrt{14} - \sqrt{16 - 4 \sqrt{7}}$.
But I still cannot simplify the initial expression by any obvious means (squaring, multiplying by conjugate etc). Any idea how to do it? Can it be done?
|
Suppose that
$$
\left(\sqrt{a}\pm\sqrt{b}\right)^2
=\overbrace{a+b\vphantom{\sqrt4}}^c\pm\overbrace{\sqrt{4ab}}^{\sqrt{d}}
$$
Then
$$
\begin{align}
c^2-d
&=a^2+2ab+b^2-4ab\\
&=(a-b)^2
\end{align}
$$
Thus,
$$
a=\frac{c+\sqrt{c^2-d}}2
\qquad\text{and}\qquad
b=\frac{c-\sqrt{c^2-d}}2
$$
If $c=4$ and $d=7$, we get
$$
4-\sqrt{7}=\left(\sqrt{\frac72}-\sqrt{\frac12}\right)^2
$$
Therefore,
$$
\begin{align}
\sqrt{14}-\sqrt{16-4\sqrt7}
&=\sqrt{14}-2\sqrt{4-\sqrt7}\\
&=\sqrt{14}-\left(\sqrt{14}-\sqrt2\right)\\
&=\sqrt2
\end{align}
$$
|
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|
Four numbers drawn from 1 to 100000 randomly, the same number may be chosen more than once. Four numbers drawn from 1 to 100000 randomly, the same number may be chosen more than once. Determine probability the last digit from multiplying that four numbers is 1 or 9.
I have tried with many cases with it but it come a big number on combination.
|
As mentioned in the comments, we may assume we only draw numbers from the set $\{0,1,2,\ldots,9\}$. This is the case because only the last digit of the number we picked matters. Let us call the random numbers $A$, $B$, $C$, $D$. If one of $A$, $B$, $C$, $D$ is even or divisible by $5$, then $ABCD$ is as well and then it will not end in a $1$ or $9$. The probability that $ A$, $B$, $C$, $D$ are all coprime to $10$ is $(\frac25)^4$ (we can only pick $1$, $3$, $7$, $9$). We can now show that the last digit of $AB$ takes these values $1$, $3$, $7$, $9$ with equal probability and that the same holds for $ABC$ and $ABCD$. In half of the cases the last digit will be $1$ or $9$. This yields a final answer of
$$
\left(\frac25\right)^4 \cdot \frac12 = 0.0128.
$$
Edit: Consider the multiplication table of $1$, $3$, $7$ and $9$ (we consider last digits only):
$$
\begin{array}{c|cccc}
\times & 1 & 3 & 7 & 9 \\ \hline
1 & {\color{green}1} & 3 & 7 & {\color{green}9} \\
3 & 3 & {\color{green}9} & {\color{green}1} & 7 \\
7 & 7 & {\color{green}1} & {\color{green}9} & 3 \\
9 & {\color{green}9} & 7 & 3 & {\color{green}1}
\end{array}
$$
Exactly half of the entries of this table is a $1$ or a $9$, so if the last digits of $X$ and $Y$ are independently equidistributed on $\{1,3,7,9\}$ then so is the last digit of $XY$.
|
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|
If $a$ is a quadratic residue of odd prime $p$, then is it the case that $a^{-1}$ is also a quadratic residue? If $a$ is a quadratic residue of odd prime $p$, then is it the case that $a^{-1}$ is also a quadratic residue ?
I notice quadratic residues of odd prime $13$ :
$$1^2\equiv 12^2\equiv 1\\2^2\equiv 11^2\equiv 4\\3^2\equiv 10^2\equiv 9\\4^2\equiv 9^2\equiv 3\\5^2\equiv 8^2\equiv 12\\6^2\equiv 7^2\equiv 10$$
$4^{-1}=10\\9^{-1}=3\\1^{-1}=1\\12^{-1}=12$
Is this true always ? How to go about proving ? thanks!
|
If $a^2 = b$ in a ring $R$ with unit, $a$ is invertible if and only if $b$ is and then $b^{-1} = (a^{-1})^2$. (Tell if this is clear to you, if not, I'll detail in an edit.) Apply this to $R = \mathbf{Z} / n \mathbf{Z}$. Apply the latter to the case $n = p$ is prime. ;-)
Detail. If $a$ is (for instance left) invertible of left inverse $c$ then $ca = 1$ so that $ca^2 = a$ which implies that $c^2 a^2 = ca = 1$, that is $c^2 b = 1$ and $b$ is invertible. If $b$ is (let's say left) invertible then $db=1$ and then $(d a) a = 1$ and $a$ is left invertible. Same for right inverse. The relation $c^2 a^2 = 1$ shows you the assertion on quadratic residue, that is, that if you are invertible and a quadratic residue, that you your inverse is also a quadratic residue.
|
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|
What am i doing wrong when solving this differential equation $$
f(x) = \frac{\frac{d^2}{dx^2}[e^y]}{\frac{d}{dx}[e^y]}
$$
Given that $f(x) = cx$
$$
\frac{c}{2}x^2 + k_1 = \ln(e^y y')
$$
$$
k_2\int e^{\frac{c}{2}x^2} dx = e^y
$$
$$
y = \ln(k_2\int e^{\frac{c}{2}x^2} dx)
$$
Therefore
$$
y'^2 - cxy' + y'' = 0
$$
$$
y' = \frac{c}{2}x \pm \sqrt{\frac{c^2}{4}- y''}
$$
$$
\frac{e^{\frac{c}{2}x^2}}{\int e^{\frac{c}{2}x^2} dx} = \frac{c}{2}x \pm \sqrt{\frac{c^2}{4}x^2- y''}
$$
Yet when $c=4$
$$
\frac{2\sqrt{\frac{2}{\pi}}e^{2x^2}}{\operatorname{erfi}(\sqrt{2}x)+c} = 2x \pm\sqrt{4x^2 + 8e^{2x^2}\frac{e^{2x^2}-\sqrt{2\pi}x\operatorname{erfi}(\sqrt{2}x)}{\pi \operatorname{erfi}(\sqrt{2}x)^2}}
$$
Is not correct, what am i doing wrong?
|
Using the very long way, assuming that $y$ is a function of $x$,$$f(x) = \frac{\frac{d^2}{dx^2}[e^y]}{\frac{d}{dx}[e^y]}$$ finally write $$y''-y'^2-y' f(x)=0$$ which eventually allows reduction order setting $z=y'$. So, if $f(x)=c x$, the solution of $z'-z^2-z f(x)=0$ is $$z=\frac{2 \sqrt{c} e^{\frac{c x^2}{2}}}{\sqrt{2 \pi } \text{erfi}\left(\frac{\sqrt{c}
x}{\sqrt{2}}\right)+2 \sqrt{c} k_1} $$ and finally $$y=\log \left(\sqrt{2 \pi } \text{erfi}\left(\frac{\sqrt{c} x}{\sqrt{2}}\right)+2
\sqrt{c} k_1\right)+k_2$$ The solution can be checked since $$\frac{d^2}{dx^2}[e^y]=2 c^{3/2} x e^{\frac{c x^2}{2}+k_2}$$ $$\frac{d}{dx}[e^y]=2 \sqrt{c} e^{\frac{c x^2}{2}+k_2}$$ $$\frac{\frac{d^2}{dx^2}[e^y]}{\frac{d}{dx}[e^y]}=cx=f(x)$$
Using the short way (as suggested by chu), define $z=\frac{d}{dx}[e^y]$ and the equation write $$\frac{z'}{z}=cx$$ so $\log(z)=\frac c2 x^2+k_1$, that is to say $z=k_1 e^{\frac c2 x^2}$ and then the gaussian integral and the final result.
|
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|
Explaining that $1 \cdot 3 \cdot 5 \dotsm (2n+1) = 1 \cdot 3 \cdot 5 \dotsm (2n-1)(2n+1)$ I have a few students that are having trouble understanding that
$$1 \cdot 3 \cdot 5 \dotsm (2n+1) = 1 \cdot 3 \cdot 5 \dotsm (2n-1)(2n+1),$$
specifically that
$$\frac{1 \cdot 3 \cdot 5 \dotsm (2n+1)}{1 \cdot 3 \cdot 5 \dotsm (2n-1)} = 2n+1.$$
I've tried explaining it a few different ways, but it didn't seem to go over very well. What would be a good way to explain this?
|
I would suggest put some various number
for example n=3 , and force them to write numbers in two rows
then simplify
$$n=3 \to 2n+1=2(3)+1=7 \to1.3.5.7 \\2n-1=2(3)-1=5 \to 1.3.5\\ \frac{1.3.5...(2n+1)}{1.3.5...(2n-1)}=\frac{1.3.5.7}{1.3.5}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove $2^{n+1} * 2^{n+1} = (2^n*2^n)+(2^n*2^n)+(2^n*2^n)+(2^n*2^n)$ Below diagram is used as part of a proof of induction to prove that $E$ a way to tile a $2^n * 2^n$ region with square missing :
What is the proof that $2^{n+1} * 2^{n+1}$ = $(2^n*2^n)+(2^n*2^n)+(2^n*2^n)+(2^n*2^n)$ ?
|
In general, $a^m\cdot a^n=a^{m+n}$, so $a^n\cdot a^n=a^{n+n}=a^{2n}$. Thus
$$
a^n\cdot a^n+a^n\cdot a^n+a^n\cdot a^n+a^n\cdot a^n=4a^n\cdot a^n=4a^{2n}
$$
Similarly, $a^{n+1}\cdot a^{n+1}=a^{2n+2}$. Now let $a=2$; the former expression is
$$
4\cdot2^{2n}=2^2\cdot2^{2n}=2^{2n+2}
$$
|
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Find all integral solutions to $a+b+c=abc$. Find all integral solutions of the equation $a+b+c=abc$.
Is $\{a,b,c\}=\{1,2,3\}$ the only solution? I've tried by taking $a,b,c=1,2,3$.
|
The solutions with $abc=0$ obviously are of form $0,b,-b$. And since $a,b,c$ is a solution iff $-a,-b,-c$ is, we find all solutions with $abc\neq 0$ by finding all with $abc>0$. We show the only such solution is $1,2,3$.
If $abc>0$ then at least one of $a,b,c$ is $>0$ and the other two are alike in sign. But if that sign was negative then the sum would be less than the greatest of $a,b,c$ while the product would be greater, so this would not give a solution. So if $abc>0$ then $a,b,c$ are each $>0$.
If $a=b$, apply the quadratic formula to find $a$ in the equation $2a+c=a^2c$. You find $a$ is a positive integer iff $c=0$ which is a case we already know.
So we can assume $0<a<b<c$.
For fixed $0<a<b$ consider
$$f_{ab}(c) = abc - a-b-c=(ab-1)c-(a+b)$$
as a linear function of $c$. We want an integer solution $f_{ab}(c)=0$ with $c>2$. The slope of $f_{ab}$ is $ab-1$ which is $>0$, so there is at most one $c$, and indeed for $a=1,b=2$ there is the known solution $c=3$. When $a=1$ and $b>2$ the smallest value of $c$ we need consider is $c=b+1$. For that value of $c$ the function is already strictly positive and since it has positive slope there is no solution with any greater value of $c$.
When $a>1$ then again $b>2$, the smallest value of $c$ we need consider is $c=b+1$ and for that value of $c$ the function is already strictly positive and since it has positive slope there is no solution.
|
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If $\frac {1}{2+a} + \frac {1}{2+b} + \frac {1}{2+c} = 1$, prove $\sqrt{ab} + \sqrt{ac} + \sqrt{bc} \leq 3$ Let $a,b,c$ be non-negative numbers such that $$\frac {1}{2+a} + \frac {1}{2+b} + \frac {1}{2+c} = 1.$$
Prove that $ \sqrt{ab} + \sqrt{ac} + \sqrt{bc} \leq 3 $.
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The condition gives that there are $\alpha\geq0$, $\beta\geq0$ and $\gamma\geq0$ such that $\alpha+\beta+\gamma=\pi$ for which $\sqrt{ab}=2\cos\gamma$, $\sqrt{ac}=2\cos\beta$ and $\sqrt{bc}=2\cos\alpha$.
Hence, we need to prove that $\cos\alpha+\cos\beta+\cos\gamma\leq\frac{3}{2}$, which is obvious.
|
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|
Show $x^2+4x+18\equiv 0\pmod{49}$ has no solution My method was to just complete the square:
$x^2+4x+18\equiv 0\pmod{49}$
$(x+2)^2\equiv -14\pmod{49}$
$x+2\equiv \sqrt{35}\pmod{49}$
So $x\equiv\sqrt{35}-2\pmod{49}$, which has no real solutions.
I feel that this may be too elementary, is this the correct way to solve this?
|
Indeed, it is wrong. $49 \mid (x+2)^2+14$ implies $7\mid (x+2)^2$, and in particular $7\mid x+2$. But this means that $49\mid (x+2)^2$, and by difference $49$ should divide $14$ too, which is false.
|
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|
Is it possible to write Catalan's product for e as a product of a sequence? Catalan found a product for $e$:
$$e=\dfrac{2}{1}\left(\dfrac{4}{3}\right)^{\frac{1}{2}}\left(\dfrac{6\cdot 8}{5\cdot 7}\right)^{\frac{1}{4}}\left(\dfrac{10\cdot 12\cdot 14\cdot 16}{9\cdot 11\cdot 13\cdot 15}\right)^{\frac{1}{8}}\cdots$$
Is it possible to write this as an infinite product in the form:
$$e = \prod_{n=0}^\infty a_n$$
I know that it would likely be some sequence of factorials to the $\frac{1}{2^n}$ power, but I cannot think of what the sequence would be.
Thanks!
|
Look at that fourth term:
$$\left(\frac{10\cdot 12\cdot 14\cdot 16}{9\cdot 11\cdot 13\cdot 15}\right)^{1/8}$$
Factor out the $2$s in the numerator:
$$10\cdot 12\cdot 14\cdot 16=2^4\cdot5\cdot6\cdot7\cdot 8=2^4\frac{8!}{4!}$$
The denominator is $$\begin{align}9\cdot 11\cdot 13\cdot 15 &= \frac{16!}{2^8\cdot 8!\cdot (1\cdot 3\cdot 5\cdot 7)}\\
1\cdot 3\cdot 5\cdot 7&=\frac{8!}{2^4 4!}
\end{align}$$
So:
$$9\cdot 11\cdot 13\cdot 15 = \frac{16!\cdot 2^4\cdot 4!}{2^8\cdot 8!\cdot 8!}=\binom{16}{8}\frac{4!}{2^4}$$
So $$a_4^{8}=\frac{2^4\frac{8!}{4!}}{\binom{16}{8}\frac{4!}{2^4}}=\frac{2^8\binom{8}{4}}{\binom{16}{8}}$$
of $$a_4=2\left(\frac{\binom{8}{4}}{\binom{16}{8}}\right)^{1/8}$$
Thus, for $n>1$, we might guess:
$$a_n =2\left(\frac{\binom{2^{n-1}}{2^{n-2}}}{\binom{2^n}{2^{n-1}}}\right)^{1/2^{n-1}}$$
Somewhat tricky, because $n=1$ doesn't work.
If you want to expand it as factorials:
$$a_n==2\left(\frac{\left(2^{n-1}!\right)^3}{2^n!\left(2^{n-2}!\right)^2}\right)^{1/2^{n-1}}$$
If you define $$b_k=2\left(\frac{\left((2k)!\right)^3}{(4k)!(k!)^2}\right)^{\frac{1}{2k}}$$
Then $a_n=b_{2^{n-2}}$ for $n\geq2$ and $a_1=1$.
|
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|
Find all possible values of $ a^3 + b^3$ if $a^2+b^2=ab=4$. Find all possible values of $a^3 + b^3$ if $a^2+b^2=ab=4$.
From $a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)(4-4)=(a+b)0$. Then we know $a^3+b^3=0$.
If $a=b=0$, it is conflict with $a^2+b^2=ab=4$.
If $a\neq0$ and $b\neq0$, then $a$ and $b$ should be one positive and one negative. This contradict with $ab$=4.
I don't know the solution.
Any help please.
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There are no solutions in the real numbers.
Plotting $a^2+b^2=4$ on a graph, we get a circle radius 2. Plotting $ab=4$ on the same graph, we get a hyperbola which goes through $(2,2)$ (with the negative side going through $(-2,-2)$) which does not intersect the circle.
A tag of "recreational mathematics" wouldn't usually cause me to go looking for complex solutions unless they were apparent from the setup. However Lythia's answer correctly identifies that the options in the complex plane are restricted to $a$ and $b$ choosing from the $6^{th}$ roots of $-64$. Since $ab=4$, $a$ and $b$ must be complex conjugates. The $a^2+b^2=4$ requirement additionally means that $a$ and $b$ cannot be $\pm 2i$, giving the four answer pairs of $a=\pm \sqrt 3\pm i$ and $b=a^*$.
Allowing complex $a$ and $b$ rescues the question's original analysis and reinstates the answer of $a^3+b^3 = 0$.
|
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|
Solve $\int\frac{x}{\sqrt{x^2-6x}}dx$ I need to solve the following integral
$$\int\frac{x}{\sqrt{x^2-6x}}dx$$
I started by completing the square,
$$x^2-6x=(x-3)^2-9$$
Then I defined the substitution variables..
$$(x-3)^2=9\sec^2\theta$$
$$(x-3)=3\sec\theta$$
$$dx=3\sec\theta\tan\theta$$
$$\theta=arcsec(\frac{x-3}{3})$$
Here are my solving steps
$$\int\frac{x}{\sqrt{(x-3)^2-9}}dx = 3\int\frac{(3\sec\theta+3)\sec\theta\tan\theta}{\sqrt{9(\sec^2\theta-1)}}d\theta$$
$$=\int\frac{(3\sec\theta+3)\sec\theta\tan\theta}{\tan\theta}d\theta$$
$$=\int(3\sec\theta+3)\sec\theta\tan\theta$$
$$=3\int\sec^2\theta\tan\theta d\theta + 3\int\sec\theta\tan\theta d\theta$$
$$u = \sec\theta, du=\sec\theta\tan\theta d\theta$$
$$=3\int udu + 3\sec\theta$$
$$=\frac{3\sec\theta}{2}+3\sec\theta+C$$
$$=\frac{3\sec(arcsec(\frac{x-3}{3}))}{2}+3\sec(arcsec(\frac{x-3}{3}))+C$$
$$=\frac{3(\frac{x-3}{3})}{2}+3(\frac{x-3}{3})$$
$$=\frac{x-3}{2}+x-3+C$$
However, the expected answer is
$$\int\frac{x}{\sqrt{x^2-6x}}dx=\sqrt{x^2-6x}+3\ln\bigg(\frac{x-3}{3}+\frac{\sqrt{x^2-6x}}{3}\bigg)$$
What did I misunderstood?
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I'll give you an answer that is completely free from all this trigonometric nonsense, that I personally think is cleaner and easier. We wish to compute
$$\int \frac{x}{\sqrt{x^2-6x}}\,\mathrm{d}x.$$
Consider first rewriting it as
$$\int \frac{x}{\sqrt{x^2-6x}}\,\mathrm{d}x=\frac{1}{2}\int \frac{2x-6}{\sqrt{x^2-6x}}\,\mathrm{d}x+3\int \frac{\mathrm{d}x}{\sqrt{x^2-6x}}\,.$$
Now for the first integral, letting $u=x^2-6x$ we get that
$$\frac{1}{2}\int \frac{2x-6}{\sqrt{x^2-6x}}\,\mathrm{d}x=\int\frac{\mathrm{d}u}{2\sqrt{u}}=\sqrt{u}+A=\sqrt{x^2-6x}+A.$$
This takes care of the first integral. Now for the second integral, consider the Euler substitution
$$\begin{cases}
t=x+\sqrt{x^2-6x},\\
\mathrm{d}t=\frac{x+\sqrt{x^2-6x}-3}{\sqrt{x^2-6x}}\mathrm{d}x.
\end{cases}$$
This yields that
$$3\int \frac{1}{\sqrt{x^2-6x}}\,\mathrm{d}x=3\int \frac{\mathrm{d}t}{t-3}=3\ln\lvert t-3\rvert +D=3\ln\lvert x+\sqrt{x^2-6x}-3\rvert+B.$$
Combining this the answer becomes
$$\int \frac{x}{\sqrt{x^2-6x}}\,\mathrm{d}x=\sqrt{x^2-6x}+3\ln\lvert x+\sqrt{x^2-6x}-3\rvert+C.$$
|
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How to solve $\int\sqrt{1+x\sqrt{x^2+2}}dx$ I need to solve
$$\int\sqrt{1+x\sqrt{x^2+2}}dx$$
I've chosen the substitution variables
$$u=\sqrt{x^2+2}$$
$$du=\frac{x}{\sqrt{x^2+2}}$$
However, I am completly stuck at
$$\int\sqrt{1+xu} dx$$
Which let me believe I've chosen wrong substitution variables.
I've then tried letting $u=x^2+2$ or simply $u=x$, but it does not help me at all solving it.
Would someone please give me an hint on this ?
Thanks.
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$x=\sqrt{2}\tan{u},dx=\sqrt{2}\sec^2{u}du,\sqrt{1+x\sqrt{x^2+2}}dx=\sqrt{2+4\tan{u}|\sec{u}|}\sec^2{u}du=\dfrac{\sqrt{2\cos^2{u} \pm 4\sin{u}}}{|\cos{u}|\cos^2{u}}du=\pm\dfrac{\sqrt{2\cos^2{u} \pm 4\sin{u}}}{cos^4{u}}d\sin{u}=\pm\dfrac{\sqrt{2(1-v^2) \pm 4v}}{(1-v^2)^2}dv ,v=\sin{u}$
consider case "+",$2(1-v^2)+4v=2(2-(v-1)^2)=2(2-y^2),y=v-1$
$\dfrac{\sqrt{2(1-v^2) +4v}}{(1-v^2)^2}dv=\dfrac{\sqrt{4-2y^2}}{(y(y+2))^2}dy$
$\dfrac{1}{y^2(y+2)^2}=\dfrac{1}{4(y+1)}\left(\dfrac{1}{y^2}-\dfrac{1}{(y+2)^2}\right)=\dfrac{1}{4y}\left(\dfrac{1}{y}-\dfrac{1}{(y+1)}\right)-\dfrac{1}{4(y+2)}\left(\dfrac{1}{y+1}-\dfrac{1}{(y+2)}\right)$
note :
$\dfrac{\sqrt{4-2y^2}}{y^2},\dfrac{\sqrt{4-2y^2}}{y}...$can be solved,so the problem can be solved.
|
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|
$\sup$ and $\inf$ of this set These are exercises from my textbook, and I am not sure if the solutions are correct or not.
Given a set $B = \{\frac{n}{2n+1} : n \in \mathbb{N} \}$
Find the $\sup$ and $\inf$ of $B$, and maxima and minima if they exist.
The solutions give $\sup B = \frac{1}{2}$ and $\inf B = \frac{1}{3}$
I am confused by this result. The $\sup$ has to be the least upper bound, $\frac{1}{2} \in B$ and $\frac{1}{3}$ is also in $B$. So how can we claim that $\sup B = \frac{1}{2}$ when $\frac{1}{3} > \frac{1}{2}$
Is this a typo?
Additionally, shouldn't $\inf B = \frac{1}{2}$ since as $n \to \infty$, $\frac{n}{2n+1} \to \frac{1}{2}$
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First $\frac{1}{3} < \frac{1}{2}$.
Second $B$ is bounded above by $\lim \frac{n}{2n+1} = \frac{1}{2}$. Then $\sup B \leq \frac{1}{2}$. Take any $\epsilon > 0$ then there exists $N \in \mathbb N$ such that
$$n > N \implies |x_n - \frac{1}{2}| < \epsilon $$
then there exists $x_k = \frac{k}{2k + 1} \in B$, such that $$\frac{1}{2} - \epsilon < x_k \leq \epsilon$$ then $\sup B = \frac{1}{2}$.
Now $\inf B = \frac{1}{3}$ because when $n = 1$ you have $\frac{1}{2 \dot \ 1 + 1} = \frac{1}{3}$. Then $B$ is bounded below by $\frac{1}{3}$, try to show that in fact $\inf B = \frac{1}{3} $.
|
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$ \int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}dx$ How do we compute this integral ?
$$ \int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}dx$$
I have tried partial fraction but it is quite hard to factorize the denominator. Any help is appreciated.
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This is not a final answer but it is too long for a comment.
Beside the elegant solution given by Dr.MV using the residue theorem, there is something I found interesting in the problem. $$\frac{1+x^6}{1-x^2+x^4-x^6+x^8}=(1+x^2+x^6+x^8)\sum_{k=0}^{\infty}(-1)^k x^{10k}$$ So $$I=\int_{0}^1\frac{1+x^6}{1-x^2+x^4-x^6+x^8}\,dx=\sum_{k=0}^{\infty}\int_{0}^1 (-1)^k(1+x^2+x^6+x^8)x^{10k}\,dx$$ which leads to $$I=\sum_{k=0}^{\infty}(-1)^k \left(\frac{1}{10 k+1}+\frac{1}{10 k+3}+\frac{1}{10 k+7}+\frac{1}{10
k+9}\right)$$ and $I$ has a very nice and simple expression.
|
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|
Find the Derivative and simplify (write w/o negative exponents and factor if possible) y=cos^3(x^2+2) I am currently practicing on an old college calculus final. Would like to know if my work is correct.
Find the derivative for f(x) = x^1/2 + (3/x^2)-2 and y=cos^3(x^2+2) . Thanks for your help.
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You got one part right, but there are a few mistakes. For example, the function $\frac{3}{x^2}-2$ is equivalently written as $3x^{-2}-2$. By the power rule, the derivative would be $$(-2)3x^{-2-1} = -6x^{-3} = \frac{-6}{x^3}$$ Your derivative of $x^{1/2}$ is correct.
The derivative of $\cos^3(x^3+2)$ is wrong for a few reasons. First, you are missing one "layer" in your chain rule. Second, it is the cosine argument being raised to the $3$rd power, not the quantity $(x^3+2)$ inside the argument of cosine. You can think of it as $$\left[\cos(x^3+2) \right]^3 = \cos(x^3+2)\cdot \cos(x^3+2) \cdot \cos(x^3+2)$$ By the power rule, you should get $$\frac{d}{dx}\left[\cos(x^3+2) \right]^3 = 3\left[\cos(x^3+2) \right]^2 \cdot \left[\frac{d}{dx} \cos(x^3+2)\right]$$ In differentiating $\frac{d}{dx}\cos(x^3+2)$ you should get $$-\sin(x^3+2) \cdot \left[\frac{d}{dx}(x^3+2)\right] = -\sin(x^3+2)\cdot 3x^2$$ So all together, $$\frac{d}{dx}\left[\cos(x^3+2) \right]^3 = -9x^2\left[\cos(x^3+2) \right]^2\sin(x^3+2)$$
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Solution of $y''+xy=0$ The differential equation $y''+xy=0$ is given.
Find the solution of the differential equation, using the power series method.
That's what I have tried:
We are looking for a solution of the form $y(x)= \sum_{n=0}^{\infty} a_n x^n$ with radius of convergence of the power series $R>0$.
Then:
$$y'(x)= \sum_{n=1}^{\infty} n a_n x^{n-1}= \sum_{n=0}^{\infty} (n+1) a_{n+1} x^n$$
$$y''(x)= \sum_{n=1}^{\infty} (n+1) n a_{n+1} x^{n-1}= \sum_{n=0}^{\infty} (n+2) (n+1) a_{n+2} x^n$$
Thus:
$$\sum_{n=0}^{\infty} (n+2) (n+1) a_{n+2} x^n+ x \sum_{n=0}^{\infty} a_n x^n=0 \\ \Rightarrow \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n+ \sum_{n=0}^{\infty} a_n x^{n+1}=0 \\ \Rightarrow \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n+ \sum_{n=1}^{\infty} a_{n-1} x^n=0 \\ \Rightarrow 2a_2+\sum_{n=1}^{\infty} \left[ (n+2) (n+1) a_{n+2}+ a_{n-1}\right] x^n=0$$
So it has to hold:
$$a_2=0 \\ (n+2) (n+1) a_{n+2}+a_{n-1}=0, \forall n=1,2,3, \dots$$
For $n=1$: $3 \cdot 2 \cdot a_3+ a_0=0 \Rightarrow a_3=-\frac{a_0}{6}$
For $n=2$: $4 \cdot 3 \cdot a_4+a_1=0 \Rightarrow a_4=-\frac{a_1}{12}$
For $n=3$: $5 \cdot 4 \cdot a_5+a_2=0 \Rightarrow a_5=0$
For $n=4$: $6 \cdot 5 \cdot a_6+a_3=0 \Rightarrow 30 a_6-\frac{a_0}{6}=0 \Rightarrow a_6=\frac{a_0}{6 \cdot 30}=\frac{a_0}{180}$
For $n=5$: $7 \cdot 6 \cdot a_7+ a_4=0 \Rightarrow 7 \cdot 6 \cdot a_7-\frac{a_1}{12}=0 \Rightarrow a_7=\frac{a_1}{12 \cdot 42}$
Is it right so far? If so, how could we find a general formula for the coefficients $a_n$?
EDIT: Will it be as follows:
$$a_{3k+2}=0$$
$$a_{3k}=(-1)^k \frac{a_0}{(3k)!} \prod_{i=0}^{k-1} (3i+1)$$
$$a_{3k+1}=(-1)^k \frac{a_1}{(3k+1)!} \prod_{i=0}^{k-1} (3i+2)$$
If so, then do we have to write seperately the formula for the coefficients of $x^0, x^1$, because otherwhise the sum would be from $0$ to $-1$ ?
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$$y''+xy=0$$
$$y=\sum_{n=0}^{\infty} c_n x^n, y'=\sum_{n=1}^{\infty} nc_n x^{n-1}, y''=\sum_{n=2}^{\infty} n(n-1)c_n x^{n-2}$$
$$\therefore y''+xy=0=\underbrace{\sum_{n=2}^{\infty} n(n-1)c_n x^{n-2}}_{k=n-2\Rightarrow n=k+2}+\underbrace{\sum_{n=0}^{\infty} c_n x^{n+1}}_{k=n+1\Rightarrow n=k-1}=\sum_{k=0}^{\infty} (k+2)(k+1)c_{k+2} x^{k}+\sum_{k=1}^{\infty} c_{k-1} x^k$$
$$=2c_2+\sum_{k=1}^{\infty} (k+2)(k+1)c_{k+2} x^{k}+\sum_{k=1}^{\infty} c_{k-1} x^k=2c_2+\sum_{k=1}^{\infty} [(k+2)(k+1)c_{k+2}+ c_{k-1}] x^k=0$$
Thus,
$$2c_2=0\Rightarrow c_2=0$$
$$(k+2)(k+1)c_{k+2}+ c_{k-1}=0\Rightarrow c_{k+2}=-\frac{c_{k-1}}{(k+2)(k+1)}\forall k=1,2,3,\ldots$$
Choosing $c_0=1$ and $c_1=0$, we find
$$c_2=0,c_3=-1/6,c_4=0, c_5=0, c_6=1/180\ldots$$
and so on.
Choosing $c_0=0$ and $c_1=1$, we find
$$c_2=0,c_3=0,c_4=-1/12, c_5=0, c_6=0,c_7=1/504\ldots$$
and so on.
Thus, the two solutions are
$$y_1=1-\frac{1}{6}x^3+\frac{1}{180}x^6+\ldots$$
$$y_2=x-\frac{1}{12}x^4+\frac{1}{504}x^7+\dots$$
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Derivative Chain Rule $f(x) = (3x^2+2)^2 (x^2 -5x)^3$ I'm learning chain rule in derivative,
and I don't understand this in the example.
$$f(x) = (3x^2+2)^2 (x^2 -5x)^3\\$$
\begin{align}
f'(x)&= (3x^2 +2)^2[3(x^2 -5x)^2(2x-5)] + (x^2-5x)^3[2(3x^2+2)(6x)]
\\&=3(3x^2 +2)(x^2-5x)^2[(3x^2+2)(2x-5)+4x(x^2-5x)]
\\&= 3(3x^2 +2)(x^2-5x)^2[6x^3-15x^2+4x-10+4x^3-20x^2]
\\&=3(3x^2+2)(x^2-5x)^2(10x^3-35x^2+4x-10)
\end{align}
I understand the first line, because they just applied the chain rule,
but I don't understand the second line. I think they are simplifying it, but still I don't understand.
Could anyone explain what happened there? or add extra steps so it's easier??
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\begin{align}
f'(x) &= (3x^2 +2)^2[3(x^2 -5x)^2(2x-5)] + (x^2-5x)^3[2(3x+2)(3)]\\
&=3(3x^2+2)(x^2-5x)^2[(2x-5)-(x^2-5x)(2)] \\
&=3(3x^2+2)(x^2-5x)^2[2x-5-4x^2+10x] \\
&=3(3x^2+2)(x^2-5x)^2(-4x^2+12x-5)
\end{align}
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1200300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Find the Inverse Laplace Transforms Find the inverse Laplace transform of:
$$\frac{3s+5}{s(s^2+9)}$$
Workings:
$\frac{3s+5}{s(s^2+9)}$
$= \frac{3s}{s(s^2+9} + \frac{5}{s(s^2+9)}$
$ = \frac{3}{s^2+9} + \frac{5}{s}\frac{1}{s^2+9}$
$ = \sin(3t) + \frac{5}{s}\frac{1}{s^2+9}$
Now I'm not to sure on what to do.
Any help will be appreciated.
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The convolution theorem is, found here,
\begin{align}
\mathcal{L}^{-1} \{f(s)g(s)\} = \int_{0}^{t} f(t-u) g(u) \, du
\end{align}
In the case here $f(s) = 1/s$ which is the transform of $1$ and $g(s)$ being the transform of $\sin$. Now
\begin{align}
\mathcal{L}^{-1}\{ \frac{1}{s ( s^{2} + a^{2})} \} &= \frac{1}{a} \, \int_{0}^{t} (1) \, \sin(a u) \, du = - \frac{1}{a} \, \left[ \frac{\cos(au)}{a} \right]_{0}^{t} \\
&= - \frac{1}{a^{2}} ( \cos(at) - 1)
\end{align}
With all this it is seen that:
\begin{align}
\mathcal{L}^{-1}\{ \frac{3 s + 5}{ s (s^{2} + 9) } \} &= \mathcal{L}^{-1}\{ \frac{3}{ s^{2} + 9 } \} + \mathcal{L}^{-1}\{ \frac{5}{ s (s^{2} + 9) } \} \\
&= \sin(3t) + \frac{5(1 - \cos(3t))}{9}
\end{align}
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1201187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.