312 Chapter 5 Higher Dimensions and Other Coordinates
When the differentiations in Eq. (1) are carried out and the equation is multi-
plied byr,itbecomes
r^2 R′′=α(α− 1 )c 0 rα+(α+ 1 )αc 1 rα+^1 +(α+ 2 )(α+ 1 )c 2 rα+^2 +···
+(α+k)(α+k− 1 )ckrα+k+···
rR′= αc 0 rα +(α+ 1 )c 1 rα+^1 +(α+ 2 )c 2 rα+^2 +···
+(α+k)ckrα+k+···
−μ^2 R=−μ^2 c 0 rα −μ^2 c 1 rα+^1 −μ^2 c 2 rα+^2 −···
−μ^2 ckrα+k+···
λ^2 r^2 R= λ^2 c 0 rα+^2 +···
+λ^2 ck− 2 rα+k+···The expression forλ^2 r^2 Ris jogged to the right to make like powers ofrline up
vertically. Note that the lowest power ofrpresent inλ^2 r^2 Risrα+^2.
Now we add the tableau vertically. The sum of the left-hand sides is, accord-
ing to the differential equation, equal to zero. Therefore
0 =c 0(
α^2 −μ^2)
rα+c 1[
(α+ 1 )^2 −μ^2]
rα+^1
+[
c 2(
(α+ 2 )^2 −μ^2)
+λ^2 c 0]
rα+^2
+···+[
ck(
(α+k)^2 −μ^2)
+λ^2 ck− 2]
rα+k+···.Each term in this power series must be zero in order for the equality to hold.
Therefore, the coefficient of each term must be zero:
c 0(
α^2 −μ^2)
= 0 ,
c 1(
(α+ 1 )^2 −μ^2)
= 0 ,
..
.
ck(
(α+k)^2 −μ^2)
+λ^2 ck− 2 = 0 , k≥ 2.
As a bookkeeping agreement, we takec 0 =0. Thusα=±μ. Let us study the
caseα=μ≥0. The second equation becomes
c 1(
(μ+ 1 )^2 −μ^2)
= 0
and this impliesc 1 =0. Now, in general the relation
ck=− λ(^2) ck− 2
(μ+k)^2 −μ^2
=−λ^2 ck−^2
k( 2 μ+k)
, k≥ 2 , (3)
says thatckcan be found fromck− 2 .Inparticular,wefind
c 2 =− λ
2
2 ( 2 μ+ 2 )c^0 ,
c 4 =− λ
2
4 ( 2 μ+ 4 )
c 2 = λ
4
2 · 4 ·( 2 μ+ 2 )( 2 μ+ 4 )
c 0 ,