I am trying to prove that if a function $f(x)$ can be written as a power series in the form \begin{equation} f(x)=\sum_{n=0}^{\infty}c_n(x-x_0)^n \end{equation} such that $|x-x_0|<r$, then \begin{equation} f(A)=\sum_{n=0}^{\infty}c_n(A-x_0I)^n \end{equation} as long as $|x_k-x_0|<r$ for all eigenvalues $x_k$ of $A$.
I have written $f(A)=Zf(J)Z^{-1}=Z\text{diag}(f(J_1),f(J_2),...,f(J_s))Z^{-1}$ where $J_k$ are Jordan blocks and then $f(J_k)=\begin{pmatrix} f(x_k) & f'(x_k) & \cdots & \frac{f^{(m_k-1)}(x_k)}{(m_k-1)!} \\ & f(x_k) & \ddots & \vdots \\ & & \ddots & f'(x_k) \\ & & & f(x_k) \end{pmatrix}$. After this I have substituted for $f(x_k), f'(x_k)$ etc. the power series and am trying to show $f(J_k)=\sum_{n=0}^\infty c_nJ_k^n$.
From there I can say $f(A)=Z\text{diag}(\sum_{n=0}^\infty c_nJ_1^n,...,\sum_{n=0}^\infty c_nJ_s^n)Z^{-1}=\sum_{n=0}^\infty c_n(Z\text{diag}(\sum_{n=0}^\infty J_1^n,...,\sum_{n=0}^\infty J_s^n)Z^{-1})=\sum_{n=0}^\infty c_nA^n$.
Does this make any sense? If so, how do I complete the proof, if not, then how do I prove this?
Thanks