The exercise [1, p. 86 ex. 3.16] asks to prove that if the eigenfunctions and eigenvalues of an operator $\widehat{A}$ are $\{\phi_{n}\}$ and $\{a_{n}\}$, respectively ($\ensuremath{\widehat{A}}\phi_{n}=a_{n}\phi_{n}$) then the eigenfunctions of a function $f(x)$ having an expansion of the form:
 $f(x)=\sum\limits^{\infty}_{l=0}b_{l}x^{l}$ (1)

will also be $\phi_n$  with corresponding eigenvalues $f(a_n)$ , $n=0,1...$ . That is $f(\widehat{A})\phi_n=f(a_n)\phi_n$. read the conclusion >