how to find this analytic function satisfying such condition

Question

Decide whether there exists analytic $ f$ in $\mathbb{C}$ such that $f(n) = \cos(\sqrt{n})$ for all $n \in \mathbb{N}$.

I tried to raise this issue by Taylor's expansion, but I could not find a consistent result. Can someone help me?

Answer 1

The well-known Taylor series for the complex cosine is $$ \cos (z) = \sum_{k=0}^\infty \frac{(-1)^k}{(2k)!} z^{2k} $$ In particular, $$ \cos (\sqrt n) = \sum_{k=0}^\infty \frac{(-1)^k}{(2k)!} n^k $$ for all $n \in \Bbb N$. This suggests to define $f: \Bbb C \to \Bbb C$ as $$ f(z) = \sum_{k=0}^\infty \frac{(-1)^k}{(2k)!} z^k $$ It remains to show (which should not be too difficult) that the radius of convergence of that series is $\infty$, to that $f$ is an entire function with the desired property.

More generally, for any even entire function $g$ there is an entire function $f$ such that $f(z^2) = g(z)$ for all $z \in \Bbb C$.