APPENDIX i 
To prove that 
er =m aed 
Rn=3, 24 rin—y! C2) 
is a solution to the difference equation 
n i 
MD os Ja oy uu (25) 
Substituting (22) to (25) 
(n-+1)! 7=dhl-™ 1 nl T2=m 1 mi t=nti-m 1 
gurl > ri(ntl—r)! 2X2" <0 ri(n—r)! 2X2" =5 AG@= 
Multiplying by 2°*'/n!, bringing all terms on a common denominator, and combining the first 
and third terms, one obtains 
r=n+l1—m r r=n—™M n+1—r < 
2 rinti—n! << ACG 
or 
1 freon ate l) ae 
(n—m)\(m)! 3 ri(n+1—r) 
0 (I) 
Equation (1) is satisfied for any values of n and m, as will be proven by the method of mathe- 
matical induction. First, it will be shown to hold for n—m=0, n—m=1, and n—m=2. Then it 
will be shown that if we assume (I) to hold for n—m=k, it must ‘also hold for n—m=k-+1, and 
hence for any value of n—m. 
As n—m=0, m=n: 
1 n+l1 nya 
Olm! Olnm+1)!. n! nn! 
Asn—m=1, m=n—1: 
le (oar  @=@apl) eee lem 
NG] Mecsono Im! (n—1)! ao om GS 
As n—m=2, m=n—2: 
1 Lg WO Bee ee er ell 6 
Tel am" ow PIGS t SG=0n (=n Tei Jes TGs. 
As n—m=k, m=n—k: 
1 Sh 2n = (al) 
lab 2 ri@ti—r)—° ay) 
As n—m=k+1, m=n—k—1: 
1 rektl Qr—(n-+1) _ n—k =k Wr—(n+1) , 2k+1)—(+1) __ 
@4)i@—k—pit Pa rivti—ni FI) Na— ky es AGl=alt CLieaaT 
n—2k—1 n—2k—1 
C2DIG=O! ELDIG=pI (in view of Equation II) =0 
g. €. de 
37 
