Interpolating an infinite sequence ?

[JmsNxn]

Interpolating is actually pretty easy. It's when you ask for a functional equation that it's difficult.

Assume \(a_n \to \infty\) and \(b_n\to\infty\) and we want to find \(f(b_n) = a_n\). Define a Weierstrass function \(W(z)\), such that \(W(b_n) = 0\). Then define:

\[
f(z) = W(z) \sum_{n=0}^\infty  \frac{a_n}{W'(b_n)(z-b_n)}\\
\]

You can choose \(W\) such that the series converges, and that's pretty much it.

This is an exercise in John B Conway's complex analysis, if you're looking for a source.

hmm

Is sin(x) = W(x) valid ?

then we get 

\[
f(z) = sin(z) \sum_{n=0}^\infty  \frac{a_n}{cos(2 \pi n)(z-2 \pi n)}\\
\]

I guess I made a mistake there ...

You'd need to choose a specific zero function depending on \(a_n\). Such that we have \(W'(b_n) \) is large enough to force the series to converge. In your cause, you would need to find a function with zeroes at \(n\) but has a derivative at \(n\) which causes the series to converge. For example, use:

\[
f(z) = A(z)\sin(2 \pi z) \sum_{n=0}^\infty \frac{a_n}{2\pi A(n) (z-n)}\\
\]

This satisfies:

\[
f(n) = a_n\\
\]

And you can force convergence of the series by letting  \(A(n)\) be as large as possible. So for example \(A(z) = e^z\) works.