Mittag-Leffler’s Expansion Theorem

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Theorem

Let $f$ be a meromorphic function with only simple poles continuous, or with a removable singularity, at $0$.

Let $X$ be the set of poles of $f$.

For $N \in \N$, let $C_N$ be a circle, centred at the origin, of radius $R_N$, where $R_N \to \infty$ as $N \to \infty$, such that $\partial C_N$ contains no poles of $f$ for any $N$.

Let $M > 0$ be a real number independent of $N$ such that for all $z \in \partial C_N$, $\cmod {\map f z} < M$ , for all $N \in \N$.


Then:

$\displaystyle \map f z = \map f 0 + \sum_{n \mathop \in X} \Res f n \paren {\frac 1 {z - n} + \frac 1 n}$

where:

$\Res f n$ is the residue of $f$ at $n$
$z$ is not a pole of $f$
$\displaystyle \map f 0 = \lim_{z \mathop \to 0} \map f z$ if $f$ has a removable singularity at $0$.


Proof

Let $\zeta \in \C \setminus X$.

Then:

$\displaystyle \frac {\map f z} {z - \zeta}$

has simple poles for $z \in X \cup \set \zeta$.

Let $X_N$ be the set of poles contained within $C_N$.

Then:

\(\displaystyle \frac 1 {2 \pi i} \oint_{\partial C_N} \frac {\map f z} {z - \zeta} \rd z\) \(=\) \(\displaystyle \Res {\frac {\map f z} {z - \zeta} } \zeta + \sum_{n \mathop \in X_N} \Res {\frac {\map f z} {z - \zeta} } n\) Residue Theorem
\(\displaystyle \) \(=\) \(\displaystyle \lim_{z \mathop \to \zeta} \paren {\frac {\paren {z - \zeta} \map f z} {z - \zeta} } + \sum_{n \mathop \in X_N} \paren {\lim_{z \mathop \to n} \frac {\paren {z - n} \map f z} {z - \zeta} }\) Residue at Simple Pole
\(\displaystyle \) \(=\) \(\displaystyle \lim_{z \mathop \to \zeta} \map f z + \sum_{n \mathop \in X_N} \paren {\lim_{z \mathop \to n} \paren {\paren {z - n} \map f z } \cdot \lim_{z \mathop \to n} \frac 1 {z - \zeta} }\) Product Rule for Limits
\(\displaystyle \) \(=\) \(\displaystyle \map f \zeta + \sum_{n \mathop \in X_N} \frac {\Res f n} {n - \zeta}\) $f$ is continuous at $\zeta$, Residue at Simple Pole

Setting $\zeta = 0$, we obtain:

$\displaystyle \frac 1 {2 \pi i} \oint_{\partial C_N} \frac {\map f z} z \rd z = \map f 0 + \sum_{n \mathop \in X_N} \frac {\Res f n} n$

So:

$\displaystyle \frac 1 {2 \pi i} \oint_{\partial C_N} \map f z \paren {\frac 1 {z - \zeta} - \frac 1 z} \rd z = \map f \zeta + \sum_{n \in X_N} {\Res f n} \paren {\frac 1 {n - \zeta} - \frac 1 n} - \map f 0$

It remains to show that the integral on the left hand side vanishes as $N \to \infty$.

We have:

\(\displaystyle \cmod {\frac 1 {2 \pi i} \oint_{\partial C_N} \map f z \paren {\frac 1 {z - \zeta} - \frac 1 z} \rd z}\) \(=\) \(\displaystyle \cmod {\frac \zeta {2 \pi i} \oint_{\partial C_N} \frac {\map f z} {z \paren {z - \zeta} } \rd z}\)
\(\displaystyle \) \(\le\) \(\displaystyle \frac {\cmod \zeta} {2 \pi} \cdot \frac M {R_N \paren {R_N - \cmod \zeta} } \cdot 2 \pi R_N\) Triangle Inequality for Contour Integrals, Reverse Triangle Inequality, noting that $\cmod z$ for $z \in \partial C_N$
\(\displaystyle \) \(\sim\) \(\displaystyle \frac 1 {R_N}\)
\(\displaystyle \) \(\to\) \(\displaystyle 0\) as $N \to \infty$, $R_N \to \infty$

Letting $N \to \infty$ gives:

$\displaystyle 0 = \map f \zeta + \sum_{n \in X} {\Res f n} \paren {\frac 1 {n - \zeta} - \frac 1 n} - \map f 0$

Giving:

$\displaystyle \map f \zeta = \map f 0 + \sum_{n \mathop \in X} \Res f n \paren {\frac 1 {\zeta - n} + \frac 1 n}$

$\blacksquare$


Source of Name

This entry was named for Magnus Gustaf Mittag-Leffler.


Sources