I am trying to solve a problem where I am asked to compute the Taylor expansion series of the function $\sin\left(\frac{1}{1-z}\right)$ around $z=0$. Now, I know that to find the coefficients of the series I can use the Cauchy Integral formula, but here is my doubt: the Taylor series for $\sin(z)$ is valid when $z$ is closed to $0$, and in my case when $z\sim 0$ I have $\sin\left(\frac{1}{1-z}\right) \sim \sin(1)$.
Does everything work the same? Why?
Best Answer
You can use Bell polynomials and the Taylor series of $\frac{1}{1-z}$ and $\sin z$, so we have : $$\frac{1}{1-z}=\sum_{n=0}^\infty z^n,\ \sin z=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$ Then we have : $$\sin \biggl(\frac{1}{1-z}\biggr)=\sum_{n=0}^\infty\frac{\sum_{k=0}^n \alpha_kB_{n,k}(0!,1!,...,(n-k+1)!)}{n!}z^n$$ Where : $$\alpha_{2n}=0,\\ \alpha_{2n+1}=(-1)^n\frac{1}{(2n+1)!}$$ For more information see the link https://en.wikipedia.org/wiki/Bell_polynomials.