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Why does 11/3+1/51/7... converge to Pi/4? In Latex,
\[\sum_{k=0}^{\infty}\frac{(1)^k}{2k+1}=\frac{\pi}{4}\]
Don't give me a copy/paste of what Wikipedia says though, it skips a couple steps and I don't understand!
 one year ago
 one year ago
Why does 11/3+1/51/7... converge to Pi/4? In Latex, \[\sum_{k=0}^{\infty}\frac{(1)^k}{2k+1}=\frac{\pi}{4}\] Don't give me a copy/paste of what Wikipedia says though, it skips a couple steps and I don't understand!
 one year ago
 one year ago

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mukushlaBest ResponseYou've already chosen the best response.4
sure u know that\[\arctan x=\sum_{n=0}^{\infty} (1)^n \frac{x^{2n+1}}{2n+1}\]
 one year ago

mukushlaBest ResponseYou've already chosen the best response.4
\[\frac{1}{1+t^2}=1t^2+t^4t^6+t^8...=\sum_{n=0}^{\infty} (1)^n t^{2n} \ \ \ \ \ \ \ t<1\]integrate both sides from \(0\) to \(x\) for \(x<1\)
 one year ago

vf321Best ResponseYou've already chosen the best response.1
Wow should have caught that more easily. It seems so clear when you see it. But go on with the proof....
 one year ago

mukushlaBest ResponseYou've already chosen the best response.4
after integrating the geometric series u will get\[\arctan x=\sum_{n=0}^{\infty} (1)^n \frac{x^{2n+1}}{2n+1}\]let \(x=1\) u will have\[\arctan 1=\frac{\pi}{4}=\sum_{n=0}^{\infty} \frac{(1)^n }{2n+1}\]
 one year ago

mukushlaBest ResponseYou've already chosen the best response.4
but this is not a complete proof
 one year ago

vf321Best ResponseYou've already chosen the best response.1
Yeah but you can't just let x=1 cause that series only holds true for \[x<1\]
 one year ago

vf321Best ResponseYou've already chosen the best response.1
So really the question is how Wikipedia got that first substitution of arctanx with the series that had a larger radius of convergence. I get the rest.
 one year ago

mukushlaBest ResponseYou've already chosen the best response.4
here we must use Abel's theorem that im not comfortable with
 one year ago

vf321Best ResponseYou've already chosen the best response.1
Well can we try to reason this out? Clearly the introduction of \[\frac{(1)^{n+1}x^{2n+2}}{1+x^2}\]in the summation \[\frac{1}{1+x^2}=\sum_{k=0}^n(1)^nx^{2k}+\frac{(1)^{n+1}x^{2n+2}}{1+x^2}\]Had something to do with it...
 one year ago

vf321Best ResponseYou've already chosen the best response.1
My guess is that Wikipedia made that sum finite (with n on top instead of infy), and so that messy term at the end is just a correction factor for when x=1 in the mclauren series 1/(1+x^2). Then, taking the limit n>infty later guarentees that the new series models arctan x, and, after integration, the correction factor approaches 0 at infity.
 one year ago

vf321Best ResponseYou've already chosen the best response.1
So integration actually expanded our radius of convergence (as I understand it)
 one year ago

vf321Best ResponseYou've already chosen the best response.1
If I'm not mistaken, then it looks something like this: \[\frac{1}{1+x^2}=\sum_{k=0}^n (1)^kx^{2k}+R\]where\[R=\{_{0,x<1}^{f(x, k), x=1}\]We know that there must be some existing f(x, k) for finite series. We also know, because of the radius of convergencIe for a geometric series, that \[\lim_{n\rightarrow\infty}\sum_{k=0}^nf(x, k)=0\]If and only if x<1, and for x=1, the same sum diverges.
 one year ago

vf321Best ResponseYou've already chosen the best response.1
However, integrating both sides of my first equation in the previous reply gets us: \[arctan(x)=\sum_{k=0}^n\int(1)^kx^{2k}dx+\int Rdx\] Taking the limit n>infty of both sides, we find that, due to the properties of f(x, k), \[\lim_{n\rightarrow\infty}\int R dx=0\] For 1 < x <= 1. Then your proof works.
 one year ago

vf321Best ResponseYou've already chosen the best response.1
EDIT: Last limit should be: \[\lim_{n\rightarrow\infty}\sum_{k=0}^n\int R dx=0\]
 one year ago

vf321Best ResponseYou've already chosen the best response.1
The whole thing is interesting, though. It leads me to another question, that I'll post separately (about whether or not such a function f(x, k) can exist with those properties.
 one year ago
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