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 >