Does for any sequence $(a_n)$ exist a power series $f$ such that $f(i)=a_i$?

Question

Consider the sequence of real numbers $(a_n)$. Does there exist power series $f(x)=\sum_{i=0}^\infty c_ix^i$, where $c_i \in \mathbb R$, such that $f(k)=a_k$ or $f(kx_0)=a_k$ for some $x_0>0$?

I know that the answer is true for holomorfic function $f$ (according to Analytic "Lagrange" interpolation for a countably infinite set of points?), but is it true for $f$ is power series or even polynomial?

@Dzoooks $a_i$ is the coefficient of power series, whereas $x_i$ is the element of sequence I try to interpolate. — Tyrone Ward, Jul 02 '18 at 23:13
But you also use $x$ as the variable in your power series. You should clean up your notation a bit. — Dzoooks, Jul 02 '18 at 23:17
It's certainly impossible to find a single polynomial (of finite degree) whose values are $f(n) = F_n$ (since $\Delta^k F_n$ is not identically zero for any $k$). — Daniel Schepler, Jul 02 '18 at 23:18
Also, such a power series would have to have radius of convergence $\infty$, like $e^x$ for example. Your $c_i$'s would have to decay rapidly. — Dzoooks, Jul 02 '18 at 23:19

score 2 · Answer 1 · answered Jul 02 '18 at 23:59

Consider a Newton-style interpolation formula: $$f(x) = b_0 + b_1 x (x/1)^{n_1} + b_2 x (x-1) (x/2)^{n_2} + b_3 x (x-1) (x-2) (x/3)^{n_3} + \cdots.$$

By recursion, we can find $b_0$ which guarantees $f(0) = a_0$, then find $b_1$ which guarantees $f(1) = a_1$ and choose $n_1$ according to some criteria below, then find $b_2$ which guarantees $f(2) = a_2$ and choose $n_2$ by the below criteria, etc. Furthermore, within this process, we can choose $n_1, n_2, n_3, \ldots$ such that none of the terms in the expansion "overlap" in terms of producing multiple coefficients for any $x^i$; and we can also choose $n_1, n_2, n_3, \ldots$ such that the coefficients of the resulting power series will satisfy the requirements of the Root Test and give an infinite radius of convergence. (In particular, if one of the terms of the expansion of $b_i x(x-1)(x-2)\cdots(x-i+1)$ is $C x^m$ then multiplying by $(x/i)^{n_i}$ gives $C/i^{n_i} x^{m+n_i}$, so the root test term would give $|C|^{1/(m+n_i)} / i^{n_i/(m+n_i)}$ which approaches $1/i$ as $n_i \to \infty$, so by choosing $n_i$ sufficiently large, we can ensure all of these terms are less than $2/i$.)

score 1 · Answer 2 · answered Jul 02 '18 at 23:58

It is not possible to find a polynomial (as mentioned in the comments) but you can find an entire function withe stated property. A much stronger result is proved in Rudin's RCA. See "An Interpolation Problem" in the chapter on "zeros of Holomorphic Functions". [ Take $A=\{0,1,2...\}$ and $m(k)=0$ for all $k$ in that theorem].

Does for any sequence $(a_n)$ exist a power series $f$ such that $f(i)=a_i$?

2 Answers2