Can any function on naturals be interpolated to a smooth function on reals?

Question

Let $f : \mathbb{N} \rightarrow \mathbb{N}$ be an arbitrary function from naturals to naturals. Is it always possible to find a function $g : \mathbb{R} \rightarrow \mathbb{R}$ such that

1. for any $n \in \mathbb{N}$, we have $f(n) = g(n)$, and
2. $g \in C^\infty$?

I'm asking because I was trying to prove a result about the ratios of functions from naturals to naturals and it occurred to me that if I could always interpolate to get back smooth functions from integers to integers, I could conceivably use l'Hopital's rule to resolve the limits.

Thanks!

Answer 1

Whittaker–Shannon interpolation, using the $sinc$ function, achieves that. http://en.wikipedia.org/wiki/Whittaker%E2%80%93Shannon_interpolation_formula $$g(x)=\sum_{n=-\infty}^{+\infty}f_n\frac{\sin\pi(x-n)}{\pi(x-n)}.$$

Answer 2

First, one needs an analytic function such that $f(0)=1$ and $f(n)=n$ for $n\in\mathbb Z \setminus\{0\}$. The sinc function can be useful for that: $$\operatorname{sinc}x = \frac{\sin x}x, ~~ x\in \mathbb R\setminus\{0\}$$ $$\operatorname{sinc}0 = \lim_{x\to 0}\frac{\sin x}x = 1$$

Note that $\operatorname{sinc}0=1$and $\mathrm{sinc} (n\pi) = 0$ for $n\in\mathbb Z \setminus\{0\}$, as needed. Therefore, one could construct $$ g(x) = \sum_{n\in\mathbb N} \operatorname{sinc}[(n-x)\pi] f(n) $$ and have $g(n) = f(n)$ for all $n \in \mathbb N$.

However, this would not in general even converge for any choice $f(n)$, and even if it converges, it might not have all the derivatives (i.e. the derivatives might not converge). In order to fix this, the function used to construct $g(x)$ should be modified to fall sufficiently fast. Just replacing this function with a different one which would be the same for every $n$ can work for a wider class of functions $f(n)$, but it will not work in general, regardless of the choice. The modification should depend on individual values $f(n)$. Something like this:

$$ g(x) = \sum_{n\in\mathbb N} \operatorname{sinc}[(n-x)\pi] e^{-f(n)^2(n-x)^2} f(n) $$

That way, no matter how absurdly fast $f(n)$ grows (e.g. something that needs to be written using Knuth's up-arrow notation or Conway chained arrow notation or worse, like the TREE function), $g(x)$ will converge everywhere and it will even be in $C^\omega$.

Answer 3

Given a sequence of points $z_n$ in an open subset $\Omega$ of $\mathbb{C}$ without an accumulation point, and $\alpha_n$ a sequence of values, there exists a holomorphic function $\phi$ on $\Omega$ with $\phi(z_n)= \alpha_n$, for all $n$. Now, if $z_n$ and $\alpha_n$ are real, we can take $\phi$ real on $\Omega\cap \mathbb{R}$ ( take $\frac{\phi(z) + \bar \phi(\bar z)}{2}$).
This is a standard result in complex analysis.