[math-fun] Roger-Ramanujan as matrix product for contfrac
As I suspected the matrix product for contfracs is helpful: Write convergents P_k, Q_k of a0 + b1/(a1 + b2/(a2 + b3/(a3 + ... ))) as matrix product:
[ P P ] [ P P ] [ a 1 ] ___k [ a 1 ] [ k k-1 ] [ k-1 k-2 ] [ k ] | | [ j ] [ ] = [ ] [ ] = | | [ ] [ Q Q ] [ Q Q ] [ b 0 ] | |j=0 [ b 0 ] [ k k-1 ] [ k-1 k-2 ] [ k ] [ j ]
(fxtbook p.720 gives transposed version). Now we have (set b_k = a*q^k and a_k = 1)
inf /===\ | | [ 1 1 ] [ V V ] (for a==1) [ G G ] | | [ ] = [ ] = [ ] k = 0 [ a*q^k 0 ] [ U U ] [ H G ]
where G(q)=sum(n=0,N,q^(n^2)/prod(k=1,n,1-q^k)); /* g.f. of A003114 */ H(q)=sum(n=0,N,q^(n^2+n)/prod(k=1,n,1-q^k)); /* g.f. of A003106 */ U(a,q) = a * sum(n=0,N, a^n*q^(n^2+n) / prod(k=1,n, 1-q^k) ); \\ = H(q) for a==1 \\ Note I kept factor a for consistency with prior postings V(a,q) = sum(n=0,N, a^n*q^(n^2) / prod(k=1,n, 1-q^k) ); \\ = G(q) for a==1 V(a,q) / U(a,q) = 1/a * ( 1+a*q/(1+a*q^2/(1+a*q^3/(1+a*q^4/(1+a*q^5/(1+...))))) ) (We certainly can get rid of the ugly factor 1/a, but my brain is not working anymore today.) And for a==1 this is the Roger-Ramanujan contfrac G(q) / H(q) = 1+q/(1+q^2/(1+q^3/(1+q^4/(1+...)))) see https://oeis.org/A003823 Now it's officially beer o'clock.
participants (1)
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Joerg Arndt