A FORTRAN program to compute the continued fraction coefficients for the series 

 given by eq (72) is given in appendix C. This program can be easily modified to determine 

 the other set. 



Two Proofs 



In this section proofs will be given of two facts used in the previous section. Follow- 

 ing this, the number of terms required, the accuracy, and similar topics will be discussed. 

 To prove that the first 2N+ 1 terms of the quotient A^/Bn are equal to the same terms 

 when N is a larger integer, use long division on eq (79) and (80) to obtain 



An/Bn = An_i/Bn_i + an iA^_2 Bn_i - A^.j Bn_2)/(Bn Bn-i) • (84) 



If the first quotient on the right is to have terms equal to the quotient on the left up 

 through term 2N- 1 , the remainder must have no terms with y to a higher power than 

 -(2N-1). The final divisor, Bjsj Bjsj.j, contains y to the (2N- 1) and lower powers. There- 

 fore, the proof is complete if the numerator of the remainder is a constant. To show this, 

 use eq (79) and (80) to evaluate Bj^f-l and Ajsj-l ; it can be shown that 



An-2 ^N-1 - An-1 ^N-2 = "^N-1 ^^N-3 ^N-2 ~ ^N-2 ^N-S^ 



= (-l)NaN_iaN_2-..ai . 



The right-hand product is obtained by repeatedly applying the middle result. The product 

 of a's is a constant, completing the proof. 



The second proof required is that in the quotient of Ajsj/B^, C2N (the coefficient of 

 y-2N) wi]i contain no aj or bj to higher than term N and C2N+1 (the coefficient of y~2N-l) 

 will involve no aj to higher than term N + 1 and no h[ to higher than term N. 



The first part is intuitively obvious. Since from the preceding proof C2N will be the 

 same when derived from the ratio A^/B^ for any x as long as it is N or greater, we need 

 consider only the case where x is N. But since from eq (79) and (80) Ajsj and Bj^f contain no 

 a's or b's of greater than term N, C2N cannot contain a's or b's of higher terms. 



By the same argument C2N+1 can contain no a's or b's to higher terms than N+ 1. 

 There remains to be proven only that bj^i+j cannot exist in C2N+1 or that its coefficient, 

 which we will call E, is zero. Applying eq (82) for N+ 1 and j = N gives 



N+l 



%+1+N " " Z %+l,N+l-iCN+H-N-i • 

 i=l 



E, the coefficient of b^+i in this expression from eq (83), takes the form 



N+1 



--1 



%,N+l-i^2N-i-l-i 



i=l 



31 



