Hints and answers

16.7 HINTS AND ANSWERS
(c) Show that the corresponding non-terminating series solutions Sm (z) have as
their first few terms
9
1
S0 (z) = a0 z + z 3 + z 5 + · · · ,
3!
5!
1 2
3
S1 (z) = a0 1 − z − z 4 − · · · ,
2!
4!
3 3 15 5
S2 (z) = a0 z − z − z − · · · ,
3!
5!
9 2 45 4
S3 (z) = a0 1 − z + z + · · · .
2!
4!
16.16
Obtain the recurrence relations for
the solution of Legendre’s equation (18.1) in
inverse powers of z, i.e. set y(z) =
an z σ−n , with a0 = 0. Deduce that, if is an
integer, then the series with σ = will terminate and hence converge for all z,
whilst the series with σ = −( + 1) does not terminate and hence converges only
for |z| > 1.
16.7 Hints and answers
16.1
16.3
16.5
16.7
16.9
16.11
Note that z = 0 is an ordinary point of the equation.
For σ = 0, an+2 /an = [n(n + 2) − λ]/[(n + 1)(n + 2)] and, correspondingly, for
σ = 1, U2 (z) = a0 (1 − 4z 2 ) and U3 (z) = a0 (z − 2z 3 ).
σ = 0 and 3; a6m /a0 = (−1)m /(2m)! and a6m /a0 = (−1)m /(2m + 1)!, respectively.
y1 (z) = a0 cos z 3 and y2 (z) = a0 sin z 3 . The Wronskian is ±3a20 z 2 = 0.
(b) an+1 /an = [ ( + 1) − n(n + 1) ]/[ 2(n + 1)2 ].
(c) R = 2, equal to the distance between z = 1 and the closest singularity at
z = −1.
(−1)n z 2n
A typical term in the series for y(σ, z) is
.
[ (σ + 2)(σ + 4) · · · (σ + 2n) ]2
The origin is an ordinary point. Determine the constant of integration by examining the behaviour
z of the related functions for small x.
y2 (z) = (exp z 2 ) 0 exp(−x2 ) dx.
Repeated roots σ = 2.
∞
"
(n + 1)(−2z)n+2 ! a
y(z) = az 2 − 4az 3 + 6bz 3 +
+ b [ln z + g(n)] ,
n!
4
n=2
where
g(n) =
16.13
16.15
1
1
1
1
− −
− · · · − − 2.
n+1 n n−1
2
The transformed equation is xy + 2y + y = 0; an = (−1)n (n + 1)−1 (n!)−2 a0 ;
du/dz = A[ y1 (z) ]−2 .
(a) (i) an+2 = [an (n2 − m2 )]/[(n + 2)(n + 1)],
(ii) an+2 = {an [(n + 1)2 − m2 ]}/[(n + 3)(n + 2)]; (b) 1, z, 2z 2 − 1, 4z 3 − 3z.
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