382 ON THE PRODUCT OF ANY TWO LAPLACE'S COEFFICIENTS 



Similarly in the expression for D m P n xD p P q , when q=p+\, the co- 

 efficient of (2n + 2q-4:r+l)D" t+ "P n+ll _ 2r is 



_. 



' r! (q-r)\ (n-r}\ (2n + 2q-2r + 1)! 



NOTE. In the last expression the quantity in the square brackets 

 may be stated in either of the following forms : 



[(n-m+l)2q-2r(2n + 2q-2r+l)'] = [(n-m-2r+l)2q-2r(2n-2r+l)'], 

 or = [{(n + q-2r) - (m+j))} 2q - 2r (2n - 2r+ 1)], 



or = [{(n + q)-(m+2>)}2q-2r{2(n + q)-2r+l}'], 



or =[(j-i)(j+i+l)-(n+m)(n-m+l)'], 



where j = n + q 2r and i=m,+p. 



6. From the equation 



(2n + 1) pB m P n = (n - m + 1 ) D m P n+l + (n + m) D n P n .^ 

 we get, by putting n+p 2r for n and m+p for m, 



(2n + 2p- 4r + 1) p.D"+*P n+p _ tr = (n-m-2r+l) D m+1 >P n , p _, r+l 



+ (n + m + 2p- 2r) rP+P n + p _ w _,. 



If we put n+p 2r+l for n and m + p for m, we get 



(2n + 2p - 4r + 3) /tZ> w+ *P B+1 ,_, r+1 - (n -m- 2r + 2) D m+ P n+p _ w+1 



+ (n + m + 2p-2r+l) D m+p P n+p _^. 



The successive equations may be obtained from these two formulae by 

 changing r into r 1, r 2, &c. in the two formulae alternately. Thus 

 putting rl for r in the last formula, we get 



(2n + 2p- 4r + 7) nD m+ P n+p _, r+3 = (n-m- 2r + 4) D"^P^ p _^ t 



+ (n + m + 2p- 2r + 3) D m+p P n+p _ w+ ,. 



Again, putting p + 1 for n and p for m in the above equation, trans- 

 posing and multiplying the result by D m P n . we get 



2D m P n x D'P p+t = (2p + 3) ^D m P n x D*P V+1 - (2p + 1) D m P n x D P P P 



