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I have a complex analysis question that I am stuck on.

Let $\sum_{n=0}^{\infty}a_{n}z^{n}$ be the Taylor series expansion at 0 for $(1-z)^{-a}$ where $a>0$. Using the generalized Cauchy integral Formula, obtain an estimate for the coefficient $a_{n}$.

I know the Cauchy integral Formula says for an analytic function $f$ and a regular closed curve $\gamma$ with $z\in\gamma$ then $ f^{(k)}(z)=\frac{k!}{2\pi i}\int_{\gamma}\frac{f(w)}{(w-z)^{k+1}}dw $. So I see that $a_{n}=\frac{f^{(n)}(0)}{n!}=\frac{1}{2\pi i}\int_{\gamma}\frac{f(w)}{(w)}dw$.

Additionally, I see that the distance is 1 between 0 the singularity at $z=1,$ so that $\gamma$ is at most of radius 1. However, I am stuck on how to proceed. I want to plug in for my particular function $(1-z)^{-a}$ but I don't know exactly how to do it. Can an exact expression for $a_n$ be attained or just an estimate?

Any help would be appreciated.

share|improve this question
    
Does $a\in \mathbb{Z}$? –  Nameless Dec 8 '12 at 21:16
    
Good question. No, a is any real number. If a is an integer than the coefficients can be found quite easily by differentiating the geometric series. –  Daniel Dec 8 '12 at 21:34
    
Do you know the formula $|a_n| \leq \frac{1}{r^n} \cdot \max \{|f(\xi)|; |\xi| \leq r\}$ where $0<r<1$...? –  saz Dec 9 '12 at 9:41
    
Yes. I'm tempted to use that formula. However this function is unbounded near 0. Doesn't this present a problem. –  Daniel Dec 10 '12 at 0:28

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