Elementary function $f:\mathbb{R}\to \mathbb{R}$ so that $f(0)=0, f(n)=2^n$ for $n\in \mathbb{N}$

Question

Is there any elementary function $f:\mathbb{R}\to \mathbb{R}$ such that $f(0)=0$ and $f(n)=2^n$ for every positive interger $n$?

By elementary functions, I mean functions that are sum, product, inverse, or composition of exponential, polynomial, rational, trigonometric functions.

https://en.wikipedia.org/wiki/Elementary_function

Would you be happy with using the sinc function ? – charmd Dec 03 '20 at 07:19 — charmd, Dec 03 '20 at 07:19
how about $$f(n)=\sum_{k=0}^{n}\binom{n}{k}$$ – Albus Dumbledore Dec 03 '20 at 07:25 — Albus Dumbledore, Dec 03 '20 at 07:25

score 3 · Accepted Answer · answered Dec 03 '20 at 07:27

The function $f(x) = \exp(\log(2)x)$ satisfies $f(n) = 2^n$ for all integers $n \ge 1$ and $f(0)=1$.

To get what you want, we could try to find a function $g$ such that $g(0)=1$ and $g(n)=0$ for all integers $n \ge 1$ (and then take $f-g$). Such a function $g$ is easy to find: just take $g(x) = \sin\big(\frac{\pi}{2} 2^x\big)$.

Then $(f-g)(0) = 1-\sin(\frac{\pi}{2})=0$ and for $n \ge 1$, $(f-g)(n) = 2^n - \sin(\pi 2^{n-1}) = 2^n$.

So $x \mapsto 2^x - \sin(\pi 2^{x-1})$ is a solution.

Elementary function $f:\mathbb{R}\to \mathbb{R}$ so that $f(0)=0, f(n)=2^n$ for $n\in \mathbb{N}$

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