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Math Help - Power Series

  1. #1
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    Power Series

    Im having trouble with this power series, I know it converges absolutely but im having trouble getting there.


    \sum_{n=0}^{\infty}\frac{x^{2n+1}}{n!}


    any help appreciated, Thanks.
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  2. #2
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    Quote Originally Posted by Itsygo View Post
    Im having trouble with this power series, I know it converges absolutely but im having trouble getting there.


    \sum_{n=0}^{\infty}\frac{x^{2n+1}}{n!}


    any help appreciated, Thanks.
    Are you trying to prove it converges absolutely?

    If so, you need to show that \lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right| < 1.


    So in your case a_n = \frac{x^{2n + 1}}{n!} and a_{n + 1} = \frac{x^{2n + 3}}{(n + 1)!}.


    Thus \left|\frac{a_{n + 1}}{a_n}\right| = \left|\frac{x^{2n + 3}}{(n + 1)!}\cdot \frac{n!}{x^{2n + 1}}\right|

     = \frac{x^2}{n + 1}.


    What happens as n \to \infty?
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    awesome thanks for the quick post, ya sorry I needed to prove a.c.

    it eventually gets infinity small < 1
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  4. #4
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    Quote Originally Posted by Itsygo View Post
    Im having trouble with this power series, I know it converges absolutely but im having trouble getting there.


    \sum_{n=0}^{\infty}\frac{x^{2n+1}}{n!}


    any help appreciated, Thanks.
    In fact, you can determine what that sum is equal to. It is well known that e^x= \sum_{n=0}^\infty \frac{x^n}{n!}. Here you have \sum_{n=0}^\infty \frac{x^{2n+1}}{n!}= \sum_{n=0}^\infty x\frac{(x^2)^n}{n!}= x e^{x^2}
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    In fact, you can determine what that sum is equal to. It is well known that e^x= \sum_{n=0}^\infty \frac{x^n}{n!}. Here you have \sum_{n=0}^\infty \frac{x^{2n+1}}{n!}= \sum_{n=0}^\infty x\frac{(x^2)^n}{n!}= x e^{x^2}
    If this power series isn't known to you (which it really should be) we can find it slighly differently. Suppose we just want to find the explicity function described by f(x)=\sum_{n=0}^{\infty}\frac{x^{2n}}{n!} notice that we may (take for granted here that you can...there is criterion it needs to meet but it does...so) differentate to obtain f'(x)=\sum_{n=1}^{\infty}\frac{2nx^{2n-1}}{n!}=\sum_{n=0}^{\infty}\frac{2(n+1)x^{2(n+1)-1}}{(n+1)!}=\sum_{n=0}^{\infty}\frac{2x^{2n+1}}{n!  }=2xf(x). Therefore, f'(x)=2xf(x)\implies \frac{f'(x)}{f(x)}=2x\implies \ln\left(f(x)\right)=x^2+C\implies f(x)=C_1e^{x^2}. And seeing that f(0)=1 we can conclude that C_1=1\implies f(x)=e^{x^2} therefore \sum_{n=0}^{\infty}\frac{x^{2n+1}}{n!}=xf(x)=xe^{x  ^2}.
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