(13) 



196 Mr. R. Hargreaves on a Diffraction Problem 



These last give 



P 2p =(-l) p 2 vV J 2P+ |M cos r, } 



Q 2p = (-l) p 2 x^7rj 2p+h {r) sin r, 



iVi=(~i) p+1 2 V^J 2P+ i(^ sin ^ 



Q2 P +i= (~1) P 2 v/7rJ 2f+ f(r) cos r. j 



It will be sufficient to sketch the argument by which the 

 results in (13) were reached, as I have since found they are 

 particular cases of formulae given in Nielsen *. Since 



2(P P cos r + Q p sin r) cos (2p + l)co, 

 v 

 and 2(P p sinr — Q p cos r) cos (2p-\-l)<o 

 p 



IT 



are solutions of (y 2 + l)/=0, and a>= ■= — j-, we conjecture 



that one of the brackets with p as index will vanish and the 

 other be proportional to J p+A , since the function of type 

 J_ i is not admissible. If the expression for J +i in terms 

 of sin r and cos r with polynomial coefficients is used, an 

 expression for J 2 +i (r) cos r with polynomials multiplying 

 sin 2r and 1 + cos 2r results, which gives for coefficient of 

 the general term a finite series. This is identified with that 

 required for (13) by means of 



„0 P =S(-1)' C C . . . (14) 



q = n + p+ln + q + lp-\-qq 



which can be readily proved by a repeated use of 



n C r = C — C and n C n = C , 



n+lr+l n r+l n+1 n+l 



so applied that each step raises by 1 the value of the prefix. 

 From (13) follow 



P(/a)cosr+Q(p) sin r=2 v^2( — l)*J 2p+ i(r) cos (ty+l)a>, 



v 

 and 



-P(p) sin r-Q(p) cosr = 2\/n2,(-iy+ 1 J 2p+3 (r)cos(±p + 3) l 

 and then (8) gives 

 <j)(r, y) = cos (y + kYt ) [JJikr) cos co — J 5 cos 5 co + . . . ] ") 



/t \ *' * r (16) 



— sin(^ + kYt) [J s cos 3co — J x ccs 7&> +...]. J 



* Nielsen, Cylinder Functioned p. 20. Write v=2p+% and v=2p+l 

 in (5) and (6) to obtain the four results. 



