I have the given problem. Let $f(z)$ be an entire function such that $\int_\gamma\frac{f^{(n)}(z)dz}{z}=\frac{1}{n}$, where $(n)$ denotes the $n$-th derivative and $\gamma=|z|=3$ counterclockwise. Determine the Taylor series for $f(z)$ centered at $z=0$ (assume that $f(0)=0)$.
Taylor series is:
\begin{equation}
f(x)=f(x_0)+\frac{f'(x_0)(x-x_0)^1}{1!}+\frac{f''(x_0)(x-x_0)^2}{2!}+\ldots+\frac{f^{(n)}(x_0)(x-x_0)^n}{n!}+\ldots
\end{equation}
But how do we derive the Taylor series from here?
Thanks
Best Answer
You have$$(\forall n\in\Bbb N):\frac1n=\int_\gamma\frac{f^{(n)}(z)}z\,\mathrm dz=2\pi if^{(n)}(0),$$and therefore$$\left(\forall n\in\Bbb N\right):\frac{f^{(n)}(0)}{n!}=-\frac i{2\pi n.n!}.$$So, the Taylor series of $f$ centered at $0$ is$$-\sum_{n=1}^\infty\frac{iz^n}{2\pi n.n!}.$$