Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
vf321
Group Title
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!
 2 years ago
 2 years ago
vf321 Group Title
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!
 2 years ago
 2 years ago

This Question is Closed

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

mukushla Group TitleBest ResponseYou've already chosen the best response.4
for \(x<1\)
 2 years ago

mukushla Group TitleBest 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 \(x<1\)
 2 years ago

vf321 Group TitleBest 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....
 2 years ago

mukushla Group TitleBest 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\[\arctan 1=\frac{\pi}{4}=\sum_{n=0}^{\infty} \frac{(1)^n }{2n+1}\]
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.4
but this is not a complete proof
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
Yeah but you can't just let x=1 cause that series only holds true for \[x<1\]
 2 years ago

vf321 Group TitleBest 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.
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.4
here we must use Abel's theorem that im not comfortable with
 2 years ago

vf321 Group TitleBest 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...
 2 years ago

vf321 Group TitleBest 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.
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
So integration actually expanded our radius of convergence (as I understand it)
 2 years ago

vf321 Group TitleBest 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.
 2 years ago

vf321 Group TitleBest 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.
 2 years ago

vf321 Group TitleBest ResponseYou've already chosen the best response.1
EDIT: Last limit should be: \[\lim_{n\rightarrow\infty}\sum_{k=0}^n\int R dx=0\]
 2 years ago

vf321 Group TitleBest 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.
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.