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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.

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

2 Answers 2

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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?

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  • $\begingroup$ I'm giving it a look. It is every possible that this is a valid answer. $\endgroup$ Oct 13, 2017 at 15:22
  • $\begingroup$ $1/ \Gamma(-x + 2)$ does the trick. Thank you very much, sir. $\endgroup$ Oct 13, 2017 at 15:26
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Arguably a better answer:

$$f(x) = \begin{cases} 1 & x = 0 \\ \frac{\sin(\pi x)}{\pi x} & x\neq 0\end{cases}.$$

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.

$$f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n\pi^{2n}x^{2n}}{(2n+1)!}.$$

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