Hypergeometric Function

Author

Eric W. Weisstein
August 24, 2008

This notebook downloaded from http://mathworld.wolfram.com/notebooks/SpecialFunctions/HypergeometricFunction.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/HypergeometricFunction.html.

©2008 Wolfram Research, Inc. except for portions noted otherwise

Linear Transformation Formulas

Equal@@@Table[{Hypergeometric2F1[a,b,c,z],
(((1-z))^(c-a-b))Hypergeometric2F1[c-a,c-b,c,z],
(((1-z))^(-a))Hypergeometric2F1[a,c-b,c,(z/(z-1))],
(((1-z))^(-b))Hypergeometric2F1[b,c-a,c,(z/(z-1))]}/.Thread[{a,b,c,z}->Table[Random[Complex,5(I+1){-1,1},20],{4}]],{100}]//Union

Differential Equation

-(((a b) y(z))/(z (1-z)))+(((c-(a+b+1) z) y'(z))/(z (1-z)))+y''(z)==0

DSolve[z(1-z)y''[z]+(c-(a+b+1)z)y'[z]-a b y[z]==0,y[z],z]

eqn=ExpandAll[z(1-z)y''[z]+(c-(a+b+1)z)y'[z]-a b y[z]/.y->Function[z,Sum[A[n]z^n,{n,0,∞}]]]

eqn2=eqn/.{(z^(e1_.))(∑_(n=0)^∞)l_ (z^(e2_))A[n] :>(∑_(n=0)^∞)l (z^(e1+e2))A[n]}/.{(∑_(n=0)^∞)l_ (z^(-1+k_))A[n]:>(∑_(n=0)^∞)(l (z^(-1+k))A[n]/.k->(n+1))}

FullSimplify[eqn2/.Verbatim[Sum][l__]:>First[{l}]]

RSolve[A[n+1]==(((n+a)(n+b))/((n+1)(n+c)))A[n],A[n],n]

FullSimplify[((C[1] Gamma[1+c] Gamma[a+n] Gamma[b+n])/(Gamma[1+a] Gamma[1+b] Gamma[1+n] Gamma[c+n]))/.n->0]

FullSimplify[((a b)/c)(( Gamma[1+c] Gamma[a+n] Gamma[b+n])/(Gamma[1+a] Gamma[1+b] Gamma[1+n] Gamma[c+n]))]

<<DiscreteMath`RSolve`

SeriesTerm[Hypergeometric2F1[a,b,c,z],{z,0,n}]

Table[((Gamma[c] Gamma[a+n] Gamma[b+n])/(Gamma[a] Gamma[b] Gamma[1+n] Gamma[c+n])),{n,0,3}]//FullSimplify

Sum[((Gamma[c] Gamma[a+n] Gamma[b+n])/(Gamma[a] Gamma[b] Gamma[1+n] Gamma[c+n]))(z^n),{n,0,∞}]//FullSimplify

2nd Solution

y''[z]+((c-(a+b+1)z)/(z(1-z)))y'[z]-((a b)/(z(1-z)))y[z]==0

FullSimplify[Integrate[((c-(a+b+1)z)/(z(1-z))),z]]

FullSimplify[Exp[-((1+a+b-c) Log[-1+z]+c Log[z])]]

Hypergeometric2F1[a,b,c,z] Integrate[(((((-1+z))^(-1-a-b+c)) (z^(-c)))/((Hypergeometric2F1[a,b,c,z])^2)),z]

Integer Arguments

Flatten[Table[{{i,j,k},Hypergeometric2F1[i,j,k,z]//FullSimplify},{i,3},{j,3},{k,2}],2]//ColumnForm

Log Identity

Log[1+z]+Sum[(((((-1))^n)(z^n))/n),{n,N-1}]

Rational Values

Rational Values

M. Trott, Aug 5, 2002

{#,N[#,50]}&[Hypergeometric2F1[1/3,2/3,5/6,27/32]==8/5]//TraditionalForm

Zucker and Joyce (2001)

{#,N[#,50]}&[Hypergeometric2F1[1/4,1/2,3/4,80/81]==9/5]//TraditionalForm

Quadratic Surds

{#,N[#,50]}&[Hypergeometric2F1[1/8,3/8,1/2,2400/2401]==2/3Sqrt[7]]//TraditionalForm

Others

Zucker and Joyce (2001)

{#,N[#,50]}&[Hypergeometric2F1[1/12,5/12,1/2,1323/1331]==3/4(11)^(1/4)]//TraditionalForm

Zucker and Joyce (2001)

{#,N[#,50]}&[Hypergeometric2F1[1/12,5/12,1/2,121/125]==(((2^(1/6))(15^(1/4)))/(4Sqrt[π]))(((Gamma[1/3])^3)/((Gamma[1/4])^2))(1+Sqrt[3])]//TraditionalForm

Zucker and Joyce (2003)

{#,N[#,50]}&[Hypergeometric2F1[1/6,1/2,2/3,125/128]==4 (2^(1/6))/3]//TraditionalForm

Glasser's Identity

Glasser[k_,x_]:={HypergeometricPFQ[Range[k]/(k+1),Append[Range[2,k-1]/k,(k+1)/k],(x(1-x^k)/(k/(1+k)^(1+1/k)))^k],1/(1-x^k)}

Solve[1/(1-x^3)==27/26,x]

{#,N[Subtract@@#,100]}&[Glasser[3,1/3]]

With[{k=2},
PFQ[Range[k]/(k+1),Append[Range[2,k-1]/k,(k+1)/k],(x(1-x^k)/(k/(1+k)^(1+1/k)))^k]]

FunctionExpand[Glasser[2,x]]

Subtract@@@Table[{HypergeometricPFQ[{(1/3),(2/3)},{(3/2)},(27/4) (x^2) (((1-(x^2)))^2)],(1/(1-(x^2)))}/.x->Random[Real,{0,1/Sqrt[3]},50],{20}]//Chop

Table[(k+1)^(-1/k),{k,10}]

N[%]

Solve[(x(1-x^k)/(k/(1+k)^(1+1/k)))^k==1,x]

Gamma Identity

FullSimplify[HypergeometricPFQ[{a,b,c},{d,e},2]-Pi^2(d e-(a+1)(b+1)(c+1)+a b c)/(Cos[d Pi]Cos[e Pi]-Cos[a Pi]Cos[b Pi]Cos[c Pi])*Gamma[d]/(Gamma[d-a]Gamma[d-b]Gamma[d-c])*
Gamma[e]/(Gamma[e-a]Gamma[e-b]Gamma[e-c])
/. e->a+b+c-d+2]

FullSimplify[% /. c->0]

FullSimplify[%14 /. c->1]

Series

About 0

SeriesCoefficient[Hypergeometric2F1[a,b,c,z],{z,0,n}]//InputForm

FullSimplify[%]

Sums

(∑_(n=0)^∞)(((x^n) (Γ(γ) Γ(n+α) Γ(n+β)))/(n! Γ(α) Γ(β) Γ(n+γ)))==(Subscript[, 2])(Subscript[F, 1])(β,α;γ;x)

Sum[((Gamma[n+α]Gamma[n+β]Gamma[γ])/(n!Gamma[α]Gamma[β]Gamma[n+γ]))(x^n),{n,0,∞}]

(∑_(n=0)^∞)(((z^n) ((Subscript[(a), n]) (Subscript[(b), n])))/(n! (Subscript[(c), n])))==(Subscript[, 2])(Subscript[F, 1])(a,b;c;z)

Sum[((Pochhammer[a,n]Pochhammer[b,n])/(Pochhammer[c,n]))((z^n)/(n!)),{n,0,∞}]