446 
but by (7) 
J(t,x,k,1) = —k fG@,2+1,k—1,0) 
Therefore, 
(8) a«f(t,v,k,0) = f(t+1,2,k,0) + k f(t.c+1,k—1,0), Pe MR 2 ee 
In (5) setting ¢ = 0: 
m 
(9) es fe a! f(—m,x,k,n-+-m—i) = Shi 70) 
i=0\ 7 
eT IO oe ee : m — 1, 2, 3, 
Now S(k,n) vanishes if k > n; therefore f(—m,z,k,n) satisfies the linear homo 
geneous difference equation of order m: 
m , ; 
(10) BS F | x f(—m,2,k,n+m—i) = 0. 
=0 
ice — 0 al eee Di — Nee tee ae 
of which the characteristic equation is 
(eg) =.0 
whence the complete solution is 
(11) f(—m,z,k,n) = (co tem+...... + Cm—1 a) (—2)" 
m= 1,23 ....}n=0,1,2, 2... . kl; not forash 
however, the equation (10) itself will give f(—m,vz,k,n) for 
For m = 1, we have 
f(—1,2,k,n) = co (—2)" = 06 E23, eee k. 
and setting n = 0, we determine 
Co = f(—1,2,k,0). 
setting ¢ = —1 in (8) P 
f(—1,2,k,0) : [S(,0) + kf(—1,2-+1,k—1,0)] 
ll 
— 
— —= when 4 — 0 
re 
= f(—1,2+1,k—-1,0) k=1,2,3... 
