Let's say I have a sequence $s_n$ of numbers, and I want a series $a_i$ which computes the sequence; that is

$\sum_{i=0}^\infty a_i n^i = s_n$

Clearly $a_0 = s_0$, but after that I am stuck. I need $\sum_{i=0}^\infty a_i = s_1$, so I'm not sure how I should go about picking the $a_1$ -- it seems that I need to choose all the $a_i$ "simultaneously".

Are there any methods for doing this? I am most interested in computable methods, though I suspect it may not be a computable problem in general. I am interested in theoretical results about this as well.

  • $\begingroup$ What do you mean by computable? If you truncate the $s_n$ sequence this is just polynomial interpolation. $\endgroup$ – MR_Q Mar 24 at 4:28
  • $\begingroup$ @MR_Q yeah but I don't want to truncate. I mean can I compute (exactly, or perhaps to arbitrary closeness) each $a_i$ from a finite amount of $s_i$. $\endgroup$ – luqui Mar 24 at 5:08

More generally, for any sequences $p_j$ and $s_j$ of complex numbers such that $p_j$ has no limit points, there exist entire functions $f$ such that $f(p_j) = s_j$ for all $j$. See e.g. Rudin, "Real and Complex Analysis", Theorem 15.15. You then want the Maclaurin series of $f$.

EDIT: One way to construct such an $f$ in your case is as follows. Let $ g_n(z)$ be an entire function with $g_n(n) = (-1)^n s_n/\pi$ and $|g_n(z)| \le 1/n$ for $|z| \le n/2$. For example, you might take $g_0(z) = s_1/\pi$ and $g_n(z) = (-1)^n s_n (z/n)^{k_n} \pi$ for $n \ge 1$, where $k_n > \log_2(n |a_n|/\pi)$. Let $g(z) = \sum_{n=0}^\infty \frac{g_n(z)}{z-n}$. This series converges to a meromorphic function with simple poles at the nonnegative integers, having residue $(-1)^n s_n/\pi$ at $z=n$. Then $ f(z) = \sin(\pi z) g(z)$ is entire with $f(n) = s_n$ for all nonnegative integers $n$.

Note that to compute $a_j$ we need only consider the $g_n$ where $k_n \le j$, and we can arrange it so $k_n \ge n$.


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