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I have a function defined only by it's taylor series: $$f(x) = \sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!}x^k$$

Obviously, integer derivatives can be defined as $$\frac{d^n}{dx^n} f(x) = \sum_{k=0}^\infty \frac{f^{(k)}(0)}{(k-n)!}x^{k-n}$$

However, this completely fails if $n$ is not an integer and $x=0$; every term vanishes, which is obviously incorrect. Is there a better way to take a fractional derivative of a Taylor series?

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Why do you say every term vanishes? The fractional derivative of $x^n$ exists and is nonzero. – Cameron Williams Nov 19 '13 at 5:07
$0^n$ is zero if $n>0$. And as I said, I'm obviously not taking it correctly. – fivex Nov 19 '13 at 5:16
Oh I see your problem now. You're taking the derivative and then evaluating at $0$? – Cameron Williams Nov 19 '13 at 5:37
Yes, that is exactly what I'm trying to do. – fivex Nov 19 '13 at 5:44
I think the issue is that when you take the $\frac{3}{2}$ derivative of, say, $x$, you don't get a zero value. The $\frac{3}{2}$ derivative of $x$ is defined to be $\frac{\Gamma(1+1)}{\Gamma(1-\frac{3}{2}+1)}x^{1-\frac{3}{2}} = \frac{\Gamma(2)}{\Gamma(\frac{1}{2})}x^{-\frac{1}{2}} = \frac{1}{\sqrt{\pi}}x^{-\frac{1}{2}}$ since $\Gamma(n+1) = n!$ if $n$ is an integer. – Cameron Williams Nov 19 '13 at 6:09

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