# Analytic function such that $f(1) = 1$ and $\forall n \in \mathbb{N}, n \neq 1 f(n) = 0$

Those there exist a function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(1) = 1$ and $\forall n \in \mathbb{N}, n \neq 1 , f(n) = 0$ and such that $f$ is analytic?

I have been thinking that I could use some composition of trigonometric functions.

Using $\sin{(g(x))}$ I would need a function $g(x)$ such that at the naturals, $g(x)$ was a multiple of $\pi$. I thought a linear function such as $g(x) = \pi x$ could do, but of course, this isn't the case, since at $x=1$, I would have $f(1)=0$ too.

I'm not sure whether such a function even exists, but my intuition says there isn't really any reason for such a function not to exist even being analytic.

• Try the series for sinc Oct 13, 2017 at 15:22
• @CameronWilliams You should give this as an answer. Not sure why you mention "series" though.
– zhw.
Oct 13, 2017 at 15:34
• @zhw. done. Reason I said "series" was because it's clearer that it is analytic that way. Oct 13, 2017 at 18:56
• @CameronWilliams OK, I guess I was assuming everyone knows $\sin (u)/u$ is analytic.
– zhw.
Oct 13, 2017 at 19:11

On the complex plane, the inverse of the Gamma function $f(z)=1/\Gamma(z)$ is an analytic function everywhere and has the property that it vanishes at $0$, $-1$, $-2$, $-3$ etc., but nowhere else. It is also real on the real line. Can you adapt it to suit your purposes?
• $1/ \Gamma(-x + 2)$ does the trick. Thank you very much, sir. Oct 13, 2017 at 15:26
$$f(x) = \begin{cases} 1 & x = 0 \\ \frac{\sin(\pi x)}{\pi x} & x\neq 0\end{cases}.$$
It is not a priori obvious that this function is analytic, but if you look at the following series, it is not hard to see that $f$ agrees with it.
$$f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n\pi^{2n}x^{2n}}{(2n+1)!}.$$