Entire function vanishing at $n+\frac{1}{n}$ for $n\geq 1$.

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.?]

Solution 1:

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...