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It was a problem: does there exists an entire function which vanishes at $n+\frac{1}{n}$ for all $n\in\mathbb{N}$?

Since the set $\left\{n+\frac{1}{n}\right\}_{n\geq 1}$ has no limit point in $\mathbb{C}$, by Weierstrass theorem, there exist such function.

Question: Can we produce an entire function (non-zero) which vanishes on above set using the well-known functions $\sin z, \cos z, e^z$, polynomial etc.? [In other words, can we write our required function as a composition of $\sin, \cos, e$, polynomials etc.?]

rubik
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1 Answers1

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If you don't mind a bunch of extraneous zeroes, you can take $$ f(x)=\cos(\pi\sqrt{x^2-4})-\cos(\pi x) $$ which is entire because $\cos x$ is even and entire, and so $\cos\sqrt{x}$ is entire.

Using the product-to-sum formula, we have $$ f(x)=2\sin\left(\pi \frac{x+\sqrt{x^2-4}}{2}\right)\sin\left(\pi \frac{x-\sqrt{x^2-4}}{2}\right) $$ and the first factor in this product is already zero at each of the points $n+\frac{1}{n}$: if $x=n+\frac{1}{n}$ and $x > 1$, then $\sqrt{x^2-4}=n-\frac{1}{n}$, and so the argument being passed to $\sin$ is a multiple of $\pi$.

Note however that this will also vanish when $x=n+\frac{1}{n}$ for $n$ a negative integer (because of the second factor). So if you're looking for something that could have come out of the Weierstrass theorem (i.e., something with precisely your original zero set), this isn't it...

Micah
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  • that doesn't seem to be entire though – Albert Feb 23 '15 at 20:02
  • @Glougloubarbaki: Why do you say that? – Micah Feb 23 '15 at 20:02
  • because of the square root. I doesn't seem to be holomorphic near $x=2$ to me – Albert Feb 23 '15 at 20:10
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    @Glougloubarbaki Note that $\cos$ is even, so the square root only appears under an even power. Being explicit, $\cos(\pi \sqrt{x^2-4}) = \sum_{n=0}^\infty \frac{(-1)^n \pi^{2n} (x^2-4)^n}{(2n)!}$. It is straightforward, if tedious, to rearrange this into an ordinary power series which you will find converges globally. – Ian Feb 23 '15 at 20:12
  • @Glougloubarbaki More intuitively (but in a way which is much harder to generalize), $\sqrt{z}$ is basically bad because it has a sign ambiguity in general. An even function collapses this ambiguity. – Ian Feb 23 '15 at 20:15
  • ok, I rescind my objection :) – Albert Feb 23 '15 at 20:22
  • This is awesome. How did you find the function? Also, to limit it to $x \geq 2$, cannot we substitute $\left(\sqrt x\right)^2$ in place of $x$? – rubik Feb 23 '15 at 22:35
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    @rubik: $g(x)=\frac{x+\sqrt{x^2-4}}{2}$ is the inverse function to $n+1/n$, so it seemed natural to look for things involving $\sin(\pi g(x))$. Unfortunately $\sin(\pi g(x))$ on its own is not entire, so I played around until I found a factor I could multiply it by that would straighten out the branch points at $\pm 2$. – Micah Feb 24 '15 at 05:24
  • @rubik: If you substitute $\sqrt{x}^2$ for $x$, I think it stops being entire again... – Micah Feb 24 '15 at 05:26
  • @Micah: It makes sense! Thank you! – rubik Feb 24 '15 at 08:41