Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let {an} be a sequence of complex numbers indexed by the positive integers. Does there always exist an analytic function f such that f(n) = an for n=1,2,...? If not, are there any simple necessary or sufficient conditions for the existence of such f? This analytic function should be defined on some connected domain in the complex plane containing the positive integers.

To make this concrete, consider Ackermann's function, which is defined recursively: first define the sequence of functions Ak, k=1,2,..., as

A1(n) = 2n,
Ak(1) = 2, Ak(n) = Ak-1(Ak(n-1)),

and then define Ackermann's function as the diagonal A(n) = An(n) for n >= 1. Does there exist an analytic function f such that f(n) = A(n) for n = 1,2,...?

Actually, the individual functions Ak are interesting as well. A1(n) = 2n, as given above; A2(n) = 2n, and A3(n) = 2^(2^(...2^2)) (with n twos in the expression). Obviously, A1 and A2 have analytic extensions. According to Wikipedia (which uses a slightly different definition and notation), analytic extensions of A3 or Ak for any other k aren't known, but from the language, it isn't clear whether the existence of an extension is itself in question, or whether one simply hasn't been found yet. Also, it doesn't say anything about the diagonal A(n) (unless I missed it).

There are many other obvious sequences that don't seem to have obvious analytic extensions, like the prime-counting function (just to name one!). As far as my knowledge is concerned, this seems more like the rule than the exception. My knowledge here is admittedly very limited, though, so anything at all that you can share will probably teach me something.

share|improve this question
add comment

3 Answers

up vote 14 down vote accepted

It's a standard theorem in complex analysis that if z_n is a sequence that goes to infinity, there is an entire function taking any prescribed values at the z_n. There is a function f vanishing to order 1 at each z_n (for z_n=n, you could take f(z)=\sin \pi z), and then consider \sum_n a_nf(z)/(f'(z_n)(z-z_n)). This may not converge, but you can tweak it by multiplying each term by something that is 1 at z_n (eg, exp(c_n(z-z_n)) for c_n chosen appropriately) to make it converge.

(I don't know off the top of my head how to choose the c_n; this is copied from Exercise 1 on page 197 of Ahlfors's Complex Analysis.)

EDIT: It's easy to show that such c_n exist. If you write b_n=a_n/(z_n f'(z_n)), then for any fixed z, the terms of the sum will be approximately b_n exp(c_n z_n) for n large. You can obviously pick c_n so that this converges.

share|improve this answer
1  
Ah, cool. Ironically, this also turns out to be an exercise in my old complex analysis textbook! Its construction, which they name the "Pringsheim interpolation formula," is very similar to yours. –  Darsh Ranjan Oct 28 '09 at 1:21
add comment

If I remember correctly you can get even a stronger result. Let a_n, b_n and c_n be sequences of complex numbers and i_n be a sequence of natural numbers. Then I believe there exists a meromorphic function that takes the value b_n on a_n for all n, and has a pole of order i_n at c_n for all n. Assuming that the a_n and c_n are disjoint and have no accumulation point. It's been a while since I thought about complex analysis but I seem to remember learning this.

share|improve this answer
add comment

Actually, there is an even stronger result, often called the interpolation theorem, which follows from a well known theorem of Mittag-Leffler :

Let (z_n) be sequence of complex numbers with no accumulation point. For each n, let l(n) be any integer greater or equal to 1 and (a_nk) (0 <= k <= l(n) ) complex numbers. Then there exists an entire function g(z) such that

g^(k)(z_n) = a_nk

for every n>=1 and every 0 <= k <= l(n)

That is, you can fix values for the derivative at the z_j's.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.