Expand \(f(z) = z^2\cos\qty{z\over 3}\) about \(z=0\).
Solution
\[ \cos\qty{z\over3} =\sum_{k=0}^\infty(-1)^k{(z/3)^{2k}\over(2k)!}, \] so \[ z^2\cos\qty{z\over3} =\sum_{k=0}^\infty(-1)^k{z^{2k+2}\over 3^{2k}(2k)!} =z^2-{z^4\over 2!\,3^2}+{z^6\over 4!\,3^4}-\cdots. \] This is an ordinary Taylor series; there are no negative powers.