[math-fun] Linear/linear CF closed form
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 + --- . . . 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
Bill, first part garbled a bit. did you mean Hypergeometric1F1Regularized[(d+e)/d,(a(a+b)+d)/a^2,d e/a^2]/ a /Hypergeometric1F1Regularized[e/d,(a b+d)/a^2,d/a^2] ?? and the Mma4.1 equivalent of the CF formula would be Fold[Function[n,d n+e][#2]+Function[n,a n+b][#2]/#1&,Infinity,Reverse@Table[k,{k,0,big}]] but then I get, to sufficient accuracy, Fold[Function[n,2 n+3] [#2]+Function[n,5 n+7] [#2]/#1&,Infinity,Reverse@Table[k,{k,0,64}]]~N~48 gives 4.09416383356161184979137013700204211345271113443 while Hypergeometric1F1Regularized[(d+e)/d,(a(a+b)+d)/a^2,d e /a^2]/ a /Hypergeometric1F1Regularized[e/d,(a b+d)/a^2,d/a^2]/. Thread[{a,b,d,e}->{5,7,2,3}] ~N~48 gives 0.158710559426226163598984126912885264985530397927 hmmm, Wouter. =============================== This email is confidential and intended solely for the use of the individual to whom it is addressed. If you are not the intended recipient, be advised that you have received this email in error and that any use, dissemination, forwarding, printing, or copying of this email is strictly prohibited. You are explicitly requested to notify the sender of this email that the intended recipient was not reached.
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
participants (2)
-
Bill Gosper -
Meeussen Wouter (bkarnd)