546 Chapter^8
which is precisely the product of the two series
00
1 +La~ Zn
n=l n.andHence the use of the symbolic method is justified as far as any linear
combination or product of two series of the form (8.5-1) is concerned.
Examples 1. Consider f(z) = z/(ez - 1) for z -:/:- 0 and f(O) =
limz-.o z/(ez - 1) = 1. The function thus defined is analytic in C ex-
cept at the points Zk = 2k7ri (k = ±1, ±2, ... ). Hence it has a series
expansion in powers of z valid for lzl < 271". To determine the coefficients
of the expansion, let
--Z = 1 + B1z + -B2 z^2 + .. · • + -Bn z n + .. · ( 8.5-3 )
ez -1 2! n!By applying the symbolic method we have
which gives
Z :::: e(B+l)z _ eBzso that
_z_ :eBz
ez -1(B+1)^2 -B^2 :::0, (B+1)^3 -B^3 :::o,
Simplifying and returning to the subindex notation, we get
2B1 + 1 = 0, 3B2 + 3B1 + 1 = 0, 4Ba + 6B2 + 4B1 + 1 = O,... ' (n ~ l)Bn + (n ~ l)Bn-1 +, .. + (n ~ l)B1+1=0
From these equations the numbers B 1 , B 2 , • • • can easily be calculated.
We find
B 3 = 0,B 9 =0,
1
B4 = -
30, B 5 = 0,
5
B10 =
66, B 11 = 0,1B6 = 42'
It is easy to prove that all numbers Bn with odd index greater than 1 are
zero. In fact, substituting B 1 = -^1 / 2 in (8.5-3), we have
z 1 z ez + 1 z z
--+ -z = - --= - coth -
ez - 1 2 2 ez - 1 2 2