[math-fun] three term recurrence ⇏ CF
Given F[c_] := HypergeometricPFQ[{a, b, c}, {d, e + c}, 1], there are many ways to get the three term recurrence r[c] == -(((1 + c - d) (c + e) (1 + c + e))/((1 + c) (1 - a + c + e) (1 - b + c + e) (-(((1 + c + e) (1 + a (-1 + b - c) + 3 c - b (1 + c) + 2 e + (2 c - d) (c + e)))/ ((1 + c) (1 - a + c + e) (1 - b + c + e))) + r[1 + c]))) with r[c]:=F[c+1]/F[c]. (E.g. A&S or DLMF 15, or the 3x3 matrices in the "3F2 Rosetta stone" (after j->j+k), or Annihilator in HolonomicFunctions.m, or my CONTIGUATE function that uses the extra parameters feature of the indefinite summation algorithm.) Unfortunately, r[c] ≠ ContinuedFractionK[-(((1 - d + k) (e + k) (1 + e + k))/((1 + k) (1 - a + e + k) (1 - b + e + k))), -(((1 + e + k) (1 + 2 e + a (-1 + b - k) + 3 k - b (1 + k) + (e + k) (-d + 2 k)))/ ((1 + k) (1 - a + e + k) (1 - b + e + k))), {k, c, ∞}] Brutally specializing a=b=d=1, and changing s(k) = k*(k+e-1)^2 r(k)/(k+e) gives s[k]=k*(k+e-1), satisfying the recurrence s[k] == -((k^2 (-1 + e + k)^2)/(-2 k^2 - e (1 + 2 k) + s[1 + k])) , which is also obviously satisfied by ContinuedFractionK[-k^2 (-1 + e + k)^2, -2 k^2 - e (1 + 2 k), {k, c, ∞}]= (c (-1 + c + e) (PolyGamma[0, 1 + c] - PolyGamma[0, c + e]))/ (PolyGamma[0, c] - PolyGamma[0, -1 + c + e]) The general solution to the recurrence is s[k] -> 2 - e - 4 k + 2 e k + 2 k^2 - ((-1 + k) (-2 + e + k) (-1 + e + C[1] (HarmonicNumber[-2 + k] + PolyGamma[0, e] - PolyGamma[0, -2 + e + k])))/(-1 + e + EulerGamma C[1] + C[1] (PolyGamma[0, e] + PolyGamma[0, k] - PolyGamma[0, -1 + e + k]))}} which requires C[1]->0 for k*(k+e-1) and C[1] -> (1 - e)/(EulerGamma + PolyGamma[0, e]) for the correct value of the CF. Such an exotic C[1] makes it unclear how to use the original 3F2 to evaluate the original CF. ? --rwg
participants (1)
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Bill Gosper