Atomic Structure Theory

262 Solutions
From the above, it follows
2 · l ···2 · 1
Il =
(2l +1)···5 · 3 2=
2l l!
3 · 5 ···(2l +1) 2
l 2
2 l!
= 2 .
(2l + 1)!
Therefore,
1
c = =
Il
1
2l
(2l + 1)!
.
l! 2
1.3 maple program to generate first 10 Legendre polynomials using Rodrigues’
formula:
for l from 0 to 9 do
if l>0 then p := expand(diff((x^2-1)^l,x$l)/(2^l*l!))
else p :=1 fi;
print(P[l]=p);
od;
mathematica program,
P = {1, Table[Expand[D[(x^2 - 1)^l, {x, l}]/(2^l l!)],
{l, 1, 9}]} // Flatten // TableForm
1.4 mathematica program to generate first 10 Legendre polynomials using
the recurrence relation:
For[P[0] = 1; P[1] = x; l = 2, l < 11,
P[l] = Expand[((2l - 1)x P[l - 1] - (l - 1)P[l - 2])/l];
Print["P[", l, "]=", P[l]]; l++]
1.5 maple routine to generate P m
l (x) forl≤4and1≤m≤ l:
for l from 1 to 4 do
for m from 1 to l do
q:=(sqrt(1-x^2))^m*
diff(expand(diff((x^2-1)^l,x$l)/(2^l * l!)),x$m);
print(P[l,m]= q);
od;
od;
1.6 mathematica routine to find j6(x) using the upward recurrence relation:
j[0] = Sin[x]/x;
j[1] = Sin[x]/x^2 - Cos[x]/x;
Do[j[n + 1] = Expand[(2n + 1) j[n]/x - j[n - 1]], {n, 1, 5}]
j[6]