16.7 HINTS AND ANSWERS
(c) Show that the corresponding non-terminating series solutionsSm(z) have as
their first few termsS 0 (z)=a 0(
z+1
3!
z^3 +9
5!
z^5 +···)
,
S 1 (z)=a 0(
1 −
1
2!
z^2 −3
4!
z^4 −···)
,
S 2 (z)=a 0(
z−3
3!
z^3 −15
5!
z^5 −···)
,
S 3 (z)=a 0(
1 −
9
2!
z^2 +45
4!
z^4 +···)
.
16.16 Obtain the recurrence relations for the solution of Legendre’s equation (18.1) in
inversepowers ofz,i.e.sety(z)=
∑
anzσ−n,witha 0 = 0. Deduce that, ifis an
integer, then the series withσ=will terminate and hence converge for allz,
whilst the series withσ=−(+ 1) does not terminate and hence converges only
for|z|> 1.16.7 Hints and answers16.1 Note thatz= 0 is an ordinary point of the equation.
Forσ=0,an+2/an=[n(n+2)−λ]/[(n+1)(n+ 2)] and, correspondingly, for
σ=1,U 2 (z)=a 0 (1− 4 z^2 )andU 3 (z)=a 0 (z− 2 z^3 ).
16.3 σ=0and3;a 6 m/a 0 =(−1)m/(2m)! anda 6 m/a 0 =(−1)m/(2m+ 1)!, respectively.
y 1 (z)=a 0 cosz^3 andy 2 (z)=a 0 sinz^3 .TheWronskianis± 3 a^20 z^2 =0.
16.5 (b)an+1/an=[(+1)−n(n+1)]/[2(n+1)^2 ].
(c)R= 2, equal to the distance betweenz= 1 and the closest singularity at
z=− 1.
16.7 A typical term in the series fory(σ, z)is
(−1)nz^2 n
[(σ+2)(σ+4)···(σ+2n)]^2.
16.9 The origin is an ordinary point. Determine the constant of integration by exam-
ining the behaviour of the related functions for smallx.
y 2 (z)=(expz^2 )
∫z
0 exp(−x(^2) )dx.
16.11 Repeated rootsσ=2.
y(z)=az^2 − 4 az^3 +6bz^3 +
∑∞
n=2(n+1)(− 2 z)n+2
n!{a4+b[lnz+g(n)]}
,
whereg(n)=1
n+1−
1
n−
1
n− 1−···−
1
2
− 2.
16.13 The transformed equation isxy′′+2y′+y=0;an=(−1)n(n+1)−^1 (n!)−^2 a 0 ;
du/dz=A[y 1 (z)]−^2.
16.15 (a) (i)an+2=[an(n^2 −m^2 )]/[(n+2)(n+ 1)],
(ii)an+2={an[(n+1)^2 −m^2 ]}/[(n+3)(n+ 2)]; (b) 1,z,2z^2 −1, 4z^3 − 3 z.