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Math Help - intergration

  1. #1
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    intergration

    Show that by expanding 1/(1+x^4) as a geometric series and integrating termwise.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by bigb View Post
    Show that by expanding 1/(1+x^4) as a geometric series and integrating termwise.
    We know that \forall{x}\in(0,1)~\frac{1}{1+x^4}=\sum_{n=0}^{\in  fty}(-1)^nx^{4n} so \begin{aligned}\forall{x}\in(0,1)~\frac{1+x^2}{1+x  ^4}&=(1+x^2)\sum_{n=0}^{\infty}(-1)^nx^{4n}\\<br />
&=\sum_{n=0}^{\infty}(-1)^nx^{4n}+\sum_{n=0}^{\infty}(-1)^nx^{4n+2}\end{aligned}

    So \begin{aligned}\int_0^1\frac{1+x^2}{1+x^4}&=\int_0  ^1\bigg[\sum_{n=0}^{\infty}(-1)^nx^{4n}+\sum_{n=0}^{\infty}(-1)^nx^{4n+2}\bigg]dx\\<br />
&=\sum_{n=0}^{\infty}\frac{(-1)^n}{4n+1}+\sum_{n=0}^{\infty}\frac{(-1)^n}{4n+3}\\<br />
&=\frac{\sqrt{2}\pi}{4}\end{aligned}

    Now what I think you meant to say was this

    \forall{x}\in(0,1)~\frac{1}{1+x^2}=\sum_{n=0}^{\in  fty}(-1)^nx^{2n}

    So \begin{aligned}\int_0^1\frac{dx}{1+x^2}&=\int_0^1\  sum_{n=0}^{\infty}(-1)^nx^{2n}dx\\<br />
&=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}<br />
\end{aligned}

    But note that \int_0^1\frac{dx}{1+x^2}=\arctan(1)=\frac{\pi}{4}

    So \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}

    Also note that the interchanging of all the summations and integrals was due to the integrability of a power seris for all points on the interior of its interval of convergence.
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  3. #3
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    Quote Originally Posted by Mathstud28 View Post
    We know that \forall{x}\in(0,1)~\frac{1}{1+x^4}=\sum_{n=0}^{\in  fty}(-1)^nx^{4n} so \begin{aligned}\forall{x}\in(0,1)~\frac{1+x^2}{1+x  ^4}&=(1+x^2)\sum_{n=0}^{\infty}(-1)^nx^{4n}\\<br />
&=\sum_{n=0}^{\infty}(-1)^nx^{4n}+\sum_{n=0}^{\infty}(-1)^nx^{4n+2}\end{aligned}

    So \begin{aligned}\int_0^1\frac{1+x^2}{1+x^4}&=\int_0  ^1\bigg[\sum_{n=0}^{\infty}(-1)^nx^{4n}+\sum_{n=0}^{\infty}(-1)^nx^{4n+2}\bigg]dx\\<br />
&=\sum_{n=0}^{\infty}\frac{(-1)^n}{4n+1}+\sum_{n=0}^{\infty}\frac{(-1)^n}{4n+3}\\<br />
&=\frac{\sqrt{2}\pi}{4}\end{aligned}

    Now what I think you meant to say was this

    \forall{x}\in(0,1)~\frac{1}{1+x^2}=\sum_{n=0}^{\in  fty}(-1)^nx^{2n}

    So \begin{aligned}\int_0^1\frac{dx}{1+x^2}&=\int_0^1\  sum_{n=0}^{\infty}(-1)^nx^{2n}dx\\<br />
&=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}<br />
\end{aligned}

    But note that \int_0^1\frac{dx}{1+x^2}=\arctan(1)=\frac{\pi}{4}

    So \sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}

    Also note that the interchanging of all the summations and integrals was due to the integrability of a power seris for all points on the interior of its interval of convergence.
    I am suppose to show in a three step process that the

    and this was the first step part a)
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by bigb View Post
    I am suppose to show in a three step process that the

    and this was the first step part a)
    Hmm, but this just simply is not true?

    But for the sum look here http://www.mathhelpforum.com/math-he...415-post2.html and note then that your sum should be equal to \frac{1}{16}\bigg[\frac{\pi}{2\frac{1}{4}\sin\left(\frac{\pi}{4}\rig  ht)}-\frac{1}{2\left(\frac{1}{4}\right)^2}\bigg]=\frac{\sqrt{2}\pi-4}{8} which is what your first equation indicates.
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  5. #5
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    Quote Originally Posted by Mathstud28 View Post
    Hmm, but this just simply is not true?

    But for the sum look here http://www.mathhelpforum.com/math-he...415-post2.html and note then that your sum should be equal to \frac{1}{16}\bigg[\frac{\pi}{2\frac{1}{4}\sin\left(\frac{\pi}{4}\rig  ht)}-\frac{1}{2\left(\frac{1}{4}\right)^2}\bigg]=\frac{\sqrt{2}\pi-4}{8} which is what your first equation indicates.
    What do you think is simply not true??

    This portion of your first post is my second part of this problem, where i gotta show that the integral from my first post is equal to sqrt(2)*pi/4 like you have shown.

    We know that \forall{x}\in(0,1)~\frac{1}{1+x^4}=\sum_{n=0}^{\in  fty}(-1)^nx^{4n} so \begin{aligned}\forall{x}\in(0,1)~\frac{1+x^2}{1+x  ^4}&=(1+x^2)\sum_{n=0}^{\infty}(-1)^nx^{4n}\\<br />
&=\sum_{n=0}^{\infty}(-1)^nx^{4n}+\sum_{n=0}^{\infty}(-1)^nx^{4n+2}\end{aligned}

    So \begin{aligned}\int_0^1\frac{1+x^2}{1+x^4}&=\int_0  ^1\bigg[\sum_{n=0}^{\infty}(-1)^nx^{4n}+\sum_{n=0}^{\infty}(-1)^nx^{4n+2}\bigg]dx\\<br />
&=\sum_{n=0}^{\infty}\frac{(-1)^n}{4n+1}+\sum_{n=0}^{\infty}\frac{(-1)^n}{4n+3}\\<br />
&=\frac{\sqrt{2}\pi}{4}\end{aligned}

    is there a way to conclude from here to prove
    Last edited by bigb; November 22nd 2008 at 01:07 PM.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by bigb View Post
    What do you think is simply not true??

    This portion of your first post is my second part of this problem, where i gotta show that the integral from my first post is equal to sqrt(2)*pi/4 like you have shown.

    We know that \forall{x}\in(0,1)~\frac{1}{1+x^4}=\sum_{n=0}^{\in  fty}(-1)^nx^{4n} so \begin{aligned}\forall{x}\in(0,1)~\frac{1+x^2}{1+x  ^4}&=(1+x^2)\sum_{n=0}^{\infty}(-1)^nx^{4n}\\<br />
&=\sum_{n=0}^{\infty}(-1)^nx^{4n}+\sum_{n=0}^{\infty}(-1)^nx^{4n+2}\end{aligned}

    So \begin{aligned}\int_0^1\frac{1+x^2}{1+x^4}&=\int_0  ^1\bigg[\sum_{n=0}^{\infty}(-1)^nx^{4n}+\sum_{n=0}^{\infty}(-1)^nx^{4n+2}\bigg]dx\\<br />
&=\sum_{n=0}^{\infty}\frac{(-1)^n}{4n+1}+\sum_{n=0}^{\infty}\frac{(-1)^n}{4n+3}\\<br />
&=\frac{\sqrt{2}\pi}{4}\end{aligned}

    is there a way to conclude from here to prove
    What is untrue is that \int_0^1\frac{1+x^2}{1+x^4}dx\ne\sum_{n=0}^{\infty  }\frac{(-1)^n}{2n+1}=\int_0^1\frac{dx}{1+x^2}

    And try not to quote it like that, it is hard to discern what I wrote and what you wrote.
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