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Evaluate $\int_0^1 \ln^2{\left(x^4+x^2+1\right)} \, \mathrm{d}x$
Evaluate $$\int_0^1 \ln^2{\left(x^4+x^2+1\right)} \, \mathrm{d}x$$
First thing I saw is $x^4+x^2+1=(x^2+x+1)(x^2-x+1)$ so the integral is the same as:
\begin{gather*}
\int_0^1\ln^2{\left(x^2+x+1\right)} \, \mathrm{d}x
+ \int_0^1\ln^2{\left(x^2-x+1\right)} \, \mathrm{d}x \\
+ 2\int_0^1\ln{\left(x^2+x+1\right)}\ln{\left(x^2-x+1\right)} \, \mathrm{d}x
\end{gather*}
I dont know if this helps though. The last integral above seems to be the hardest, but still I don't even know how to evaluate first 2 integrals (perhaps Feynman's method)?
Source: https://tieba.baidu.com/p/4794735082
|
I would maybe try:
$$x^4+x^2+1=(x^2+\frac12)^2+\frac34$$
and so:
$$\int_0^1\ln^2(x^4+x^2+1)dx=\int_0^1\ln^2\left[(x^2+\frac12)^2+\frac{\sqrt{3}}{2}^2\right]dx$$
Now we know that:
$$\tan^2u+1=\sec^2u$$
so by letting:
$$\left[\frac{2}{\sqrt{3}}\left(x^2+\frac12\right)\right]=\tan(u)$$
$$\frac{4}{\sqrt{3}}xdx=\sec^2udu$$
we get:
$$I=\frac{\sqrt{3}}4\int\limits_{\frac\pi6}^{\frac\pi3}\ln^2\left[\frac32\tan^2u+\frac32\right]\left(\frac{\sqrt{3}\tan(u)-1}{2}\right)^{-1/2}\sec^2(u)du$$
and whilst the inside of the natural log simplifies nicely the rest is still rather ugly so I'm unsure if I can get a nice result from it but its worth a try :)
|
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How to prove that $\frac{\cos(x)-\cos(2x)}{\sin(x)+\sin(2x)} = \frac{1-\cos(x)}{\sin(x)}$ in a simpler way. EDIT: Preferably a LHS = RHS proof, where you work on one side only then yield the other side.
My way is as follows:
Prove: $\frac{\cos(x)-\cos(2x)}{\sin(x)+\sin(2x)} = \frac{1-\cos(x)}{\sin(x)}$
I use the fact that $\cos(2x)=2\cos^2(x)-1, \sin(2x)=2\sin(x)\cos(x)$
(1) LHS = $\frac{\cos(x)-2\cos^2(x)+1}{\sin(x)(1+2\cos(x))}$
(2) Thus it would suffice to simply prove that $\frac{\cos(x)-2\cos^2(x)+1}{1+2\cos(x)}=1-\cos(x)$
(3) Then I just used simple algebra by letting $u = \cos(x)$ then factorising and simplifying.
(4) Since that equals $1-\cos(x)$ then the LHS = $\frac{1-\cos(x)}{\sin(x)} = $ RHS.
Firstly, on the practice exam, we pretty much only had maximum 2-2.5 minutes to prove this, and this took me some trial and error figuring out which double angle formula to use for cos(2x).
This probably took me 5 minutes just experimenting, and on the final exam there is no way I can spend that long.
What is the better way to do this?
EDIT: I also proved it by multiplying the numerator and denominator by $1-\cos(x)$, since I saw it on the RHS. This worked a lot better, but is that a legitimate proof?
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$$\frac{\cos x-\cos2x}{\sin x+\sin 2x } = \frac{1-\cos x }{\sin x}\iff \sin x\cos x-\sin x\cos2x=\sin x-\sin x\cos x+$$
$$+\sin2x-\sin2x\cos x\iff \color{red}{\sin x\cos 2x}+\sin x+\sin2x-\color{red}{\sin2x\cos x}-2\sin x\cos x=0\iff$$
$$\color{red}{\sin(-x)}+\sin x=0$$
and we're done by the double implications all through (and assuming the first expression is well defined, of course)
Check all the cancellations are correct and check all the trigonometric identities used above.
Another way: We begin with the left side, again: assuming it is well defined
$$\frac{\cos x-\cos2x}{\sin x+\sin2x}\stackrel{\cos2x=2\cos^2x-1\\\sin2x=2\sin x\cos x}=\frac{\cos x-2\cos^2x+1}{\sin x(1+2\cos x)}\stackrel{-2t^2+t+1=-(2t+1)(t-1)}=$$
$$=\require{cancel}-\frac{\cancel{(2\cos x+1)}(\cos x-1)}{\sin x\cancel{(1+2\cos x)}}\stackrel{\cdot\frac{\cos x+1}{\cos x+1}}=-\frac{\overbrace{(\cos^2x-1)}^{=-\sin^2x}}{\sin x(\cos x+1)}=$$
$$=-\frac{(-\sin x)}{(\cos x+1)}=\frac{\sin x}{\cos x+1}$$
Finally, we show that last right side equals the right side of the original equation:
$$\frac{\sin x}{\cos x+1}\cdot\frac{\cos x-1}{\cos x-1}=\frac{\sin x(\cos x-1)}{\underbrace{\cos^2x-1}_{=-\sin^2x}}=-\frac{\cos x-1}{\sin x}=\frac{1-\cos x}{\sin x}$$
|
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Evaluate $\int_0^{\pi/2} \frac{\arctan{\left(\frac{2\sin{x}}{2\cos{x}-1}\right)}\sin{\left(\frac{x}{2}\right)}}{\sqrt{\cos{x}}} \, \mathrm{d}x$ Evaluate: $$\int_0^{\frac{\pi}{2}} \frac{\arctan{\left(\frac{2\sin{x}}{2\cos{x}-1}\right)}\sin{\left(\frac{x}{2}\right)}}{\sqrt{\cos{x}}} \, \mathrm{d}x$$
I believe there is a "nice" closed form solution but Wolfram is too weak. These arctan integrals are so tricky! I sense a substitution like $\sin{\frac{x}{2}}$ because of arctan argument and $\sqrt{\cos{x}}$ but I just cant get it. Any ideas or tips please.
Source: https://tieba.baidu.com/p/4794735082 (Exercise 3.1.22).
|
$$\int_0^\frac{\pi}{2}\arctan\left(\frac{2\sin x}{2\cos x -1}\right)\frac{\sin\left(\frac{x}{2}\right)}{\sqrt{\cos x}}dx \overset{\tan\frac{x}{2}\to x}= 2\int_0^1 \frac{x\arctan\left(\frac{4x}{1-3x^2}\right)}{\sqrt{1-x^2}(1+x^2)}dx$$
$$=2\underbrace{\int_0^1 \frac{x(\arctan x + \arctan(3x))}{\sqrt{1-x^2}(1+x^2)}dx}_{\mathcal J} - 2\pi \underbrace{\int_\frac{1}{\sqrt 3}^1 \frac{x}{\sqrt{1-x^2}(1+x^2)}dx}_{\large\frac{1}{2\sqrt2}\ln(2+\sqrt 3)}
$$
The second integral appears since for $x>\frac{1}{\sqrt 3}$ we have $x \cdot 3x>1$.
Now in order to deal with $\mathcal J$ we can differentiate under the integral sign, considering:
$$\mathcal J(a)=\int_0^1\frac{x(\arctan(a x)+\arctan(3 x))}{\sqrt{1-x^2}(1+x)}dx $$
$$\Rightarrow \mathcal J'(a) = \int_0^1 \frac{x^2}{\sqrt{1-x^2}(1+x^2)(1+a^2 x^2)}dx
\overset{\frac{1}{x}\to \sqrt{1+x^2}}=\int_0^\infty \frac{dx}{(2+x^2)(1+a^2+x^2)}$$
$$=\frac{2}{1-a^2}\int_0^\infty \left(\frac{1}{1+a^2+x^2}-\frac{1}{2+x^2}\right)dx=\frac{\pi}{2(1-a^2)}\left(\frac{1}{\sqrt{1+a^2}}-\frac{1}{\sqrt 2}\right)$$
We are looking to find $\mathcal J(1)=\mathcal J$, but $\mathcal J(-3)=0$ therefore our integral is:
$$\mathcal J=\frac{\pi}{2}\int_{-3}^1 \frac{1}{1-a^2}\left(\frac{1}{\sqrt{1+a^2}}-\frac{1}{\sqrt 2}\right)da\overset{a=\frac{1+x}{1-x}}=\frac{\pi}{4\sqrt 2}\int_0^2\frac{1}{x}\left(1-\frac{1-x}{\sqrt{1+x^2}}\right)dx$$
$$=\frac{\pi}{4\sqrt 2}\left(\ln\left(x+\sqrt{1+x^2}\right)+\ln\left(1+\sqrt{1+x^2}\right)\right)\bigg|_0^2=\frac{\pi}{\sqrt 2} \ln \varphi,\ \varphi =\frac{1+\sqrt 5}{2}$$
Therefore:
$$\boxed{\int_0^\frac{\pi}{2}\arctan\left(\frac{2\sin x}{2\cos x -1}\right)\frac{\sin\left(\frac{x}{2}\right)}{\sqrt{\cos x}}dx=\sqrt 2 \pi \ln\varphi-\frac{\pi}{\sqrt 2}\ln(2+\sqrt 3)}$$
|
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Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .
Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .
What I tried: In some step I messed up with this problem and so I think I am getting my answer wrong, so please correct me.
We have $x^2 - 3x + 2$ = $(x - 1)(x - 2)$
and I can see $(x - 1)^2 \equiv 1$ $($mod $x - 2)$ . We also have :-
$$\frac{(x - 1)^{100}}{(x - 1)(x - 2)} = \frac{(x - 1)^{99}}{(x - 2)}.$$
We have :- $(x - 1)^{98} \equiv 1$ $($mod $x - 2).$ $\rightarrow (x - 1)^{99} \equiv (x - 1)$ $($mod $x - 2)$. Now for the case of $(x - 2)^{200}$ we have :-
$$\frac{(x - 2)^{200}}{(x - 1)(x - 2)} = \frac{(x - 2)^{199}}{(x - 1)}.$$
We have :- $(x - 2) \equiv (-1)$ $($mod $x - 1)$ $\rightarrow (x - 2)^{199} \equiv (-1)$ $($mod $x - 1)$.
Adding all these up we have :- $(x - 1)^{100} + (x - 2)^{200} \equiv (x - 2)$ $($mod $x² - 3x + 2)$ .
On checking my answer with wolfram alpha , I found the remainder to be $1$, so I messed up in some step .
Can anyone help me?
|
The remainder of $(x - 1)^{100}$ divided by $(x - 1)(x - 2)$ will be $(x - 1) (2 - 1)^{99} = x - 1$. The remainder of $(x - 2)^{200}$ divided by $(x - 1)(x - 2)$ will be $(x - 2)(1 - 2)^{199} = 2 - x$ Therefore, the total remainder will be 1.
|
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Distance from plane to origin via Lagrange multipliers Let's say I have to find the least distance between origin and the plane $$x-2y-2z = 3$$
I know distance from origin to any $x-y-z$ plane is $\sqrt{x^2 + y^2 + z^2}$
so the constraint will be $$g(x, y, z) = x - 2y - 2z -3$$
however, what will my $f(x,y,z)$ be? Why is it possible to remove the sqrt and make it such that
$$
f(x,y,z) = x^2 + y^2 + z^2?
$$
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By C-S $$\sqrt{x^2+y^2+z^2}=\frac{1}{3}\sqrt{(1^2+(-2)^2+(-2)^2)(x^2+y^2+z^2)}\geq\frac{1}{3}(x-2y-2z)=1.$$
The equality occurs for $(x,y,z)||(1,-2,-2)$ and $x-2y-2z=3,$ id est, occurs,
which says that we got a minimal value.
|
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If $y= \frac1{1+x} + \frac2{1+x^2} + \frac4{1+x^4} + \dots + \frac{2^n}{1+x^{2^n}} $, then find $\frac{\mathrm dy}{\mathrm dx}$.
If $y= \dfrac1{1+x} + \dfrac2{1+x^2} + \dfrac4{1+x^4} + \dots + \dfrac{2^n}{1+x^{2^n}} $, then find $\dfrac{\mathrm dy}{\mathrm dx}$.
I am stuck up in this question. I tried taking log on both sides and generate some simplified expression but was unable to do so. Any help would be highly appreciated. thanks
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Hint:
$$\dfrac1{1+y}+\dfrac1{1-y}=\dfrac2{1-y^2}$$
Set $y=x,x^2,x^4,\cdots,x^{2^n}$ to find
$$y+\dfrac1{1+x}=\dfrac{2^n}{1-x^{2^{n+1}}}$$
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If $xyz=32$, find the minimal value of If $xyz=32;x,y,z>0$, find the minimal value of $f(x,y,z)=x^2+4xy+4y^2+2z^2$
I tried to do by $A.M.\geq M.G.$:
$\frac{x^2+4y^2+2z^2}{2}\geq\sqrt{8x^2y^2z^2}\to x^2+4y^2+2z^2\geq32$
But how can I maximaze 4xy?
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$$\dfrac{a\cdot\dfrac{x^2}a+b\cdot\dfrac{4xy}b+c\cdot\dfrac{4y^2}c+d\cdot\dfrac{2z^2}d}{a+b+c+d}$$
$$\ge\sqrt[a+b+c+d]{\left(\dfrac{x^2}a\right)^a\left(\dfrac{4xy}b\right)^b\left(\dfrac{4y^2}c\right)^c\left(\dfrac{2z^2}d\right)^d}$$
$\left(\dfrac{x^2}a\right)^a\left(\dfrac{4xy}b\right)^b\left(\dfrac{4y^2}c\right)^c\left(\dfrac{2z^2}d\right)^d=\dfrac{x^{2a+b}y^{b+2c}z^{2d}}{\cdots}$
Set $2d=b+2c=2a+b=4$
WLOG $b=2, 2a=2c=4-b=?$
|
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$(1−x^n) = (1−x)(1 + x + x^2 +\cdots + x^{n−1})$ Is this expression generally true?
$$(1−x^n) = (1−x)(1 + x + \cdots + x^{n−1})$$
The closest Identity I could find in "Mathematical Handbook of Formulas and Tables", Schaum's Outline is:
$$x^{2n+1} - y^{2n+1} = (x-y)(x^{2n} +x^{2n-1}y + x^{2n+2}y^2 + \cdots +y^{2n})$$
Which I could almost believe is the same thing if I set x=1.
$$1 - y^{2n+1} = (1-y)(1 +y + y^2 + \cdots + y^{2n})$$
let $N=2n+1$
$$1 - y^N = (1-y)(1 +y + y^2 + \cdots + y^{N-1})$$
But doesn't this restrict N to be an odd number? I was just asking because i saw this identity being used in a case where N could be an even or odd number, which made me think its generally true. I'm just not sure how to prove it.
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Here are the formulæ I learnt in med-school:
$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\dots+xy^{n-2}+y^{n-1}),$$
in other words, it is $x-y$ times the sum of all homogeneous monomials in $x$ and $y$, with total degree $n-1$.
The sum of two powers requires an odd exponent:
$$x^{2n+1}+y^{2n+1}=(x+y)(x^{2n}-x^{2n-1}y+x^{2n-2}y^2-\dots+x^2y^{2n-2}-xy^{2n-1}+y^{2n}).$$
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The coefficient of $x^{n}$ in the expansion of $(2-3 x) /(1-3 x+$ $\left.2 x^{2}\right)$ is The coefficient of $x^{n}$ in the expansion of $\frac{(2-3 x)}{(1-3 x+\left.2 x^{2}\right)}$ is
$(a) \quad(-3)^{n}-(2)^{n / 2-1}$
(b) $2^{n}+1$
$(c) 3(2)^{n / 2-1}-2(3)^{n}$
(d) None of the foregoing numbers.
Now, $1-3 x+2 x^{2}=(1-x)(1-2 x)$
Now, $(1-x)^{-1}$
$=1+x+x^{2}+x^{3}+\ldots$
Now, $(1-2 x)^{-1}$
$=1+2 x+(2 x)^{2}+(2 x)^{3}+\ldots \ldots$
Coefficient of $x^{n}$ in $(1-x)^{-1}(1-2 x)^{-1}=2^{n}+2^{n-1}+2^{n-2}+\ldots .+2+1=$
$2^{n+1}-1$
What to do next??
Any shortcut or objective approach for this type of problems would be highly appreciated.
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Your approach is kind of brute-force, but you can already get the answer using $2(2^{n+1}-1)-3(2^{(n+1)-1}-1) = (4-3)2^n-2+3=(b).$
Note that there’s a $-1$ in the exponent of the term multiplied with $-3$ because to get an $x^n$ term, when the first term is $x^1$, you need the $x^{n-1}$ term.
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Does $\lim_{n\to \infty} \sum_{k=1}^n\ln\left(1-\frac{x^2\sin^2k}{2n}\right)$ exist? Let $x \in \mathbb{R}.$ Is is true that the following limit exists : $$\lim_{n \to \infty} \sum_{k=1}^n\ln\left(1-\frac{x^2\sin^2k}{2n}\right)$$ What is the value of this limit?
I tried the Integral test for convergence, but nothing came out.
Any suggestions?
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We can expand $\ln\left(1-\frac{x^2\sin^2k}{2n}\right)$ as $$-\sum_{m=1}^{\infty} \frac{(\frac{x^2\sin^2k}{2n})^m}{m}$$ which converges for $\frac{x^2\sin^2k}{2n} \le 1 \to x^2 \le 2n \to |x| \le \sqrt{2n}$.
Then the summation in your question would become $$-\sum_{k=1}^n\sum_{m=1}^{\infty} \frac{(\frac{x^2\sin^2k}{2n})^m}{m}$$
Switching the order of summation, I get $$-\sum_{m=1}^{\infty} \frac{x^{2m}}{2^m n^{m-1} m} \frac{1}{n}\sum_{k=1}^n \sin^{2m}(k)$$
The limit $ \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \sin^{2m}(k)$ is equivalent to $$\frac{1}{\pi}\int_0^{\pi}\sin^{2m}(t)dt$$
So the sum can be rewritten as $$-\frac{n}{\pi} \int_0^{\pi} \sum_{m=1}^{\infty} \frac{\left(\frac{x^2 \sin^2(t)}{2n}\right)^m}{m} dt$$
The inside sum can be rewritten so that it ends up looking like the original $$\frac{n}{\pi} \int_0^{\pi} \ln\left( 1 - \frac{x^2 \sin^2(t)}{2n} \right) dt$$
This seems to match up with the original sum, although I feel like there is some easier way to convert the sum into this integral. We now want to find $$\lim_{n \to \infty} \frac{\int_0^{\pi} \ln\left( 1 - \frac{x^2 \sin^2(t)}{2n} \right) dt}{\frac{\pi}{n}}$$
This is a $\frac{0}{0}$ indeterminate form, so using L'Hôpital's rule, I get $$\lim_{n \to \infty} \frac{\int_0^{\pi} \frac{x^2 \sin^2(t)}{2n^2 \left( 1- \frac{x^2 \sin^2(t)}{2n} \right)} dt}{-\frac{\pi}{n^2}}$$
Simplifying, this becomes $$-\frac{x^{2}}{2\pi}\int_{0}^{\pi}\frac{\sin^{2}\left(t\right)}{1-\frac{x^{2}}{2n}\sin^{2}\left(t\right)}dt$$
As $n \to \infty$, $\frac{x^2}{n} \to 0$, so the final answer is $$-\frac{x^2}{2\pi} \int_0^{\pi} \sin^2(t) dt = -\frac{x^2}{2\pi} \frac{\pi}{2} = -\frac{x^2}{4}$$
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If $f(x)$ is differentiable for all real numbers, then what is the value of $\frac{a+b+c}{2}$?
If $f(x)=\begin {cases} a^2 + e^x & -\infty <x<0 \\ x+2 & 0\le x \le 3 \\ c -\frac{b^2}{x} & 3<x<\infty \end{cases}$, where $a,b,c$ are positive quantities. If $f(x)$ is differentiable for all real numbers, then value of $\frac{a+b+c}{2}$ is
Left hand derivative at $x=0$
$$Lf’(0) =\lim_{h\to 0} \frac{2 - (a^2 +e^{-h})}{h}$$
For limit to exist, $2-a^2=0 \implies a=\pm \sqrt 2$
$$L f’(0)=1$$
Right hand derivative at $x=0$
$$R f’(0) =\lim_{h\to 0} \frac{ h+2 -2}{h} =1$$
Left hand derivative at $x=3$
$$Lf’(3) =\lim_{h\to 0} \frac{5- (3-h+2)}{h}=1$$
And
$$Rf’(3) =\lim_{h\to 0} \frac{ c -\frac{b^2}{3+h}-5}{h}$$
For limit to exist, $c=h$
$$Rf’(3) =\lim_{h\to 0} \frac{b^2}{(3+h)(h)}=\infty$$
Where am I going wrong?
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In order to be differentiable everywhere, $f$ must be continuous. As $f$ is a piecewise continuous function:
To solve for $a$, in the last step using the fact that $a$ is positive
$$\lim_{x\rightarrow0^-}f(x)=\lim_{x\rightarrow0^+}f(x)$$
$$\lim_{x\rightarrow0^-}(a^2+e^x)=\lim_{x\rightarrow0^+}(2+x)$$
$$a^2+e^0=2+0$$
$$a^2=1$$
$$a=1$$
Similarly, we can solve for $c$ in terms of $b$:
$$\lim_{x\rightarrow3^-}f(x)=\lim_{x\rightarrow3^+}f(x)$$
$$\lim_{x\rightarrow3^-}(2+x)=\lim_{x\rightarrow3^+}(c-\frac{b^2}{x})$$
$$2+3=c-\frac{b^2}{3}$$
$$c=5+\frac{b^2}{3}$$
Now, we take into account that $f$ is differentiable at $x=3$ (and using the fact that $b$ is positive):
$$\lim_{x\rightarrow3^-}f'(x)=\lim_{x\rightarrow3^+}f'(x)$$
$$\lim_{x\rightarrow3^-}1=\lim_{x\rightarrow3^+}\frac{b^2}{x^2}$$
$$1=\frac{b^2}{9}$$
$$b=3$$
$$c=5+\frac{b^2}{3}=5+3=8$$
Therefore $\frac{a+b+c}{2}=\frac{1+3+8}{2}=6$
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In $\triangle ABC$, if $AB=x+2$, $BC=x$, $AC=x-2$, and $C = 120^\circ$, then $x=5$ I need help with this Law of cosines problem:
Prove that when $AB = x + 2,\ BC = x,\ AC = x - 2\;$ and $\;C = 120^\circ$ while $ABC$ is a triangle, then $x = 5$.
I need help to get to that answer. With law of cosines I got this
$$(x + 2)^2 = x^2 + (x-2)^2 -2x^2(x-2)\cos120^\circ$$
Then I solved the equation and I got that the solutions are $0,\ 1 + \sqrt{33}/2,\ 1 - \sqrt{33}/2$
I don't understand what I'm doing wrong.
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It was a silly mistake. Already fixed it.
$(x + 2)^2 = x^2 + (x-2)^2 -2x(x-2)\cos 120°$
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|
How can I show the quotient of the $k$th partial sums of $\sum\limits_{n=1}^{k} n$ and $\sum\limits_{n=1}^{k} n^2$ is $\frac{3}{2k+1}$? I've found using pen and paper that any trivial case of the sum of a sequence of integers from $1$ to $k$ divided by the sum of the squares of these integers is equal to $\frac{3}{2k+1}$
for example,
$$
\begin{split}
\frac{1+2+3}{1^{2}+2^{2}+3^{2}} &= \frac{3}{7}\\
\frac{1+2+3\ +\ 4}{1^{2}+2^{2}+3^{2}+4^{2}} &= \frac{3}{9}\\
\frac{1+2+3\ +\ 4\ +5}{1^{2}+2^{2}+3^{2}+4^{2}+5^{2}} &= \frac{3}{11}\\
\frac{1+2+3\ +\ 4\ +5+6}{1^{2}+2^{2}+3^{2}+4^{2}+5^{2}+6^{2}} &= \frac{3}{13}
\end{split}
$$
Using a computer I can verify this is true for very large values of $k$ and so I intuitively assume it is true for all values of $k$ but how do I prove this?
I've come up with the following equation to represent this as the quotient of two partial sums but don't know how to get from the left side of the equation to the right.
$$\frac{\displaystyle \sum_{n=1}^k n}{\displaystyle \sum_{n=1}^k n^2} = \frac{3}{2k+1}$$
|
There are explicit formulas for the partial sum (that can e.g. be shown by induction):
$$
\sum_{n=1}^k n = \frac{k(k+1)}{2}
$$
and
$$
\sum_{n=1}^k n^2 = \frac{k(k+1)(2k+1)}{6}.
$$
Take the ratio, and you get your formula.
|
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|
Method of summation for third order difference series: $2+12+36+80+150\dots$ I am unable to understand how one derived the formula for the $n$-th term $= an^3 +bn^2 + cn + d$, where the degree of the polynomial depends on the step at which we get a constant A.P. Here its at $2$nd step so degree $=2+1=3$.
But how do we derive this?
|
In this table, where $A_n$ is the $n^{th}$ term of your series,
$$
\begin{array}{c|c|c|c|c}
\style{font-family:inherit}{{n}} & \style{font-family:inherit}{A_n}
& \style{font-family:inherit}{B_n} & \style{font-family:inherit}{C_n} & \style{font-family:inherit}{D_n}\\\hline
0 & 2 & 10 & 14 &6\\\hline
1 & 12 & 24 & 20&6 \\\hline
2 & 36 & 44 & 26 \\\hline
3 & 80 & 70 & \\\hline
4 & 150
\end{array},
$$
$D_n=6=C_{n+1}-C_n; $ $C_n=B_{n+1}-B_n$; and $B_n=A_{n+1}-A_n$.
Therefore, $C_n=14+\sum\limits_{i=0}^{n-1}6=14+6n$;
$B_n=10+\sum\limits_{i=0}^{n-1}(14+6i)=10+14n+6\dfrac{(n-1)n}2=10+11n+3n^2$;
and $A_n=2+\sum\limits_{i=0}^{n-1}(10+11n+3n^2)=2+10n+11\dfrac{(n-1)n}2+3\dfrac{(n-1)n(2n-1)}6$
$=n^3+4n^2+5n+2.$
|
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|
On the golden ratio and even perfect numbers (Note: This post is an offshoot of this earlier MSE question.)
Here is my question in this post:
Is $I(2^{p-1}) - 1 > 1/I(2^{p-1})$ true when $I(2^{p-1}) = \sigma(2^{p-1})/2^{p-1}$ is the abundancy index of $2^{p-1}$ and $6 \neq 2^{p-1}(2^p - 1)$ is an even perfect number (with corresponding Mersenne prime $2^p - 1$)?
MY ATTEMPT
Claim For $p \geq 3$,
$$\frac{7}{4} \leq I(2^{p-1}) < 2.$$
Proof: Only the left-hand inequality is not evident (as $2^{p-1}$ is deficient, being a proper divisor of the perfect number $2^{p-1}(2^p - 1)$).
$$I(2^{p-1}) = \frac{\sigma(2^{p-1})}{2^{p-1}} = \frac{2^p - 1}{2^{p-1}} = 2 - \bigg(\frac{1}{2^{p-1}}\bigg).$$
But since $6 \neq 2^{p-1}(2^p - 1)$, then $p \geq 3$, which implies that
$$2^{p-1} \geq 4 \implies \frac{1}{2^{p-1}} \leq \frac{1}{4} \implies 2 - \bigg(\frac{1}{2^{p-1}}\bigg) \geq 2 - \frac{1}{4} = \frac{7}{4}.$$
QED
Checking now whether this inequality is satisfied:
$$I(2^{p-1}) - 1 > \frac{1}{I(2^{p-1})}$$
For $p \geq 3$:
$$I(2^{p-1}) - 1 \geq \frac{3}{4} > \frac{4}{7} \geq \frac{1}{I(2^{p-1})}.$$
For general $p$:
$$I(2^{p-1}) - 1 = 1 - \bigg(\frac{1}{2^{p-1}}\bigg) = \frac{2^{p-1} - 1}{2^{p-1}}$$
$$\frac{1}{I(2^{p-1})} = \frac{2^{p-1}}{2^p - 1}$$
$$\bigg(I(2^{p-1}) - 1\bigg) - \frac{1}{I(2^{p-1})} = \frac{2^{p-1} - 1}{2^{p-1}} - \frac{2^{p-1}}{2^p - 1} > 0 \text{ when } p \geq 3.$$
Note that, since $I(2^{p-1})=x$ satisfies the inequality
$$I(2^{p-1}) - 1 > \frac{1}{I(2^{p-1})} \iff x - 1 > \frac{1}{x} \implies x^2 - x - 1 > 0 \text{ since } x > 1 > 0 \implies x > \frac{1 + \sqrt{5}}{2} = \varphi \approx 1.618,$$
which is trivial compared to the inequality
$$I(2^{p-1})=x \geq \frac{7}{4} = 1.75.$$
Finally, notice that $6 = 2^{2 - 1} \cdot (2^2 - 1)$ was excluded in this analysis because it is squarefree.
|
You have written a proof for the follwing claim :
Claim For $p \geq 3$, $\frac{7}{4} \leq I(2^{p-1}).$
$$2^{p-1} \geq 4 \implies \frac{1}{2^{p-1}} \leq \frac{1}{4} \implies 2 - \bigg(\frac{1}{2^{p-1}}\bigg) \geq 2 - \frac{1}{4} = \frac{7}{4}.$$
I've found no errors here.
(One can also say "Since $I(2^{p-1})$ is increasing, we have $I(2^{p-1})\ge I(2^{3-1})=\frac 74$".)
After this, you have written
For $p \geq 3$:$$I(2^{p-1}) - 1 \geq \frac{3}{4} > \frac{4}{7} \geq \frac{1}{I(2^{p-1})}.$$
I've found no errors here.
In conclusion, I think that you have correctly proved the inequality.
|
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|
How can you approach $\int_0^{\pi/2} x\frac{\ln(\cos x)}{\sin x}dx$ Here is a new challenging problem:
Show that
$$I=\int_0^{\pi/2} x\frac{\ln(\cos x)}{\sin x}dx=2\ln(2)G-\frac{\pi}{8}\ln^2(2)-\frac{5\pi^3}{32}+4\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}$$
My attempt:
With Weierstrass substitution we have
$$I=2\int_0^1\frac{\arctan x}{x}\ln\left(\frac{1-x^2}{1+x^2}\right)dx\overset{x\to \frac{1-x}{1+x}}{=}4\int_0^1\frac{\frac{\pi}{4}-\arctan x}{1-x^2}\ln\left(\frac{2x}{1+x^2}\right)dx$$
$$=\pi\underbrace{\int_0^1\frac{1}{1-x^2}\ln\left(\frac{2x}{1+x^2}\right)dx}_{I_1}-4\underbrace{\int_0^1\frac{\arctan x}{1-x^2}\ln\left(\frac{2x}{1+x^2}\right)dx}_{I_2}$$
By setting $x\to \frac{1-x}{1+x}$ in the first integral we have
$$I_1=\frac12\int_0^1\frac{1}{x}\ln\left(\frac{1-x^2}{1+x^2}\right)dx$$
$$=\frac14\int_0^1\frac{1}{x}\ln\left(\frac{1-x}{1+x}\right)dx=\frac14\left[-\text{Li}_2(x)+\text{Li}_2(-x)\right]_0^1=-\frac38\zeta(2)$$
For the second integral, write $\frac{1}{1-x^2}=\frac{1}{2(1-x)}+\frac{1}{2(1+x)}$
$$I_2=\frac12\int_0^1\frac{\arctan x}{1-x}\ln\left(\frac{2x}{1+x^2}\right)dx+\frac12\int_0^1\frac{\arctan x}{1+x}\ln\left(\frac{2x}{1+x^2}\right)dx$$
The first integral is very similar to this one
$$\int_0^1\frac{\arctan\left(x\right)}{1-x}\,
\ln\left(\frac{2x^2}{1+x^2}\right)\,\mathrm{d}x =
-\frac{\pi}{16}\ln^{2}\left(2\right) -
\frac{11}{192}\,\pi^{3} +
2\Im\left\{%
\text{Li}_{3}\left(\frac{1 + \mathrm{i}}{2}\right)\right\}$$
So we are left with only $\int_0^1\frac{\arctan x\ln(1+x^2)}{1+x}dx$ as $\int_0^1\frac{\arctan x\ln x}{1+x}dx$ is already nicely calculated by FDP here. Any idea?
I noticed that if we use $x\to\frac{1-x}{1+x}$ in $\int_0^1\frac{\arctan x\ln(1+x^2)}{1+x}dx$ we will have a nice symmerty but still some annoying integrals appear.
In $I$, I also tried the Fourier series of $\ln(\cos x)$ but I stopped at $\int_0^{\pi/2} \frac{x\cos(2nx)}{\sin x}dx$. I would like to see different approaches if possible.
Thank you.
|
$$ \int_0^1 \frac{\arctan x \ln(1+x^2)}{1+x} dx=\frac{\pi}{16}\ln^{2}\left(2\right) -
\frac{11}{192}\,\pi^{3} +
2\Im\left\{%
\text{Li}_{3}\left(\frac{1 + \mathrm{i}}{2}\right)\right\}+{G\ln2}$$
$$\int_0^1\frac{\arctan x\ln(\frac{2x}{1+x^2})}{1-x}dx=\frac{\pi^3}{192}-\dfrac{G\ln 2}{2}$$
$$\int_0^1\frac{\arctan x\ln(\frac{2x}{1+x^2})}{1+x}dx=\frac{\pi}{16}\ln^{2}\left(2\right) +
\frac{\pi^3}{24} -
2\Im\left\{%}
\text{Li}_{3}\left(\frac{1 + \mathrm{i}}{2}\right)\right\}-\dfrac{G\ln 2}{2}$$
|
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|
Complex Matrix is Orthogonal if and only if... Let D be a 2x2 matrix with entries in the complex numbers. Prove that D is orthogonal if and only if, it is of the form:
\begin{pmatrix}
a & -b\\
b & a
\end{pmatrix}
or
\begin{pmatrix}
a & b\\
b & -a
\end{pmatrix}
Proof. I've already proved that if D is equal to those forms then, it implies that D is an orthogonal matrix. But how can I prove this? If D is orthogonal, then it must be of the form:
\begin{pmatrix}
a & -b\\
b & a
\end{pmatrix}
or
\begin{pmatrix}
a & b\\
b & -a
\end{pmatrix}
Update: a and b must satisfy that $a^2+b^2=1$
|
If $A= \begin{pmatrix}a & b \\
c & d\end{pmatrix}$
then we get from $A^TA= I$ that $a^2+c^2=1= b^2+d^2$ and also $ab+cd = 0$. Then $ab = -cd$ so squaring it gives $a^2b^2 = c^2d^2$. I'll let you finish the rest.
|
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|
For $\triangle ABC$, show that $ac\cos B+ab\cos C-bc\cos A-a^2 \le \frac{c^2}{8\cos^2(90^\circ-C)}$
Triangle $\triangle ABC$ has sides $a$, $b$, and $c$, and circumradius $R$. Prove that
$$ac \cos B + ab \cos C - bc \cos A - a^2 \le \frac{c^2}{8\cos^2(90^\circ - C)}$$
When does equality occur?
I came across this question in a different forum and I thought it was interesting. I made a bit of progress but not much: I changed $R^2$ to the fraction in the inequality. I think that there is probably another use of Law of Sines or Law of Cosines but I can't find one.
Edit: A lot of people have questions about if the problem is right; here is the original problem:
Triangle $\triangle ABC$ has sides $a$, $b$, and $c$, and circumradius $R$. Prove that
$b^2 + c^2 - a^2 \ge -R^2$
When does equality occur?
|
Answer to the second question (equality).
Triangle $ABC$ has sides $a$, $b$, and $c$, corresponding angles
$\alpha,\beta,\gamma$, semiperimeter $\rho$, inradius $r$ and
circumradius $R$. Prove that
\begin{align} R^2-a^2+b^2+c^2\ge0\tag{1}\label{1}. \end{align}
When does equality occur?
By dividing \eqref{1} by $R^2$, we have
\begin{align}
1-4\sin^2\alpha+4\sin^2\beta+4\sin^2\gamma&\ge0
\tag{2}\label{2}
.
\end{align}
It's easy to verify that \eqref{2} becomes an equality
for $\alpha=120^\circ,\beta=\gamma=30^\circ$.
In other words,
\eqref{1} becomes an equality
for an isosceles triangle with $\alpha=120^\circ$.
|
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|
Evaluate $\int e^{2x}(7+e^x)^{1/2}\,dx$ $\int e^{2x}(7+e^x)^{1/2}\,dx$
Let $u=7+e^x \rightarrow du=e^xdx$
So the integral becomes:
$\int u^{\frac{3}{2}}-7u^{\frac{1}{2}}du$ and so our answer is
$\frac{2}{5}(7+e^x)^{\frac{5}{2}}-\frac{14}{3}(7+e^x)^{\frac{3}{2}}+C$
But this is not what wolfram says. Did I make a mistake?
|
You could have made life simpler getting rid of the radical
$$u=\sqrt{7+e^x}\implies x=\log \left(u^2-7\right)\implies dx=\frac{2 u}{u^2-7}\,du$$
$$e^{2x}\sqrt{7+e^x}\,dx=2u^2(u^2-7)\,du$$
$$\int e^{2x}\sqrt{7+e^x}\,dx=2\int (u^4-7u^2)\,du=\frac25u^5-\frac 73 u^3+ C$$
Back to $x$
$$\int e^{2x}\sqrt{7+e^x}\,dx=\frac{2}{15} \sqrt{e^x+7} \left(3 e^{2 x}+7 e^x-98\right)+C$$
|
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|
How many ways are there to distribute 6 passengers into three different hotels? 6 individuals want to go to 3 different hotels such that each hotel can select zero through 6 people all states are possible. From the passenger's angle, we know that there are $3^6=729$ different ways to do this task. But from the perspective of hotels, how can solve be this problem?
We know that hotel 1 can get 6's and hotel 2,3 zero's i.e. $(6,0,0)$ OR hotel 1 5's and hotel 2 1's and hotel 3 nothing $(5,1,0)$ OR hotel 1 3's hotel 2 2's and hotel 3 1's $(3,2,1)$ OR so on. But this counting method does not yield the correct answer.
Thanks in advance for your help.
|
The number of ways to choose $a$ people for the first hotel, $b$ for the second hotel, and $c$ for the third hotel, with $a+b+c=6$, is the multinomial coefficient
$$\binom{6}{a,b,c}= \frac{6!}{a! b! c!}$$
so the total number of possible arrangements is
$$\sum_{a+b+c = 6} \binom{6}{a,b,c}$$
where the summation is over all integer triples $(a,b,c)$ with $a+b+c = 6$ and $a,b,c \ge 0$. We could work this out, but there is a shortcut.
By the multinomial theorem,
$$(x+y+z)^6 = \sum_{a+b+c = 6} \binom{6}{a,b,c} x^a y^b z^c$$
where, as before, the summation is over all integer triples $(a,b,c)$ with $a+b+c = 6$ and $a,b,c \ge 0$. Now let $x=y=z=1$, and we have
$$3^6 = \sum_{a+b+c = 6} \binom{6}{a,b,c}$$
which reproduces the previous answer of $3^6 = 729$.
|
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|
Limit using Taylor expansion : which term do we expand? I want to check the limit $\displaystyle{\lim_{n\rightarrow +\infty}\text{exp}\left (\frac{n}{2}\ln\left (\frac{n^2-2n+1}{n^2+1}\right )\right )}$ using the Taylor expansion.
I have done the following:
$$\lim_{n\rightarrow +\infty}\text{exp}\left (\frac{n}{2}\ln\left (\frac{n^2-2n+1}{n^2+1}\right )\right )=\lim_{n\rightarrow +\infty}\text{exp}\left (\ln\left (\frac{n^2-2n+1}{n^2+1}\right )^{\frac{n}{2}}\right )=\lim_{n\rightarrow +\infty}\left (\frac{n^2-2n+1}{n^2+1}\right )^{\frac{n}{2}}=\lim_{n\rightarrow +\infty}\left (\frac{(n-1)^2}{n^2+1}\right )^{\frac{n}{2}}=\lim_{n\rightarrow +\infty}\frac{(n-1)^n}{\left (n^2+1\right )^{\frac{n}{2}}}$$
For which term do we have to write the Taylor expansion?
|
We have that
$$\frac{n^2-2n+1}{n^2+1}=1-\frac{2n}{n^2+1}$$
then we can start by by first order Taylor's expansion for $\log(1+x)$ to obtain
$$\ln\left (\frac{n^2-2n+1}{n^2+1}\right )=\ln\left (1-\frac{2n}{n^2+1}\right )=-\frac{2n}{n^2+1}+O\left(\frac1{n^2}\right)$$
then
$$\text{exp}\left (\frac{n}{2}\ln\left (\frac{n^2-2n+1}{n^2+1}\right )\right )=\text{exp}\left( -\frac{n^2}{n^2+1}+O\left(\frac1{n}\right)\right)\to e^{-1}$$
then in this case a first order expansion suffices.
In general there is not a method to determine a priori which order we need to expand to but after some practice it becomes relatively easy for standard limits.
|
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|
If $f(x)=\big\lfloor x\lfloor x\rfloor\big\rfloor$ for all $x\geq 0$, then for an integer $n$, solve for $x\geq 0$ such that $f(x)=n$.
Let $f(x) = \big\lfloor x \lfloor x \rfloor \big\rfloor$ for $x \ge 0.$
(a) Find all $x \ge 0$ such that $f(x) = 1.$
(b) Find all $x \ge 0$ such that $f(x) = 3.$
(c) Find all $x \ge 0$ such that $f(x) = 5.$
(d) Find the number of possible values of $f(x)$ for $0 \le x \le 10.$
Attempt.
I've solved part (a), but I'm stuck on how to solved Part (b), (c), and (d).
The answer I got for (a) is $1 \leq x < 2$. If $0\le x<1$, then $\lfloor x\rfloor=0$, so $f(x)=0\ne 1$. If $x\ge 2$, then $\lfloor x\rfloor=2$ and $x\lfloor x\rfloor \ge 4$, so $f(x)\ge 4$. Hence, when $f(x)=1$ we must have $1\le x<2$. This means $\lfloor x\rfloor=1$ so $1\le x\lfloor x\rfloor<2$ and $f(x)=1$.
Could someone help me out with the other parts of the question? Thanks! (Also, there is a hint to divide into cases based on the value of $\lfloor x \rfloor.$
but I don't exactly understand.)
|
If $x$ were allowed to be negative this would be a real pain but if $x \ge 0$ then $[x] \ge 0$
If $[x] = n$ then $n \le x < n+1$ and $n^2 \le nx \le n^2 + n$ with the second equality holding only if $n= 0$....
So if $0 \le x < 1$ then $[x[x]] = 0$.
If $x \ge 1$ then $[x] = n \ge 1$ and $n \le x < n+1$ so $1\le n^2 \le nx = x[x] < n^2 + n$ so $n^2 \le [nx]=[x[x]] < [nx]+1 \le n^2+n$.
So if $f(x) = 0$ then $x\in[0,1)$
And if $f(x) = k$ then there is an $n\in \mathbb N$ so that $n^2 \le k < n^2 + n$ and $x \in [n,n+1)$ (but there is only one such $n$.)
But if $k$ is such that there is no such $n$ (which will happen frequently if there is an $m$ so that $m^2 + m \le k < (m+1)^2$) there will be no solutions.
So $f(x) = 1$ means that $1^2 \le f(x) < 1^2 + 1$ so $1\le x < 2$. $f(x) =3$ means $n^2 \le 3< n^2 +n$ which is impossible.
$f(x) = 5$ means that $2^2 \le f(x) < 2^2 + 2$ so $2\le x < 3$.
And $0 =0$ and that occurs if $x\in [0,1)$.
And $1 \le 1 < 2$ and so if $ x\in [1,2)$ then $f(x)=1$..
And $2^2 \le 4$ and $5 < 2^2+2$ so $f(x) =4,5$ are possible if $x\in [2,3)$. If $x < 2.5$ then $x[x]=2x < 5$ and $f(x) =4$. If $4x \geq 2.5,$ then $f(x) =5$.
$3^2 \le 9,10,11 < 3^2 + 3$ so $f(x) = 9,10,11$ are possible. if $x \in [3,4)$. If so then $[x] = 3$ and $[x[x]] = 9$ if $x < 3\frac 13$ and $f(x) =10$ if $3\frac 13\le x < 3\frac 23$ and if $x \ge 3\frac 23$ then $f(x)=11$. If $x \ge 4$ then $f(x) \ge 16$.
|
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|
Find $\sum_{r=1}^{20} (-1)^r\frac{r^2+r+1}{r!}$.
Calculate $$\sum_{r=1}^{20} (-1)^r\frac{r^2+r+1}{r!}\,.$$
I broke the sum into partial fractions and after writing 3-4 terms of the sequence I could see that it cancels but I wasn't able to arrive at the exact expression. I understand that it's trivial but for whatever reason I just couldn't get it so just need a little help.
|
$\mathcal{Hint}$
After writing $r=1$ separately, use the fact that the numerator can be written as $r(r-1)+2r+1$.
So, $$(-1)^r\frac{r^2+r+1}{r!}=(-1)^r\left(\frac{1}{(r-2)!}+\frac{2}{(r-1)!}
+\frac{1}{r!}\right)$$
Can you finish now?
$\mathbf{Edit:-}$
Writing a few terms for $r=2$ onwards, we get
$$\left(\frac{1}{0!}+\frac{2}{1!}+\frac{1}{2!}\right)-\left(\frac{1}{1!}+\frac{2}{2!}+\frac{1}{3!}\right)+\left(\frac{1}{2!}+\frac{2}{3!}+\frac{1}{4!}\right)-...+\left(\frac{1}{16!}+\frac{2}{17!}+\frac{1}{18!}\right)-\left(\frac{1}{17!}+\frac{2}{18!}+\frac{1}{19!}\right)+\left(\frac{1}{18!}+\frac{2}{19!}+\frac{1}{20!}\right)
$$
Notice how every term with denominator $2!,3!,...,18!$ completely cancel...Numerators of both $1$ and $19!$ are $2-1=1$ ...and we have the first and last terms $0!$ and $20$ as it is...
So your final answer is $\frac{1}{0!}+\frac{1}{1!}+\frac{1}{19!}+\frac{1}{20!}-3$.
|
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|
Can $\int_{0}^{2\pi} \frac{d\theta}{\sqrt{R^2+x^2-2Rx\cos\theta}},$ where $R$ and $x$ are positive constants, be solved using substitution? While I was finding the potential at a point in the plane of a uniformly charged ring, I got the following integral as the solution.
$$\int_{0}^{2\pi} \frac{d\theta}{\sqrt{R^2+x^2-2Rx\cos\theta}},$$
where $R$ and $x$ are positive constants.
Can this integral be solved using substitution?
|
$$\frac{1}{\sqrt{R^2+x^2-2 R x \cos (\theta )}}=\frac{1}{\sqrt{R^2+x^2}}\frac{1}{\sqrt{1-k\cos (\theta )}}$$ with $k=\frac{2Rx}{R^2+x^2}$.
$$\int \frac{d\theta}{\sqrt{1-k\cos (\theta )}}=\frac 2 {\sqrt{{1-k}}} F\left(\frac{\theta }{2}|\frac{2 k}{k-1}\right)$$
$$\int_0^{2\pi} \frac{d\theta}{\sqrt{1-k\cos (\theta )}}=\frac 4 {\sqrt{{1-k}}}K\left(\frac{2 k}{k-1}\right)$$
If $k$ is small, you can use the expansion
$$\frac 4 {\sqrt{{1-k}}}K\left(\frac{2 k}{k-1}\right)=2\pi \left(1+\frac{3 k^2}{16}+\frac{105 k^4}{1024}+\frac{1155 k^6}{16384} \right)+O\left(k^8\right)$$
If you want a much better approximation, you could use the Padé approximant
$$\frac 4 {\sqrt{{1-k}}}K\left(\frac{2 k}{k-1}\right)=2\pi\frac{1-\frac{497 }{576}k^2+\frac{3835}{36864}k^4 } {1-\frac{605 }{576}k^2+\frac{7315 }{36864}k^4 }$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Quadratic form reduction on the $n$-dimensional complex space $\mathbb{C}^n$ How to reduce the quadratic form
$$\sum_{1 \leq k < l \leq n}(k + il)x_kx_l$$
to the canonical form on the $n$-dimensional complex space $\mathbb{C}^n$?
I started with its symmetric matrix
\begin{align*}
S = \frac{1}{2}\begin{pmatrix}
0 & 1 + 2i & 1 + 3i & \cdots & 1 + ni \\
1 + 2i & 0 & 2 + 3i & \cdots & 2 + ni \\
1 + 3i & 2 + 3i & 0 & \cdots & 3 + n i \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 + ni & 2 + ni & 3 + ni & \cdots & 0
\end{pmatrix},
\end{align*}
and then tried to reduce it to the diagonal matrix following the algorithmic recipe, i.e., tried reducing it by eliminating the off diagonal elements row/column after row/column: the first step is to clear the first two rows/columns by calculating $P^TSP$, where
\begin{align*}
P = \begin{pmatrix}
1 & -1 & 0 & \cdots & 0 \\
1 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 1
\end{pmatrix}
\begin{pmatrix}
1 & 0 & -\frac{1 + 3i}{1 + 2i} & \cdots & -\frac{1 + ni}{1 + 2i} \\
0 & 1 & \frac{1}{2(1 + 2i)} & \cdots & \frac{1}{2(1 + 2i)} \\
0 & 0 & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & \cdots & 1
\end{pmatrix}.
\end{align*}
The resultant matrix $P^TSP := (s_{ij})$ has already quite complicated entries:
\begin{align*}
& s_{kk} = \frac{k^2 - 2 - 3ki}{1 + 2i}, \; k = 3, 4, \ldots, n; \\
& s_{kl} = \frac{k - 2l + 2kl - 4 - li}{2(1 + 2i)}, \; l > k,
\end{align*}
making the subsequent calculation notoriously nasty, so I am still inconclusive.
Is my attempt on the right track to solve this problem? I think there probably should be a better and cleaner way.
|
This might be helpful
If we define
$$
\Phi_1:=\sum_{1\leq k<l\leq n}(k+il)x_kx_l
$$
and
$$
\epsilon_{k,l}=\left\{\begin{array}{cc}
1\textrm{ , if }k<l\\
i\textrm{ , if }k>l\\
0\textrm{ , else }
\end{array}\right\}\textrm{, }i=\sqrt{-1}
$$
then
$$
\Phi_1=\sum^{n}_{k,l=1}\epsilon_{kl}kx_kx_l=\frac{1+i}{1-i}\sum_{k<l}(k+l)x_kx_l-\frac{2i}{1-i}\sum^{n}_{k<l}kx_kx_l
$$
|
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|
Proving $\sum_{cyc}\frac{(a-1)(c+1)}{1+bc+c}\geq 0$ for positive $a$, $b$, $c$ with $abc=1$. I recently saw the following inequality,
$$\frac{(a-1)(c+1)}{1+bc+c}+\frac{(b-1)(a+1)}{1+ca+a} + \frac{(c-1)(b+1)}{1+ab+b} \geq 0 \tag1$$
for all $abc=1$ and $a,b,c \in \mathbb{R}_+ \setminus \{0 \}$.
To prove this inequality, I thought to try to look at each term, by knowing that $abc=1$, and as such we could express the three terms of our inequalities in terms of three/two variables and eliminate the denominator.
First, we shall look at the expression:
$$\frac{(a-1)(c+1)}{1+bc+c} = \frac{(a-1)(\frac{1}{ab}+1)}{1+ b \frac{1}{ab}+\frac{1}{ab}} = \frac{\frac{(a-1)(ab+1)}{ab}}{1+ \frac{1}{a}+\frac{1}{ab}} = \frac{(a-1)(ab+1)}{ab(1+ \frac{1}{a}+\frac{1}{ab})} = \frac{(a-1)(ab+1)}{ab+ {b}+{1}}$$
One can see that the denominator now is the same as the denominator of the third term of our inequality. Thus, we shall also bring the denominator of the second term to the form $ab+b+1$. We have
$$\frac{(b-1)(a+1)}{1+ca+a}=\frac{(b-1)(a+1)}{1+\frac{1}{ab}a+a} = \frac{(b-1)(a+1)}{1+\frac{1}{b}+a} = \frac{(b-1)(a+1)b}{b(1+\frac{1}{b}+a)} = \frac{(b-1)(ab+b)}{b+1+ab} = \frac{(b-1)(ab+b)}{ab+b+1}$$
Now we will sum our terms now with the same denominator,
\begin{align*}
I &= \frac{(a-1)(ab+1)}{ab+ {b}+{1}} + \frac{(b-1)(ab+b)}{ab+b+1} +\frac{(c-1)(b+1)}{1+ab+b} \\ &= \frac{(a-1)(ab+1)+(b-1)(ab+b)+(c-1)(b+1)}{ab+ {b}+{1}}
\end{align*}
with
$$I=\sum_{cyc}\frac{(a-1)(c+1)}{1+bc+c} $$
Thus
$$I = \frac{a^2b+a-ab-1+ab^2+b^2-ab-b+bc+c-b-1}{ab+ {b}+{1}}$$
So
$$I = \frac{a^2b+ab^2 -2(ab+b+1)+b^2+bc+a+c}{ab+ {b}+{1}}$$
Now we must prove that
$$\frac{a^2b+ab^2 -2(ab+b+1)+b^2+bc+a+c}{ab+ {b}+{1}} \geq 0$$
Since $a,b,c$ are positive real numbers, $ab+b+1 \in \mathbb{R}_+^*$, so we now are left to prove
$$a^2b+ab^2 -2(ab+b+1)+b^2+bc+a+c \geq 0$$
$$a^2b+ab^2+b^2+bc+a+c \geq 2(ab+b+1)$$
Note. Here, AM-GM would be a good idea to use. I am currently thinking about a way to use it.
If anybody has a solution or a hint, it'd be much appreciated.
|
Now, use AM-GM:
$$ab^2+a\geq2ab,$$
$$b^2+1\geq2b$$ and $$a^2b+bc+c\geq3\sqrt[3]{a^2b^2c^2}=3.$$
After summing we'll get your last inequality.
|
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|
What is $3^{99} \pmod{100}$? I saw this post on how to solve $3^{123}\pmod{100}$ using Euler's Totient Theorem.
How about for $3^{99}\pmod{100}$? It seems more complicated because applying Euler's Totient Theorem gets us $3^{40}\equiv 1\pmod{100}$. This means $3^{80}\equiv 1 \pmod{100}$, which isn't enough, because we still need to find $3^{19}\pmod{100}$.
Now, when the terms are listed, a pattern does emerge.
$$\begin{array}{|c|c|c|c|}
\hline
3 & 9 & 27 & 81 \\ \hline
43 & 29 & 87 & 61 \\ \hline
83 & 49 & 47 & 41\\ \hline
23 & 69 & 07 & 21 \\ \hline
03 & 89 & 67 & 01 \\ \hline
\end{array}$$
And $3^{19}$ ends in $67$.
But how can I find this in other methods, besides bashing? Is there some sort of theorem that I can use?
|
We have $ 3^4 \equiv 81 \ (\text{mod} \ 100), \\ 3^5=243 \equiv 43 \ (\text{mod} \ 100),\\ 3^{10}=43^2 =1849 \equiv 49 \ (\text{mod} \ 100).$
Then $ \ 3^{20} =49^2=2401 \equiv 1 \ (\text{mod} \ 100) \Rightarrow 3^{80} \equiv 1 \ (\text{mod} \ 100).$
Also $3^9=3^4 \cdot 3^5 = 81 \cdot 43=3483 \equiv 83 \ (\text{mod} \ 100).$
Thus,$$ 3^{99}=3^9 \cdot.3^{10} \cdot 3^{80}=83 \cdot 49 \cdot 1=4067 \equiv 67 \ (\text{mod} \ 100) .$$
Hence $3^{99} \equiv {\color{blue}{67}} \ (\text{mod} \ 100)$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding intersection line question without the $z$ axis. What is the intersection line of these two planes:
$$ P_1 : 3x+4y+7z=5$$
$$P_2: 2x-5y=8$$
I will first multiply one by $4$ and the second by $5$ just to get rid of $y$:
$$15x+20y+35z -25 = 0$$
$$8x-20y-32=0$$
Adding those two, in order to get the intersection plane:
$$23x + 35z -57 = 0$$
Now we want to find $2$ points in order to get the line:
So first let's substitute a random value, say $x=1$ :
$$23 +35z -57 = 0$$
$$z = \frac{34}{35}$$
Now checking what is $y$ by checking it on our plane:
$$ 3 \cdot 1 + 4y + \frac{7 \cdot 34}{35} -5 = 0 $$
$$y = - \frac{6}{5}$$
So we have one point:
$$ (1, - \frac{6}{5}, \frac{34}{35} )$$
Now for the second point:
I substituted : $ z = 0$
$$23x = 57 \Rightarrow x = \frac{57}{23}$$
And to get $y$ I subs. it in the inters. plane:
$$3 \cdot \frac{57}{23} + 4y - 5 =0 $$
$$y = 5 - 3 \cdot \frac{53}{23} = - \frac{14}{23}$$
And so we have our second point:
$$(\frac{57}{23} , - \frac{14}{23}, 0)$$
We will now take the difference to get the direction vector:
$$ ( \frac{57}{23} - 1 , - \frac{14}{23} + \frac{6}{5} , - \frac{34}{35}) = (\frac{34}{23}, \frac{68}{115}, - \frac{34}{35})$$
To get that our intersection line is (starting at the first or second point we found) and the "direction" vector :
$$(1 , - \frac{6}{5}, \frac{34}{35}) + t ( \frac{34}{23}, \frac{68}{115}, -\frac{34}{35})$$
This is not the right answer as GeoGebra says:
This is clearly not the intersecting line...
I would appreciate your help, I really checked myself dozens of times... thanks!
|
$P_1 : 3x+4y+7z=5$
$P_2: 2x-5y=8$
You can either find the cross-product of the normal vector of the two planes and a point on the intersection to find the equation of the line or you can do as below -
For $z = t$,
$3x+4y=5-7t$ and $2x-5y=8$
Equating the two, you get -
$x = \dfrac {57-35t}{23}, y = \dfrac{-14-14t}{23}, z = t$
The parameterized equation of the line will be,
$(x,y,z) = (\dfrac{57}{23},-\dfrac {14}{23}, 0)+(-35,-14,23)t$
If you do the cross product of the normal vectors of the planes, again you will get the direction vector of the line as ($-35,-14,23)$. You can find a point on the line assuming $z = 0$. You get the same equation.
|
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|
Find $\operatorname{Var}(Y)$ when $f_{X}(x)=3x^{2}I_{\{0Let $X$ be a random variable with the following pdf:
$$
f_{X}(x)=3x^{2}I_{\{0<x<1\}}
$$
let $Y$ be a random variable so $\left(Y\mid X=x\right)\sim \operatorname{Uni}[96x,100x]$. Calculate $\operatorname{Var}(Y)$.
What I did: If $(Y\mid X=xt)\sim \operatorname{Uni}[96x,100x]$ then $f_{Y\mid X}(y\mid x)=\frac{1}{4x}$ and so:
$$
f_{X,Y}(x,y)=f_{Y\mid X}(y\mid x)\cdot f_{X}(x)=\frac{1}{4x}\cdot3x^{2}I_{\{0<x<1\}}=\frac{3}{4}xI_{\{0<x<1\}}
$$
Let's find $f_{Y}(y)$:
$$
f_{Y}(y)=\int_{-\infty}^{\infty}f_{X,Y}(x,y)\,dx=\int_{-\infty}^{\infty}\frac{3}{4}xI_{\{0<x<1\}}dx=\frac{3}{4}\int_{0}^{1}x\,dx=\frac{3}{8}
$$
Let's find $E\left(Y\right)$:
$$
E\left(Y\right)=\int_{-\infty}^{\infty}y\cdot f_{Y}(y)\,dy=\frac{3}{8}\int_{0}^{1}y\,dy=\frac{3}{16}
$$
Lets find $E\left(Y^2\right)$:
$$
E\left(Y^{2}\right)=\int_{-\infty}^{\infty}y^{2}\cdot f_{Y}(y)\,dy=\frac{3}{8}\int_{0}^{1}y^{2}\,dy=\frac{1}{8}
$$
Then we get:
$$
\operatorname{Var}(Y)=\frac{1}{8}-\left(\frac{3}{16}\right)^2=\frac{23}{256}
$$
I'm not even close to the answer $360.95$. Some questions:
*
*When should I write the indicator $I$? I'm a bit confused about what does it represent.
*Where is my mistake? I guess that the integral limits that were used in calculating $E(Y)$ and $E(Y^2)$ ($0$ to $1$) are not correct. What should they be?
|
The entire calculation is unnecessary. Rather, apply the law of total variance.
Note for all positive integers $k$, $$\operatorname{E}[X^k] = \int_{x=0}^1 3x^{k+2} \, dx = \frac{3}{k+3}.$$ Then $$\begin{align}
\operatorname{Var}[Y] &= \operatorname{Var}[\operatorname{E}[Y \mid X]] + \operatorname{E}[\operatorname{Var}[Y \mid X]] \\
&= \operatorname{Var}\left[\frac{96X + 100X}{2}\right] + \operatorname{E}\left[\frac{(100X - 96X)^2}{12}\right] \\
&= \operatorname{Var}[98X] + \operatorname{E}[\tfrac{4}{3}X^2] \\
&= 98^2 \operatorname{Var}[X] + \frac{4}{3}\operatorname{E}[X^2] \\
&= (98^2 + \tfrac{4}{3}) \operatorname{E}[X^2] - 98^2 \operatorname{E}[X]^2 \\
&= \left(98^2 + \frac{4}{3}\right) \frac{3}{5} - 98^2 \frac{9}{16} \\
&= \frac{7219}{20}.
\end{align}$$
If you must do it the hard way, we observe that the full joint density is $$f_{X,Y}(x,y) = \frac{3}{4}x \mathbb 1(0 \le x \le 1) \mathbb 1(96x \le y \le 100x),$$ since you cannot ignore the fact that $Y$ given $X$ is supported on an interval that is a function of $X$. Thus the support is a triangle; the unconditional variance of $Y$ is then computed directly as
$$\begin{align}\operatorname{E}[Y^k]
&= \int_{x=0}^1 \int_{y=96x}^{100x} y^k \frac{3}{4}x \, dy \, dx \\
&= \int_{x=0}^1 \frac{3}{4} x \left[\frac{y^{k+1}}{k+1}\right]_{y=96x}^{100x} \, dx \\
&= \frac{3(4^k)(25^{k+1} - 24^{k+1})}{k+1} \int_{x=0}^1 x^{k+2} \, dx \\
&= \frac{3(4^k)(25^{k+1} - 24^{k+1})}{(k+1)(k+3)}.
\end{align}$$
Substituting $k = 1$ and $k = 2$ gives
$$\operatorname{Var}[Y] = \frac{28816}{5} - \left(\frac{147}{2}\right)^2 = \frac{7219}{20}.$$
|
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|
Maximum of $(\sin{x} + \sqrt3)(\cos{x}+1)$ The problem is
Maximum of $(\sin{x} + \sqrt3) (\cos{x}+1)$
I tried to set $\sin(t) = y$, $\cos(t) = x$, so that the problem turned out to be finding
Max $(y + \sqrt3)(x + 1)$
s.t. $x^2 + y^2 =1 $
I tried Lagrange multipliers, but it turns out to be a quartic equation.
I also tried a linear transformation such that the problem becomes
Max $x^2 - y^2$
s.t. $(x + \frac{\sqrt3+1}{2})^2 + (y - \frac{\sqrt3-1}{2})^2 = 1$
Since $x^2-y^2$ is saddle surface, the answer is still not trivial.
Can you help me find the way out?
|
We can proceed by tangent half angle identities to obtain
$$f(t)=\left(\frac{2t}{1+t^2} + \sqrt3\right) \left(\frac{1-t^2}{1+t^2} + 1\right) \implies f'(t)=-4\frac{\sqrt 3t^3+3t^2+\sqrt 3t-1}{(1+t^2)^3}$$
and since
$$g(t)=\sqrt 3t^3+3t^2+\sqrt 3t-1 \implies g'(t)=3\sqrt 3t^2+6t+\sqrt 3=3\sqrt 3\left(t+\frac{\sqrt 3}{3}\right)^2\ge 0$$
we have that $f'(t)$ has only one root which corresponds to the maximum for $f(t)$.
Now since
$$g'(t)=3\sqrt 3\left(t+\frac{\sqrt 3}{3}\right)^2 \implies g(t)=\sqrt 3\left(t+\frac{\sqrt 3}{3}\right)^3+c$$
expanding we find $c=-\frac43$ therefore the root is given by
$$\sqrt 3\left(t+\frac{\sqrt 3}{3}\right)^3=\frac43 \implies t\approx 0.339$$
|
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|
Solve $I^{\prime}(a)=\int_{0}^{\pi} \frac{2 a-2 \cos x}{1-2 a \cos x+a^{2}} d x$ I tried to compute $I^{\prime}(a)=\int_{0}^{\pi} \frac{2 a-2 \cos x}{1-2 a \cos x+a^{2}} d x$
\begin{equation}\begin{aligned}
&\int_{0}^{\pi} \frac{2 a-2 \cos x}{1-2 a \cos x+a^{2}} d x&\\
&=\int_{0}^{\pi} \frac{2(a-1)+4 \sin ^{2} \frac{x}{2}}{(a-1)^{2}+4 a \cdot \sin ^{2} \frac{x}{2}} d x\\
&=2 \int_{0}^{\frac{\pi}{2}} \frac{2(a-1)+4 \sin ^{2} x}{(a-1)^{2}+4 \operatorname{asin}^{2} x} d x\\
&=\frac{2}{a} \int_{0}^{\frac{\pi}{2}}\left(1+\frac{2 a(a-1)-(a-1)^{2}}{(a-1)^{2}+4 a \sin ^{2} x}\right) d x\\
&=\frac{\pi}{a}+\frac{2}{a} \int_{0}^{\frac{\pi}{2}} \frac{a^{2}-1}{(a-1)^{2}+4 a \sin ^{2} x} d x\\
&=\frac{\pi}{a}+\frac{2}{a} \int_{0}^{\frac{\pi}{2}} \frac{\frac{a+1}{a-1} \sec ^{2} x}{1+\left(\frac{a+1}{a-1}\right)^{2} \tan ^{2} x} d x\\
&=\frac{\pi}{a}+\frac{2}{a} \int_{0}^{\infty} \frac{1}{1+u^{2}}d u\\
&=\frac{2 \pi}{a}
\end{aligned}\end{equation}
But I can't find my mistake.
|
This result $I=2\pi/a$ is true only if $a>1$, for a<1 $I=0$.
When $a<1$, $t=\frac{a+1}{a-1} \tan x \implies t=0 (x=0) ~\text{but}~ t=-\infty (x=\pi/2)$. Then
$$I=\left(1+\frac{|a-1|}{a-1}\right)\frac{\pi}{a}$$
When $a=1$ the orininal integral is $\pi$.
|
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|
How to evaluate this integral $\int^{\infty}_{-\infty} \frac{t^{2}}{2s^{2}} e^{-(t+ \frac{t^{2}}{2s^{2}} )} dt$ If the question was $\int^{\infty}_{-\infty} e^{-(t+ \frac{t^{2}}{2s^{2}} )} dt$, I'd have gone with completing the square of $(t+ \frac{t^{2}}{2s^{2}} )$ and using the normal density trick, but I am stumped with the additional factor. Integration by parts doesn't work either. Please help.
|
Note
\begin{align}
I &= \int^{\infty}_{-\infty} \frac{t^{2}}{2s^{2}} e^{-(t+ \frac{t^{2}}{2s^{2}} )} dt
= \int^{\infty}_{-\infty} \frac{t^{2}}{2s^{2}} e^{-\frac1{2s^2}(t+ s^2 )^2 +\frac12s^2}dt\\
& \overset{x=t+s^2}=\frac{e^{\frac{s^2}{2}}}{2s^2} \int^{\infty}_{-\infty} (x-s^2)^2 e^{-\frac{x^2}{2s^2}} dx
= \frac{e^{\frac{s^2}{2}}}{2s^2} \int^{\infty}_{-\infty} (x^2+s^4)e^{-\frac{x^2}{2s^2}} dx
\end{align}
Let $a=\frac1{2s^2}$ and use
$$J(a)=\int^{\infty}_{-\infty} e^{-ax^2}dx= \sqrt{\frac\pi a},\>\>\>\int^{\infty}_{-\infty} x^2e^{-ax^2}dx=-J’(a) = \frac{\sqrt\pi}{2a^{3/2}}$$
to obtain
$$I= \frac{e^{\frac{s^2}{2}}}{2s^2}
\left( \sqrt{2\pi}s^3 + \sqrt{2\pi} s^5\right)=\sqrt{\frac\pi2} e^{\frac{s^2}{2}}s(1+s^2)
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
Find the $26^{th}$ digit of a $50$ digit number divisible by $13$. $N$ is a $50$ digit number (in the decimal scale). All digits except the $26^{th}$ digit (from the left)
are $1$. If $N$ is divisible by $13$, find the $26^{th}$ digit.
This question was asked in RMO $1990$ and is very similar to this question and the same as this question but it is not solved by the approach used by me whereas I want to verify my approach.
My approach:
Suppose $N=111\cdots a\cdots111$ and $N\equiv 0\pmod {13}$
Now $N=10^{49}+10^{48}+\ldots+a10^{24}+\ldots+10+1=(10^{49}+10^{48}\ldots+10+1)+(a-1)10^{24}$
$N=\dfrac{10^{50}-1}{9}+(a-1)10^{24}$
Now $10^{12}\equiv 1\pmod {13}\Rightarrow 10^{24}\equiv 1\pmod {13}$ by fermat's little theorem.
Thus $(a-1)10^{24}\equiv (a-1) \pmod{13}\Rightarrow \dfrac{10^{50}-1}{9}\equiv 1-a\pmod{13}$ since $N\equiv 0\pmod{13}$
$10^{24}\equiv 1\pmod{13}\Rightarrow 10^{48}\equiv 1\pmod{13}$ or $10^{50}-1\equiv -5 \pmod{13}$
Now $10^{50}-1\equiv -5\pmod {13}\Rightarrow 9(1-a)\equiv -5\pmod{13}$
$a=3$ clearly satisfies the above conditions
$\therefore$ The $26^{th}$ digit from the left must be $3$.
Please suggest what is incorrect in this solution and advice for alternative solutions.
THANKS
|
After your editions, your approach is correct. Here's an alternative one:
The number $N$ consists of $24$ ones followed by the two digits $1a$ (the $2$-digit number $10+a$) followed by another $24$ ones, so with the number $M$ consisting of $24$ ones, $M:=\sum_{k=0}^{23}10^k=\frac{10^{24}-1}{9}$, we have
$$N=M\cdot10^{24+2}+(10+a)\cdot10^{24}+M$$
Since $13$ is a prime, from Fermat's Little Theorem we know that $10^{12}\equiv1\pmod{13}$, and it follows that $13\mid(10^{12}-1)(10^{12}+1)=10^{24}-1=9M \Rightarrow 13\mid9
\lor 13\mid M$. Obviously, $13\nmid 9$, so $13\mid M$.
Now, if $13\mid N$, it follows that $13\mid (10+a)\cdot10^{24}$, and since $13\nmid10^{24}$, it must be $13\mid10+a$. Since $0\le a\le9$, it must be $a=3$.
|
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|
Find the power series of $\frac{3x+4}{x+1}$ around $x=1$. I'm trying to find the power series of
$$
\frac{3x+4}{x+1}
$$
around $x=1$.
My idea was to use the equation
$$
\left(\sum_{n\ge0}a_n (x-x_0)^n\right)\left(\sum_{k\ge0}b_k (x-x_0)^k\right) = \sum_{n\ge0}\left(\sum_{k=0}^n a_k b_{n-k} \right)(x-x_0)^n \tag{1}
$$
taken from here, to solve this problem.
My attempt
I can write the numerator of the expression as
$$
3x +4 = 3(x-1)+7 = \sum_{n\ge0}a_n (x-1)^n
$$
where
$$
a_n=\begin{cases}
7 & n=0 \\
3 & n=1 \\
0 & n>1
\end{cases}
$$
On the other hand, given $|x-1| <2$ we get
\begin{align*}
\frac{1}{x+1} = \frac{1}{2} \cdot \frac{1}{1-\left(\frac{1-x}{2}\right)} = \frac{1}{2} \sum_{k\ge0} \left(\frac{1-x}{2}\right)^n = \sum_{k\ge0}\underbrace{\frac{-1}{(-2)^{k+1}}}_{\color{blue}{b_k}} (x-1)^k
\end{align*}
And then, using equation $(1)$ we get
\begin{align*}
\frac{3x+4}{x+1} &= \left(\sum_{n\ge0}a_n (x-1)^n\right)\left(\sum_{k\ge0}b_k (x-1)^k\right)\\
&= \sum_{n\ge0}\left(\sum_{k=0}^n a_k b_{n-k} \right)(x-1)^n\\
&= \sum_{n\ge0}\left(a_0 b_n + a_1 b_{n-1} \right)(x-1)^n\\
&= \sum_{n\ge0}\left(7 \frac{-1}{(-2)^{n+1}} + 3 \frac{-1}{(-2)^{n}} \right)(x-1)^n\\
&= \sum_{n\ge0}\left( \frac{-7 +3(-1)(-2)}{(-2)^{n+1}} \right)(x-1)^n\\
&= \sum_{n\ge0} \frac{-1}{(-2)^{n+1}} (x-1)^n\\
\end{align*}
And this seems to imply that $\frac{3x+4}{x+1} = \frac{1}{x+1}$, at least for $|x-1| <2$, which is clearly not true.
I've gone over the steps, but I don't see where my mistake is. Could anyone tell me where my solution went wrong? Thank you!
|
You know that all power series that converge to the function $f$ agree. So you can just write for $u = x - 1$:
$\begin{align*}
\frac{3 x + 4}{x + 1}
&= \frac{3 u + 7}{u + 2} \\
&= 3 + \frac{2}{1 + u / 2} \\
&= 3 + 2 \sum_{n \ge 0} (-1)^n \left( \frac{u}{2} \right)^n \\
&= 3 + \sum_{n \ge 0} \frac{(-1)^n}{2^{n - 1}} (x - 1)^n
\end{align*}$
|
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|
Sums of powers of two.. with one restriction. For a positive integer $n,$ let $a_n$ denote the number of ways of representing $n$ as a sum of powers of 2, where each power of 2 appears at most three times, and the order of the terms does not matter. For example, $a_8 = 5$ because of the representations
\begin{align*}
8 &= 8 \\
&= 4 + 4 \\
&= 4 + 2 + 2 \\
&= 4 + 2 + 1 + 1 \\
&= 2 + 2 + 2 + 1 + 1.
\end{align*}(The representation $1 + 2 + 1 + 4$ is the same as $4 + 2 + 1 + 1.$) Compute $a_{1000}.$
|
Let's prove @Oldnboy's conjecture with a hint by @JyrkiLahtonen to a technique the OP might not know well. Restate $n=\sum_jc_j2^j$, with $0\le c_j\le3$, as $x^n=\prod_jx^{c_j2^j}$, so $a_n$ is the $x^n$ coefficient in$$\begin{align}\prod_{j\ge0}(1+x^{2^j}+x^{2\cdot2^j}+x^{3\cdot2^j})&=\prod_j\frac{1-x^{4\cdot2^j}}{1-x^{2^j}}\\&=\frac{1}{(1-x)(1-x^2)}\\&=(1+x+x^2+\cdots)(1+x^2+x^4+\cdots).\end{align}$$This is just the number of integers from $0$ to $n$ inclusive with $n$'s parity, i.e. $\lfloor\tfrac{n}{2}\rfloor+1$ as claimed.
We can use this technique if $3$ is replaced with another Mersenne number. Otherwise, it's a bit tricky because the product doesn't telescope. I've asked a question about further generalization.
|
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|
Find all real solutions $x$ for the equation $x^{1/2} − (2−2x)^{1/2} = 1$ This is what the answer says:
Note that the equation can be rewritten as $\sqrt{x} − \sqrt{2 − 2x} = 1$,
and the existence of such real $x$ implies that $x$ is larger than or equal to $0$ and $x$ is less than or equal to $1$, since we implicitly assume that $\sqrt{x}$ and $\sqrt{2 − 2x}$ are real as well. Then
$\sqrt{x} − \sqrt{2 − 2x} < \sqrt{1} = 1 \text{ if } x < 1$, and
$\sqrt{x} − \sqrt{2 − 2x} = 1 \text{ if } x = 1$.
Thus, $x = 1$ is the only real solutions of the equation $\sqrt{x} − \sqrt{2 − 2x} = 1$.
I don't understand how
$\sqrt{x} − \sqrt{2 − 2x} < \sqrt{1} = 1 \text{ if } x < 1$.
I understand everything else.
Can someone please explain? Thank you.
|
Observe that
$$\sqrt{x} - \sqrt{2-2x} = 1 \implies x = 1 + 2-2x + 2\sqrt{2-2x} \implies 3x - 3 = \sqrt{2-2x}$$ $$ \implies 9(x^2 - 2x + 1) = 2-2x \implies 9x^2 - 16x +7 = 0 \implies (x -1)(9x - 7) = 0 $$
Thus $ x = 1, \frac{7}{9} $. Now, $$ \sqrt{\frac{7}{9}}-\sqrt{2-\frac{14}{9}} \neq 1 $$
Thus $x=1 $ is the solution.
|
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|
Proving $\sum_{i=1}^n (1-\frac{1}{(i+1)^2}) = \frac{n+2}{2n+2}$ using induction. My textbook has the following question:
Prove the follwing statement using induction for all natural numbers $n$
$$(1- \frac{1}{4})+(1- \frac{1}{9})+.......+(1- \frac{1}{(n+1)^2})=\frac{n+2}{2n+2}$$
So, I check both the sides for $n=1$. In that case LHS=RHS=$\frac{3}{4}$.
Now I assume the statement to be true for $n=k$ which gives
$$
\sum_{i=1}^k (1-\frac{1}{(i+1)^2}) = \frac{k+2}{2k+2}
$$
Now I evaluate the original statement for $n=k+1$ which leaves us with the
$$ LHS= \frac{3k^3+16k^2+26k+14}{2(k+1)(k+2)^2} $$
And we have to prove this LHS to be equal to RHS which is
$$
RHS= \frac{k+3}{2k+4}
$$
But these (new) LHS and RHS don't seem to be equal. And hence I am not able to complete the proof.
How should I proceed?
A solution without induction is also welcome.
Book: Comprehensive Algebra VOL-1
Author: Vinay Kumar
Publisher: McGraw Hill Education.
|
For $n=2$ we have
$$\sum_{i=1}^{2}(1-\frac{1}{(i+1)^{2}})=(1-\frac{1}{4})+(1-\frac{1}{9})=\frac{59}{36}$$ but $$\frac{n+2}{2n+2}\big|_{n=2}=\frac{2+2}{2(2)+2}=\frac{2}{3}\neq\frac{59}{36}.$$
|
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|
If $A,B,C$ are collinear, prove $\vec{A}\times \vec{B} + \vec{B}\times \vec{C} + \vec{C}\times \vec{A} =\begin{pmatrix} 0 \\ 0\\ 0 \end{pmatrix}$ Prove that if $A, B$ and $C$ are collinear, then
$\overrightarrow{A}\times \overrightarrow{B} + \overrightarrow{B}\times \overrightarrow{C} + \overrightarrow{C}\times \overrightarrow{A} =\begin{pmatrix} 0 \\ 0\\ 0 \end{pmatrix}$.
So far, I know that $\vec{AB}, \vec{BC}, \vec{AC}$ are scalar multiples and their cross product is zero. I'm not sure how to apply this to prove $\overrightarrow{A}\times \overrightarrow{B} + \overrightarrow{B}\times \overrightarrow{C} + \overrightarrow{C}\times \overrightarrow{A}$ though.
|
Setting up some vector identities:
$$
\overrightarrow{AB} = \overrightarrow{B} -\overrightarrow{A} \\
\overrightarrow{A} \times \overrightarrow{B} = - \overrightarrow{B} \times \overrightarrow{A} \\
\overrightarrow{A} \times \overrightarrow{A} = \overrightarrow{0}
$$
Proof:
$$
\begin{align}
&\overrightarrow{A}\times \overrightarrow{B} + \overrightarrow{B}\times \overrightarrow{C} + \overrightarrow{C}\times \overrightarrow{A} \\
&= \overrightarrow{A} \times \overrightarrow{B} - \overrightarrow{A} \times \overrightarrow{C} + \overrightarrow{B} \times \overrightarrow{C} \\
&= \overrightarrow{A} \times (\overrightarrow{B} - \overrightarrow{C}) + \overrightarrow{B} \times \overrightarrow{C} \\
&= (\overrightarrow{B} - \overrightarrow{AB}) \times \overrightarrow{CB} + \overrightarrow{B} \times \overrightarrow{C} \\
&= \overrightarrow{B} \times \overrightarrow{CB} - \overrightarrow{AB} \times \overrightarrow{CB} + \overrightarrow{B} \times \overrightarrow{C} \\
&= \overrightarrow{B} \times (\overrightarrow{CB} + \overrightarrow{C}) - \overrightarrow{0} \\
&= \overrightarrow{B} \times (\overrightarrow{B} - \overrightarrow{C} + \overrightarrow{C}) \\
&= \overrightarrow{0}
\end{align}
$$
|
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|
Solving $z^4=(2+3i)^4$ To solve the equation, I calculated right side:
$z^4=(2+3i)^4=(-5+12i)^2=-119-120 i$
And then I get the correct answer:
$z_k=\underbrace{\sqrt[8]{119^2+120^2}}_{\sqrt{13}} \times Cis(\cfrac{\pi+\tan^{-1}(\frac{120}{119})}{4}+\cfrac{k \pi}{2}), k=0,1,2,3$
But, I am looking for a way to solve the equation $z^4=(2+3i)^4$ without expanding the right side. So I tried :
$z={ \left| r \right| }e^{i \theta}$
$r^4e^{4 \theta i}=(\sqrt{13} e^{(2k\pi+\tan ^{-1}(\frac{3}{2}))i})^4$
$r=\sqrt{13}$
$4\theta=4 \times {(2k\pi+\tan ^{-1}(\frac{3}{2}))}$
$\theta=2k\pi+\tan ^{-1}(\frac{3}{2})$
But I calculated the value of $\theta$ wrongly. How can I fix it?
|
If $z^4=(2+3i)^4$ then $Z^4 = 1$ where $Z = \frac{z}{2+3i}$.
Hence the solutions set is
$$\{(2+3i), -(2+3i), i(2+3i), -i(2+3i)\}=\\ \{\sqrt{13} e^{i \phi},\sqrt{13} e^{i (\phi + \pi)},\sqrt{13} e^{i (\phi + \pi/2)},\sqrt{13}e^{i (\phi - \pi/2)}\}$$
where $\phi$ is such that $\cos \phi = \frac{2}{\sqrt{13}}, \sin \phi =\frac{3}{\sqrt{13}}$.
|
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|
To Prove $\frac{1}{b}+\frac{1}{c}+\frac{1}{a} > \sqrt{a}+\sqrt{b}+\sqrt{c}$ The three sides of a triangle are $a,b,c$, the area of the triangle is $0.25$, the radius of the circumcircle is $1$.
Prove that $1/b+1/c+1/a > \sqrt{a}+\sqrt{b}+\sqrt{c}$.
what I've tried:
$$\frac{1}{4} = \frac{1}{2}ab\sin C \Rightarrow ab=\frac{1}{2}\sin C \\c=2\sin C \Rightarrow \frac{1}{c}=\frac{1}{2}*\sin C $$
so, $$\frac{1}{c}=ab \Rightarrow abc=1 \Rightarrow \sqrt{abc}=1$$
now the problem becomes
$$ab+bc+ac > \frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ac}},$$
with $0<a\leq b\leq c\leq 2$, and $a+b>c$.
But even so I don't know how to prove it.
Any help or hint is appreciated. Thank you.:)
|
By your work $abc=1$.
Let $a=x^2$, $b=y^2$ and $c=z^2$, where $x$, $y$ and $z$ are positives.
Thus, $$xyz=1$$ and we need to prove that:
$$\sum_{cyc}\frac{1}{x^2}>x+y+z$$ or
$$\sum_{cyc}(x^2y^2-x^2yz)>0$$ or
$$\sum_{cyc}z^2(x-y)^2>0,$$
which is true because $x=y=z$ is impossible.
Indeed, let $x=y=z$.
Thus, $a=b=c=1$ and $R=\frac{1}{\sqrt3},$ which is a contradiction.
|
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|
Simplify $(1+\sqrt{3}) \cdot \sqrt{2-\sqrt{3}}$ Can someone help me simplify $(1+\sqrt{3})\times\sqrt{2-\sqrt{3}}$?
The end result is $\sqrt{2}$, however, I honestly do not know how to get there using my current skills.
I asked a teacher/tutor and he proposed setting the expression equal to X and working backwards, squaring both sides so that:
$$X^2 = (1 + \sqrt{3})^2 \cdot (2 - \sqrt{3}) =(4+2\sqrt{3})(2-\sqrt{3}) =8-2\sqrt{3}^2 = 2 \require{cancel}$$
$$\Rightarrow X = \sqrt{2}$$
My main question is:
*
*What are the steps to simplify this without setting equal to X?
*I tried watching a youtube video, but had no success - is difference of squares applicable here?
Thanks!
|
This is what the OP means by the 'setting equal to X' method.
Let $x = (1+\sqrt{3}) \cdot \sqrt{2-\sqrt{3}}$, and thus:
$$x^2 = (1 + \sqrt{3})^2 \cdot (2 - \sqrt{3}) =(4+2\sqrt{3})(2-\sqrt{3}) =8-2\sqrt{3}^2 = 2 \require{cancel}$$
$$\Rightarrow x = \cancel{-\sqrt{2}}, \sqrt{2}$$
The reason why we have crossed out $-\sqrt{2}$ is because $1 + \sqrt{3} > 1 > 0$, and $\sqrt{2 - \sqrt{3}} > 0$, thus $x = (1 + \sqrt{3})(\sqrt{2 - \sqrt{3}}) > 0$. Hence $x = \sqrt{2}$ is the only possible value.
|
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For any odd prime $p$, a quadratic is solvable mod $p^2$ if it is solvable mod $p$, and $p$ does not divide the discriminant and leading coefficient.
Let $p$ be an odd prime number and $a$, $b$ and $c$ integers such that $p$ does not divide $a$, and does not divide $D=b^2-4ac$, and such that
$$ax^2+bx+c\equiv0\pmod{p},$$
is solvable. Prove that in this case $ay^2 + by + c \equiv 0 \mod p^2$ is solvable too.
Well, the only thing that I was able to do is get $D$ from quadratic equation:
\begin{eqnarray*}
ax^2 + bx + c &=& a(x^2 + bx/a) + c\\
& =& a(x^2 + 2\cdot b/2a + (b/2a)^2) - a(b/2a)^2 + c\\
&=&a(x + b/2a)^2 - \left(\frac{b^2 - 4ac}{4a}\right)\\
&=& a(x + b/2a)^2 - \left(\frac{D}{4a}\right)
\end{eqnarray*}
Since $D$ is not divisible by $p$, $D/4a$ is not divisible by $p$ too. So in order to make this equation true remainders of both $\frac{D}{4a}$ and $a(x + b/2a)^2$ should be equal to each other. Let us say that it is $r_1$.
Okay, then let $a(x + b/2a)^2$ be $p\cdot k + r_1$ and $D/4a = p \cdot m + r_1$. But can't move on with this proof. I tried several paths but all of them didn't lead me anywhere.
Your help will be much appreciated!
|
Let $p$ be an odd prime number and $a$, $b$ and $c$ integers such that $p\nmid a$ and $p\nmid D$, where $D=b^2-4ac$. Let $x$ be an integer such that
$$ax^2+bx+c\equiv0\pmod{p},\tag{1}$$
so $ax^2+bx+c=mp$ for some integer $m$.
Exercise: Show that $p$ does not divide $2ax+b$, because $p\nmid a$ and $p\nmid D$ and $p$ is odd.
If $2ax+b\equiv0\pmod{b}$ then because $p$ is odd and $p\nmid a$ we have $$x\equiv-b(2a)^{-1}\pmod{p}.$$ Plugging this into $(1)$ yields \begin{eqnarray*} 0&\equiv&a(-b(2a)^{-1})^2+b(-b(2a)^{-1})+c\\ &\equiv&(2a)^{-2}(ab^2-2ab^2+4a^2c)\\ &\equiv&(2a)^{-2}(-aD), \end{eqnarray*} contradicting the fact that $p\nmid a$ and $p\nmid D$ and $p$ is odd.
Now if $y$ is an integer such that
$$ay^2+by+c\equiv0\pmod{p}^2,\tag{2}$$
then reducing mod $p$ shows that $y$ is a solution to $(1)$. So it makes sense to suppose that $y=x+kp$ for some integer $k$. Then
\begin{eqnarray*}
ay^2+by+c&=&a(x+kp)^2+b(x+kp)+c\\
&=&(ax^2+bx+c)+(2ax+b)kp+ak^2p^2\\
&=&((2ax+b)k+m)p+ak^2p^2
\end{eqnarray*}
We see that $y$ satisfies $(2)$ if and only if
$$(2ax+b)k+m\equiv0\pmod{p}.$$
Now because $2ax+b\not\equiv0\pmod{p}$ we are be done by taking
$$k\equiv-m(2ax+b)^{-1}\pmod{p}.$$
|
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|
Find the values of $x$ for which $x^{12}-x^9+x^4-x+1>0$
Find the values of $x$ for which $x^{12}-x^9+x^4-x+1>0$.
I tried to substitute some basic values like $-1,0,1$ and try to find the roots of the function but couldn't.
Then I graphed the function on desmos and this is the graph.
So from this, we can say that $x^{12}-x^9+x^4-x+1>0$ for all values of $x$.
But I want to know how to find the required values of $x$ without graphing
|
The sum of square form
$$2(x^{12}-x^9+x^4-x+1)$$
$$=x^6(x^3-1)^2+\left(x^6-\frac{1}{2}\right)^2+2\left(x^2-\frac{1}{4}\right)^2+(x-1)^2+\frac{5}{8}>0.$$
|
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|
use mathematical induction to show that $n^3 + 5n$ is divisible by $3$ for all $n\ge1$ What I have so far
Base: $n^3 + 5n$
Let $n=1$
$$
1^3 + 5(1) = 6
$$
$6$ is divisible by $3$
Induction step: $(k+1)^3 + 5(k+1)$
$(k^3 + 3k^2 + 8k + 6)$ is divisible by $3$
I kind of get lost after this point. For starters, how do I prove that this isn't applicable for any number less than $1$? Also, where do I go after this? Thank you!
|
In the induction step, we assume that it holds for $n = k$.
That means that $\exists a \in \mathbb{Z} : k^3 + 5k = 3a$.
Then, for $(k+1)$, we get
$$
\begin{split}
(k+1)^3 + 5(k+1)
&= k^3 + 3k^2 + 8k + 6 \\
&= (k^3 + 5k) + 3k^2 + 3k + 6 \\
&= 3a + 3k^2 + 3k + 6 \\
&= 3(a + k^2 + k + 2).
\end{split}
$$
So, if $k^3 + 5k$ is a multiple of $3$, then $(k+1)^3 + 5(k+1)$ must also be a multiple of $3$.
|
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|
Is there a solution to $2^a+2^b = 10^c+10^d$, with $0 \leq a < b$ and $0 \leq c < d$? This question arose on the code golf StackExchange:
Is there a solution to $2^a+2^b = 10^c+10^d$, with $0 \leq a < b$ and $0 \leq c < d$?
In other terms: is there an integer that looks like $\color{blue}{1000...001000...}$ in both binary and decimal?
I feel like there probably isn't, but I can't think of a simple counterargument.
(A computer search by one of the commenters suggests there is no such integer up to $10^{100000}$.)
|
NOT AN ANSWER. Commentary too long to fit in a Comment
*
*The equation: $2^a+2^b=10^c+10^d;\ 0\le a<b,\ 0\le c<d$. If $b>a,\ d>c$ were not already stated, it is provable that $a\ne b,\ c\ne d$ from which we could validly assume WLOG $b>a,\ d>c$.
*Although $a=0$ and $c=0$ are contemplated by the question as stated, neither $a$ nor $c$ can actually be $0$. If only one of them is, then we have an odd sum equal to an even sum. If both of them are, then by subtracting $1$ from each side we obtain $2^b=10^d$ which is impossible.
*$b-a=4k-2$. This follows from $2^a+2^b\equiv 0 \bmod 5$. Positive integer powers of $2$ modulo $5$ cycle in order through $2,4,3,1$. Sums of powers of $2$ that are $\equiv 0 \bmod 5$ arise from $2^{4m+1}+2^{4n+3}\text{ or }2^{4m}+2^{4n+2}$, and the differences of the exponents have the stated form.
*$a=c$. The original equation can be rearranged to $2^a(2^{b-a}+1)=2^c5^c(10^{d-c}+1)$. Plainly, there are $a$ factors of $2$ on the LHS and $c$ factors of $2$ on the RHS, so $a=c$.
*$a$ is even. From point 4 we obtain $2^a+2^b=10^a+10^d \Rightarrow 1+2^{b-a}=5^a(1+10^{d-a})$. We look at this equation $\bmod 3$. $1+2^{b-a}\not \equiv 1 \bmod 3$, and $(1+10^{d-a})\equiv 2 \bmod 3$. This requires $5^a \not \equiv 2 \bmod 3$ which in turn requires $a$ is even. This impacts point 3 by restricting the exponents to $(a,b)=(4m,4n+2)$ up to order. So $b$ is also even.
*$d$ is even. From points 4 and 5, $c$ is even. $10\equiv -1 \bmod 11$ so $10^c+10^d \equiv 1+(-1)^d \equiv 0,2 \bmod 11$; $0$ if $d$ is odd and $2$ if $d$ is even. $2^{4m}+2^{4n+2}\not \equiv 0 \bmod 11$. There are instances where $2^{4m}+2^{4n+2} \equiv 2 \bmod 11$. So if there are solutions to the equation, $d$ is even.
This is as far as I have gotten; I will update if I get further.
|
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|
Transformation of roots of a polynomial Suppose there is a cubic polynomial in x with roots A,B,C and another cubic polynomial (in t) with roots
1/(A-1), 1/(B-1), 1/(C-1)
is to be found.
My text book mentions two ways of doing this. One is to use vieta's relations, which is time consuming.
Another way mentioned is to let t=1/(x-1), make x the subject, and then substitute and simplify. However, I cant really wrap my mind around this. Just because this is the relation between the roots doesnt mean that all values of x and t are in thay relation, does it?
Moreover, if I try this method with more generality i.e trying to find a polynomial with roots f(A),f(B),f(C) where f(u) is some invertible function like e^u, it appears to fail.
Could someone please elaborate on the proper method of tackling such questions.
|
Let $A$, $B$ and $C$ be roots of the polynomial $x^3+ax^2+bx+c.$
Thus, $$A+B+C=-a,$$ $$AB+AC+BC=b,$$ $$ABC=-c$$ and we obtain:
$$\sum_{cyc}\frac{1}{A-1}=\frac{\sum\limits_{cyc}(A-1)(B-1)}{\prod\limits_{cyc}(A-1)}=\frac{b+2a+3}{-c-b-a-1},$$ which gives a coefficient before $x^2$ in new polynomial:
$$\frac{2a+b+3}{a+b+c+1}.$$
Also, $$\sum_{cyc}\frac{1}{(A-1)(B-1)}=\frac{\sum\limits_{cyc}(A-1)}{\prod\limits_{cyc}(A-1)}=\frac{-a-3}{-a-b-c-1}=\frac{a+3}{a+b+c+1}$$ and
$$\prod_{cyc}\frac{1}{A-1}=\frac{1}{-a-b-c-1}$$ and we got the answer:
$$x^3+\frac{2a+b+3}{a+b+c+1}x^2+\frac{a+3}{a+b+c+1}x+\frac{1}{a+b+c+1}.$$
We used the Viete's theorem and this is a right method for solving your problem.
Also, you can see that by using a cyclic summation it turns out easy enough.
I think the second method takes more time.
Let $x=\frac{1}{y-1}$, where $y$ is a root if the polynomial $x^3+ax^2+bx+c.$
Thus, since $y=\frac{x+1}{x},$ we obtain:
$$\left(\frac{x+1}{x}\right)^3+a\left(\frac{x+1}{x}\right)^2+b\left(\frac{x+1}{x}\right)+c,$$ which gives the same result, but with a bit of more computations.
|
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How to solve such a system of quadratic equations: $x^2+y^2-xy=a^2, x^2+z^2-xz=b^2, y^2+z^2-yz=c^2$ I don't know how to solve this this system:
$$x^2+y^2-xy=a^2\\x^2+z^2-xz=b^2\\y^2+z^2-yz=c^2$$ The system of quadratic equation in symmetry form has many geometric meaning,
this seems to be a triangular pyramid, three adjacent angles with a common point are $\frac{\pi}{3}$, and opposite to this point is a triangle with $a,b,c$.
But this geometric meaning makes no sense to me, it can't simplify the problem. I don't have other idea about it.
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\begin{align}
(y+z-x)(y-z)&=y^2-z^2-xy+xz=a^2-b^2\\
(x+y-z)(x-y)&=x^2-y^2-zx+zy=b^2-c^2\\
(z+x-y)(z-x)&=z^2-x^2-yz+yx=c^2-a^2
\end{align}
Let $u:=y-z$, $v:=y+z$, $\alpha^2:=a^2-b^2$, $\beta^2:=b^2-c^2$, $\gamma^2:=c^2-a^2$, then
\begin{align}
(v-x)u&=\alpha^2\\
(x+u)(u+v-2x)&=-2\beta^2\\
(x-u)(v-u-2x)&=2\gamma^2
\end{align}
Eliminating $x=v-\frac{\alpha^2}{u}$, (unless $\alpha=0=u$) \begin{align}
(uv+u^2-\alpha^2)(u^2+uv-2uv+2\alpha^2)&=-2\beta^2u^2\\
(uv-u^2-\alpha^2)(uv-u^2-2uv+2\alpha^2)&=2\gamma^2u^2\\
\therefore\ (u^2+uv-\alpha^2)(u^2-uv+2\alpha^2)&=-2\beta^2u^2\\
(u^2-uv+\alpha^2)(u^2+uv-2\alpha^2)&=2\gamma^2u^2\\
\therefore\ (u^2+\tfrac{1}{2}\alpha^2)^2-(uv-\tfrac{3}{2}\alpha^2)^2&=-2\beta^2u^2\\
(u^2-\tfrac{1}{2}\alpha^2)^2-(uv-\tfrac{3}{2}\alpha^2)^2&=2\gamma^2u^2
\end{align}
These two equations are actually the same: $$ (u^2-\tfrac{1}{2}\alpha^2)^2-(u^2+\tfrac{1}{2}\alpha^2)^2=-2\alpha^2u^2=2(\beta^2+\gamma^2)u^2$$
So we get $$v=\left(\tfrac{3}{2}\alpha^2\pm\sqrt{(u^2+\tfrac{1}{2}\alpha^2)^2+2\beta^2u^2}\right)/u$$
It seems we've lost one degree of freedom. This is due to taking differences of the equations to start with. So reintroduce the $y,z$ equation as $$(u+v)^2+(u-v)^2+(u+v)(u-v)=4c^2$$ $$3u^2+v^2=4c^2$$ Substituting $v$ in this equation gives a quartic in $u^2$, which can be solved.
Hence find $v$, $y$, $z$, $x$ in turn.
Note: There may be 8 or 4 or 0 real solutions.
|
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|
Comparing $(2+\frac{1}{2})(3+\frac{1}{3})(4+\frac{1}{4})(5+\frac{1}{5})$ with $(2+\frac{1}{5})(3+\frac{1}{4})(4+\frac{1}{3})(5+\frac{1}{2})$ This question appeared in one of the national exams (MCQs) in Saudi Arabia.
In this exam;
*
*Using calculators is not allowed,
*The student have $72$ seconds on average to answer one question.
PROBLEM:
Compare $a=(2+\frac{1}{2})(3+\frac{1}{3})(4+\frac{1}{4})(5+\frac{1}{5})$ with $b=(2+\frac{1}{5})(3+\frac{1}{4})(4+\frac{1}{3})(5+\frac{1}{2})$.
CHOICES:
A) $a>b$
B) $a<b$
C) $a=b$
D) Given information is not enough
Using algebra to evaluate each expression is easy, and the correct choice is $A$, but that will take a long time.
Any suggestion to solve this problem in a short time? THANKS.
|
We have that
$$a=\left(2+\frac{1}{2}\right)\left(3+\frac{1}{3}\right)\left(4+\frac{1}{4}\right)\left(5+\frac{1}{5}\right)=$$
$$=\frac12\left(4+1\right)\frac13\left(9+1\right)\frac14\left(16+1\right)\frac15\left(25+1\right)=$$
$$=\frac{5\cdot 10\cdot17\cdot26}{120}$$
and similarly
$$b=\left(2+\frac{1}{5}\right)\left(3+\frac{1}{4}\right)\left(4+\frac{1}{3}\right)\left(5+\frac{1}{2}\right)=\frac{11\cdot 13\cdot13\cdot11}{120}$$
with
$$5\cdot 10\cdot17\cdot26 > 11\cdot 13\cdot13\cdot11$$
$$5\cdot 10\cdot 17\cdot 2 > 11\cdot 13\cdot11$$
$$100 \cdot 17 > 121 \cdot 13$$
|
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|
Find $[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+....+[\sqrt{2019}]$
Let [$x$] denote the greatest integer less than or equal to $x$. Find the value of $$[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+....+[\sqrt{2019}]$$
I was able to find out the general pattern in the series i.e. $3(1)+ 5(2)+ 7(3)+\ldots+87(43)$ but was unable to proceed further.
|
Hint 1: $[\sqrt{n^2}], [\sqrt{n^2 +1}], .........[\sqrt{(n+1)^2 - 1}]$ all equal the same the same thing: $n$.
Hint 2: $(n+1)^2 - 1= n^2 + 2n$ so from $[\sqrt{n^2}]$ to $[\sqrt{(n+1)^2 - 1}]=[\sqrt{n^2 + 2n}]$ there are $2n+1$ terms.
Hint 3: So we have $\color{blue}{[\sqrt {1}] + [\sqrt 2]+ [\sqrt 3]} + \color{purple}{[\sqrt 4] + ....+[\sqrt 8]} +\color{orange}{[\sqrt 9]+ ....+[\sqrt{15}]}.....+\color{green}{[\sqrt M^2]+...... + [\sqrt {(M+1)^2-1}]} + \color{red}{[\sqrt {(M+1)^2}] + ...... + \sqrt{2019}}$ where $(M+1)^2 \le 2019 < (M+2)^2$.
============
Solution:
$M+1= 44$ and $44^2 = 1936$ so so we have
$\color{blue}{1 + 1+ 1} + \color{purple}{2 + ....+2} +\color{orange}{3+ ....+3}.....+\color{green}{43+...... + 43} + \color{red}{44 + ...... + 44}$
$\color{blue}{1\cdot (2\cdot 1 + 1)} + \color{purple}{2\cdot (2\cdot 2 + 1)} +\color{orange}{3\cdot (2\cdot 3 + 1)}.....+\color{green}{43\cdot (2\cdot 43 + 1)} + \color{red}{44\times(2019-1936+1)}=$
$\sum_{k=1}^{43}[k(2k+1)] + 44\cdot 84=$
$2\sum_{k=1}^{43} k^2+\sum_{k=1}^{43} k + 3696$
..... and that's it. Apply $\sum_{k=1}^n k = \frac {n(n+1)}2$ and $\sum_{k=1}^n k^2 = \frac {n(n+1)(2n+1)}6$ and you are done.
|
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|
Power series $\ \frac{x}{(x-2)(x-3)}$, radius of convergence I got a little trouble with this function while expanding this to power series:
$$\ f(x) = \frac{x}{(x-2)(x-3)} $$
After partial fraction decomposition:
$$\ f(x) = \frac{3}{x-3} - \frac{2}{x-2} $$
I can see here geometric series and rewrite as the sums:
$$\ \frac{1}{1-t} = \sum^{\infty}_{n=0} t^n $$
$$\ f(x) = 3 \frac{1}{\frac{1}{3}x-1} - 2 \frac{1}{1-\frac{1}{2}x} $$
I can set $ t = \frac{1}{2}x $, but what can I do with $ \frac{1}{3}x $? The signs don't match in the denominator to the formula.
As a radius of convergence, I would choose $-2<x<2$ as the smaller range for the sum of the series
EDIT: I have tried this - with your suggestion:
$$\ f(x) = \frac{3}{3} \frac{1}{\frac{1}{3}x-1} - \frac{2}{2} \frac{1}{1-\frac{1}{2}x} $$
$$\ f(x) = \frac{3}{3} \frac{1}{-(1-\frac{1}{3}x)} - \frac{2}{2} \frac{1}{1-\frac{1}{2}x} $$
$$\ f(x) = -\frac{3}{3} \frac{1}{-(1-\frac{1}{3}x)} - \frac{2}{2} \frac{1}{1-\frac{1}{2}x} $$
$$\ -(\sum^{\infty}_{n=0} (\frac{1}{2}x)^n + \sum^{\infty}_{n=0} (\frac{1}{3}x)^n) $$
$$\ \min\{2,3\} = 2 ~~\text{radius of convergence} $$
Is it right?
|
Hint:
We can write $$\dfrac3{x-3}=-\dfrac1{1-\dfrac x3}$$ right?
So, we need $|x|<$min$(2,3)$
|
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|
Proving inequality using Taylor polynomial of arctan I have to prove that $ x - \frac{x^3}{3} < \arctan(x) < x - \frac{x^3}{6} $ is true if $ 0 < x \leq 1 $
I tried to convert the second inequality into the one I'm trying to prove. Since $ \arctan(x) $ is a monotonically increasing function, applying it to each part of the second inequality should not change the relationships:
$$ 0 < x \leq 1 \Rightarrow \arctan(0) < \arctan(x) \leq \arctan(1) $$
Then I calculated the 3rd degree Taylor polynomial of $ \arctan(x) $ centered at $ x = 0 $ and at $ x = 1 $ to approximate the left and right sides of the inequality:
$$ \arctan(0) \approx x - \frac{x^3}{3} $$
$$ \arctan(1) \approx \frac{\pi}{4} + \frac{1}{2} (x-1) + \frac{1}{2} (x-1)^2 + \frac{1}{3} (x-1)^3$$
So it's evident that $ x - \frac{x^3}{3} < \arctan(x) $ is correct. But I'm stuck trying to prove that $ \arctan(x) < x - \frac{x^3}{6} $
|
We have that
$$f(x)=\arctan x- x+\frac16 x^3 \implies f'(x)=\frac{x^2(x^2-1)}{2(x^2+1)}\le 0$$
with $f(0)=0$.
|
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|
Hint to prove $\sin^4(x) + \cos^4(x) = \frac{3 + \cos(4x)}{4}$ Could go from LHS to RHS by adding zero but I need to know how to do this WITHOUT knowing the half-angle formula. So from RHS to LHS, you an expand $\cos4x$ twice. I get as close as
$$\frac{ \cos^4x + \sin^4x + 3(1 - 2\sin^2x\cos^2x)}{4}$$
|
To check, you can see if the LHS equals the RHS by using
$$\cos^2 x = \dfrac {\cos 2x + 1}{2} \\ \sin^2 x = \dfrac {1- \cos 2x}{2}$$
Squaring both sides
$$\cos^4 x = \dfrac 14(\cos^2 2x + 2 \cos 2x +1) \\ \sin^4 x = \dfrac 14(1-2 \cos 2x + \cos^2 2x)$$
Adding we get
$$\cos^4 x + \sin^4 x = \left(\dfrac 14 {\cos^2 2x} + \dfrac 14 \right) +\left(\dfrac 14 + \dfrac 14 {\cos^2 2x}\right) \\ \cos^4 x + \sin^4 x = \dfrac 12 ({1+\cos^2 2x}) $$
Since $2 \cos^2 2x - 1 = \cos 4x$, $\cos^2 2x = \dfrac 12 ({\cos 4x +1})$ so
$$\cos^4 x + \sin^4 x = \dfrac 12 (1 + \dfrac 12 (\cos 4x + 1)) \\ \cos^4 x + \sin^4 x = \dfrac 12 + \dfrac 14 (\cos 4x + 1)$$ or $$\mathbf {\cos^4 x + \sin^4 x = \dfrac {3 + \cos 4x}{4}}$$
|
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|
Integral of $x^2 \sqrt{1 + x^2}$ How may one evaluate $\int x^2 \sqrt{x^2 + 1}\ dx$? I tried parts by integrating $\sqrt{x^2 + 1}$ but that seems to lead me down a rabbit hole of endless computations.
|
Substitute $x=\tan(u)$ then we have $$\int x^2 \sqrt{x^2 + 1}\ dx$$
$$=\int \tan^2(u)\sec^3(u)du=\int(\sec^2(u)-1)\sec^{3}(u)du$$
$$=\int \sec^5(u)du-\int \sec^3(u)du.$$
Now applying the reduction formula $$I_{n}=\int \sec^{n}(x)dx=\frac{1}{n-1}\tan(x)\sec^{n-2}(x)+\frac{n-2}{n-1}I_{n-2}$$ gives $$=\int \sec^5(u)du-\int \sec^3(u)du$$
$$=\frac{1}{4}\tan(u)\sec^3(u)+\frac{3}{4}\int\sec^3(u)du-\int\sec^3(u)du$$
$$=\frac{1}{4}\tan(u)\sec^3{u}-\frac{1}{4}\big[\frac{1}{2}\tan(x)\sec(x)+\frac{1}{2}\int\sec(u)du\big]$$
$$=\frac{1}{4}\tan(u)\sec^3(u)-\frac{1}{8}\tan(u)\sec(u)-\frac{1}{8}\ln|\tan(u)+\sec(u)|+C$$
Finally since $u=\arctan(x)$, we have $\tan(u)=x$ and $\sec(u)=\sqrt{x^2+1},$ thus
$$\int x^2 \sqrt{x^2 + 1}\ dx=\frac{1}{4}x(x^2+1)^{\frac{3}{2}}-\frac{1}{8}x\sqrt{x^2+1}-\frac{1}{8}\ln|x+\sqrt{x^2+1}|+C.$$
|
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|
reduce a differential equation $y'+\dfrac{x}{y}=0$ I want to reduce a differential equation.
$y'+\dfrac{x}{y}=0$
I reduced this. But my answer don't much with Wolfram alpha. Please tell me what is wrong.
$y'=-\dfrac{x}{y}=-\dfrac{1}{\left( \dfrac{y}{x}\right) }$
When I put, $u=\dfrac{y}{x}$,
$\dfrac{dy}{dx}=-\dfrac{1}{u}$
differentiate both sides with respect to x
$\dfrac{du}{dx}=-\dfrac{1+u^{2}}{ux}$
$\dfrac{u}{1+u^{2}}\dfrac{du}{dx}=-\dfrac{1}{x}$
Integrate both sides with respect to x
$\dfrac{1}{2}\log \left| u^{2}+1\right| =-\log \left| x\right| +C'$
Organize the formula
$u^{2}+1=\dfrac{C}{x}$
$\begin{aligned}\dfrac{y^{2}}{x}+1=\dfrac{C}{x}\\ y=\pm \sqrt{Cx-x^{2}}\end{aligned}$
|
Not sure exactly what our OP langhtorn means by "reduction", or how the calculations presented in the text of the question itself work, but here is my solution:
$y' + \dfrac{x}{y} = 0; \tag 1$
multiply through by $y$:
$yy' + x = 0; \tag 2$
$\dfrac{1}{2}(y^2)' + x = 0; \tag 3$
$(y^2)' = -2x; \tag 4$
integrate:
$y^2 = -x^2 + C = C - x^2; \tag 5$
then
$y = \pm \sqrt{C - x^2} = \pm (C - x^2)^{1/2}. \tag 6$
Check:
$y' = \pm \dfrac{1}{2}(C - x^2)^{-1/2}(-2x) = \pm (-x(C - x^2)^{-1/2})$
$= \pm (-\dfrac{x}{(C - x^2)^{1/2}}) = -\dfrac{x}{\pm (C - x^2)^{1/2}} = -\dfrac{x}{y}. \checkmark \tag 7$
In light of this verified solution, I am skeptical of our OPs calculations, but as of this writing have not scrutinized them carefully enough to find an error.
|
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|
How to solve this integral $I = \int\dfrac{\cos^3x}{\sin x + \cos x}dx$? $\displaystyle\int\dfrac{\cos^3x}{\sin x + \cos x}dx$
I added $J =\displaystyle \int\dfrac{\sin^3x}{\sin x + \cos x}dx$
then $I + J = \displaystyle\int\dfrac{\cos^3x + \sin^3x}{\sin x + \cos x}dx = x + \dfrac{1}{2}\cos2x + C$
but I can't find how to solve $I-J$
And is that the true way to solve it?
Please help!
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If you use $\tan(x)=t$, you end with
$$I = \int\dfrac{\cos^3(x)}{\sin (x) + \cos (x)}dx=\int \frac{dt}{(t+1) \left(t^2+1\right)^2}$$ Using partial fraction decomposition
$$\frac{1}{(t+1) \left(t^2+1\right)^2}=\frac{1-t}{4 \left(t^2+1\right)}+\frac{1-t}{2 \left(t^2+1\right)^2}+\frac{1}{4
(t+1)}$$ does not seem too bad.
|
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Mathematical Induction Proof for $(\sum_{k=1}^{n}a^2_k)(\sum_{k=1}^{n}b^2_k) \geq (\sum_{k=1}^{n}a_kb_k)^2$ Mathematical Induction Proof $$(\sum_{k=1}^{n}a^2_k)(\sum_{k=1}^{n}b^2_k) \geq (\sum_{k=1}^{n}a_kb_k)^2$$
Base case: $n = 2$, $$(a^2_1+a^2_2)(b^2_1+b^2_2) \geq (a_1b_1
+ a_2b_2)^2$$
$$(a^2_1+a^2_2)(b^2_1+b^2_2) - (a_1b_1
+ a_2b_2)^2 \geq 0$$
$$(a_1b_2-a_2b_1)\geq 0$$
Let $$\sum_{i=1}^{k}a^2_i = A^2, \sum_{i=1}^{k}b^2_i = B^2,$$
$$(A^2 + a^2_{k+1})(B^2 + b^2_{k+1}) =$$
$$(A^2B^2 + a^2_{k+1}b^2_{k+1} + A^2b^2_{k+1} + B^2a^2_{k+1})$$
This step is where I don't understand:
$$(A^2B^2 + a^2_{k+1}b^2_{k+1} + A^2b^2_{k+1} + B^2a^2_{k+1}) \geq (AB + a_{k+1}b_{k+1})^2 \geq (\sum_{k=1}^{n}a_kb_k)^2 $$
I know that $$(AB + a_{k+1}b_{k+1})^2 = A^2B^2 + a^2_{k+1}b^2_{k+1} + 2ABa_{k+1} b_{k+1}$$
and that $$A^2b^2_{k+1} + B^2a^2_{k+1} \geq 0$$
hence,
$$A^2b^2_{k+1} + B^2a^2_{k+1} \geq 2ABa_{k+1} b_{k+1}$$
hence, $$(A^2B^2 + a^2_{k+1}b^2_{k+1} + B^2a^2_{k+1} + A^2b^2_{k+1}) \geq (AB + a_{k+1}b_{k+1})^2 $$
but why $$(AB + a_{k+1}b_{k+1})^2 \geq (\sum_{k=1}^{n}a_kb_k)^2 $$?
|
By the assumption of the induction twice and by AM-GM we obtain:
$$\sum_{i=1}^{n+1}a_i^2\sum_{i=1}^{n+1}b_i^2=\sum_{i=1}^{n}a_i^2\sum_{i=1}^{n}b_i^2+a_{n+1}^2\sum_{i=1}^nb_i^2+b_{n+1}^2\sum_{i=1}^na_i^2+a_{n+1}^2b_{n+1}^2\geq$$
$$\geq\left(\sum_{i=1}^na_ib_i\right)^2+2\sqrt{a_{n+1}^2b_{n+1}^2\sum_{i=1}^na_i^2\sum_{i=1}^nb_i^2}+a_{n+1}^2b_{n+1}^2\geq$$
$$\geq\left(\sum_{i=1}^na_ib_i\right)^2+2a_{n+1}b_{n+1}\sum_{i=1}^na_ib_i+a_{n+1}^2b_{n+1}^2=\left(\sum_{i=1}^{n+1}a_ib_i\right)^2.$$
|
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Rationalize the denominator of $2\over{2 - \sqrt[4]{2}}$?
Rationalize the denominator of $2\over{2 - \sqrt[4]{2}}$.
Here's my progress. Let $x = \sqrt[4]{2}$. Then our expression can be written as $x^4/(x^4 - x)$, which simplifies to $x^3/(x^3 - 1)$. Multiply top and bottom by $(x^3 + 1)$ to get $x^3(x^3 + 1)/(x^6 - 1)$. Multiply top and bottom by $(x^6 + 1)$ to get$$x^3(x^3 + 1)(x^6 + 1)/(x^{12} - 1) = x^3(x^3 + 1)(x^6 + 1)/7 = {1\over7}(8 + 4\sqrt[4]{2} + 2 \sqrt{2} + 2^{3/4}).$$However, Wolfram Alpha also tells me that we can write this as$${1\over{14}}\Big(16 + 4\sqrt{2} + 7\sqrt{{{64}\over{49}} + {{72{\sqrt2}}\over{49}}}\Big)$$But how do I derive that? Seems impossible!
|
$$\frac{2}{2-\sqrt[4]2}=\frac{\sqrt[4]{8}}{\sqrt[4]{8}-1}$$ and use $$(x-1)(x+1)(x^2+1)=x^4-1$$ for $x=\sqrt[4]{8}.$
Now, about the WA's bang.
$$7\sqrt{\frac{64}{49}+\frac{72\sqrt2}{49}}=\sqrt{64+72\sqrt2}=\sqrt{8\sqrt2(4\sqrt2+9)}=$$
$$=2\sqrt[4]8\sqrt{(2\sqrt2)^2+4\sqrt2+1}=2\sqrt[4]8(2\sqrt2+1)$$ and we made denesting.
|
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|
Limit of the finite series $\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3}$ The problem is to find the limit of:
$$\ \lim_{n\to\infty}\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3}$$
A the series is finite, it looks as if it would be required to find the sum of the series - however, I have to find the limit. It resembles for me:
$$\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ (k+n)^2+n^2-kn }{(k+n)^3-2kn^2-2k^2n}=\\
\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ (k+n)^2+n(n-k) }{(k+n)^3-2kn(n+k)}=\\
\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ (k+n)^2+n(n-k) }{(k+n)((k+n)^2-2kn)}$$
but I don't know what to do next and how to solve it. I would appreciate your help.
Edit: Does it has something in common with Riemann sum?
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Thanks to Peter Franek I would try to calculate the limit.
The sum can be rewritten as:
$$\ \sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3} = \sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} (\frac{n}{k^2+n^2}+\frac{1}{k+n}) $$
$$\ a_n=\frac{n}{k^2+n^2} = \frac{n}{n^2}\cdot\frac{1}{\frac{k^2}{n^2}+1}$$
$$\ b_n=\frac{1}{k+n} = \frac{1}{n}\cdot\frac{1}{\frac{k}{n}+1} $$
Let us consider $f(x) = \frac{1}{x^2+1} $ and $g(x)=\frac{1}{x+1}$
$$\lim_{n\to\infty}a_n=\int_0^1f(x)dx=\int_0^1\frac{1}{x^2+1}dx=\arctan(x)|_0^1=\frac{\pi}{4} \\\lim_{n\to\infty}b_n=\int_0^1g(x)dx=\int_0^1\frac{1}{x+1}dx = \int_0^1\frac{c'(x)}{c(x)}=\ln(x+1)|_0^1=\ln(2)$$
As the element $\sqrt(n)$ becomes unsignificant for $n\to\infty$ as $\frac{\sqrt(n)}{n}\to0$, the limit is:
$$ \lim_{n\to\infty}\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} (\frac{n}{k^2+n^2}+\frac{1}{k+n}) = \frac{\pi}{4}+\ln(2)$$
|
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|
In trapezoid $ABCD$, $AB \parallel CD$ , $AB = 4$ cm and $CD = 10$ cm.
In trapezoid $ABCD$, $AB \parallel CD$ , $AB = 4$ cm and $CD = 10$ cm. Suppose the lines $AD$ and $BC$ intersect at right angles and the lines $AC$ and $BD$ when extended at point $Q$ form an angle of $45^\circ$. Compute the area of $ABCD$.
What I Tried :- Here is the picture :-
Now to find the area of $ABCD$, I just need to find its height, but I cannot find it.
I can see that $\Delta AOB \sim \Delta COD$. So :-
$$\frac{AB}{CD} = \frac{AO}{OD} = \frac{BO}{OC} = \frac{2}{5}$$
So I assumed $AO = 2x$ , $BO = 2y$ , $CO = 5y$ , $DO = 5x$.
Now in $\Delta AOB$, by Pythagorean Theorem :-
$AO^2 + OB^2 = AB^2$
$\rightarrow 4x^2 + 4y^2 = 16$
$\rightarrow x^2 + y^2 = 4$
Also $\Delta QAB \sim \Delta QDC$. So:-
$$\frac{QA}{AC} = \frac{QB}{BD}$$
I get $AC$ and $BD$ by Pythagorean Theorem again, which gives me :-
$$\frac{QA}{\sqrt{4x^2 + 25y^2}} = \frac{QB}{\sqrt{25x^2 + 4y^2}}$$
I don't know how to proceed next, as this result only gives me that $\left(\frac{QA}{OB}\right)^2 = \frac{21y^2 + 16}{21x^2 + 16}$ . Also I couldn't think of any way to use the $45^\circ$ angle, except that I can figure out that the triangle is cyclic.
Can anyone help?
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By your work $$(2x)^2+(2y)^2=4^2,$$ which gives $$x^2+y^2=4.$$
Also, $$AD=\sqrt{DO^2+AO^2}=\sqrt{25y^2+4x^2}$$ and
$$BC=\sqrt{25x^2+4y^2}.$$
Now, let $PABC$ be parallelogram.
Thus, $P\in DC$, $AP=BC$, $$DP=DC-PC=10-4=6$$ and $$\measuredangle DAP=\measuredangle Q=45^{\circ}.$$ Thus, by the law of cosines for $\Delta DAP$ we obtain:
$$\frac{25y^2+4x^2+25x^2+4y^2-36}{2\sqrt{(25x^2+4y^2)(25y^2+4x^2)}}=\frac{1}{\sqrt2}$$ or
$$\frac{29\cdot4-36}{\sqrt2}=\sqrt{(25x^2+4y^2)(25y^2+4x^2)}$$ or
$$3200=641x^2y^2+100(x^4+y^4)$$ or
$$3200=441x^2y^2+100(x^2+y^2)^2,$$ which gives
$$xy=\frac{40}{21}.$$
Id est, $$S_{ABCD}=\frac{1}{2}AC\cdot DB=\frac{1}{2}\cdot7x\cdot7y=\frac{140}{3}.$$
|
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Number of right isosceles triangles that can be formed with points lying on the curve $8x^3+y^3+6xy=1$ Number of right isosceles triangles that can be formed with points lying on the curve $$8x^3+y^3+6xy=1$$
MY ATTEMPT :
We have,
$$8x^3+y^3+6xy=1$$
adding both the sides $$6xy^2+12x^2$$ and simplifying we get ,
$$y^2+y(1-2x)+4x^2+2x+1=0$$
after this I got struck pls help me out with this question
Answer given is 3
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Since $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc),$$ we see that our equation it's
$$(2x+y-1)(4x^2+y^2+1-2xy+2x+y)=0,$$ which gives $$2x+y-1=0$$ or
$$(2x-y)^2+(2x+1)^2+(y+1)^2=0.$$
The second gives a point $A\left(-\frac{1}{2},-1\right)$ and two other points are placed on the line $2x+y-1=0$.
Thus, the vertex of the right angle of the triangle may be in three these points.
Easy to see that all these possibilities occur.
Indeed, let $AD$ be a perpendicular to the line $l:2x+y-1=0$ and $\{B,C\}\subset l$ such that $D$ be a mid-point of $BC$ and $AD=DB=DC.$
Thus, $\Delta ABC$, $\Delta ADB$ and $\Delta ADC$ they are right-angled isosceles triangles.
|
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|
Solving for $\lambda$ and $\mu$ in $5\sin(x) = \lambda(2\sin(x)-3) + \mu 2\cos(x)$
Solve for $\lambda$ and $\mu$ in the following equation:
$$5\sin(x) = \lambda(2\sin(x)-3) + \mu 2\cos(x)$$
I have tried:
$$
5\sin(x) = 2\lambda \sin(x) -3\lambda+2\mu \cos(x)
$$
Comparing the coefficient of $\sin x$:
$$
2\lambda = 5
$$
$$
\lambda = \frac{5}{2}
$$
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It's $$(2\lambda-5)\sin{x}+2\mu\cos{x}=3\lambda$$ or since the case $2\lambda-5=\mu=0$ is impossible,
$$\frac{2\lambda-5}{\sqrt{(2\lambda-5)^2+4\mu^2}}\sin{x}+\frac{2\mu}{\sqrt{(2\lambda-5)^2+4\mu^2}}\cos{x}=\frac{3\lambda}{\sqrt{(2\lambda-5)^2+4\mu^2}}.$$
Now, take $$\frac{2\lambda-5}{\sqrt{(2\lambda-5)^2+4\mu^2}}=\cos\alpha$$ and
$$\frac{2\mu}{\sqrt{(2\lambda-5)^2+4\mu^2}}=\sin\alpha.$$
Can you end it now?
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|
solving $\tan^{-1}(x+1)+\tan^{-1}x+\tan^{-1}(x-1)=\tan^{-1}3$ solve $$\tan^{-1}(x+1)+\tan^{-1}x+\tan^{-1}(x-1)=\tan^{-1}3$$.
we have the formula;
let $$\tan^{-1}(\frac{x+y}{1-xy})=a$$ :when $xy<1$ $$\tan^{-1}x+\tan^{-1}y= a$$
when $x>0,y>0,xy>1$ $$\tan^{-1}x+\tan^{-1}y= \pi+a$$
when $x<0,y<0,xy>1$ $$\tan^{-1}x+\tan^{-1}y= -\pi+a$$
My try
$$\tan^{-1}(x+1)+\tan^{-1}(x-1)=\tan^{-1}3-\tan^{-1}x$$
when $x^2-1<1$ and $3x<1$or when $-\sqrt{2}<x<1/3...(1)$:
$$\tan^{-1}(\frac{2x}{2-x^2})=\tan^{-1}(\frac{3-x}{1+3x})$$
$$\frac{2x}{2-x^2}=\frac{3-x}{1+3x}$$solving and neglecting the ones coming out of the bound defined by $x$ in (1).
we have $x=-1$.
I know i have to do other cases when $xy>1$ . But suprisingly the answer given in book is only $x=-1$ and the solution also does not take the other cases.Why are they neglecting other cases. Morover i feel doing so many cases is very less efficient .How do i tackle this question easily?.
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$$\tan \left(\arctan x+\arctan(x+1)\right)=\frac{-2 x-1}{x^2+x-1}$$
$$\tan \left(\arctan 3-\arctan(x-1)\right)=\frac{4-x}{3 x-2}$$
provided that $\left(x (x+1)<1\right)\land \left(3 (x-1)<1\right)$ that is
$\frac{1}{2} \left(-\sqrt{5}-1\right)<x<\frac{1}{2} \left(\sqrt{5}-1\right)\land x\ne -1$
The equation becomes
$$\frac{-2 x-1}{x^2+x-1}=\frac{4-x}{3 x-2}$$
expanding and reordering
$$x^3-9 x^2-4 x+6=0$$
factoring
$$(x+1) \left(x^2-10 x+6\right)=0$$
which gives the solutions
$$x=-1,x= 5-\sqrt{19},x= 5+\sqrt{19}$$
$x=-1$ is discarded because the given equation is not verified
$$\arctan (-1+1)+\arctan(-1) +\arctan(-1-1)\ne\arctan 3$$
only $x=5-\sqrt{19}$ satisfies the constraint above therefore it is the only solution.
|
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Choose two sequences from a set such that the average of one sequence is larger than the other by 4th decimal point
Choose two sequences $a_n, b_m$ from $\{ \frac{1}{3}, \frac{2}{3}, 1, \frac{4}{3}, \frac{5}{3}, 2 \}$. Calculate their average $\bar{a_n}, \bar{b_m}$. Turned out that $\bar{a_n}$ is the same with $\bar{b_m}$ until the 4th decimal point (for example $\bar{a_n} = 0.xxxy$ and $\bar{b_m} = 0.xxxz$. Of course $\bar{a_n}$ can be $1$ too)
What's the minimum value of $\max(n,m)$.
If we only have two integers, $1, 2$, the results look simple. You need to have 1001 (with 1000 1's and 1 2's). But what if we got $\{ \frac{1}{3}, \frac{2}{3}, 1, \frac{4}{3}, \frac{5}{3}, 2 \}$?
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Fill in the gaps as needed.
*
*The terms of the sequences are $ \{ \frac{1}{3},\frac{2}{3},\frac{3}{3},\frac{4}{3},\frac{5}{3},\frac{6}{3} \}$. So, we can write $ \bar{a} = \frac{ A}{ 3n}$, where $ n \leq A \leq 6n$. Likewise, $ \bar{b} = \frac{ B}{ 3m}$, where $ m \leq A \leq 6m$.
*$ \frac{A}{3n} \neq \frac{B}{3m} \Rightarrow Am-Bn \neq 0 $.
*$ | \bar{a} - \bar{b} | < \frac{1}{1000} \Rightarrow 1000 | Am - Bn | < 3 mn$.
*If $|Am-Bn| = 1$, then $1000 < 3mn \Rightarrow \max(m, n) \geq 19$.
*If $\max(m, n) = n = 19$, then $m \geq \frac{1000}{3n} \Rightarrow m \geq 18$. Notice that $|19A - 19B| = 1 $ has no solutions, so we must have $m = 18$ only.
*Solving for $18A - 19B = 1$, we get solutions $A = 19K + 18, B = 18K + 17$. However, verify that none of these lead to actual solutions.
*Likewise for $18A - 19B = -1$, we don't get solutions.
*Apart from the trial and error, that there are no solutions in these cases can be explained as such: We have $\frac{|Am-Bn|} { 3mn } = \frac{1}{3}\times 19 \times 18 \approx 0.000974\ldots$, so this means that we either $\frac{A}{n} $ or $\frac {B}{m}$ will have to look like $0.xxx00, 0.xxx01, 0.xxx02, 0.xxx03$ in order for the other term to have the form $0.xxxz$. This is extremely restrictive (though we'd still have to check the various numbers to ensure that it isn't of this form).
*If $ \max(m, n) = n = 20$, then we have $ m \geq\frac{1000}{3n} \Rightarrow m\geq 17$. So let's consider $(m, n) = (17,20)$ in a similar manner. Solving $20A - 17B = 1$ gives us $A = 17k + 6, B = 20k + 7$. Because $\frac{27}{20} = 0.45000\ldots$, we know that this will work, namely $\frac{ 23}{17} = 0.45098\ldots$.
Hence, $\min (\max(m, n) ) = 20$.
Notes:
*
*As it turns out, selecting terms from $ \{ \frac{1}{3},\frac{2}{3},\frac{3}{3},\frac{4}{3},\frac{5}{3},\frac{6}{3} \}$ is a distractor. We only needed $\{ \frac{1}{3}, \frac{2}{3} \}$.
*To test your understanding of what is happening, figure out what happens with $(m, n) = (18, 20)$ (no solutions, why?), and $(m, n) = (19,20)$ (has solutions, which are?).
*We lucked out somewhat with the testing. If we wanted them to be the same until the 5th decimal point, what would the answer be?
|
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|
Use mathematical induction to prove that (n+2)(n+3)(n+7) is divisible by 6. Use mathematical induction to prove that $q(n)=(n+2)(n+3)(n+7)$ is divisible by $6$.
I have already proved the base case at n=1. I need help on the second part to prove $n=k+1$.
What I did: $(n+2)(n+3)(n+7)=6P$
\begin{align*}
((k+2)+1)&((k+1)+3)((k+1)+7) = (k+3)(k+4)(k+8)\\
= &(k+3)[(k+2)+2][(k+7)+1]\\
= &[(k+3)(k+2)+(2)(k+3)][(k+7)+1]\\
= &(k+2)(k+3)(k+7)+2(k+3)(k+7)+(k+2)(k+3)+2(k+3)\\
= &6P+2k^2+20k+42+k^2+5k+6+2k+6\\
= &6P+3k^2+27k+54\\
= &6p+3(k^2+9k+18)
\end{align*}
I'm not sure what to do, my proof turned out to be divisible by 3 instead of 6. Please let me know how I can move forward with this. Thank you!
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Hint: You are done, because $k^2+9k+18$ is always even, so that you have divisibility by $2\cdot 3=6$.
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|
Using the number e to find the limit of a sequence I have been trying to solve this problem for the past day and I just can't get my head around it, I know I need to use Euler's formula but how do I proceed after simplifying it.
$$
B_n = \left( \frac{n^2 -1}{n^2 +2} \right)^{2n^2-3}
= \left( \frac{n^2 +2 -3 }{n^2 +2} \right)^{2n^2-3}
= \left(1 + \frac{-3 }{n^2 +2} \right)^{2n^2-3}
$$
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Set $y=n^2+2$, to get:
$$
\lim_{n\to\infty} \left(1 + \frac{-3 }{n^2 +2} \right)^{2n^2-3}= \lim_{y\to\infty} \left(1 + \frac{-3 }{y} \right)^{2y-7}\\
=\lim_{y\to\infty} \bigg(\left(1 + \frac{-3 }{y} \right)^y\bigg)^2\left(1 + \frac{-3 }{y} \right)^{-7}\\=(e^{-3})^2\cdot 1=e^{-6}
$$
|
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If $b^2+c^2+bc=3$ then $b+c\leq 2$ Suppose that $b,c\geq 0$ such that $b^2+c^2+bc=3$. Prove that $b+c\leq 2$.
I tried to do that by contradiction but I failed.
Indeed, if $b+c>2$ then $b^2+2bc+c^2>4$ then $(b^2+bc+c^2)+bc>4$. Hence $3+bc>4$ or equivalently $bc>1$.
|
Let $b+c=k\iff b=k-c$
$$3=b^2+bc+c^2=(k-c)^2+(k-c)c+c^2$$
$$\iff c^2-kc+k^2-3=0$$ which is a quadratic equation in $c$
As $c$ is real, the discriminant must be $\ge0$
i.e., $$(-k)^2\ge4(k^2-3)\iff k^2\le4\iff-2\le k\le2$$
|
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|
$x+1$ is always invertible in $\Bbb Z_{x^3}$.
Claim: For any $x > 1$ and $x \in \Bbb N$, $x+1$ is always invertible in $\Bbb Z_{x^3}$.
Proof: We know that $x^3 +1 = (x+1)(x^2-x+1)$. Since $x > 1$ and $x \in \Bbb N$, we have $x^2-x+1 >0$ and $(x+1)(x^2-x+1) = x^3 +1 \equiv 1 \mod x^3$. Hence $x+1$ is always invertible in $\Bbb Z_{x^3}$.
Is the proof correct?
|
Actually you have proved that, for any positive integer $x$, $x+1$ is invertible modulo $x^3$.
Indeed, $(x+1)(x^2-x+1)=x^3+1\equiv1\pmod{x^3}$.
No need to check that $x^2-x+1>0$ (which it is, and it may be just remarked). If you find $u\in\mathbb{Z}$ such that $xu\equiv 1\pmod{x^3}$, then you also find $v>0$ such that $xv\equiv1\pmod{x^3}$: just keep adding $x^3$ to $u$ until you get past $0$.
A few more notes.
The $x>1$ clause is just to avoid congruences modulo $1$, which however don't pose relevant problems, do they?
By the way, I'd also state that the restriction for $x$ to be a positive integer can be replaced by “$x$ is an integer”. Even $x=0$ is good, because congruence modulo $0$ is just equality and $0+1=1$ is certainly invertible modulo $0$.
|
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How to solve this recursion which is not homogenous I have the following recursion
$$a_n = \frac{1}{4}a_{n-1}+\frac{1}{4}(\frac{2}{3})^{n-1}$$
I've tried first to solve the homogeneous equation (shifting by one)
$$(E - \frac{1}{4})a_n = 0$$
where $Ea_n = a_{n+1}$ is the shift operator. The only solution to this equation is $E=\frac{1}{4}$. Now I thought that for a non-homogeneous equation, where the term $d(n)$ does not depend on the underlying recursion has the form $d(n) = k\mu^n$ and $\mu$ is not a root of the homogeneous equation, the solution is given by
$$a_n = \frac{k\mu^n}{\Phi(\mu)}$$
where $\Phi$ is the characteristic equation of the homogeneous one. In my case $d(n) = \frac{1}{4}\frac{2}{3}^{n}$, so $k=\frac{1}{4}$ and $\mu = \frac{2}{3}$. Thus the solution should be given by
$$a_n = \frac{\frac{1}{4}\frac{2}{3}^n}{\frac{2}{3}-\frac{1}{4}}=\frac{\frac{1}{4}\frac{2}{3}^n}{\frac{5}{12}}=\frac{3}{5}\frac{2}{3}^n$$
However, the solution should be $$\frac{3}{5}\frac{2}{3}^n-\frac{3}{5}\frac{1}{4}^n$$. What did I wrong?
Note: the question arises from another problem, see here
|
The recurrent equation is
\begin{align}
a_n-\dfrac{1}{4}a_{n-1}=\dfrac{1}{4}\left(\dfrac{2}{3}\right)^{n-1}, n=1,2,\ldots.
\end{align}
Solve the homogeneous equation,
$$a_n-\dfrac{1}{4}a_{n-1}=0.$$
The characteristic equation is
$$r-\dfrac{1}{4}=0$$
which gives
$$r=\dfrac{1}{4}.$$
The solution of homogeneous equation is
$$a_n^{(c)}=C\left(\dfrac{1}{4}\right)^n.$$
Now, we solve non-homogenous equation.
Let the particular solution is
$$a_n^{(p)}=A\left(\dfrac{2}{3}\right)^{n-1}.$$
Substituting particular solution to recurrent equation gives
\begin{align}
A\left(\dfrac{2}{3}\right)^{n-1}-\dfrac{1}{4}A\left(\dfrac{2}{3}\right)^{n-2}=\dfrac{1}{4}\left(\dfrac{2}{3}\right)^{n-1}, n=1,2,\ldots.
\end{align}
Now, we have
\begin{alignat}{2}
&&
A\left(\dfrac{2}{3}\right)^{n-1}-\dfrac{3}{8}A\left(\dfrac{2}{3}\right)^{n-1}&=\dfrac{1}{4}\left(\dfrac{2}{3}\right)^{n-1}, n=1,2,\ldots.\\
\iff\quad
&&
\dfrac{5}{8}A\left(\dfrac{2}{3}\right)^{n-1}&=\dfrac{1}{4}\left(\dfrac{2}{3}\right)^{n-1}, n=1,2,\ldots.
\end{alignat}
Now we have
\begin{alignat}{2}
&&
\dfrac{5}{8}A&=\dfrac{1}{4}\\
\iff\quad
&&
A&=\dfrac{2}{5}.
\end{alignat}
So, the particular solution is
$$a_n^{(p)}=\dfrac{2}{5}\left(\dfrac{2}{3}\right)^{n-1}.$$
So, the solution of recurrent equation is
\begin{alignat}{2}
&&
a_n&=a_n^{(c)}+a_n^{(p)}\\
\iff\quad
&&
a_n&=C\left(\dfrac{1}{4}\right)^n+\dfrac{2}{5}\left(\dfrac{2}{3}\right)^{n-1}\\
\iff\quad
&&
a_n&=C\left(\dfrac{1}{4}\right)^n+\dfrac{3}{5}\left(\dfrac{2}{3}\right)^{n}.
\end{alignat}
Related to this question: Markov chain probability state question,
the initial condition is $a_1=\dfrac{1}{4}$.
We find constant $C$ as below
\begin{alignat}{2}
&&
a_n&=C\left(\dfrac{1}{4}\right)^n+\dfrac{3}{5}\left(\dfrac{2}{3}\right)^{n}\\
\iff\quad
&&
a_1&=C\left(\dfrac{1}{4}\right)+\dfrac{3}{5}\left(\dfrac{2}{3}\right)=\dfrac{1}{4}
\\
\iff\quad
&&
\dfrac{1}{4}C&=\dfrac{1}{4}-\dfrac{2}{5}=-\dfrac{3}{20}\\
\iff\quad
&&
C&=-\dfrac{3}{5}
\end{alignat}
So, the solution is
$$
a_n=-\dfrac{3}{5}\left(\dfrac{1}{4}\right)^n+\dfrac{3}{5}\left(\dfrac{2}{3}\right)^{n}.
$$
|
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|
Proving that every positive integer is of the form $x^2+y^2-5z^2$
Prove that every positive integer can be written as
$$x^2+y^2-5z^2$$
with $x$, $y$ and $z$ are non-zero integers.
I made the following observations
*
*if a number is congruent to 0,1,2 mod 4 than it can easily be expressed in this by taking z to be zero , as for the case when z is non zero I am not sure.
*if the number is congruent to 3 mod 4 than (x,y )have to even and z has to be odd all other cases dont work, the opposite is true if the number is congruent 2 mod 4
*to generalize for all types of integers mod 4 the parity of the numbers(x,y,z) that will satisfy are given below(I will denote even as 0 and odd as 1) and (x,y) can obviously be interchanged, therefore
-if 0 mod 4 then (0,0,0) and (0,1,1)
-if 1 mod 4 then (0,1,0) and (1,1,1)
-if 2 mod 4 then (1,1,0)
-if 3 mod 4 then (0,0,1)
so I tried to write a number congruent to 3 mod 4 as follows
$$x^2+y^2-5z^2$$=
$$(2a)^2+(2b)^2-5(2c+1)^2$$
$$4(a^2+b^2-5c^2-5c-1)^2-1$$
|
$$2=1^2+9^2-5\cdot4^2$$
$$4=20^2+3^2-5\cdot9^2$$
If $n \ge 3$ then $$2n=(n-2)^2+(2n-1)^2-5(n-1)^2$$
$$1=10^2+9^2-5\cdot6^2$$
If $n \ge 2$ then $$2n-1=n^2+(2n-2)^2-5(n-1)^2$$
|
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|
Find the equation of the circle which touches the curve $x^2+xy-y^2=4$ at $(2,2)$ and the line $3x-y+6=0$ For the first curve, the slope of the tangent is $3$
Let the circle be $x^2+y^2+2gx+2fy+c=0$
Then
$$\frac{2+g}{2+f} =-3$$
$$g+3f+8=0$$
Also the equation of tangent to a circle is
$$y=mx \pm r\sqrt{1+m^2}$$
From $y=3x+6$
$m=3$ and $6=r\sqrt{10}$
$$\implies g^2+f^2 -c=\frac{18}{5}$$
Also the circle passes through $(2,2)$, so
$$4g+4f+8+c=0$$
Solving all these equations, I didn’t get the right answer.
What is wrong in my method, and is there a shorter way to solve it?
|
You have used the equation for tangent to a circle centred at origin but you don't know the centre is at the origin.
The tangent to $x^2+xy-y^2=4$ at $A(2,2)$ is parallel to the line $y=3x+6$. Find the point $B$ on $y=3x+6$ so that $AB$ has slope $-1/3$ i.e. $AB$ is perpendicular to the tangent and the line. The centre of the circle will lie on the midpoint of $AB$ and the radius will be the distance between $AB$ divided by $2$.
You will find $B(-1,3)$ giving the equation $(x-1/2)^2+(y-5/2)^2=2.5$.
|
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|
Proving $\frac{7 + 2b}{1 + a} + \frac{7 + 2c}{1 + b} + \frac{7 + 2a}{1 + c} \geqslant \frac{69}{4}$. Here's the inequality
For positive variables, if $a+b+c=1$, prove that
$$
\frac{7 + 2b}{1 + a} +
\frac{7 + 2c}{1 + b} +
\frac{7 + 2a}{1 + c} \geqslant
\frac{69}{4}
$$
Here equality occurs for $a=b=c=\frac{1}{3}$ which is not-so-usual, so I decide to write the inequality as
$$
\frac{21 + 2q}{3 + p} +
\frac{21 + 2r}{3 + q} +
\frac{21 + 2p}{3 + r} \geqslant
\frac{69}{4}
$$Where the constraint now is $p + q + r = 3$ and the equality occurs for $p = q = r = 1$. Now we are left to proving that just
$$
2\sum_{cyc}{\frac{q}{3 + p}} +
21\sum_{cyc}{\frac{1}{3 + p}} \geqslant \frac{69}{4}
$$
Now it is sufficient to prove that
$$
\sum_{cyc}\frac{q}{3 + p}\geqslant\frac{3}{4} \quad \textrm{and} \quad \sum_{cyc}\frac{1}{3 + p} \geqslant \frac{3}{4}
$$ The second is true but I can't prove that the first is true.
|
By C-S $$\sum_{cyc}\frac{7+2b}{1+a}=\sum_{cyc}\left(\frac{7+2b}{1+a}-\frac{7}{2}\right)+\frac{21}{2}=\frac{1}{2}\sum_{cyc}\frac{4b+7(1-a)}{1+a}+\frac{21}{2}=$$
$$=\frac{1}{2}\sum_{cyc}\frac{11b+7c}{1+a}+\frac{21}{2}=\frac{1}{2}\sum_{cyc}\frac{(11b+7c)^2}{(11b+7c)(1+a)}+\frac{21}{2}\geq$$
$$\geq\frac{1}{2}\frac{324(a+b+c)^2}{\sum\limits_{cyc}(11b+7c)(1+a)}+\frac{21}{2}=\frac{162(a+b+c)^2}{18(a+b+c)^2+18(ab+ac+bc)}+\frac{21}{2}\geq$$
$$\geq\frac{162(a+b+c)^2}{18(a+b+c)^2+6(a+b+c)^2}+\frac{21}{2}=\frac{69}{4}.$$
|
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|
Difference between anti-derivative and indefinite integral My teacher gave me the following integral to evaluate:
$$\int \frac{x^2}{(x\sin(x)+\cos(x))^2}dx$$
After half an hour of uselessly fumbling around with trig identities I gave up and plugged it into an integral calculator: https://www.integral-calculator.com/. However I am confused: it displayed ANTIDERIVATIVE COMPUTED BY MAXIMA as $$-\dfrac{\left(2x^2-2\right)\sin\left(2x\right)+4x\cos\left(2x\right)}{\left(x^2+1\right)\sin^2\left(2x\right)+4x\sin\left(2x\right)+\left(x^2+1\right)\cos^2\left(2x\right)+\left(2-2x^2\right)\cos\left(2x\right)+x^2+1}+C$$ and I pressed the simplify button to obtain $$-\dfrac{\left(x^2-1\right)\cos\left(x\right)\sin\left(x\right)+2x\cos^2\left(x\right)-x}{\left(x^2-1\right)\sin^2\left(x\right)+2x\cos\left(x\right)\sin\left(x\right)+1}+C$$ However the "MANUALLY" COMPUTED ANTIDERIVATIVE displayed the following $$\dfrac{\sin\left(x\right)-x\cos\left(x\right)}{x\sin\left(x\right)+\cos\left(x\right)}+C$$ Which was computed by the following method:
$$\int \frac{x^2}{(x\sin(x)+\cos(x))^2}dx= \int \Bigg(\frac{x\sin(x)}{x\sin(x)+cos(x)}-\frac{x\cos(x)(\sin(x)-x\cos(x))}{(x\sin(x)+\cos(x))^2}\Bigg) dx$$
Using integration by parts
$$ \int \frac{x\cos(x)(\sin(x)-x\cos(x))}{(x\sin(x)+\cos(x))^2} dx= \dfrac{\sin\left(x\right)-x\cos\left(x\right)}{x\sin\left(x\right)+\cos\left(x\right)}+ \int\frac{x\sin(x)}{x\sin(x)+cos(x)}dx$$
$$\Rightarrow \int \frac{x^2}{(x\sin(x)+\cos(x))^2}dx=\dfrac{\sin\left(x\right)-x\cos\left(x\right)}{x\sin\left(x\right)+\cos\left(x\right)}+ \int\frac{x\sin(x)}{x\sin(x)+cos(x)}dx-\int\frac{x\sin(x)}{x\sin(x)+cos(x)}dx=\dfrac{\sin\left(x\right)-x\cos\left(x\right)}{x\sin\left(x\right)+\cos\left(x\right)}+C $$
My question is: why do I get different results from computing the anti-derivative and the indefinite integral? I simplified the anti-derivative so shouldn't it be simplified to the indefinite integral above? Are these two equations equal? Are functions for anti-derivatives and indefinite integrals vastly different? Any help will be appreciated
|
For example, because $$\begin{align*}&(x^2-1)\sin^2x+2x\sin{x}\cos{x}+1\\&=(x^2-1)\sin^2x+2x\sin{x}\cos{x}+\sin^2x+\cos^2x\\&=x^2\sin^2x+2x\sin{x}\cos{x}+\cos^2x\\&=(x\sin{x}+\cos{x})^2.\end{align*}$$
Now, what does happen in the numerator?
We have the following:
$$\begin{align*}&(x^2-1)\cos{x}\sin{x}+2x\cos^2x-x\\&=(x^2-1)\cos{x}\sin{x}+2x\cos^2x-x\sin^2x-x\cos^2x\\&=x\cos^2x+(x^2-1)\cos{x}\sin{x}-x\sin^2x\\&=(x\sin{x}+\cos{x})(x\cos{x}-\sin{x}).\end{align*}$$
I hope now it's clear.
|
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|
Find the equation of line $AP$ in triangle $ABC$ where $P$ will make $PA=PB=PC$ I was recently doing COMC (Canadian Opening Mathematic Challenge) past year exams. Here is another question that I want to ask!
Here is the question:
Triangle $ABC$ has its sides determinded in the following way: side $AB$ by line $3x-2y+3 = 0$, side $BC$ by line $x+y-14 = 0$, and side $AC$ by line $y=3$. If the point $P$ is chosen so that $PA=PB=PC$, determine the equation of the line containing $A$ and $P$
So I successfully figured out the three coordinates of the triangle $ABC$, which are (1,3), (5,9), (11,3). Now I can't find the point $P$ that makes $PA=PB=PC$. Hope someone would like to help me!
Thank you very much!
This question is from COMC 1998 Part B Question 1.
|
In other words, the point $P$
is the center of the circumscribed circle,
and
the coordinates of the circumcenter
is known to be found as
\begin{align}
P&=
\frac{a^2(b^2+c^2-a^2)\cdot A+b^2(a^2+c^2-b^2)\cdot B+c^2(b^2+a^2-c^2)\cdot C}
{a^2(b^2+c^2-a^2)+b^2(a^2+c^2-b^2)+c^2(b^2+a^2-c^2)}
\tag{1}\label{1}
,
\end{align}
where $a=|BC|$, $b=|AC|$, $c=|AB|$.
For $A=(1,3),\ B=(5,9),\ C=(11,3)$
the coordinates of the circumcenter
is therefore $P=(6,4)$,
and the question is simplified to:
find the equation of the line through
the two distinct points with known coordinates.
|
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|
Find all values of the real parameter $a$ for which the equation $4x^4+(8+4a)x^3+(a^2+8a+4)x^2+(a^3+8)x+a^2=0$ has only real roots
Find all values of the real parameter a for which the equation $$4x^4+(8+4a)x^3+(a^2+8a+4)x^2+(a^3+8)x+a^2=0$$ has only real roots.
Obviously as soon as you factor this equation to $$(4x^2+8x+a^2)(x^2+ax+1)=0$$ then you have finished.
However, how am I supposed to think of factoring the equation in such a manor? That you are supposed to factor it, is obvious, but how can you find out which are its factors, aside from blind luck?
|
Observing and categorizing the coefficients can be effective here.
Rewrite the equation into
$$(4x^4+8x^3+a^2x^2)+(4ax^3+8ax^2+a^3x)+(4x^2+8x+a^2)=0\\\Longrightarrow x^2(4x^2+8x+a^2)+ax(4x^2+8x+a^2)+(4x^2+8x+a^2)=0\\\Longrightarrow (x^2+ax+1)(4x^2+8x+a^2)=0$$
Or in this way:
$$(4x^4+4ax^3+4x^2)+(8x^3+8ax^2+8x)+(a^2x^2+a^3x+a^2)=0\\\Longrightarrow 4x^2(x^2+ax+1)+8x(x^2+ax+1)+a^2(x^2+ax+1)=0\\\Longrightarrow (4x^2+8x+a^2)(x^2+ax+1)=0$$
Does this help? I'm not sure if you are asking for a general way or just for this one problem.
|
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|
Solving a functional equation with derivatives? Am I doing this correctly?
Given
\begin{align}
f(x) = \frac{x^n}{1+x^n} - kf(x-1)
\end{align}
, then evaluating its the derivative
\begin{align}
\frac{df}{dx}=\frac{nx^{n-1}}{(1+x^n)^2}-k\frac{df}{dx}(x-1).
\end{align}
Let $t=x-1$ so $x=t+1$ and $dx=dt$. Thus,
\begin{align}
\frac{df}{dt} &= \frac{n(t+1)^{n-1}}{(1+(t+1)^n)^2} - k\frac{df}{dt}\\
\Leftrightarrow (k+1)\frac{df}{dt} &= \frac{n(t+1)^{n-1}}{(1+(t+1)^n)^2}\\
\Leftrightarrow \frac{df}{dt} &= \biggl(\frac{n}{k+1}\biggr)\frac{(t+1)^{n-1}}{(1+(t+1)^n)^2} \\
\Leftrightarrow \frac{df}{dt} &= \biggl(\frac{n}{k+1}\biggr)\frac{(x)^{n-1}}{(1+(x)^n)^2}\\
&= \biggl(\frac{1}{k+1}\biggr)\frac{nx^{n-1}}{(1+x^n)^2}\\
\Leftrightarrow f(x) &= \biggl(\frac{1}{k+1}\biggr)\frac{x^n}{1+x^n}
\end{align}
EDIT
As pointed out in the comments. My solution is incorrect. Can anyone please help me solve this?
|
Formally a solution is given by
$$
f_0(x) = \sum_{j = 0}^\infty (-k)^j\frac{(x-j)^n}{1+(x-j)^n}
$$
One can expect convergence if $|k| < 1$. Mathematica can evaluate this for certain choices of $k$ and $n$. For example, if $k = 1/2, \, n = 2$, then
$$
f_0(x) = \frac{1}{6} \left(4+3 i \left(-\frac{1}{2}\right)^{x-i}
\left(\left(-\frac{1}{2}\right)^{2 i}
B_{-\frac{1}{2}}(-x-i,0)-B_{-\frac{1}{2}}(i-x,0)\right)\right)
$$
where $B$ is the Beta function. For $n = 1$, we obtain
$$
f_0(x) = \left(-k\right){}^{x+1} B_{-k}(-x,-1)-x \Phi
\left(-k,1,-x-1\right)
$$
where $\Phi$ is the Lerch transcendent.
In other cases, the answer may be expressed in terms of hypergeometric functions.
|
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|
On the quantity $I(q^k) + I(n^2)$ where $q^k n^2$ is an odd perfect number with special prime $q$ The topic of odd perfect numbers likely needs no introduction.
In what follows, we let $\sigma(x)$ denote the sum of divisors of the positive integer $x$. Let
$$D(x) = 2x - \sigma(x)$$
denote the deficiency of $x$, and let
$$s(x) = \sigma(x) - x$$
denote the sum of aliquot/proper divisors of $x$. Finally, denote the abundancy index of $x$ by
$$I(x) = \frac{\sigma(x)}{x}.$$
Euler proved that a hypothetical odd perfect number must have the form $N = q^k n^2$ where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Since $q$ is prime and $N$ is perfect, we obtain
$$\frac{q+1}{q} \leq I(q^k) < \frac{q}{q-1}$$
so that we get
$$\frac{2(q-1)}{q} < I(n^2) = \frac{2}{I(q^k)} \leq \frac{2q}{q+1}.$$
Using the identity
$$\frac{D(x)}{x} = 2 - I(x)$$
we obtain the bounds
$${q^k}\bigg(\frac{q-2}{q-1}\bigg) < D(q^k) \leq {q^{k-1}}\bigg(q-1\bigg)$$
and
$$\frac{2n^2}{q+1} \leq D(n^2) < \frac{2n^2}{q}.$$
This implies that
$${2q^k n^2}\cdot\bigg(\frac{q-2}{(q-1)(q+1)}\bigg) < D(q^k)D(n^2) < {2q^k n^2}\cdot\bigg(\frac{q-1}{q^2}\bigg).$$
Dividing both sides of the last inequality by $2q^k n^2$, we get
$$\frac{q-2}{(q-1)(q+1)} < \frac{D(q^k)D(n^2)}{2q^k n^2} < \frac{q-1}{q^2}.$$
But we know that
$$D(q^k)D(n^2)=2s(q^k)s(n^2),$$
which can be verified by a direct, brute-force computation.
Thus, the fraction in the middle of the last inequality simplifies to
$$\frac{D(q^k)D(n^2)}{2q^k n^2}=\bigg(\frac{s(q^k)}{q^k}\bigg)\bigg(\frac{s(n^2)}{n^2}\bigg)=\bigg(I(q^k) - 1\bigg)\bigg(I(n^2) - 1\bigg) = 3 - \bigg(I(q^k) + I(n^2)\bigg).$$
We therefore finally have the bounds
$$3 - \bigg(\frac{q-1}{q^2}\bigg) < I(q^k) + I(n^2) < 3 - \bigg(\frac{q-2}{(q-1)(q+1)}\bigg)$$
which does not improve on the known bounds
$$3 - \bigg(\frac{q-2}{q(q-1)}\bigg) < I(q^k) + I(n^2) \leq 3 - \bigg(\frac{q-1}{q(q+1)}\bigg).$$
(See this paper for a proof.)
Here are my:
QUESTIONS Is it possible to improve on the bounds for $D(q^k)$ and $D(n^2)$ (where $q^k n^2$ is an odd perfect number with special prime $q$) to hopefully produce stronger bounds for $I(q^k) + I(n^2)$? If so, how could this be done?
|
Let me try to work backwards from
$$3 - \bigg(\frac{q-2}{q(q-1)}\bigg) < I(q^k) + I(n^2) \leq 3 - \bigg(\frac{q-1}{q(q+1)}\bigg).$$
This can be rewritten as
$$\frac{q-1}{q(q+1)} \leq 3 - \bigg(I(q^k) + I(n^2)\bigg) = \frac{D(q^k)D(n^2)}{2q^k n^2} < \frac{q-2}{q(q-1)}.$$
We also have
$$\frac{2n^2}{q+1} \leq D(n^2) < \frac{2n^2}{q},$$
which we can rewrite as
$$q < \frac{2n^2}{D(n^2)} \leq q + 1.$$
We therefore obtain
$$\frac{q-1}{q+1} = \frac{q(q-1)}{q(q+1)} < \bigg(\frac{q-1}{q(q+1)}\bigg)\cdot\bigg(\frac{2n^2}{D(n^2)}\bigg) \leq \frac{D(q^k)}{q^k} < \bigg(\frac{q-2}{q(q-1)}\bigg)\cdot\bigg(\frac{2n^2}{D(n^2)}\bigg) \leq \frac{(q+1)(q-2)}{q(q-1)},$$
which implies that
$$\frac{q-1}{q+1} < 2 - I(q^k) < \frac{(q+1)(q-2)}{q(q-1)},$$
which simplifies to
$$\frac{q^2 - q + 2}{q^2 - q} = 2 - \bigg(\frac{(q+1)(q-2)}{q(q-1)}\bigg) < I(q^k) < 2 - \bigg(\frac{q-1}{q+1}\bigg) = \frac{q+3}{q+1}.$$
Note that both bounds do not improve on the currently known
$$\frac{q+1}{q} \leq I(q^k) < \frac{q}{q-1}.$$
|
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|
How to find the dot product using the law of cosines I'm working with the following problem:
We have a triangle with sides $AB=3$ and $BC=2$, the angle $ABC$ is 60 degrees. Find the dot product $AC \cdotp AB$
Since we don't actually know the side $AC$ my first step is to calculate this side via the law of cosines.
$$AC^2=AB^2 +BC^2 -2AB\cdot BC\cos x$$
$$\implies AC^2=3^2 +2^2 -2\cdot2\cdot3\cos 60^\circ$$
$$\implies AC^2=9 +4 -12\cdot0.5$$
$$\implies AC^2=13-6=7$$
$$\implies AC=\sqrt 7.$$
My next step is to calculate the angle $BAC$; we should be able to use the law of cosines here as well:
$$BC^2=AB^2+AC^2-2\cdot AB\cdot AC\cos x$$
$$\implies4=9+(\sqrt 7)^2 -2\cdot3\sqrt 7\cos x$$
$$\implies4=9+7 -6\sqrt 7\cos x$$
$$\implies-2=-6\sqrt 7\cos x$$
$$\implies\frac{1}{3}=\sqrt 7\cos x$$
$$\implies\frac{1}{3\sqrt 7}=\cos x.$$
If we want to calculate our dot product using only the vector lengths, we would use the fact that $A\cdotp B=|A||B|\cos x$, which in this case would mean that:
$$AB \cdot AC=3\sqrt7\frac{1}{3\sqrt 7}=1,$$
which is quite wrong since the answer is supposed to be $12$.
Can someone please tell me where I've made a mistake?
In the proposed solution another method seem to be used. The textbook claims that $AB \cdotp AC = AB \cdotp (AB + BC)=AB \cdotp AB + AB \cdotp BC = |AB|^2 + |AB||BC|cos(60)=9+6*\frac{1}{2}=12$
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$$\cos\measuredangle BAC=\frac{3^2+(\sqrt7)^2-2^2}{2\cdot3\cdot\sqrt7}=\frac{2}{\sqrt7},$$
which gives $$\vec{AC}\cdot\vec{AB}=3\cdot\sqrt7\cdot\frac{2}{\sqrt7}$$
In your solution this statement is wrong: $-2=-6\sqrt7\cos{x}$.
It should be $$-12=-6\sqrt7\cos{x}.$$
About you last adding.
It should be
$$\vec{AB}\cdotp\vec{AC}=\vec{AB}\cdotp (\vec{AB} + \vec{BC})=\vec{AB}\cdotp\vec{AB}+\vec{AB}\cdotp\vec{BC} =$$
$$=|AB|^2+ |\vec{AB}||\vec{BC}|\cos120^{\circ}=9-6\cdot\frac{1}{2}=6$$
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Need help with complex equation: |z−i|+|z+i|=2 I am trying to solve this equation: |z−i|+|z+i|=2 and don't know how to do it. This what I have:
$$\sqrt{(x+1)^2+y^2}+ \sqrt{(x-1)^2+y^2} = 2 /^2$$
$$(x+1)^2+y^2+(x-1)^2+y^2 + 2\sqrt{[(x+1)^2+y^2][(x-1)^2+y^2]} = 4$$
$$x^2 +2x + 1+y^2+x^2-2x + 1+y^2 + 2\sqrt{[(x^2 +2x + 1)+y^2][(x^2-2x + 1)+y^2]}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{[(x^2 +2x + 1)+y^2][(x^2-2x + 1)+y^2]}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{(x^2 +2x + 1)(x^2-2x + 1)+x^2y^2-2xy^2+y^2+x^2y^2-2xy^2+y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{(x^2 +2x + 1)(x^2-2x + 1)+2x^2y^2-4xy^2+2y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{x^4+2x^3+x^2-2x^3+4x^2-2x+x^2+2x+1+2x^2y^2-4xy^2+2y^2+y^4}=4$$
$$2x^2+2y^2 + 2 + 2\sqrt{x^4+6x^2+1+2x^2y^2-4xy^2+2y^2+y^4}=4$$
And I have no idea what to do next.
Any help would be much appreciated
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Triangle inequality tells us $|z+i| + |z-i| \ge |(z+i)-(z-i)| = |2i| = 2$ with equality holding if and only if geometrically of $0$ is colinear and between $z+i$ and $z-i$.
The only values colinear between $z + i$ and $z-i$ must all have the same real value so the real part as $z$ has and if $0$ is one such number, the real part of $z$ must be $0$.
If $Im(z) = b$ then $z = bi$ and we have $|bi+i| + |bi - i| = |b+1|+|b-1|=2$. By cases we can verify then this will hold if and only if $b \in [-1,1]$.
(If $b > 1$ then $|b+1| > 0$ and $|b-1| > 2$ and if $b < -1$ then $|b+1| > 2$ and $|b-1| > 0$; but if $-1 < b < 1$ then $|b+1| = b+1$ and $|b-1|= 1-b$ and $|b+1|+|b-1| = (b+1)+(1-b)=2$.)
So $z$ can be any $bi; -1 \le b \le 1$.
======
Alternatively (in hindsight) note:
If $z = a + bi$ then $|z\pm i| = \sqrt{a^2 + (b\pm 1)^2}$ so
$|z+i| +|z-i|$
$=\sqrt {a^2 + (b+ 1)^2} + \sqrt {a^2 + (b-1)^2}$
$[*]\ge\sqrt {(b+1)^2} +\sqrt {(b-1)^2}$
$ =|b+1|+|b-1|$
$=\begin{cases}(b+1)+(b-1)=2b>2& b> 1\\ (b+1)+(1-b)=2 &-1\le b \le 1\\(-b-1)+(1-b)=-2b>2 & b < -1\end{cases}$
$[**] \ge 2$
with equality holding if and only if $a = 0$ and $-1 \le b \le 1$.
[*] Equality holding if and only if $a = 0$.
[**] Equality holding if and only if $-1 \le b \le 1$.
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Using De Moivre to show $\tan6\theta=\frac{6\tan\theta-20\tan^3\theta+6\tan^5\theta}{1-15\tan^2\theta+15\tan^4\theta-\tan^6\theta}$
Use the De Moivre Theorem to show that
$$\tan6\theta=\frac{6\tan\theta-20\tan^3\theta+6\tan^5\theta}{1-15\tan^2\theta+15\tan^4\theta-\tan^6\theta}$$
I got this question on my exam today and had no idea how to do it.
There must be a connection to complex numbers since its the topic its connected to. I only learnt how to use the De Moivre theorem with complex numbers in polar form so I was completely lost here.
Sorry if its a dumb question :<.
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By De Moivre's theorem we have
$$\cos(6\theta)+i\sin(6\theta)=(\cos(\theta)+i\sin(\theta))^6$$
$$=-\sin^6(\theta) + \cos^6(\theta) + 6 i \sin(\theta) \cos^5(\theta) - 15 \sin^2(\theta) \cos^4(\theta) - 20 i \sin^3(\theta) \cos^3(\theta) + 15 \sin^4(\theta) \cos^2(\theta) + 6 i \sin^5(\theta) \cos(\theta)$$ by the Binomial expansion. Then comparing the real and imaginary parts we have:
$$\cos(6\theta)=-\sin^6(\theta) + \cos^6(\theta)- 15 \sin^2(\theta) \cos^4(\theta) + 15 \sin^4(\theta) \cos^2(\theta)$$
$$\sin(6\theta)= 6 \sin(\theta) \cos^5(\theta) - 20 \sin^3(\theta) \cos^3(\theta) + 6 \sin^5(\theta) \cos(\theta)$$
So we have $$\tan(5\theta)=\frac{\sin(6\theta)}{\cos(6\theta)}$$
$$=\frac{6 \sin(\theta) \cos^5(\theta) - 20 \sin^3(\theta) \cos^3(\theta) + 6 \sin^5(\theta) \cos(\theta)}{-\sin^6(\theta) + \cos^6(\theta)- 15 \sin^2(\theta) \cos^4(\theta) + 15 \sin^4(\theta) \cos^2(\theta)}$$
$$=\frac{6\tan\theta-20\tan^3\theta+6\tan^5\theta}{1-15\tan^2\theta+15\tan^4\theta-\tan^6\theta}$$
after dividing through by $\cos^6(\theta)$
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Vectors : finding scalar $\mu$ in $\overrightarrow a = \mu \overrightarrow b + 4\overrightarrow c $
Non-zero vectors $\overrightarrow a ,\overrightarrow b \& \overrightarrow c $ satisfy $\overrightarrow a .\overrightarrow b = 0$, $\left( {\overrightarrow b - \overrightarrow a } \right).\left( {\overrightarrow b + \overrightarrow c } \right) = 0$ and $2\left| {\overrightarrow b + \overrightarrow c } \right| = \left| {\overrightarrow b - \overrightarrow a } \right|$. If $\overrightarrow a = \mu \overrightarrow b + 4\overrightarrow c $, then find all possible values of $\mu $.
My approach is as follows
Given $\left( {\overrightarrow b - \overrightarrow a } \right).\left( {\overrightarrow b + \overrightarrow c } \right) = 0\& 2\left| {\overrightarrow b + \overrightarrow c } \right| = \left| {\overrightarrow b - \overrightarrow a } \right|$
$\overrightarrow u = \left( {\overrightarrow b - \overrightarrow a } \right);\overrightarrow v = \left( {\overrightarrow b + \overrightarrow c } \right)$
$2\left| {\overrightarrow v } \right| = \left| {\overrightarrow u } \right| \Rightarrow 4\left( {{{\left| {\overrightarrow b } \right|}^2} + {{\left| {\overrightarrow c } \right|}^2} + 2\overrightarrow b .\overrightarrow c } \right) = \left( {{{\left| {\overrightarrow b } \right|}^2} + {{\left| {\overrightarrow a } \right|}^2} - 2\overrightarrow a .\overrightarrow c } \right)$
$\overrightarrow a = \mu \overrightarrow b + 4\overrightarrow c \Rightarrow \overrightarrow b - \overrightarrow u = \mu \overrightarrow b + 4\left( {\overrightarrow v - \overrightarrow b } \right)$
$ \Rightarrow \overrightarrow b .\overrightarrow v - \overrightarrow u .\overrightarrow v = \mu \overrightarrow b .\overrightarrow v + 4\left( {\overrightarrow v .\overrightarrow v - \overrightarrow b .\overrightarrow v } \right) \Rightarrow 5\overrightarrow b .\overrightarrow v = \mu \overrightarrow b .\overrightarrow v + 4{\left| {\overrightarrow v } \right|^2}$\
$ \Rightarrow 5\overrightarrow b .\left( {\overrightarrow b + \overrightarrow c } \right) = \mu \overrightarrow b .\left( {\overrightarrow b + \overrightarrow c } \right) + 4{\left| {\overrightarrow v } \right|^2} \Rightarrow 5\overrightarrow b .\overrightarrow b + 5\overrightarrow b .\overrightarrow c = \mu \overrightarrow b .\overrightarrow b + \mu \overrightarrow b .\overrightarrow c + 4{\left| {\overrightarrow b } \right|^2} + 4{\left| {\overrightarrow c } \right|^2} + 8\overrightarrow b .\overrightarrow c $
Not able to proceed from here onward
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Let $\vec a=t \vec i$, $\vec b= s \vec j$ then $\vec c=(t\vec i-\mu s \vec j)/4$.
$$2|\vec b+ \vec c|=|\vec a- \vec b| \implies |s \vec j+(t/4) \vec i-(s\mu/4)\vec j|=|t \vec i- s\vec j|$$
$$\implies (4-\mu)^2s^2+t^2=t^2+s^2\implies (\mu^2-8\mu+12)s^2=3t^2~~~(1)$$
Next $$(\vec b-\vec c).(\vec a- \vec b)=0 \implies t^2=(4-\mu) s^2~~~(2)$$
From (1) and (2) equate $t^2/s^2$, to get
$$3(4-\mu)=\mu^2-8\mu+12 \implies \mu^2-5\mu=0 \implies \mu =0, 5$$
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Why is $\sum_{k=1}^{\infty}\left(\frac{1}{3}\right)^{k}=\frac{1/3}{2/3}$ $\sum_{k=1}^{\infty}\left(\frac{1}{3}\right)^{k}=\frac{1/3}{2/3}$ what theorem or algebra leads to this equality?
EDIT: The sum should have been infinite.
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Sum should be infinite, because for finite sum you should get:
$$\sum_{k=0}^{n}(\frac{1}{3})^k=1+\frac{1}{3}+\dots+(\frac{1}{3})^n=\frac{1-(\frac{1}{3})^{n+1}}{1-\frac{1}{3}},$$ where we used formula $1+q+\dots+q^n=\frac{1-q^{n+1}}{1-q}$ for $q=\frac{1}{3}$.
Therefore,
$$\sum_{k=1}^{\infty}\left(\frac{1}{3}\right)^{k}=\frac{1}{3}\sum_{k=0}^\infty \left(\frac{1}{3}\right)^{k}=\frac{1}{3}\lim_{k\to\infty}\sum_{i=0}^k (\frac{1}{3})^i=\frac{1}{3}\lim_{k\to\infty}\frac{1-(\frac{1}{3})^{k+1}}{1-\frac{1}{3}}=\frac{1}{3}\frac{1}{\frac{2}{3}}=\frac{1}{2},$$
since
$$\lim_{k\to\infty} (\frac{1}{3})^{k+1}=0.$$
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Prove by induction that $3^{4n+2}+1$ is divisible by $5$ when $n \ge 0.$
Prove by induction that $3^{4n+2}+1$ is divisible by $5$ when $n \ge 0.$
(1) When $n=0$ we have that $3^2+1 = 10$ which is divisible by $5$ clearly.
(2) Assuming that the condition hold for $n=k.$
(3) Proving that it holds for $n=k+1$
$$3^{4(k+1) + 2} + 1 = 3^{4k + 6} + 1 = 3^4 \cdot 3^{4k+2} + 1$$
Since we assumed that $5 \mid 3^{4k+2} + 1$ we have that
$$3^4 \cdot 3^{4k+2} + 1 = 3^4 \cdot 5t, \text{ where $t \in \Bbb Z$}.$$
Thus $5 \mid 3^{4(k+1) + 2} + 1$.
Is this a valid proof? I'm not entirely sure I'm correct with this...
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Hint: $3^4 \cdot 3^{4k+2} + 1= 3^4(3^{4k+2}+1)+1-3^4=3^4(3^{4k+2}+1)-80.$
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$\lim_{n\to\infty}\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})$ I need to find $\lim_{n\to\infty}{\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})}$ without using L'Hopital's rule, derivatives or integrals.
Empirically, I know such limit exists (I used a function Grapher and checked in wolfram) and it's equal to $-\frac{1}{4}$. I noticed that
$$\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})=\sqrt{n^3} \Big(\frac{1}{\sqrt{n+1}+\sqrt{n}}-\frac{1}{\sqrt{n+1}+\sqrt{n-1}}\Big) $$
That doesn't seem to lead to $-\frac{1}{4}$ when $n\to \infty$ . I tried another form of the original expression:
$$\sqrt{n^3}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n})=2\sqrt{n^3} \Bigg(\frac{\sqrt{n^2 - 1} - n} {\sqrt{n + 1} + \sqrt{n - 1} + 2\sqrt{n}}\Bigg)$$
If I multiply by the conjugate, we obtain
$$2\sqrt{n^3} \Bigg(\frac{\sqrt{n^2 - 1} - n} {\sqrt{n + 1} + \sqrt{n - 1} + 2\sqrt{n}}\Bigg)=-\frac{2\sqrt{n^3}}{\big(\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n}\big)\big( \sqrt{n^2-1}+n\big)}$$
Now that doesn't seem to be of any use either. Any ideas?
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The last expression
$$A=-\frac{2\sqrt{n^3}}{\big(\sqrt{n+1}+\sqrt{n-1}+2\sqrt{n}\big)\big( \sqrt{n^2-1}+n\big)}$$
leads to the result.
Since $n$ is large
$$\sqrt{n+1}\sim \sqrt{n} \qquad \sqrt{n-1}\sim \sqrt{n}\qquad \sqrt{n^2-1}\sim \sqrt{n^2}=n$$
$$A \sim -\frac{2\sqrt{n^3}}{\big(\sqrt{n}+\sqrt{n}+2\sqrt{n}\big)\big( n+n\big)}=-\frac{2n\sqrt{n}}{\big(4\sqrt{n}\big)\big( 2n\big)}=-\frac 14$$
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Distance between the curves from a particular point The distance between the points $P(u,v)$ and the curve $x^2+4x+y^2=0$ is the same as the distance between the points $P(u,v)$ and $M(2,0)$. If $u$ and $v$ satisfy the relation $u^2-\frac{v^2}{q}=1$, then $q$ is greater than or equal to
(A) 1
(B) 2
(C) 3
(4) 4
My approach is as follow Curve $x^2+4x+y^2$ represent a circle with centre $(-2,0)$ and radius $2$.
The parametric equation of the circle is $-2cos\theta+2$ and $2sin\theta$
$D^2=(u-2)^2+v^2$
Also $D^2=(u+2cos\theta-2)^2+(v-2sin\theta)^2$
How we will proceed from here
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Distance $D_1$ of P(u,v) from the circle is $D_1=\sqrt{(u+2)^2+v^2}-2=\sqrt{F}-2$
and $D_2=\sqrt{(u-2)+v^2}=\sqrt{G}$, we require
$$D_1=D_2 \implies \sqrt{(u+2)^2+v^2}-2=\sqrt{(u-2)+v^2} \implies \sqrt{F}-\sqrt{G}=2~~~(1)$$ We also have $$F-G=8u \implies \sqrt{F}+\sqrt{G}=4u~~~~(2)$$
From (1) and (2), we have $$\sqrt{F}=1+2u \implies u^2+4u+4+qu^2-q=1+4u+4u^2 \implies (q-3) u^2=(q-3)\implies q=3$$
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Is the absolute value on matrices well-defined with respect to PSD cone? Consider for two matrices $A, B$ we say $A \geq B $ if $ A - B$ is positive semidefinite.
Also, there exists a definition of absolute value for a matrix $A$ that is $|A| = \sqrt{A^\dagger A}$.
I was studying Hermitian matrices(matrices like $A$ with complex entries such that $A^\dagger = A$) and faced with a problem, is the following statement true for two arbitrary Hermitian matrices?
$$ A \geq B \geq -A \implies |A| \geq |B|$$
Note that if $A$ is a Hermitian matrix then $|A|$ is PSD and the eigenvalues of $|A|$ are equal to the absolute value of the eigenvalues of $A$.
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The statement is false. Consider
$$A=\begin{bmatrix} 6 & 0 \\ 0 & 2\end{bmatrix}, \quad B=\begin{bmatrix} 1 & 2 \\ 2 & 1\end{bmatrix}.$$
We have $$A+B = \begin{bmatrix} 7 & 2 \\ 2 & 3\end{bmatrix} \ge 0, \quad A-B = \begin{bmatrix} 5 & -2 \\ -2 & 1\end{bmatrix} \ge 0$$
so $A \ge B \ge -A$. On the other hand, we have
$$|A| = A, \quad |B| = \sqrt{B^*B} = \sqrt{\begin{bmatrix} 5 & 4 \\ 4 & 5\end{bmatrix}} = \begin{bmatrix} 2 & 1 \\ 1 & 2\end{bmatrix}$$
and therefore $$|A|-|B| = A-|B| = \begin{bmatrix} 4 & -1 \\ -1 & 0\end{bmatrix} \not\ge 0$$
so $|A| \ge |B|$ is false.
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Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+3)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$ without induction? I've been trying to solve the following problem:
Show that $1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}$ using the identity $1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$.
I had the following idea, we write:
$$1^2+2^2+\dots + (2n+1)^2=\frac{(2n+1)((2n+1)+1)(2(2n+1)+1)}{6}=k$$
Let's pretend the identity we want to prove is true, then:
$$1^2+3^2+ \dots +(2n+1)^2=\frac{(n+1)(2n+1)(2n+1)}{3}=j$$
We take then $k-x=j$ and solve for $x$. If the given identities are true, $x$ must be the sum of $2^2+4^2+\dots+(2n)^2$, and we have that
$$x= \frac{2n (n+1) (2 n+1)}{3} $$
We still don't know that $x=2^2+4^2+\dots+(2n)^2$ but that can be easily proved by induction. I'd like to know: Is there some "neater" way that doesn't involve induction?
Despite the tag, I'd like to see an induction-free demonstration. I chose that tag because I couldn't think of anything better to choose.
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$ 1^2+2^2+\dots + n^2=\frac{n(n+1)(2n+1)}{6}$ multiplying this by $ 2^2 $ gives the sum $ 2^2+4^2+\dots + (2n)^2 = \frac{2n(n+1)(2n+1)}{3} $ and subtracting this from the given identity results int the required equality.
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Tips finding the gcd of $(n^2-3n-1, n-4)$
I need tips on how to find the $gcd(n^2-3n-1, n-4)$ for any $n \in \mathbb{N}$.
I tried the two following methods (but got stuck):
Since I need to find the $gcd(n^2-3n-1, n-4)$, then there is a $q \in \mathbb{Z}$ which is the divisor of both $(n^3-3n-1)$ and $(n-4)$, giving each of them an answer of $q_1,q_2 ∈ \mathbb{Z}$ accordingly,
therefore, it is possible to write them as follow:
$n^2-3n-1 = q*q_1$
$n-4 = q*q_2$
Because $q,q_1,q_2 ∈ \mathbb{Z}$, multiplying them with each other produces another Integer (same with adding to them an Integer), so
$n^2-3n = q_3$ , $q_3 ∈ \mathbb{Z}$
$n = q_4$ , $q_4 ∈ \mathbb{Z}$
With this method, I got stuck here (didn't find a way how to continue).
Another method I've tried is:
According to the Euclidean Algorithm,
$n^2-3n-1 = (n-4)*q + r, q,r ∈ \mathbb{Z}$
Use $q=n$,
$n^2-3n-1 = (n-4)*n+r$ => $n-1=r$
$n^2-4n=(n-1)q+r$ (while $q=n$) => $-3n=r$
$n^2-n=-3n*n+r$ => $4n^2-n=4$
and so on...but also got stuck..
I'd appreciate any tips on how to continue with what I've tried or if I should try a different approach
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Bear in mind that $\gcd(a,b) = \gcd(a\pm kb, b)$ for all integers $k$.
So if $n^3 - 3n -1 = P(n)(n-4) + R(n)$ then $\gcd(n^2 -3n-1, n-4) = \gcd(n^2-3n-1 -P(n)(n-4), n-r) = \gcd(R(n), n-4)$.
And if we use synthetic division
$n^2 -3n -1 = n^2(n-4) +4n -3n -1=$
$n(n-4) +n-1=$
$n(n-4) +(n-4) +4-1=$
$n(n-4) + (n-4) + 3=$
$(n +1)(n-4) + 3$.
....
So $\gcd(n^2-3n-1,n-4)= \gcd((n+1)(n-4) + 3, n-4)=\gcd(3,n-4)=\gcd(3,n-4 +3)=\gcd(3,n-1)$.
So if $3|n-1$ or in other words if $n\equiv 1 \pmod 3$ then $\gcd(n^2-3n-1,n-4)=\gcd(3,n-1)=3$.
ANd if $3\not \mid n-1$ or in other words if $n\not \equiv 1$ or if $n\equiv 0, 2 \pmod 3$ and $n-1\equiv 2,1$ and $n-1$ and $3$ are relatively prime then $\gcd(n^2-3n-1,n-4)=\gcd(3,n-1)=1$.
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|
Why $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$? How we can easily show that $3\mid(5^{6n+6}-2^{2n+2})$ for all natural $n$. These conditions continue $3\mid(5^{6n+5}-2^{2n+3})$ and $3\mid(5^{6n+4}-2^{2n+4})$ and $3\mid(5^{6n+3}-2^{2n+1})$ and $3\mid(5^{6n+2}-2^{2n+2})$ and $3\mid(5^{6n+1}-2^{2n+1})$.
|
we have $5^2 \equiv 1\ (mod\ 3) \Rightarrow 5^{2(3n+3)} \equiv 1\ (mod\ 3) \Rightarrow 5^{6n+6} \equiv 1\ (mod\ 3) $
and we have $2^2 \equiv 1\ (mod\ 3) \Rightarrow 2^{2(n+1)} \equiv 1\ (mod\ 3) \Rightarrow 2^{2n+2} \equiv 1\ (mod\ 3) $ Thus $$5^{6n+6}-2^{2n+2} \equiv 0\ (mod\ 3)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find MacLaurin Series and radius of convergence $$\sin{2z^2}$$
$$\sin{z} = \sum_{n=0}^{\infty}(-1)^n \frac{z^{2n+1}}{(2n+1)!}$$
Let's replace z with $2z^2$ in the MacLaurin series:
$$\sin{2z^2} = \sum_{n=0}^{\infty}(-1)^n \frac{(2z^2)^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} (-1)^n
\frac{2^{2n+1}}{(2n+1)!} z^{4n+2}$$
$$= z^2 - \frac{2^3}{3!}z^6 + \frac{2^5}{5!}z^{10} -+ ... = z^2 - \frac{4}{3}z^6 + \frac{4}{15}z^{10} -+...$$
But what about the radius of convergence? How do I find this?
|
Isn’t it quite easy? Start with the standard Laurent series
$$
\sin(z)=\sum_{n=0}^\infty(-1)^n\frac{z^{2n+1}}{(2n+1)!}\,,
$$
make your substitution $z\to2z^2$, and get
$$
\sin(2z^2)=\sum_{n=0}^\infty(-1)^n\frac{2^{2n+1}z^{4n+2}}{(2n+1)!}\,,
$$
in which your $a_n$-term is $(-1)^n2^{2n+1}z^{4n+2}/(2n+1)!$.
Now form $\mid a_{n+1}/a_n\mid$ to get
$$
\left|\frac {a_{n+1}}{a_n}\right|=\frac{2^{2n+3}z^{4n+6}\big/(2n+3)!}{2^{2n+1}z^{4n+2}\big/(2n+1)!}
=\frac{(2n+1)!}{(2n+3)!}\frac{2^{2n+3}z^{4n+6}}{2^{2n+1}z^{4n+2}}=\frac{4z^4}{(2n+2)(2n+3)}\,.
$$
That rightmost thing clearly goes to zero, no matter what $z$ is, so your radius is infinite, as I’m sure you expected.
|
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|
Evaluate the triple integral $\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$ using spherical coordinates How to evaluate triple integral $$\iiint\limits_E\frac{yz\,dx\,dy\,dz}{x^2+y^2+z^2}$$
when $E$ is bounded by $x^2+y^2+z^2-x=0$?
I know that spherical coordinates mean that $$x=r\sin\theta\cos\varphi,\quad y=r\sin\theta\sin\varphi,\quad z=r\cos\theta$$
and this function in spherical coordinates is
\begin{align*}
&\iiint\limits_E\frac{yzdxdydz}{x^2+y^2+z^2} = \iiint\limits_E\frac{r^2\sin\theta}{r^2\sin^2\theta\cos^2\varphi + r^2\sin^2\theta\sin^2\varphi + r^2\cos^2\theta}drd\theta d\varphi = \\ &\iiint\limits_E\frac{r^2\sin\theta}{r^2(\sin^2\theta\cos^2\varphi + \sin^2\theta\sin^2\varphi + \cos^2\theta)}drd\theta d\varphi = \iiint\limits_E\frac{\sin\theta}{\sin^2\theta\cos^2\varphi + \sin^2\theta\sin^2\varphi + \cos^2\theta}drd\theta d\varphi
\end{align*}
but I don't know how to write $E$ as set and convert it to spherical coordinates, and also what happens with this function after conversion. Triple integrals is now topic for me and I have never used spherical coordinates before, so I would be grateful if anyone can help me with this.
|
$x^2+y^2+z^2-x=0 \implies (x-\frac{1}{2})^2 + y^2 + z^2 = (\frac{1}{2})^2$
So it is a sphere with radius $\frac{1}{2}$ centered at $(\frac{1}{2},0,0)$.
In spherical coordinates, $x = \rho \cos \theta \sin \phi, \, y = \rho \sin \theta \sin \phi, z = \rho \cos \phi \,$ where $\theta$ is the azimuthal angle and $\phi$ is the polar angle (just opposite of your convention).
$x^2 + y^2 + z^2 = \rho^2$ so substituting in the equation of our sphere,
$\rho^2 = \rho \cos \theta \sin \phi \implies \rho = \cos \theta \sin \phi$.
Now the part that you have to be careful about is the bounds of azimuthal angle. Please consider a circle in XY plane with center at $(\frac{1}{2},0)$ and radius of $\frac{1}{2}$ and equation of $r = \cos \theta$. $\theta$ varies betweeen -$\frac{\pi}{2}$ and $\frac{\pi}{2}$ (range of $\pi$ instead of $2 \pi$ for a sphere centered at the origin). It is same for this sphere.
So the integral becomes
$I = \displaystyle \int_{-\pi/2}^{\pi/2} \int_{0}^{\pi} \int_{0}^{\cos\theta \sin\phi} \rho^2 \sin \theta \sin^2 \phi \cos \phi \, d\rho \, d\phi \, d\theta$
And based on the given function, its integral above $XY$ plane and below will cancel each other.
If I was integrating over the part of the sphere only in the first octant, the integral will be
$I = \displaystyle \int_{0}^{\pi/2} \int_{0}^{\pi/2} \int_{0}^{\cos\theta \sin\phi} \rho^2 \sin \theta \sin^2 \phi \cos \phi \, d\rho \, d\phi \, d\theta = \frac{1}{72}$
|
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|
Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$? Does $\sqrt{1+\sqrt{2}}$ belong to $\mathbb{Q}(\sqrt{2})$?
We know the answer is of the form $ a + b \sqrt{2}$. Since $(a + b\sqrt{2})^2 = a^2 + 2ab\sqrt{2} + 2b^2 = 1 + \sqrt{2}$, the system we need to solve is
\begin{align*}
2ab &= 1 \\
a^2 + 2b^2 &= 1
\end{align*}
We write $b = \frac{1}{2a}$ and substitute this in the second equation.
\begin{align*}
a^2 + 2\left(\frac{1}{2a}\right)^2 &= 1 \\
a^2 + \frac{1}{2a^2} &= 1 \\
2a^4 - 2a^2 + 1 &= 0
\end{align*}
Let $z = a^2$, so $z^2 = a^4$. The equation is then
\begin{equation}
2z^2 - 2z + 1 = 0
\end{equation}
Using the quadratic formula we find $z = \frac{1 \pm i}{2}$. This worked out when checked. Thus $a = \sqrt{\frac{1 \pm i}{2}}$.
We then find $b$ using $a$ in our original system of equations.
\begin{align*}
\frac{1 \pm i}{2} + 2b^2 &= 1 \\
1 \pm i + 4b^2 &= 2 \\
\pm i + 4b^2 &= 1 \\
4b^2 &= 1 \pm i \\
2b &= \sqrt{1 \pm i} \\
b &= \frac{\sqrt{1 \pm i}}{2} \\
\end{align*}
Substituting $a$ and $b$ into the equation $2ab = 1$, leads to inconsistent solutions.
What do I need to reconsider? How can I improve my answer?
|
Note
\begin{align}
\sqrt{\sqrt{2}+1}&=\sqrt{(\sqrt2-1)(\sqrt2+1)} \cdot\sqrt{\sqrt{2}+1}
= \sqrt{\sqrt{2}-1} (1+\sqrt2)\\
\end{align}
with $a=b = \sqrt{\sqrt{2}-1}$, thus no rational simplification.
|
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|
Find the solution for Cauchy problem using transformation I'm learning differential equations for the first semester and I'm not sure how to even begin with the following problem:
Find the solution for Cauchy problem using transformation:
$\left\{\begin{matrix}
(1-t^2)x''-tx'+n^2x=0 & n \in \mathbb{Z}\\
y(0)=1 & y'(0)=0
\end{matrix}\right.$
Could you please help me?
|
In fact this belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0220.pdf
Let $s=\int\dfrac{dt}{\sqrt{1-t^2}}=\sin^{-1}t$ ,
Then $\dfrac{dx}{dt}=\dfrac{dx}{ds}\dfrac{ds}{dt}=\dfrac{1}{\sqrt{1-t^2}}\dfrac{dx}{ds}$
$\dfrac{d^2x}{dt^2}=\dfrac{d}{dt}\left(\dfrac{1}{\sqrt{1-t^2}}\dfrac{dx}{ds}\right)=\dfrac{1}{\sqrt{1-t^2}}\dfrac{d}{dt}\left(\dfrac{dx}{ds}\right)+\dfrac{t}{(1-t^2)^\frac{3}{2}}\dfrac{dx}{ds}=\dfrac{1}{\sqrt{1-t^2}}\dfrac{d}{ds}\left(\dfrac{dx}{ds}\right)\dfrac{ds}{dt}+\dfrac{t}{(1-t^2)^\frac{3}{2}}\dfrac{dx}{ds}=\dfrac{1}{\sqrt{1-t^2}}\dfrac{d^2x}{ds^2}\dfrac{1}{\sqrt{1-t^2}}+\dfrac{t}{(1-t^2)^\frac{3}{2}}\dfrac{dx}{ds}=\dfrac{1}{1-t^2}\dfrac{d^2x}{ds^2}+\dfrac{t}{(1-t^2)^\frac{3}{2}}\dfrac{dx}{ds}$
$\therefore(1-t^2)\left(\dfrac{1}{1-t^2}\dfrac{d^2x}{ds^2}+\dfrac{t}{(1-t^2)^\frac{3}{2}}\dfrac{dx}{ds}\right)-\dfrac{t}{\sqrt{1-t^2}}\dfrac{dx}{ds}+n^2x=0$
$\dfrac{d^2x}{ds^2}+\dfrac{t}{\sqrt{1-t^2}}\dfrac{dx}{ds}-\dfrac{t}{\sqrt{1-t^2}}\dfrac{dx}{ds}+n^2x=0$
$\dfrac{d^2x}{ds^2}+n^2x=0$
$x=C_1\sin ns+C_2\cos ns$
$x=C_1\sin(n\sin^{-1}t)+C_2\cos(n\sin^{-1}t)$
$x(0)=1$ :
$C_2=1$
$\therefore x=C_1\sin(n\sin^{-1}t)+\cos(n\sin^{-1}t)$
$x'=\dfrac{C_1n\cos(n\sin^{-1}t)}{\sqrt{1-t^2}}-\dfrac{n\sin(n\sin^{-1}t)}{\sqrt{1-t^2}}$
$x'(0)=0$ :
$C_1n=0$
$C_1=0$
$\therefore x=\cos(n\sin^{-1}t)$
|
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|
Nicomachus theorem proof - what did I do wrong? The exercise asked me to proof
$1^3=1$
$2^3=3+5$
$3^3=7+9+11$
$...$
I formulate the equation as
(1)$$a^3 = \sum_{i=0}^{a-1} (a-1)a+1+2i$$
Proof base case:
$$1^3=\sum_{i=0}^{1-1} (1-1)(1)+1+2i=1$$
Proof (a+1) case:
(2) $$(a+1)^3=a^3+3a^2+3a+1$$
(3) $$(a+1)^3=\sum_{i=0}^{(a+1)-1}((a+1)-1)(a+1)+1+2i=\sum_{i=0}^{a}(a)(a+1)+1+2i$$
factor out last summation term, so number of iterations match (1):
$$\sum_{i=0}^{a-1}[(a)(a+1)+1+2i]+(a)(a+1)+1+2a=\sum_{i=0}^{a-1}[(a)(a+1)+1+2i]+a^2+3a+1$$
rearrange the summation term, so part of equation match (1):
$$\sum_{i=0}^{a-1}[(a)(a-1)+1+2i]+\sum_{i=0}^{a-1}(2a)+a^2+3a+1$$
substitute summation with (1):
(4) $$a^3+\sum_{i=0}^{a-1}(2a)+a^2+3a+1=a^3+(2a)(a-1)+a^2+3a+1=a^3+3a^2+a+1$$
From (2) I got $a^3+3a^2+3a+1$ but from (3) & (4) I got $a^3+3a^2+a+1$
My result with induction is off by 2a. Can someone please help me to understand where I did wrong?
|
Your attempt seems a little messy, so I hope you don't mind that I present an alternative approach to an inductive proof.
You want to show $\displaystyle (1+2+...+n)^2 = 1^3 + 2^3 +... + n^3$
Assume, for some $k$ that $\displaystyle (1+2+...+k)^2 = 1^3 + 2^3 +... + k^3$ as the inductive hypothesis and call the expression on the left hand side $A$.
Now, let $\displaystyle B = (1+2+...+k+ (k+1))^2$
By the difference of squares identity, $\displaystyle B-A \\= (k+1)(2(1+2+...+k) + (k+1)) = (k+1)(2(\frac 12)(k)(k+1)+(k+1)) = (k+1)(k+1)(k+1) = (k+1)^3$
Hence $\displaystyle B = A + (k+1)^3$, and using the inductive hypothesis (and showing the base case), the proof is complete.
Sometimes, using the summation notation tends to cloud the thought process.
|
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|
General proof that the minimum value of $x+y$ from $\frac{4}{11} < \frac{x}{y} < \frac{3}{8}$ is $4+11+3+8=26$ I found this problem and the solution on Twitter (translated).
$x$ and $y$ is a (positive) integer that satisfies $\frac{4}{11} < \frac{x}{y} < \frac{3}{8}$. If $y$ is the smallest number, what is $x+y$?
The given solution is $\frac{x}{y} = \frac{4+3}{11+8} = \frac{7}{19}$. Therefore, $x+y=7+19=26$.
Is the solution correct? If it is, why is the solution correct? What is the proof that it is correct? And if possible, what is the general formula?
Generally, the solution will be, given $a,b,c,d \in \mathbb{N}$ where $\frac{a}{b} < \frac {c}{d}$, $a$ and $b$ are coprimes, $c$ and $d$ are coprimes, and $a \neq b \neq c \neq d$, the minimum value of $x+y$ where $x,y \in \mathbb{N}$ and $\frac{a}{b} < \frac{x}{y} < \frac{c}{d}$ is $a+b+c+d$.
Edit: it seems the bounds are also unclear. What are the correct bounds for the general statement?
What is the proof that it's right or wrong? Thank you.
Edit: If it's possible, please explain it with only 10th-grade math or lower.
|
If I were starting out from scratch here, I'd note that:
*
*$\frac{a}{b}<\frac{x}{y}$ means $bx-ay>0$. Since we are working with integers, $bx-ay\geq1$.
*$\frac{x}{y}<\frac{c}{d}$ means $-dx+cy>0$. Since we are working with integers, $-dx+cy\geq1$.
So we have the system
$$\left\{
\begin{aligned}
bx-ay&\geq1\\
-dx+cy&\geq1
\end{aligned}
\right.$$
So it is a linear programming question. It means
$$\begin{bmatrix}b&-a\\-d&c\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\in R$$
where $R$ is the closed infinite rectangular region that has $(1,1)$ as a lower left corner.
So applying the matrix inverse,$$\begin{bmatrix}x\\y\end{bmatrix}\in \frac{1}{bc-ad}\begin{bmatrix}c&a\\d&b\end{bmatrix}R$$
Now $R=\{(u,v)|u,v\geq1\}$. So $$\begin{bmatrix}x\\y\end{bmatrix}\in \left\{\frac{1}{bc-ad}\begin{bmatrix}cu+av\\du+bv\end{bmatrix}\middle| u,v\geq1\right\}$$
Each of $a,b,c,d,u,v,bc-ad$ is positive. So the smallest that the second coordinate of the above could be is when $u$ and $v$ are at their smallest, $1$. So you will have the smallest possible value for $y$ when
$$\begin{bmatrix}x\\y\end{bmatrix}=\frac{1}{bc-ad}\begin{bmatrix}c+a\\d+b\end{bmatrix}$$
BUT we still require that $x$ and $y$ be integers. One way to guarantee this is to require that $bc-ad=1$. Then $\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}c+a\\d+b\end{bmatrix}$. And the smallest $y$ is acieved when $y=d+b$ and $x=c+a$.
So it is sufficient for $bc-ad=1$ to make it such that the minimal solution to the inequality $\frac{a}{b}<\frac{x}{y}<\frac{c}{d}$ is $y=d+b$, $x=c+a$.
For example, with $\frac{4}{11}<\frac{x}{y}<\frac{3}{8}$, the minimal solution is $y=19$, $x=7$.
Furthermore if $bc-ad\neq1$ but it happens to divide each of $d+b$ and $c+a$, that would make the minimal solution be $y=\frac{d+b}{bc-ad}$, $x=\frac{c+a}{bc-ad}$.
For example, with $\frac{3}{7}<\frac{x}{y}<\frac{5}{11}$, the minimal solution is $y=\frac{11+7}{2}=9$, $x=\frac{5+3}{2}=4$. (The fraction is $\frac{4}{9}$.)
If $bc-ad$ does not divide either $d+b$ or $c+a$, then $y=\frac{d+b}{bc-ad}$, $x=\frac{c+a}{bc-ad}$ give the minimal real solutions to the inequality, but at least one of them is not an integer. Instead, we would have to look for values of $(u,v)$ not both equal to $1$ such that $bc-ad$ divides both $du+bv$ and $cu+av$. And it's not immediately clear which values of $u$ and $v$ would lead to the minimal solution.
|
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|
$4^{n-1} {{3n}\choose{n}} \sqrt{n} \geq 3^{3n-2}, n \in \mathbb{N} \setminus \{0\}$ Induction proof, need a hint for more efective proof. I am looking for a smarter way of proving that inequality by induction.
$$4^{n-1} {{3n}\choose{n}} \sqrt{n} \geq 3^{3n-2}, n \in \mathbb{N} \setminus \{0\}$$
For $n + 1$:
$$4^{n} {{3n + 3}\choose{n+1}} \sqrt{n+1} \geq 3^{3n+1}$$
$$4*4^{n-1} \frac{(3n)!(3n + 1)(3n + 2)(3n + 3)}{(n)!(n+1)(2n)!(2n+1)(2n+2)} \sqrt{n+1} \geq 3^{3n+1}$$
Then I can plug my thesis into the inequality for $n + 1$:
$$4^{n-1} {{3n}\choose{n}} \sqrt{n} \geq 3^{3n-2} \implies 4^{n-1} \frac{(3n)!}{n!(2n)!} \geq \frac{3^{3n-2}}{\sqrt{n}}$$
So I get:
$$4* \frac{3^{3n-2}}{\sqrt{n}} * \frac{(3n + 1)(3n + 2)(3n + 3)}{(n+1)(2n+1)(2n+2)} \sqrt{n+1} \geq 3^{3n+1}$$
$$4*(3n + 1)(3n + 2)(3n + 3) \sqrt{n+1} \geq 27\sqrt{n} (n+1)(2n+1)(2n+2) $$
And then, if I sqare both hand sides of that ineqality (both are positive) I get 7th powers of $n$ and there is probably some smarter / easier way to prove that inequality.
|
This isn't induction.
It uses Stirling
and it is more general.
I will show that
$0.850
\lt \dfrac{\binom{3n}{n}}{\sqrt{\dfrac{3}{4\pi n}}
\left(\dfrac{27}{4}\right)^n}
\lt 1.085
$.
Since
$n! \approx \sqrt{2\pi n}(n/e)^n$,
$\begin{array}\\
\binom{an}{bn}
&=\dfrac{(an)!}{(bn)!((a-b)n!}\\
&\sim \dfrac{\sqrt{2\pi an}(an/e)^{an}}
{(\sqrt{2\pi bn}(bn/e)^{bn})(\sqrt{2\pi n(a-b)}(((a-b)n)/e)^{(a-b)n})}\\
&= \sqrt{\dfrac{2\pi an}{2\pi bn2\pi n(a-b)}}\left(\dfrac{(an)^ae^be^{a-b}}{e^a(bn)^b((a-b)n)^{a-b}}\right)^n\\
&= \sqrt{\dfrac{ a}{2\pi bn(a-b)}}\left(\dfrac{a^an^a}{b^b(a-b)^{a-b}n^a}\right)^n\\
&= \sqrt{\dfrac{ a}{2\pi bn(a-b)}}\left(\dfrac{a^a}{b^b(a-b)^{a-b}}\right)^n\\
&=r(n, a, b)\\
\end{array}
$
If $a=3, b=1$,
$\dfrac{a}{b(a-b)}
=\dfrac{3}{2}
$
and
$\dfrac{a^a}{b^b(a-b)^{a-b}}
=\dfrac{3^3}{2^2}
=\dfrac{27}{4}
=6\frac34
$
so
$\binom{3n}{n}
\sim \sqrt{\dfrac{3}{4\pi n}}
\left(\dfrac{27}{4}\right)^n
$.
Actually
(https://en.wikipedia.org/wiki/Stirling%27s_approximation),
if
$f(n)
=\dfrac{n!}{\sqrt{2\pi n}(n/e)^n}$,
then the following inequality holds:
$$1
\lt f(n)
\lt e^{1/(12n)}
\lt \frac{e}{\sqrt{2\pi}}
= 1.0844...
$$
So,
if $u < f(n) < v$,
then
$\frac{u}{v^2}
\lt \dfrac{\binom{an}{bn}}{r(n, a, b)}
\lt \frac{v}{u^2}
$.
From the simple bounds above,
we can take
$u = 1$
and
$v = \frac{e}{\sqrt{2\pi}} $,
so the bounds are
$\dfrac{2\pi}{e^2}
=0.850...
$
and
$\dfrac{e}{\sqrt{2\pi}}
=1.0844...
$.
Therefore
$0.850
\lt \dfrac{\binom{an}{bn}}{\sqrt{\dfrac{ a}{2\pi bn(a-b)}}\left(\dfrac{a^a}{b^b(a-b)^{a-b}}\right)^n}
\lt 1.085
$.
If $a=3, b=1$ then
$0.850
\lt \dfrac{\binom{3n}{n}}{\sqrt{\dfrac{3}{4\pi n}}
\left(\dfrac{27}{4}\right)^n}
\lt 1.085
$.
|
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|
$n_1^3 + n_2^3 + ... + n_k^3 \geq (n_1 +n_2 + ...+ n_k)^2$ for natural numbers $n_1 < n_2 < ... < n_k$ Prove that for natural numbers $n_1 < n_2 < ... < n_k$:
$$n_1^3 + n_2^3 + ... + n_k^3 \geq (n_1 +n_2 + ...+ n_k)^2$$
I have no idea how to do that. I would go for induction, but there is an infromation that $n_1 < n_2 < ... < n_k$ and induction would not recognise that fact.
What's funny is that I think I saw once that exercise on stack exchange but wasn't able to find it this time.
|
We can go further to even state the biconditional equality condition. The following is a quote of Proposition 3 from Sum of Cubes is Square of Sum by Barbeau and Seraj, with permission from me, one of the authors:
We prove by induction that:
For integers $a_k$, if $1 \leq a_1 < a_2 < \cdots < a_n$ then
$$a_1^3 + a_2^3 + \cdots + a_n^3 \ge (a_1 + a_2 + \cdots + a_n)^2,$$
with equality if and only if $a_k = k$ for $1 \le k \leq n$.
This is clear for $n = 1$. Assume that the proposition holds for some natural $n$; we will prove it for $n+1$. Note that $a_k \ge k$ and that $a_{n+1} - k \ge a_{n+1-k}$ for all values of $k$. We begin with the fact that
$$(a_{n+1} - n)(a_{n+1} - n - 1) \ge 0$$
with equality if and only if $a_{n+1} = n+1$. Then expanding gives
\begin{align*}
a_{n+1}^2 - a_{n+1} & \ge 2na_{n+1} - n(n+1) = 2 \left[ na_{n+1} - {{n(n+1)}\over 2} \right]\\
& = 2 \left( \sum_{k=1}^n (a_{n+1} - k) \right) \ge 2 \sum_{k=1}^n a_{n+1-k} = 2 \sum_{k=1}^n a_k,
\end{align*}
with equality if and only if $a_k = k$ for all $k$. By the induction hypothesis, we have that
\begin{align*}
\sum_{k=1}^{n+1} a_k^3 &= a_{n+1}^3 + \sum_{k=1}^n a_k^3 \ge a_{n+1}^2 + 2a_{n+1}\left(\sum_{k=1}^n a_k \right) + \left(\sum_{k=1}^n a_k \right)^2\\
& = \left(a_{n+1} + \sum_{k=1}^n a_k \right)^2 = \left( \sum_{k=1}^{n+1} a_k \right)^2,
\end{align*}
as desired.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3944929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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|
Value of integration $\int_0^1 \sin^{-1}(3x-4x^3)dx$ I have stumbled across an definite integration problem.
The question is to evaluate $$I=\int_0^1 \sin^{-1}(3x-4x^3)dx$$
I carried out the integration as this. Let $x = \sin A$ so that A = $\sin^{-1}(x)$ then
$$\begin {align} I &=\int \sin^{-1}(3\sin A-4\sin^3A)dx\\ &=\int \sin^{-1}(\sin3A)dx\\&=\int3Adx\\&=3\int \sin^{-1}(x)\\&=3[x\sin^{-1}x+\sqrt{1-x^2}] + c\end{align}$$
Then,
$$\begin{align}I&= \biggl.3[x \sin^{-1}x+\sqrt{1-x^2}]\biggl|_0^1\\&=3[(\sin^{-1}1+\sqrt{1-1^2})-(0\times \sin^{-1}0+\sqrt{1-0})]\\&=3(\frac\pi2-1)\\&=1.71238...\end{align}$$
But the correct answer is different as given by mathematica and many other ways,
$$I=\int_0^1 \sin^{-1}(3x-4x^3)dx = 0.62535...$$
I couldn't find the mistake in my process. And I don't even know which answer is correct actually? Can you please help me here?
|
Note
\begin{align}
I&=\int_0^1 \sin^{-1}(3x-4x^3)dx \>\>\>\>\>\>\>(x=\sin A)\\
&= \int_0^{\frac\pi2} \sin^{-1}(\sin3A)\cos AdA \\
&= \int_0^{\frac\pi6} \sin^{-1}(\sin3A)\cos AdA
+ \int_{\frac\pi6}^{\frac\pi2} \sin^{-1}(\sin(\pi-3A))\cos AdA \\
&= \int_0^{\frac\pi6} 3A\cos AdA
+ \int_{\frac\pi6}^{\frac\pi2} (\pi-3A)\cos AdA \\
&=\left( \frac{3\sqrt3}2-3 +\frac\pi4 \right)+ \left(\frac{3\sqrt3}2-\frac{3\pi}4\right)\\
&=3\sqrt3-3-\frac\pi2
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3945948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
What are the equations of the three straight lines represented by $x^3+bx^2y+cxy^2+y^3=0$ when $b+c=-2$? I am given that $$x^3 + bx^2y + cxy^2 + y^3 = 0$$ represents three straight lines if $b + c = -2$.
Is there a way to find the equations of the three lines separately?
I tried factorizing the equation but wasn't able to get anywhere. Is there a way to get the separate equations using partial differentiation?
|
If $b+c=-2$, you can write $b=-k$ and $c=k-2$, so
$$\begin{align}x^3+bx^2y+cxy^2+y^3&=x^3-kx^2y+(k-2)xy^2+y^3\\&=x^3-kx^2y+kxy^2-2xy^2+y^3\\&=(x^3-xy^2)-(kx^2y-kxy^2)-(xy^2-y^3)\\&=(x-y)(x(x+y)-kxy-y^2)\\&=(x-y)(x^2-(k-1)xy-y^2)\end{align}$$
You can factor the latter part by completing the square but the computation is a bit ugly.
The three lines will be:
$$x=y\\ x=y\left(\frac{k-1}2+\sqrt{1+\left(\frac{k-1}2\right)^2}\right)\\ x=y\left(\frac{k-1}2-\sqrt{1+\left(\frac{k-1}2\right)^2}\right)$$
where $k$ is just $-b$, a constant, so the entire expression in the brackets is a constant, the reciprocal of their slopes.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3947901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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|
Finding $n$ such that $(3-3w+2w^2)^{4n+3}+(2+3w-3w^2)^{4n+3}+(-3+2w+3w^2)^{4n+3}=0$ for $w\neq1$ a cube root of unity
Let
$${\left( {3 - 3\omega + 2{\omega ^2}} \right)^{4n + 3}} + {\left( {2 + 3\omega - 3{\omega ^2}} \right)^{4n + 3}} + {\left( { - 3 + 2\omega + 3{\omega ^2}} \right)^{4n + 3}}=0$$
If $\omega\ne1$ be the complex cube root of unity, then the possible values of $n$ is
(A) $1\quad$ (B) $2\quad$ (C) $3\quad$ (D) 4
My approach $1+\omega+\omega^2=0$
Let $A={\left( {3 - 3\omega + 2{\omega ^2}} \right)}$
$B={\left( {2 + 3\omega - 3{\omega ^2}} \right)} $
$C= {\left( { - 3 + 2\omega + 3{\omega ^2}} \right)}$
$B=A\omega$, $C=A\omega^2$ as $A+B+C=0$. Not able to proceed
|
Your sum is $A^{4n+3}(1+\omega^{4n+3}+\omega^{2(4n+3)})$, and $A\ne 0$. So we must have $1+\omega^{4n+3}+\omega^{2(4n+3)}=0$. And $\omega^3=1$. So this is
$$1+\omega^n+\omega^{2n}=0$$
which is true if and only if $n\equiv 1$ or $2\bmod 3$.
As to which of the four answers to select, that is a different puzzle...
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3948268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Calculate $\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} d x$ I have to calculate the integral
$$\int_{0}^{1} \frac{1+x^{2}}{1+x^{4}} d x$$
I've calculated the integral $\int_{-\infty}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x= \sqrt{2} \pi $. Then $\int_{-\infty}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x=2 \int_{0}^{+\infty} \frac{1+x^{2}}{1+x^{4}} d x$. How do I now arrive at the initial integral?
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Just another way to do it
$$\frac{1+x^{2}}{1+x^{4}}=\frac{1+x^{2}}{(x^2+i)(x^2-i)}=\frac 12\left( \frac{1+i}{x^2+i}+\frac{1-i}{x^2-i}\right)$$ and you could even continue with partial fractions.
Without any simplifications
$$\int\frac{1+x^{2}}{1+x^{4}}\, dx=\frac{2 (i+1) \tan ^{-1}\left(\frac{x}{\sqrt{i}}\right)-(i-1) \left(\log
\left(\sqrt{i}-x\right)-\log \left(\sqrt{i}+x\right)\right)}{4 \sqrt{i}}$$ Now, playing with the complex numbers,
$$\int\frac{1+x^{2}}{1+x^{4}}\, dx=\frac{\tan ^{-1}\left(1+\sqrt{2} x\right)-\tan ^{-1}\left(1-\sqrt{2}
x\right)}{\sqrt{2}}$$ REcombining the arctangents,
$$\int\frac{1+x^{2}}{1+x^{4}}\, dx=\frac 1{\sqrt{2}}\tan ^{-1}\left(\frac{\sqrt{2} x}{1-x^2}\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3949021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
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.