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Let's say I have a multiset of complex numbers $\lbrace a_1,\cdots,a_n\rbrace$ (so some of the elements may be repeated) and I would like to construct an entire function $p(z)$ with those numbers as zeroes. However, I also have a multiset of complex numbers $B = \lbrace b_1,\cdots,b_n \rbrace$ such that I wish $p(b_i) = 1$ - p is only 1 on the $b_i$'s.

It seems like trying to use Lagrange's polynomial interpolation formula gives you a polynomial with too high a degree (greater than $n$ and less than or equal to $2n$), and then there's the possibility that $p^{-1}(1) \nsubseteq B$.

I've been thinking about doing the following:

Let $g(z) = (x-a_1) \cdots (x - a_n)$, and then via Weierstrass construct an entire function $h(z)$ such that $e^{h(b_i)} = 1/g(b_i)$. Then it seems like the entire function $e^{h(z)}g(z)$ is getting somewhat closer to what I want - but then again I don't know if there are any other $\alpha$'s such that $e^{h(\alpha)}g(\alpha) = 1$ where $\alpha \notin B$.

The problem of polynomial interpolation and fitting seems very well studied; however, I can't seem to find a reference for this particular puzzle.

Thanks in advance!

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You're imposing too many conditions. The space of polynomials of degree at most $n$ has dimension $n+1$. You are trying to impose $2n$ linear conditions on that space, which when $n > 1$ is more conditions than the dimension of your space. So there will be no solution in general. – Pete L. Clark May 19 '10 at 8:14
Based on a closer reading of your question, it sounds like you are aware of what I said in my previous comment. But then I can't figure out what you're asking: of course you can interpolate by an entire function, but not by a polynomial in general. – Pete L. Clark May 19 '10 at 8:16
Ah, I guess I was not clear at all. I'm not looking for a polynomial (because of what you just said), but rather an entire function with 0's at only those places (the $a_i$'s), and 1's at those places (the $b_i$'s). I know I can construct a Weierstrass entire function with the specified zeros, but can I force the entire function to have 1's at only those places? – Henry Yuen May 19 '10 at 8:26
up vote 8 down vote accepted

If I read you right, you want an entire function that takes the values $0$ and $1$ at only finitely many (specified) points. This implies that the function must be a polynomial, by Picard's great theorem, since there will be deleted neighbourhoods of infinity where the function misses two values.

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Then, based on Pete Clark's comment above, I'm imposing too many conditions on the polynomial for it to exist (in general)? – Henry Yuen May 19 '10 at 9:08
Henry, yes you are imposing too stringent conditions. – Robin Chapman May 19 '10 at 9:16

In your statement, you do not say explicitly, whether $p$ is aloowed to have other zeros, except those in the set $A$.

If you want to construct an entire function with zeros and ones exactly prescribed, this is clearly impossible when your sets $A$ and $B$ are both finite. For the reason explained by Robin Chapman.

If you want ones to be exactly prescribed, and function having zeros on the set $A$, and perhaps other zeros, then this is possible: take $p(z)=1+(z-b_1)...(z-b_n)\exp g(z)$ and use interpolation for $g$.

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Some very nice instances of your problem (but of course not all) are solved by so-called Shabbat polynomials, ie. by polyomials such that $p^{-1}[0,1]$ is a tree and such that $\{0,1,\infty\}$ are the only critical values. Every planar tree can be realized by an (essentially unique up to affine transformation) Shabbat polynomial. You have thus a polynomial solution if your points $a_i$ and $b_i$ form a bipartition of the vertices of such a "Shabbat-tree".

Let me add that Shabbat polynomials are the simplest instances of "dessins d'enfants" defined by Grothendieck in the hope of understanding the absolute Galois group. (Suitably normalized Shabbat polynomials have algebraic coefficients and the action of the absolute Galois group preserves them and acts thus on the corresponding trees by permuting them.)

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