[warKNESs] THE ALGEBRAIC BASIS OF CERTAIN BESSEL SERIES 119 
This may be regarded as the generalization of (44), viz. 
and of rs 
Ji (2x) =2 Jo — Ji Je + Je Ts — AA 
which can be derived at once from the special formula for 
RS 
TE (2 cos 5) given by G. and M. p. 29, by putting b=x, a=0. 
§ 19. The expansions 


1+x 2v+1 5 : 
=) = 1+ (4y4+2) x +2 (Qv+1)?x?+ [4 (2-41) Gi.) 
+ 5 (2v+1)] x? + [3 (Qv+-1)*+$ (2+ 1)? | x 
pene 
=) (Lx) = 1+ (47+4) x + [8 +1)? — i Gi.) 
pari. Za 2048 519 Pt. CM UE ue d 
Tax 2 +1 ; - Le 
=) (1x)! = 14 (4r+6) x +8 . +1. v+2x? (iii.) 
+2 Q+1 . 82+ 687439 . 27+... 
1+x 2v+1 a F . 
(ce) (1-+2)°= 1+ (4v+8) «+8. (+2) —3) x” Gv.) 
+32 .~vt1l.v+2.y7+8x3+... 
etc: 
show that while the coefficients are in general of a complicated char- 
acter, there is one simple property possessed by specially selected 
terms in each of them, viz., the coefficient of x° in 
ttx\?41 NT EURE Eee 
= (1+x)*s is 2 es ETS 


To prove this property observe that the coefficient in question is a 
polynomial in v of order s. Now this polynomial must vanish for 
\ 2v+1 
Vie nus: .efor =) (1+x)? becomes, for 
these values, 
CROP) Clara ates, 0e a) 
and for each of these expressions the middle term in x’ has a zero 
coefficient. The literal part of the coefficient under discussion is, 
therefore, (v-+1) (v+2)...(v+s). We have still to find the numerical 
multiplier 2%/s ! This can be done by giving v in 
Gs)" (1+x)? TE DE | (y+ 1) (v+2). 5 .(v+s) xs, 
xX 


the special value v=0. We thus find that 
2s+1 
s!. A =coeff. of x in CER (+ En +. = 2s; 
showing that A = 2?5/s! 
