Despite the still-unresolved anomaly below, the CF = CF formula a x z + ContinuedFractionK[ a ((x + n)^2 + b) z^2, ((a - 1) (x + 1 + n) - (a + 1) y + 1) z, {n, 0,∞}] == a y z + ContinuedFractionK[ a ((y + n)^2 + b) z^2, ((a - 1) (y + 1 + n) - (a + 1) x + 1) z, {n, 0,∞}] produces mostly true results. E.g., using ContinuedFractionK[c*n^2 + d*n + e, a*n + b, {n, 0,∞}] == (2*c*e* Hypergeometric2F1[1 + (2*e)/(d + Sqrt[d^2 - 4*c*e]), 1 + (d + Sqrt[d^2 - 4*c*e])/(2*c), (1/2)*(3 + d/c + (2*b*c - a*(c + d))/(c*Sqrt[a^2 + 4*c])), 1/2 - a/(2*Sqrt[a^2 + 4*c])])/((2*b*c - a*(c + d) + Sqrt[a^2 + 4*c]*(c + d))* Hypergeometric2F1[(2*e)/(d + Sqrt[d^2 - 4*c*e]), (d + Sqrt[d^2 - 4*c*e])/(2*c), (1/2)*(1 + d/c + (2*b*c - a*(c + d))/(c*Sqrt[a^2 + 4*c])), 1/2 - a/(2*Sqrt[a^2 + 4*c])]) and changing (lotsa) variables gives this "peculiar" transformation of quotients of contiguous 2F1s: ((a-1) (b-1) Hypergeometric2F1[a, b, c, z])/ Hypergeometric2F1[a-1, b-1, c -1, z] == -(((a + b - c - 1) (c-1))/(z-1)) + ((c-a) (c-b) Hypergeometric2F1[1 - a + c, 1 - b + c, c, z])/ Hypergeometric2F1[c-a, c-b, c -1, z] Test: Simplify[Series[%, {z, 0, 3}]] SeriesData[z, 0, {}, 4, 4, 1] == 0 Integrating both sides dz and exponentiating gives the familiar 2F1 "linear" transformation: Hypergeometric2F1[a, b, c, z] == (1 - z)^(-a - b + c) Hypergeometric2F1[-a + c, -b + c, c, z] I've always wondered how to get this from path invariance (= source of the original CF identity, which is now looking pretty safe). We should get more CF identities by log differentiating the other 2F1 transformations. Note: Contrary to a recent brainbubble of mine, Hypergeometric2F1[a, b, c, z]) is *not* trivially related to Hypergeometric2F1[a-1, b-1, c -1, z] by shifting the sum one term. Bagbiting hidden n! factor. But is *is* simply related by differentiation: In[749]:= D[Hypergeometric2F1[a, b, c, z], z] Out[749]= (a b Hypergeometric2F1[1 + a, 1 + b, 1 + c, z])/c Thus the earlier transformation actually relates log derivatives. Applied to the usual arctan CF, the no-longer(?) suspect identity gives: -1 + Sqrt[z]/ArcTan[Sqrt[z]] == ContinuedFractionK[n^2 z, 1 + 2 n, {n, 1, ∞}] == (Sqrt[1 + z]-1)/2 + ContinuedFractionK[(n-1/2)^2 z, 2 n + Sqrt[1+z], {n,1 ∞}] This has almost no effect on the convergence except on the ray z<-1, where the the usual CF just thrashes, trying to make an imaginary result from real terms, while the tweaked one appears to converge everywhere, but weakly on the ray. The interval shrink ratio estimate for CF[a[n],b[n]] is r[n] ~ -1 + (b[n] (-1 + Sqrt[1 + (4 a[1 + n])/(b[n] b[1 + n])]) b[1 + n])/(2 a[1 + n]) For the new CF this gives ((1-1/n) (2 Sqrt[1 + z] -2 - z))/z+O[n^-2] whose infinite product for z<-1 telescopes gradually to 0, whereas for the usual CF, r[n]->(2 Sqrt[1 + z] -2 - z))/z+O[n^-2], whose infinite product doesn't →0. Numerical experiments on the ray also support convergence of the new CF but not the usual CF. --rwg I may have forgotten to thank DanA for spotting an off-by-1 error in my old email deriving the convergence measure, but it doesn't affect the result. On Sat, Apr 14, 2012 at 4:35 AM, Bill Gosper <billgosper@gmail.com> wrote:
http://www.tweedledum.com/rwg/stanfordn3.pdf (p 14) (Also: Gosper, Strip mining in the abandoned orefields of Nineteenth Century mathematics, in Computers in Mathematics (D. Chudnovsky & R. Jenks, eds.) pp 261-283, m. dekker, N. Y.)
"8. Continued fractions This is three N - K planes, j = −1, 0, 1, joined by two J matrices. In the top (j = 1) plane, the k direction computes a continued fraction, and on the bottom (j = −1), it’s the n direction. [...] Thus, to get a relation between two continued fractions, we must j-hop between planes before switching directions. If k runs from c to ∞ and n runs from d to ∞, and the corresponding continued fractions converge, then they are insensitive to whatever subsequent matrices accrue on the right by way of path closure. I.e., the two paths can be (j,k,n) = (0,c,d) → (1,c,d) → (1,∞,d) and (0,c,d) → (−1,c,d) → (−1,c,∞). This yields the identity [an undersimplified version of that symmetric CF identity I sent the other day]"
All of a sudden, I'm deriving false identities with this! E.g., with K and N matrices
(c85) [km(k,n),nm(k,n)]
[ 4 n 4 k 11 1 ] [ n k 13 1 ] [ --- + --- + -- - ] [ - + - + -- - - ] [ 5 5 30 5 ] [ 5 5 60 5 ] (d85) [[ ], [ ]] [ k 1 1 ] [ k 1 1 ] [ - - - - k + - ] [ - + - n + - ] [ 2 6 3 ] [ 2 6 4 ]
Check path invariance: (c81) CHECKM();
[ 0 0 ] (d81) [ ] [ 0 0 ]
Confirm convergence of a dozen N terms: (c82) (PRUD(NM(0,N),N,0,11),DFLOAT(%%[1]/%%[2]));
(d82) [- 1.12262566430759d0, - 1.12264219198252d0]
This says that as a function of the tail t, the PRUD (matrix product) computes (A t + B)/(a t +b), with A/a ~ B/b ~ -1.12262. This is the "insensitivity" claimed above. Continuing to wander in the +n and +k directions shouldn't change this value much. But tacking on 69 k steps: (c88) (PRUD(NM(0,N),N,0,11) . PRUD(KM(K,12),K,0,69),DFLOAT(%%[1]/%%[2]));
(d88) [0.30872240339868d0, 1.27956902429756d0]
This is batbleep! Not only was the value bashed, the convergence was destroyed. Note that by path invariance, it doesn't matter how we got here:
(c89) (PRUD(KM(K,0),K,0,69) . PRUD(NM(70,N),N,0,11),DFLOAT(%%[1]/%%[2]));
(d89) [0.30872240339868d0, 1.27956902429756d0]
There's nothing obviously wrong with the K matrix except for being a bit slow to converge:
(c86) (PRUD(KM(K,0),K,0,11),DFLOAT(%%[1]/%%[2]));
(d86) [1.10980568563031d0, 1.48744247508643d0]
(c87) (PRUD(KM(K,0),K,0,69),DFLOAT(%%[1]/%%[2]));
(d87) [1.62173336398064d0, 1.62175009332196d0]
Except it's not even the same sign as the N product!
There's nothing magic about 69 terms of K. Using 11 instead still fails bigtime;
(c83) (PRUD(NM(0,N),N,0,11) . PRUD(KM(K,12),K,0,11),DFLOAT(%%[1]/%%[2]));
(d83) [- 0.77240969070225d0, - 1.11033801208252d0]
(c84) (PRUD(KM(K,0),K,0,11) . PRUD(NM(12,N),N,0,11),DFLOAT(%%[1]/%%[2]));
(d84) [- 0.77240969070225d0, - 1.11033801208252d0]
Maybe the kids can help get to the bottom of this. --rwg (j-hopping to force continued fraction form is just cosmetic, and only adds to the confusion above.)