How to perform Inverse Z transform

Question

I'm trying to compute this Inverse Z Transform:

$\displaystyle X(z) = \sin\left(\frac{1}{z}\right)$

Suppose that the sequences are right handed and one sided

Any help would be appreciated.

Answer 1

I'm going to compute the Inverse Z Transform with use of the power series for sine

$$ \ X(z)=sin(1/z)\ $$ The power series for sine is $$ \ sin(x)=x+x^3/3!+x^5/5!+... =\sum_{0}^{\infty} \frac{x^{(2i+1)}}{(2i+1)!}\ $$ We changing variables $$ \ x=1/z\ $$ $$ \ sin(1/z)=\sum_{0}^{\infty} \frac{1}{(2i+1)!z^{(2i+1)}}\ $$

$$ \ x[n]=Z^{-1}{(X(z))}=\frac{1}{2\pi j}\oint sin(1/z)z^{n-1}dz = \sum_{0}^{\infty} \frac{1}{2\pi j}\oint \frac{z^{n-1}}{(2i+1)!z^{(2i+1)}}dz $$

the last integral is not zero iff it equals $n = 2i+1$. Then

$$ \ x[n]= \sum_{0}^{\infty}\frac{1}{(2i+1)!}\ $$