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Let $f(z)=\sum^{\infty}_{k=0}\frac{k^3}{3^k}z^k$. Compute the following:

a. $f^{(6)}(0)$

b. $\int_{|z|=1}\frac{f(z)}{z^4}dz$

I dont understand how to do these problems, the answer in the book is:

a. $f^{(6)}(0)=6!a_6=\frac{6!6^3}{3^6}$

b. $\int_{|z|=1}\frac{f(z)}{z^4}d=\frac{2\pi i}{3!}f^{(3)}(0)=\frac{2\pi i(3!a_3)}{3!}=2\pi iz$

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So the radius of convergence is $>1$. – 1015 Mar 5 '13 at 1:47
up vote 2 down vote accepted

For a., the expression you have for $f(z)$ is a sum of terms $a_iz^i$ The sixth derivative of each term, when evaluated at $z=0$ is $0$ except for the sixth. All the lower ones become a constant and then have a derivative taken, while all the higher ones still have a factor of $z^{i-6}$ which is zero. So the sixth derivative evaluated at zero is $6!a_6$ as claimed.

For b. look up the Cauchy integral formula. The curve is a closed circle, so you need the term in $f$ of order $z^3$ so when you divide by $z^4$ you are left with $z^{-1}$ as the integral formula requires.

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I seem to understand b now, thanks for that. but a im still confused, can you elaborate a bit more please? – Q.matin Mar 5 '13 at 1:53

a. $f(z)$ is also $\sum_{k=0}^\infty {f^{(k)}(0) \over k!}z^k$, and the Taylor expansion is unique. What do you conclude?

b. This follows from the Residue Theorem

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How did you know $\sum_{k=0}^\infty {f^{(k)}(0) \over k!}z^k$ is the Taylor expansion? – Q.matin Mar 5 '13 at 1:54

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