Let $r>0$, $\varepsilon>0$ and $\alpha>0$. Assume that $0<\varepsilon
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.)
Convergence in binomial series
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real-analysis
sequences-and-series
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0Why don't you just use Taylor formula? – 2012-07-04
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0How, precisely, is $\binom\alpha k$ defined for non-integer $\alpha$? The only reasonable way I can think of is with the Gamma function. – 2012-07-04
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0@CameronBuie: $\binom\alpha k=\alpha(\alpha-1)\cdots(\alpha-k+1)/k!$. – 2012-07-04
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0Interesting, Harald. Does that work out for a general binomial expansion? – 2012-07-04
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0@Norbert the last power series is about $0$. Calculating the derivative of $x^{\alpha}$ at $0$ you would obtain zero in the denominator. – 2012-07-04