Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)

share|improve this question
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
add comment

1 Answer 1

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

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.