Pole expansion of meromorphic functions

[LAHLH]

Hi,

I was hoping someone may be able to help me understand what Arfken is doing in sec 7.2 when he does the pole expansion of meromorphic functions (he says this proof is due to Mittag-Leffler).

So he starts off with a function f(z) that is analytic except at some isolated poles. These poles are at isolated $z=a_n$ with $0<|a_1|<|a_2|<...$ and are all simple with residues $b_n$. He then considers a series of concentric circles C_n about the origin so that C_n contains poles $a_1,a_2,...a_n$ but no others. Finally he assumes that $|f(z)|<\epsilon R_n$ where R_n is radius of C_n and $\epsilon>0$ is small constant. He says then that:

$$f(z)=f(0)+\sum_0^{\infty} b_n\{(z-a_n)^{-1}+a_n^{-1}\}$$

converges to f(z).

To prove this he says to prove this we use residue theorem to evaluate the contour integral for z inside C_n:

$$I_n:=(2\pi i)^{-1}\int_{C_n}\,\frac{f(w)}{w(w-z)}\mathrm{d}w$$

$$I_n=\sum_{m=1}^{n}\frac{b_m}{a_m(a_m-z)}+\frac{f(z)-f(0)}{z}$$

My first question is how to prove this second equality using the residue theorem? I just don't seem to be able to get it out..

 

[morphism]

Suppose we have two functions f and g that are holomorphic inside some circle $\gamma$ except for poles a_1, ..., a_n for f and b_1, ..., b_m for g. Also suppose that no a_i is equal to a b_j. Then the residue theorem yields
$$ (2\pi i)^{-1} \int_\gamma f(z)g(z) dz = \sum_{i=1}^n \text{Res}(f,a_i)g(a_i) + \sum_{j=1}^m \text{Res}(g,b_j)f(b_j). $$

In our case g(w)=1/w(w-z), which has poles at 0 and z with residues -1/z and 1/z, resp. So
$$\begin{align} (2\pi i)^{-1} \int_{C_n} \frac{f(w)}{w(w-z)} dw &= (2\pi i)^{-1} \int_{C_n} f(w) g(w) dw \\ &= \sum_{m=1}^n b_m g(a_m) + \text{Res}(g,0)f(0) + \text{Res}(g,z)f(z) & \\ &= \sum_{m=1}^n \frac{b_m}{a_m(a_m-z)} + \frac{f(z)-f(0)}{z}, \end{align}$$
as desired.

[Note that this only holds if $a_m,z\neq0$ and $z\neq a_m$ for all m; this is consistent with the requirement that the poles of f and g be distinct.]

 

[LAHLH]

Suppose we have two functions f and g that are holomorphic inside some circle $\gamma$ except for poles a_1, ..., a_n for f and b_1, ..., b_m for g. Also suppose that no a_i is equal to a b_j. Then the residue theorem yields
$$ (2\pi i)^{-1} \int_\gamma f(z)g(z) dz = \sum_{i=1}^n \text{Res}(f,a_i)g(a_i) + \sum_{j=1}^m \text{Res}(g,b_j)f(b_j). $$

In our case g(w)=1/w(w-z), which has poles at 0 and z with residues -1/z and 1/z, resp. So
$$\begin{align} (2\pi i)^{-1} \int_{C_n} \frac{f(w)}{w(w-z)} dw &= (2\pi i)^{-1} \int_{C_n} f(w) g(w) dw \\ &= \sum_{m=1}^n b_m g(a_m) + \text{Res}(g,0)f(0) + \text{Res}(g,z)f(z) & \\ &= \sum_{m=1}^n \frac{b_m}{a_m(a_m-z)} + \frac{f(z)-f(0)}{z}, \end{align}$$
as desired.

[Note that this only holds if $a_m,z\neq0$ and $z\neq a_m$ for all m; this is consistent with the requirement that the poles of f and g be distinct.]

Ah, thank you very much, very helpful!

 

[mathwonk]

"he assumes that |f(z)|<ϵRn where R_n is radius of C_n and ϵ>0 is small constant. "

how can a function with poles be bounded?

 

[morphism]

Presumably that inequality is for points z on C_n.