1000 Solved Problems in Modern Physics

72 1 Mathematical Physics

(b) Differentiate with respect tox

∂T

∂x

=s(1− 2 sx+s^2 )−

(^32)

=

∑

(1− 2 sx+s^2 )−^1 plsl+^1

=

∑

pl′sl

Multiply by (1− 2 sx+s^2 )

∑

sl+^1 pl=

∑

(sl− 2 xsl+^1 +sl+^2 )p′l

Equate coefficients ofsl+^1

pl=p′l+ 1 − 2 xpl′+pl′− 1

orpl(x)+ 2 xpl′(x)=p′l+ 1 +p′l− 1

1.82

e−

xs

1 −s

1 −s

=

∑∞

n= 0

Ln(x)sn

n!

Putx= 0

∑∞

n= 0

Ln(0)

sn

n!

=

1

1 −s

= 1 +s+s^2 +···sn+···

=

∑∞

n= 0

sn

ThereforeLn(0)=n!

1.3.11 ComplexVariables.................................

1.83 (a) Since the pole atz=2 is not interior to|z|=1, the integral equals zero
(b) Since the pole atz=2 is interior to|z+i|=3, the integral equals 2πi.

1.84 Method 1

∮

c

4 z^2 − 3 z+ 1

(z−1)^3

dz=

∮

c

4(z−1)^2 +5(z−1)+ 2

(z−1)^3

dz

= 4

∮

c

dz

z− 1

+ 5

∮

c

dz

(z−1)^2

+ 2

∮

c

dz

(z−1)^3

=4(2πi)+5(0)+6(0)= 8 πi

where we have used the result