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Let $r>0$, $\varepsilon>0$ and $\alpha>0$. Assume that $0<\varepsilon<x<r$. I want a power series in $x$ for $x^{\alpha}$. Here is my attempt.
We may assume that $r<1$. $$ x^{\alpha}=(1+(x-1))^{\alpha}=\sum_{k=0}^{\infty}\binom{\alpha}{k}(x-1)^k $$ $$ =\sum_{k=0}^{\infty}\binom{\alpha}{k}\sum_{j=0}^k \binom{k}{j}x^j(-1)^{k-j}. $$ Now I want to obtain one sum, say, $$ =\sum_{n=0}^{\infty} c_n x^n. $$ How could I achieve this and this rearranged series will be uniformly convergent in $[\varepsilon,r]$? (A "little bit" smaller interval is also satisfactory.)

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Why don't you just use Taylor formula? –  Norbert Jul 4 '12 at 15:26
    
How, precisely, is $\binom\alpha k$ defined for non-integer $\alpha$? The only reasonable way I can think of is with the Gamma function. –  Cameron Buie Jul 4 '12 at 15:29
    
@CameronBuie: $\binom\alpha k=\alpha(\alpha-1)\cdots(\alpha-k+1)/k!$. –  Harald Hanche-Olsen Jul 4 '12 at 15:32
    
Interesting, Harald. Does that work out for a general binomial expansion? –  Cameron Buie Jul 4 '12 at 15:39
    
@Norbert the last power series is about $0$. Calculating the derivative of $x^{\alpha}$ at $0$ you would obtain zero in the denominator. –  vesszabo Jul 4 '12 at 16:25
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1 Answer 1

Such a series would converge for $|x|<r$ and that is possible only for integer nonnegative values of $\alpha$.

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right, the power series is convergent in a symmetric interval about $0$, so it will convergent for $|x|<r$ for, what function?. Why only for nonnegative integer values of $\alpha$? The binomial series absolutely and uniformly converges for $|x-1|<1-\delta$ ($\delta>0$ is "small", this is the reason of the condition $0<\varepsilon<x$ and $x<r<1$. The $\varepsilon$ condition is important. –  vesszabo Jul 4 '12 at 16:37
    
@vesszabo the answer is about the series $\sum_{n=0}^{\infty} c_n x^n$. –  Andrew Jul 4 '12 at 17:35
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