380 ON THE PRODUCT OF ANY TWO LAPLACE'S COEFFICIENTS 



and similar equations until we come to 



and differentiating once more (since D P+1 P P = 0) we get 



We have also from the above equations, 



(2p+l) p.DP p = pDP p+1 + (p + l)DP p _ i; 

 or (2n + 2p - 4r + 1 ) i*DP n+1> _, r = (n +p - 2r) DP, !+l ,_, r+l + (n +p-2r+ 



We have also from equation (6) in a previous part of the paper (p. 248), 

 (2ft + 2p-4r+l) p.!?*** P u+p _ tr = (n-m-2r+l) D m+ P n ^ r+l 



Making use of the above equations we get 

 D'"P 





-- 



> r \ (p-r}\ ~ 1.8.5...(2n + 2p- 



and 





__ 



rl (p-r)l 1.3. 5...(2 



Writing down the (r+l)th and the yth term of this series we have 



p]_ 1.3.5... (2^+l)^_3 .5... (2 -2)-l) 



L > rl^-r + l)! 1 . 3. 5 ... (2n + 2p-2r + S) 



x [_(2n + 2p - 4r+l)fjiD m+ "P a+l ,_, r (p-r + l) (2 + 2p - 2r + 3) 



- (2n + 2p - 4r + 5) p.D"~ ! 'P n+p _ 2r+2 xr(2n- 2r + 1)]. 

 But we have 



(2n + 2p - 4r + 1 ) ^D m ^P n ^_ r = (n-m-2r+l) 



and 



(2n + 2p - 4r + 5) ,j.D^P n+f .^ = (n-m-2r + 3 



