To install click the Add extension button. That's it.

The source code for the WIKI 2 extension is being checked by specialists of the Mozilla Foundation, Google, and Apple. You could also do it yourself at any point in time.

4,5
Kelly Slayton
Congratulations on this excellent venture… what a great idea!
Alexander Grigorievskiy
I use WIKI 2 every day and almost forgot how the original Wikipedia looks like.
Live Statistics
English Articles
Improved in 24 Hours
Languages
Recent
Show all languages
What we do. Every page goes through several hundred of perfecting techniques; in live mode. Quite the same Wikipedia. Just better.
.
Leo
Newton
Brights
Milds

# Mittag-Leffler's theorem

In complex analysis, Mittag-Leffler's theorem concerns the existence of meromorphic functions with prescribed poles. Conversely, it can be used to express any meromorphic function as a sum of partial fractions. It is sister to the Weierstrass factorization theorem, which asserts existence of holomorphic functions with prescribed zeros.

The theorem is named after the Swedish mathematician Gösta Mittag-Leffler who published versions of the theorem in 1876 and 1884.[1][2][3]

## Theorem

Let ${\displaystyle U}$ be an open set in ${\displaystyle \mathbb {C} }$ and ${\displaystyle E\subset U}$ be a subset whose limit points, if any, occur on the boundary of ${\displaystyle U}$. For each ${\displaystyle a}$ in ${\displaystyle E}$, let ${\displaystyle p_{a}(z)}$ be a polynomial in ${\displaystyle 1/(z-a)}$ without constant coefficient, i.e. of the form

${\displaystyle p_{a}(z)=\sum _{n=1}^{N_{a}}{\frac {c_{a,n}}{(z-a)^{n}}}.}$
Then there exists a meromorphic function ${\displaystyle f}$ on ${\displaystyle U}$ whose poles are precisely the elements of ${\displaystyle E}$ and such that for each such pole ${\displaystyle a\in E}$, the function ${\displaystyle f(z)-p_{a}(z)}$ has only a removable singularity at ${\displaystyle a}$; in particular, the principal part of ${\displaystyle f}$ at ${\displaystyle a}$ is ${\displaystyle p_{a}(z)}$. Furthermore, any other meromorphic function ${\displaystyle g}$ on ${\displaystyle U}$ with these properties can be obtained as ${\displaystyle g=f+h}$, where ${\displaystyle h}$ is an arbitrary holomorphic function on ${\displaystyle U}$.

### Proof sketch

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

## Example

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

${\displaystyle p_{k}(z)={\frac {1}{z-k}}}$
and ${\displaystyle E=\mathbb {Z} ^{+}}$, Mittag-Leffler's theorem asserts (non-constructively) the existence of a meromorphic function ${\displaystyle f}$ with principal part ${\displaystyle p_{k}(z)}$ at ${\displaystyle z=k}$ for each positive integer ${\displaystyle k}$. More constructively we can let
${\displaystyle f(z)=z\sum _{k=1}^{\infty }{\frac {1}{k(z-k)}}.}$

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

## Pole expansions of meromorphic functions

Here are some examples of pole expansions of meromorphic functions:

${\displaystyle \tan(z)=\sum _{n=0}^{\infty }{\frac {8z}{(2n+1)^{2}\pi ^{2}-4z^{2}}}}$
${\displaystyle \csc(z)=\sum _{n\in \mathbb {Z} }{\frac {(-1)^{n}}{z-n\pi }}={\frac {1}{z}}+2z\sum _{n=1}^{\infty }(-1)^{n}{\frac {1}{z^{2}-(n\,\pi )^{2}}}}$
${\displaystyle \sec(z)\equiv -\csc \left(z-{\frac {\pi }{2}}\right)=\sum _{n\in \mathbb {Z} }{\frac {(-1)^{n-1}}{z-\left(n+{\frac {1}{2}}\right)\pi }}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}(2n+1)\pi }{(n+{\frac {1}{2}})^{2}\pi ^{2}-z^{2}}}}$
${\displaystyle \cot(z)\equiv {\frac {\cos(z)}{\sin(z)}}=\lim _{N\to \infty }\sum _{n=-N}^{N}{\frac {1}{z-n\pi }}={\frac {1}{z}}+2z\sum _{k=1}^{\infty }{\frac {1}{z^{2}-(k\,\pi )^{2}}}}$
${\displaystyle \csc ^{2}(z)=\sum _{n\in \mathbb {Z} }{\frac {1}{(z-n\,\pi )^{2}}}}$
${\displaystyle \sec ^{2}(z)={\frac {d}{dz}}\tan(z)=\sum _{n=0}^{\infty }{\frac {8((2n+1)^{2}\pi ^{2}+4z^{2})}{((2n+1)^{2}\pi ^{2}-4z^{2})^{2}}}}$
${\displaystyle {\frac {1}{z\sin(z)}}={\frac {1}{z^{2}}}+\sum _{n\neq 0}{\frac {(-1)^{n}}{\pi n(z-\pi n)}}={\frac {1}{z^{2}}}+\sum _{n=1}^{\infty }{(-1)^{n}}{\frac {2}{z^{2}-(n\,\pi )^{2}}}}$