1251 
The two first coefficients may be obtained directly, for 
1 
Ge 
m= fe *da=1— — 
e 
‘ 
1 1 
a, = —[, (1) — I= — 
é 
and 
The remaining coefficients are dependent on these. For putting 
« = 1 in the recurrent relation 
(re 1) pit (2) (At 1 — 2) op, (2) + paar @) = 0 (6) 
we get 
(rn + 1) Po (1) — 2n Gp (1) + 2 Gn—i (1) = 0 
and, changing n into n+ 1 
(2 + 2) papi (1) — 2 (2 + I npt 1) + (+ Ign (1) =O. 
thus, subtracting the former from the latter equation 
(n + 2) ara — (An +1) ani + na, = 0. 
6. The expansion holding for the value «=O, we must have 
Ss Bk 
0 
and remarking that «=1 is a point of discontinuity 
Daron. 
0 
To prove these equations directly we may remark that 
n i], v2 | 
= ty = Ser Be ie nn em rn (EN 
SO 
@ 1 i eee 
= a, =— — — Lim po (1). 
l é é nzo 
Now, the number ” being very large, we have 
nia? n?x? 
Pn (e= bnr + 25 65 + eae (V nz) 
and 
| —- : Ae — 2 
Lim gp (x) = Lim J, (nx) = Lim ve — cos (Vine — 3) == 0 
na 1 === 0D no My d nv 4 
therefore 
® is i 
DE 
and finally 
