Mittag-Leffler's theorem

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In complex analysis, Mittag-Leffler's theorem concerns the existence of meromorphic functions with prescribed poles. It is sister to the Weierstrass factorization theorem, which asserts existence of holomorphic functions with prescribed zeros. It is named after Gösta Mittag-Leffler.


Let D be an open set in \mathbb C and E \subset D a closed discrete subset. For each a in  E, let p_a(z) be a polynomial in 1/(z-a). There is a meromorphic function f on D such that for each a \in E, the function f(z)-p_a(z) is holomorphic at a. In particular, the principal part of f at a is p_a(z).

One possible proof outline is as follows. Notice that if  E is finite, it suffices to take  f(z) = \sum_{a \in E} p_a(z). If E is not finite, consider the finite sum  S_F(z) = \sum_{a \in F} p_a(z) where  F is a finite subset of E. While the S_F(z) may not converge as F approaches E, one may subtract well-chosen rational functions with poles outside of D (provided by Runge's theorem) without changing the principal parts of the S_F(z) and in such a way that convergence is guaranteed.


Suppose that we desire a meromorphic function with simple poles of residue 1 at all positive integers. With notation as above, letting

p_k = \frac{1}{z-k}

and E = \mathbb{Z}^+, Mittag-Leffler's theorem asserts (non-constructively) the existence of a meromorphic function f with principal part  p_k(z) at z=k for each positive integer  k. This f has the desired properties. More constructively we can let

f(z) = z\sum_{k=1}^\infty \frac{1}{k(z-k)} .

This series converges normally on  \mathbb{C} (as can be shown using the M-test) to a meromorphic function with the desired properties.

Pole expansions of meromorphic functions[edit]

Here are some examples of pole expansions of meromorphic functions:

		= \sum_{n \in \mathbb{Z}} \frac{(-1)^n}{z-n\pi}
		= \frac{1}{z}+ \sum_{n=1}^\infty (-1)^n \frac{2z}{z^2 - n^2 \pi^2}

 \cot(z) \equiv \frac{\cos (z)}{\sin (z)}
		= \sum_{n \in \mathbb{Z}} \frac{1}{z-n\pi}
		= \frac{1}{z} + \sum_{k=1}^\infty \frac{2z}{z^2 - k^2\pi^2}

  \frac{1}{\sin^2(z)} = \sum_{n \in \mathbb{Z}} \frac{1}{(z-n\pi)^2}

\frac{1}{z \sin(z)}
		= \frac{1}{z^2} + \sum_{n \neq 0}  \frac{(-1)^n}{\pi n(z-\pi n)}
		= \frac{1}{z^2} + \sum_{n=1}^\infty \frac{(-1)^n}{n\pi} \frac{2z}{z^2 - \pi^2 n^2}

See also[edit]


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