72
1 Mathematical Physics
(b) Differentiate with respect to x
∂ T
∂ x
= s(1 − 2sx + s
2 )
−
3
2
=
(1 − 2sx + s
2 )
−1 p l s
l+1
=
p
l s
l
Multiply by (1 − 2sx + s
2 )
s
l+1 p l =
(s
l
− 2xs
l+1
+ s
l+2 ) p
l
Equate coefficients of s
l+1
p l = p
l+1 − 2x p
l + p
l−1
or p l (x) + 2x p
l (x) = p
l+1 + p
l−1
1.82
e
−
xs
1−s
1 − s
=
∞
n=0
L n (x)s
n
n!
Put x = 0
∞
n=0
L n (0)
s
n
n!
=
1
1 − s
= 1 + s + s
2
+ · · · s
n
+ · · ·
=
∞
n=0
s
n
Therefore L n (0) = n!
1.3.11 Complex Variables
1.83 (a) Since the pole at z = 2 is not interior to |z| = 1, the integral equals zero
(b) Since the pole at z = 2 is interior to |z + i| = 3, the integral equals 2πi.
1.84 Method 1
c
4z
2
− 3z + 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) = 8πi
where we have used the result