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
2
$\begingroup$
real-analysis
sequences-and-series
-
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
1 Answers
4
Such a series would converge for $|x|
-
0@vesszabo the answer is about the series $\sum_{n=0}^{\infty} c_n x^n$. – 2012-07-04