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Suppose that $x_i$ and $y_i$ are sequences in $\mathbb{C}$. Can you construct a non constant entire function such that $f(x_i)=y_i$?

What happens if $x_i$ have an accumulation point? or what happens if $x_i$ and $y_i$ is an uncountable collection of points?

I have studied the book by Churchill and Brown on complex analysis (up to residues and their applications) but I seem to have never encountered a result that seems to address the following question.

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  • $\begingroup$ The question is very broad. See Abel-Goncharov problem for a start. $\endgroup$ – user64494 Jun 19 '15 at 6:08
  • $\begingroup$ @user6449 For reals if $x_i$ is a sequence with no accumulation points and $y_i$ is some sequence then intuitively we can construct a differentiable function that satisfies $f(x_i)=y_i$ can this not be said for complex numbers? $\endgroup$ – Miz Jun 19 '15 at 6:34
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If the sequence $(x_k)$ has no finite accumulation points, i.e., if $x_k\to\infty$ as $k\to\infty$, then for any $(y_k)$ there exists an entire function such that $f(x_k)=y_k$ for all $k$. This is a theorem of Weierstrass. Idea of proof:

  1. There exists an entire function $g$ with zeros at every $x_k$, also proved by Weierstrass.
  2. For each $k$, consider the Taylor series of $g$ at $x_k$: it begins with some term $c_{n_k}(z-x_k)^{n_k}$, $c_n\ne 0$. Use the Mittag-Leffler's theorem to produce a meromorphic function $h$ with pole of order $n_k$ at each $x_k$, and with leading Laurent coefficient such that $g(z)h(z)$ evaluates to $y_k$ at $x_k$ after the singularity is removed.

If $x_k$ have an accumulation point, the values $y_k$ can no longer be arbitrary. I don't know of any sensible way of expressing a consistency condition on $y_k$ that would be required for $f$ to exist.

Uncountable case is even worse: such a set will have uncountably many points of accumulation. I'm pretty sure there is nothing constructive to be said. Unless the set has a nice structure: a line or an analytic curve, in which case we are discussing the extensibility of a function defined on such a set. This depends on the radius of convergence of its Taylor series. If the radius is infinite, there is an extension to an entire function; otherwise no.

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  • $\begingroup$ I do like your logogram... and because your answers are insightful, I (as a fallible human...) would like to know "who you are", ... to have some (pathetic) arguments for "trust", etc. (I can understand if this is not viable...) $\endgroup$ – paul garrett Jun 21 '15 at 0:14

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