On Mon, Jul 26, 2010 at 2:48 PM, Bill Gosper <billgosper@gmail.com> wrote:
Poorly tested and recklessly derived, but it sure looks like
d + e a (a + b) + d d e Hypergeometric1F1Regularized[-----, -------------, --] d 2 2 a a -------------------------------------------------------- = e a b + d d a Hypergeometric1F1Regularized[-, -------, --] d 2 2 a a
ContinuedFractionK[d n + e, a n + b, {n, 0, Infinity}],
where ContinuedFractionK[num[n],den[n],{n,0,inf}] := num[0]+den[0]/(num[1]+den[1]/... , and Hypergeometric1F1Regularized[a,b,z] := 1F1/Gamma[b].
Mma seems not to know this CF formula.
Given the three-term recurrences for 1F1. the form of this result is not surprising. But starting from the recurrences, it's not obvious how to get pure polynomials, vs rational functions, in the CF terms.
This is too nice to be new:
y e gama(z + 1, y) z ----------------- = ----------------------------------- - 1 z y y z - y + --------------------------- 2 y z - y + 1 + --------------- 3 y z - y + 2 + --- . . .
Notice how, as usual, we want the "incomplete factorial function".
where gama is little ("lower") gamma := int_0^y. But then why isn't it in A&S, MathWorld, or Wikipedia? Hmm, the latter ascribes to Gauss a CF with alternating terms that might be two-for-one equivalent to the above. Also, for upper Gamma (:= int_y^oo) they give something very similar, but with quadratic numerators.
In Wolframese,
(E^y*Gamma[1 + z, 0, y])/y^z == -1 + z/(-y + z + ContinuedFractionK[n*y, n - y + z, {n, 1, Infinity}])
y E Gamma[1 + z, 0, y] --------------------- == -1 + z y
z ------------------------------------------------------------- -y + z + ContinuedFractionK[n y, n - y + z, {n, 1, Infinity}]
Good luck testing this numerically--ContinuedFractionK needs work.
These are presumably limiting cases of K(quadratic/linear) = 2F1/2F1, which I'll now pursue. --rwg
Praying that an indefensible shortcut carries over from the confluent case, ContinuedFractionK[c*n^2 + d*n + e, a*n + b, {n, 0, Infinity}] == (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])]) 2 ContinuedFractionK[c n + d n + e, a n + b, {n, 0, Infinity}] == 2 2 e d + Sqrt[d - 4 c e] 1 d 2 b c - a (c + d) 1 a 2 c e Hypergeometric2F1[1 + --------------------, 1 + --------------------, - (3 + - + -----------------), - - ----------------] 2 2 c 2 c 2 2 2 d + Sqrt[d - 4 c e] c Sqrt[a + 4 c] 2 Sqrt[a + 4 c] --------------------------------------------------------------------------------------------------------------------------------------------------------------- 2 2 2 e d + Sqrt[d - 4 c e] 1 d 2 b c - a (c + d) 1 a (2 b c - a (c + d) + Sqrt[a + 4 c] (c + d)) Hypergeometric2F1[--------------------, --------------------, - (1 + - + -----------------), - - ----------------] 2 2 c 2 c 2 2 2 d + Sqrt[d - 4 c e] c Sqrt[a + 4 c] 2 Sqrt[a + 4 c] i.e., -Log[F[s]]'[0], with F[s]:= Hypergeometric2F1[((Sqrt[d^2 - 4*c*e] + d)/(2*c)), ((2*e)/(Sqrt[d^2 - 4*c*e] + d)), ((-((a*d)/c) + 2*b + a)/(2*Sqrt[4*c + a^2])) + ((d - c)/(2*c)) + 1, 1/2 - ((2*c*s + a)/(2*Sqrt[4*c + a^2]))] E.g., In[252]:= Append[Simplify[% /. {a -> 2, b -> 3, c -> z, d -> 2*z, e -> z}],Sqrt[z]/ArcTan[Sqrt[z]] - 1] Out[247]= ContinuedFractionK[z*n^2 + 2*z*n + z, 2* n + 3, {n, 0, Infinity}] == ((z/(1 + z))^(3/2) Sqrt[1 + z]*Hypergeometric2F1[2, 2, 5/2, 1/2 - 1/(2 Sqrt[1 + z])])/( 6 ArcSin[Sqrt[1/2 - 1/(2 Sqrt[1 + z])]]) == -1 + Sqrt[z]/ArcTan[Sqrt[z]] In[248]:= Normal[Series[1/(1 + List @@ Rest[%]), {z, 0, 4}]] Out[248]= 2 3 4 2 3 4 z z z z z z z z {1 - - + -- - -- + --, 1 - - + -- - -- + --} 3 5 7 9 3 5 7 9 ContinuedFractionK[z*n^2 , 2*n+1, {n, Infinity}] fails even more spectacularly: -Sqrt[-n^2*z]*Tan[-n^2*z] I.e., it treats n^2 as a constant! Godd ole pattern-matching. More scandal: Mma can't do the Hypergeometric ODE unless the regular singular points are already 0, 1, and oo. --rwg