Inspired of course by What's a natural candidate for an analytic function that interpolates the tower function? I am minded to ask what looks to me like a more natural question: given a sequence $a_1,a_2,a_3,\ldots$ of complex numbers, is there always a holomorphic function $f$ defined on the entire complex plane, with $f(n)=a_n$ for $n=1,2,3,\ldots$? No idea what the answer is myself, but wouldn't surprise me if it were well-known and even easy.
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asked Apr 8, 2010 at 12:06
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5$\begingroup$ Sure. More generally, for any Stein space $X$ and discrete set $S$ in $X$ and effective divisor $D$ supported on $S$, the surjective map of coherent sheaves $O_X \rightarrow O_D$ has coherent kernel and so induces a surjection on global sections. So by description of $O_D(X)$ via discreteness of $S$, there exists holomorphic $f$ on $X$ whose germ at each point of $S$ has whatever "initial part of Taylor expansion" we wish. In dimension 1 can play similar game with meromorphic $f$ holomorphic outside $S$ and Laurent tails at $S$ (generalizing Mittag-Leffler theorem) $\endgroup$– BCnrdCommented Apr 8, 2010 at 12:24
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1$\begingroup$ Typo correction above: $D$ isn't a divisor when $X$ has dimension $> 1$. I meant it to be a 0-dimensional analytic space structure on $S$ (of which there are zillions of choices as "multiplicity" grows). Presumably this intent was clear. $\endgroup$– BCnrdCommented Apr 8, 2010 at 15:03
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1$\begingroup$ This question was asked twice before: mathoverflow.net/questions/2944, mathoverflow.net/questions/7328/… $\endgroup$– Jonas MeyerCommented Apr 8, 2010 at 15:47
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$\begingroup$ @Jonas: But now with a completely different solution. :) $\endgroup$– BCnrdCommented Apr 8, 2010 at 18:24
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1$\begingroup$ For further reference, a stronger result holds; c.f. Walter Rudin's Real and Complex Analysis theorem 15.13. $\endgroup$– AkerbeltzCommented Apr 1, 2021 at 13:38
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2 Answers
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This is Exercise 6, Page 26, of Knopp's Problem Book in the Theory of Functions, Volume 2: For any sequence of complex numbers $z_n$ with no finite limit point, and for any sequence of complex numbers $w_n$, there is an entire function mapping $z_n$ to $w_n$. The proof goes like this: Use the Weierstrass Factor Theorem to construct a function $W$ with simple zeros at the $z_n$. Use the Mittag-Leffler theorem to construct a function $M$ with simple poles at the $z_n$ with residues $\frac{w_n}{W'(z_n)}$. Then the function $W\cdot M$ does the job.
answered Apr 8, 2010 at 13:10
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$\begingroup$ Ah. Of course, good old Knopp would have it (+1). $\endgroup$ Commented Apr 8, 2010 at 19:19
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Probably well-known. Easy? I'd venture to guess that an expression like $$\sum_{n=1}^\infty b_n\frac{e^{c_n(z-n)}}{(n-1)!}\prod_{k=1}^{n-1}(z-k)$$ can be made to work. You'll have to pick the $b_n$ successively to make the $N$'th partial sum equal to $a_N$, and real constants $c_n$ large enough to obtain uniform convergence to the left of any fixed vertical line. E.g., so that the $n$'th term has absolute value less than $2^{-n}$ when $\operatorname{Re} z<n/2$.
answered Apr 8, 2010 at 12:27