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What's the sum of this power series? $$f_k(x)=1-\frac{x^2}{k}+\frac{x^4}{k(k+1)\cdot2!}-\frac{x^6}{k(k+1)(k+2)\cdot3!}+\ldots$$ I'm just helping someone, I'm not good at math! :\

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This is related to a Bessel function; see en.wikipedia.org/wiki/Bessel_function#Definitions –  joriki Apr 6 '11 at 18:05
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up vote 8 down vote accepted

To expand on my comment: Your function is

$$f_k(x)=(k-1)!\sum_{m=0}^\infty \frac{(-1)^m}{m!(k+m-1)!}x^{2m}\;.$$

The Bessel function of (integer) order $n$ is

$$J_n(x)=\sum_{m=0}^\infty\frac{(-1)^m}{m!(m+n)!}\left(\frac{x}{2}\right)^{2m+n}\;.$$

Thus your function is

$$f_k(x)=n!J_n(2x)x^{-n}\;,$$

with $n=k-1$.

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+1: as we agree and you had the initial idea, I can delete my answer ;-) –  Fabian Apr 6 '11 at 18:41
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@Fabian: Sorry for the duplicated effort -- this is not the first time I'm thinking there should be an "I'm working on this" indicator -- hmm, perhaps I'll put up a feature request for that on meta :-) –  joriki Apr 6 '11 at 18:43
    
I started to work on this too, seeing that it was easy to get from the Bessel function to an answer. –  Matt Groff Apr 6 '11 at 19:04
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I don't think it's a bad thing to have 'duplicate' answers. Yes, I know the annoying feeling of working on an answer, submitting it, and seeing that someone else has done mostly the same thing (or seeing the 'new answer just submitted' popup just before submitting yourself). but for the readers, I think it's a useful thing. –  Mitch Apr 6 '11 at 19:51
    
@Mitch: I think it depends on the question. In this case, there wasn't much variation in possible answers, and so Fabian's answer turned out to be not much different from a part of mine. I agree with you in cases where different answers can throw light on different aspects of the problem. –  joriki Apr 6 '11 at 20:03
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