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DLMF

Generated by Yurie`Math`DLMFDev`DLMFExport.

Index Formula
5.5.5 \(\begin{aligned}&\Gamma (z)=\frac{2^{z-1} \Gamma \left(\frac{z}{2}+\frac{1}{2}\right) \Gamma \left(\frac{z}{2}\right)}{\sqrt{\pi }}\\\end{aligned}\)
BinomialSwapA \(\begin{aligned}&\binom{a}{n}=(-1)^n \binom{-a+n-1}{n}\\\end{aligned}\)
BinomialSwapN \(\begin{aligned}&\binom{a}{n}=\binom{a}{a-n}\\\end{aligned}\)
PochhammerSwapA \(\begin{aligned}&(a)_n=(-1)^n (-a-n+1)_n\\\end{aligned}\)
5.2.8.1 \(\begin{aligned}&\boldsymbol{\text{EvenQ}[n]} \qquad (a)_n=2^n \left(\frac{a}{2}\right)_{\frac{n}{2}} \left(\frac{a+1}{2}\right)_{\frac{n}{2}}\\\end{aligned}\)
5.2.8.2 \(\begin{aligned}&\boldsymbol{\text{OddQ}[n]} \qquad (a)_n=2^n \left(\frac{a}{2}\right)_{\frac{n+1}{2}} \left(\frac{a+1}{2}\right)_{\frac{n-1}{2}}\\\end{aligned}\)
15.1.2 \(\begin{aligned}&\,_2\tilde{F}_1(a,b;c;z)=\frac{\Fpq{2}{1}{a,b}{c}{z}}{\Gamma (c)}\\\end{aligned}\)
15.4.20 \(\begin{aligned}&\boldsymbol{c-a-b>0} \qquad \Fpq{2}{1}{a,b}{c}{1}=\frac{\Gamma (c) \Gamma (c-a-b)}{\Gamma (c-a) \Gamma (c-b)}\\\end{aligned}\)
15.5.11 \(\begin{aligned}&\Fpq{2}{1}{a,b}{c}{z}=\frac{(a-c) \Fpq{2}{1}{a-1,b}{c}{z}-a (z-1) \Fpq{2}{1}{a+1,b}{c}{z}}{z (b-a)+2 a-c}\\\end{aligned}\)
15.5.12 \(\begin{aligned}&\Fpq{2}{1}{a,b}{c}{z}=\frac{a \Fpq{2}{1}{a+1,b}{c}{z}-b \Fpq{2}{1}{a,b+1}{c}{z}}{a-b}\\\end{aligned}\)
15.5.13 \(\begin{aligned}&\Fpq{2}{1}{a,b}{c}{z}=\frac{(b-c) \Fpq{2}{1}{a,b-1}{c}{z}+a (1-z) \Fpq{2}{1}{a+1,b}{c}{z}}{a+b-c}\\\end{aligned}\)
15.5.14 \(\begin{aligned}&\Fpq{2}{1}{a,b}{c}{z}=\frac{a c (1-z) \Fpq{2}{1}{a+1,b}{c}{z}-z (c-a) (c-b) \Fpq{2}{1}{a,b}{c+1}{z}}{c (a+z (b-c))}\\\end{aligned}\)
15.5.15 \(\begin{aligned}&\Fpq{2}{1}{a,b}{c}{z}=\frac{a \Fpq{2}{1}{a+1,b}{c}{z}-(c-1) \Fpq{2}{1}{a,b}{c-1}{z}}{a-c+1}\\\end{aligned}\)
15.5.16 \(\begin{aligned}&\Fpq{2}{1}{a,b}{c}{z}=\frac{c \Fpq{2}{1}{a-1,b}{c}{z}+z (b-c) \Fpq{2}{1}{a,b}{c+1}{z}}{c (1-z)}\\\end{aligned}\)
15.5.17 \(\begin{aligned}&\Fpq{2}{1}{a,b}{c}{z}=\frac{(a-c) \Fpq{2}{1}{a-1,b}{c}{z}+(c-1) (1-z) \Fpq{2}{1}{a,b}{c-1}{z}}{a+z (b-c+1)-1}\\\end{aligned}\)
15.5.18 \(\begin{aligned}&\Fpq{2}{1}{a,b}{c}{z}=\frac{(c-1) c (z-1) \Fpq{2}{1}{a,b}{c-1}{z}+z (c-a) (c-b) \Fpq{2}{1}{a,b}{c+1}{z}}{c (z (-a-b+2 c-1)-c+1)}\\\end{aligned}\)
15.8.1.1 \(\begin{aligned}&\Fpq{2}{1}{a,b}{c}{z}=(1-z)^{-a} \Fpq{2}{1}{a,c-b}{c}{\frac{z}{z-1}}\\\end{aligned}\)
15.8.1.2 \(\begin{aligned}&\Fpq{2}{1}{a,b}{c}{z}=(1-z)^{-b} \Fpq{2}{1}{c-a,b}{c}{\frac{z}{z-1}}\\\end{aligned}\)
15.8.1.3 \(\begin{aligned}&\Fpq{2}{1}{a,b}{c}{z}=(1-z)^{-a-b+c} \Fpq{2}{1}{c-a,c-b}{c}{z}\\\end{aligned}\)
15.8.2 \(\begin{aligned}&\Fpq{2}{1}{a,b}{c}{z}=\pi \Gamma (c) \csc (\pi (b-a)) \left(\frac{(-z)^{-a} \Fpq{2}{1}{a,a-c+1}{a-b+1}{\frac{1}{z}}}{\Gamma (b) \Gamma (a-b+1) \Gamma (c-a)}-\frac{(-z)^{-b} \Fpq{2}{1}{b,b-c+1}{-a+b+1}{\frac{1}{z}}}{\Gamma (a) \Gamma (-a+b+1) \Gamma (c-b)}\right)\\\end{aligned}\)
15.8.3 \(\begin{aligned}&\Fpq{2}{1}{a,b}{c}{z}=\pi \Gamma (c) \csc (\pi (b-a)) \left(\frac{(1-z)^{-a} \Fpq{2}{1}{a,c-b}{a-b+1}{\frac{1}{1-z}}}{\Gamma (b) \Gamma (a-b+1) \Gamma (c-a)}-\frac{(1-z)^{-b} \Fpq{2}{1}{b,c-a}{-a+b+1}{\frac{1}{1-z}}}{\Gamma (a) \Gamma (-a+b+1) \Gamma (c-b)}\right)\\\end{aligned}\)
15.8.4 \(\begin{aligned}&\Fpq{2}{1}{a,b}{c}{z}=\pi \Gamma (c) \csc (\pi (-a-b+c)) \left(\frac{\Fpq{2}{1}{a,b}{a+b-c+1}{1-z}}{\Gamma (c-a) \Gamma (c-b) \Gamma (a+b-c+1)}-\frac{(1-z)^{-a-b+c} \Fpq{2}{1}{c-a,c-b}{-a-b+c+1}{1-z}}{\Gamma (a) \Gamma (b) \Gamma (-a-b+c+1)}\right)\\\end{aligned}\)
15.8.5 \(\begin{aligned}&\Fpq{2}{1}{a,b}{c}{z}=\pi \Gamma (c) \csc (\pi (-a-b+c)) \left(\frac{z^{-a} \Fpq{2}{1}{a,a-c+1}{a+b-c+1}{1-\frac{1}{z}}}{\Gamma (c-a) \Gamma (c-b) \Gamma (a+b-c+1)}-\frac{z^{a-c} (1-z)^{-a-b+c} \Fpq{2}{1}{c-a,1-a}{-a-b+c+1}{1-\frac{1}{z}}}{\Gamma (a) \Gamma (b) \Gamma (-a-b+c+1)}\right)\\\end{aligned}\)
15.8.13 \(\begin{aligned}&\boldsymbol{2 b-c=0} \qquad \Fpq{2}{1}{a,b}{c}{z}=\left(1-\frac{z}{2}\right)^{-a} \Fpq{2}{1}{\frac{a}{2},\frac{1}{2}+\frac{a}{2}}{\frac{1}{2}+b}{\frac{z^2}{(2-z)^2}}\\\end{aligned}\)
15.8.14 \(\begin{aligned}&\boldsymbol{2 b-c=0} \qquad \Fpq{2}{1}{a,b}{c}{z}=(1-z)^{-\frac{a}{2}} \Fpq{2}{1}{\frac{a}{2},-\frac{a}{2}+b}{\frac{1}{2}+b}{\frac{z^2}{-4+4 z}}\\\end{aligned}\)
15.8.15 \(\begin{aligned}&\boldsymbol{a-b+1-c=0} \qquad \Fpq{2}{1}{a,b}{c}{z}=(1+z)^{-a} \Fpq{2}{1}{\frac{a}{2},\frac{1}{2}+\frac{a}{2}}{1+a-b}{\frac{4 z}{(1+z)^2}}\\\end{aligned}\)
15.8.16 \(\begin{aligned}&\boldsymbol{a-b+1-c=0} \qquad \Fpq{2}{1}{a,b}{c}{z}=(1-z)^{-a} \Fpq{2}{1}{\frac{a}{2},\frac{1}{2}+\frac{a}{2}-b}{1+a-b}{-\frac{4 z}{(1-z)^2}}\\\end{aligned}\)
15.8.17 \(\begin{aligned}&\boldsymbol{a+b+1-2 c=0} \qquad \Fpq{2}{1}{a,b}{c}{z}=(1-2 z)^{-a} \Fpq{2}{1}{\frac{a}{2},\frac{1}{2}+\frac{a}{2}}{\frac{1}{2} (1+a+b)}{\frac{4 (-1+z) z}{(1-2 z)^2}}\\\end{aligned}\)
15.8.18 \(\begin{aligned}&\boldsymbol{a+b+1-2 c=0} \qquad \Fpq{2}{1}{a,b}{c}{z}=\Fpq{2}{1}{\frac{a}{2},\frac{b}{2}}{\frac{1}{2} (1+a+b)}{4 (1-z) z}\\\end{aligned}\)
15.8.19 \(\begin{aligned}&\boldsymbol{a+b-1=0} \qquad \Fpq{2}{1}{a,b}{c}{z}=(1-2 z)^{1-a-c} (1-z)^{-1+c} \Fpq{2}{1}{\frac{a+c}{2},\frac{1}{2} (-1+a+c)}{c}{\frac{4 (-1+z) z}{(1-2 z)^2}}\\\end{aligned}\)
15.8.20 \(\begin{aligned}&\boldsymbol{a+b-1=0} \qquad \Fpq{2}{1}{a,b}{c}{z}=(1-z)^{c-1} \Fpq{2}{1}{\frac{1}{2} (-a+c),\frac{1}{2} (-1+a+c)}{c}{4 (1-z) z}\\\end{aligned}\)
15.8.21 \(\begin{aligned}&\boldsymbol{a-b+1-c=0} \qquad \Fpq{2}{1}{a,b}{c}{z}=\left(1+\sqrt{z}\right)^{-2 a} \Fpq{2}{1}{a,\frac{1}{2}+a-b}{1+2 a-2 b}{\frac{4 \sqrt{z}}{\left(1+\sqrt{z}\right)^2}}\\\end{aligned}\)
15.8.22 \(\begin{aligned}&\boldsymbol{a+b+1-2 c=0} \qquad \Fpq{2}{1}{a,b}{c}{z}=\left(\frac{-1+\sqrt{1-\frac{1}{z}}}{1+\sqrt{1-\frac{1}{z}}}\right)^a \Fpq{2}{1}{a,\frac{a+b}{2}}{a+b}{\frac{4 \sqrt{1-\frac{1}{z}}}{\left(1+\sqrt{1-\frac{1}{z}}\right)^2}}\\\end{aligned}\)
15.8.23 \(\begin{aligned}&\boldsymbol{a+b-1=0} \qquad \Fpq{2}{1}{a,b}{c}{z}=\left(-1+\sqrt{1-\frac{1}{z}}\right)^{1-a} \left(1+\sqrt{1-\frac{1}{z}}\right)^{1+a-2 c} \left(1-\frac{1}{z}\right)^{-1+c} \Fpq{2}{1}{-\frac{1}{2}+c,-a+c}{-1+2 c}{\frac{4 \sqrt{1-\frac{1}{z}}}{\left(1+\sqrt{1-\frac{1}{z}}\right)^2}}\\\end{aligned}\)
15.8.24 \(\begin{aligned}&\boldsymbol{a-b+1-c=0} \qquad \Fpq{2}{1}{a,b}{c}{z}=\frac{\left(\sqrt{\pi } (1-z)^{-a} \Gamma (1+a-b)\right) \Fpq{2}{1}{\frac{a}{2},\frac{1}{2}+\frac{a}{2}-b}{\frac{1}{2}}{\frac{(1+z)^2}{(-1+z)^2}}}{\Gamma \left(\frac{1}{2}+\frac{a}{2}\right) \Gamma \left(1+\frac{a}{2}-b\right)}-\frac{\left(2 \sqrt{\pi } (1-z)^{-1-a} (1+z) \Gamma (1+a-b)\right) \Fpq{2}{1}{\frac{1}{2}+\frac{a}{2},1+\frac{a}{2}-b}{\frac{3}{2}}{\frac{(1+z)^2}{(-1+z)^2}}}{\Gamma \left(\frac{a}{2}\right) \Gamma \left(\frac{1}{2}+\frac{a}{2}-b\right)}\\\end{aligned}\)
15.8.25 \(\begin{aligned}&\boldsymbol{a+b+1-2 c=0} \qquad \Fpq{2}{1}{a,b}{c}{z}=\frac{\left(\sqrt{\pi } \Gamma \left(\frac{1}{2} (1+a+b)\right)\right) \Fpq{2}{1}{\frac{a}{2},\frac{b}{2}}{\frac{1}{2}}{(1-2 z)^2}}{\Gamma \left(\frac{1}{2}+\frac{a}{2}\right) \Gamma \left(\frac{1}{2}+\frac{b}{2}\right)}-\frac{\left(2 \sqrt{\pi } (1-2 z) \Gamma \left(\frac{1}{2} (1+a+b)\right)\right) \Fpq{2}{1}{\frac{1}{2}+\frac{a}{2},\frac{1}{2}+\frac{b}{2}}{\frac{3}{2}}{(1-2 z)^2}}{\Gamma \left(\frac{a}{2}\right) \Gamma \left(\frac{b}{2}\right)}\\\end{aligned}\)
15.8.26 \(\begin{aligned}&\boldsymbol{a+b-1=0} \qquad \Fpq{2}{1}{a,b}{c}{z}=\frac{\left(\sqrt{\pi } (1-z)^{-1+c} \Gamma (c)\right) \Fpq{2}{1}{-\frac{a}{2}+\frac{c}{2},-\frac{1}{2}+\frac{a}{2}+\frac{c}{2}}{\frac{1}{2}}{(1-2 z)^2}}{\Gamma \left(\frac{a}{2}+\frac{c}{2}\right) \Gamma \left(\frac{1}{2} (1-a+c)\right)}-\frac{\left(2 \sqrt{\pi } (1-2 z) (1-z)^{-1+c} \Gamma (c)\right) \Fpq{2}{1}{\frac{1}{2}-\frac{a}{2}+\frac{c}{2},\frac{a}{2}+\frac{c}{2}}{\frac{3}{2}}{(1-2 z)^2}}{\Gamma \left(-\frac{a}{2}+\frac{c}{2}\right) \Gamma \left(\frac{1}{2} (-1+a+c)\right)}\\\end{aligned}\)
15.8.27 \(\begin{aligned}&\Fpq{2}{1}{a,b}{\frac{1}{2}}{z}=\frac{\Gamma \left(a+\frac{1}{2}\right) \Gamma \left(b+\frac{1}{2}\right) \left(\Fpq{2}{1}{2 a,2 b}{a+b+\frac{1}{2}}{\frac{1}{2} \left(1-\sqrt{z}\right)}+\Fpq{2}{1}{2 a,2 b}{a+b+\frac{1}{2}}{\frac{1}{2} \left(\sqrt{z}+1\right)}\right)}{2 \sqrt{\pi } \Gamma \left(a+b+\frac{1}{2}\right)}\\\end{aligned}\)
15.8.28 \(\begin{aligned}&\Fpq{2}{1}{a,b}{\frac{3}{2}}{z}=-\frac{\Gamma \left(a-\frac{1}{2}\right) \Gamma \left(b-\frac{1}{2}\right) \left(\Fpq{2}{1}{2 a-1,2 b-1}{a+b-\frac{1}{2}}{\frac{1}{2} \left(1-\sqrt{z}\right)}-\Fpq{2}{1}{2 a-1,2 b-1}{a+b-\frac{1}{2}}{\frac{1}{2} \left(\sqrt{z}+1\right)}\right)}{4 \sqrt{\pi } \sqrt{z} \Gamma \left(a+b-\frac{1}{2}\right)}\\\end{aligned}\)
16.4.14 \(\begin{aligned}&\boldsymbol{a+b+c-d-e-f+1-n=0} \qquad \Fpq{4}{3}{-n,a,b,c}{d,e,f}{1}=\frac{\left((e-a)_n (f-a)_n\right) \Fpq{4}{3}{-n,a,-b+d,-c+d}{d,1+a-e-n,1+a-f-n}{1}}{(e)_n (f)_n}\\\end{aligned}\)
14.3.1 \(\begin{aligned}&P_a^b(z)=\frac{\left(\frac{z+1}{1-z}\right)^{b/2} \Fpq{2}{1}{a+1,-a}{1-b}{\frac{1}{2}-\frac{z}{2}}}{\Gamma (1-b)}\\\end{aligned}\)
14.3.2 \(\begin{aligned}&Q_a^b(z)=\frac{\pi \left(\frac{z+1}{1-z}\right)^{b/2} \cot (\pi b) \Fpq{2}{1}{a+1,-a}{1-b}{\frac{1}{2}-\frac{z}{2}}}{2 \Gamma (1-b)}-\frac{\pi \left(\frac{1-z}{z+1}\right)^{b/2} \csc (\pi b) \Gamma (a+b+1) \Fpq{2}{1}{a+1,-a}{b+1}{\frac{1}{2}-\frac{z}{2}}}{2 \Gamma (b+1) \Gamma (a-b+1)}\\\end{aligned}\)
14.3.6 \(\begin{aligned}&P_a^b(z)=\frac{\left(\frac{z+1}{z-1}\right)^{b/2} \Fpq{2}{1}{a+1,-a}{1-b}{\frac{1}{2}-\frac{z}{2}}}{\Gamma (1-b)}\\\end{aligned}\)
14.3.7 \(\begin{aligned}&Q_a^b(z)=\frac{\sqrt{\pi } 2^{-a-1} e^{i \pi b} \left(z^2-1\right)^{b/2} z^{-a-b-1} \Gamma (a+b+1) \Fpq{2}{1}{\frac{a}{2}+\frac{b}{2}+1,\frac{a}{2}+\frac{b}{2}+\frac{1}{2}}{a+\frac{3}{2}}{\frac{1}{z^2}}}{\Gamma \left(a+\frac{3}{2}\right)}\\\end{aligned}\)
LegendreAtZ=0 \(\begin{aligned}&P_a(0)=\frac{\sqrt{\pi }}{\Gamma \left(\frac{1}{2}-\frac{a}{2}\right) \Gamma \left(\frac{a}{2}+1\right)}\\&Q_a(0)=-\frac{\sqrt{\pi } \sin \left(\frac{\pi a}{2}\right) \Gamma \left(\frac{a+1}{2}\right)}{2 \Gamma \left(\frac{a+2}{2}\right)}\\&P_a^{-1}(0)=\frac{\sqrt{\pi }}{\Gamma \left(1-\frac{a}{2}\right) \Gamma \left(\frac{a}{2}+\frac{3}{2}\right)}\\&Q_a^{-1}(0)=\frac{\sqrt{\pi } \cos \left(\frac{\pi a}{2}\right) \Gamma \left(\frac{a}{2}\right)}{4 \Gamma \left(\frac{a+3}{2}\right)}\\\end{aligned}\)
LegendreAtB=1/2 \(\begin{aligned}&P_a^{\frac{1}{2}}(z)=\frac{\left(z+i \sqrt{1-z^2}\right)^{-a-\frac{1}{2}}+\left(z+i \sqrt{1-z^2}\right)^{a+\frac{1}{2}}}{\sqrt{2 \pi } \sqrt[4]{1-z^2}}\\&Q_a^{\frac{1}{2}}(z)=-\frac{1}{2} i \left(\frac{\sqrt{\frac{\pi }{2}} \left(z+i \sqrt{1-z^2}\right)^{-a-\frac{1}{2}}}{\sqrt[4]{1-z^2}}-\frac{\sqrt{\pi } \left(z-i \sqrt{1-z^2}\right)^{-a-\frac{1}{2}}}{\sqrt{2} \sqrt[4]{1-z^2}}\right)\\&P_a^{\frac{1}{2}}(z)=\frac{\left(\sqrt{z^2-1}+z\right)^{-a-\frac{1}{2}}+\left(\sqrt{z^2-1}+z\right)^{a+\frac{1}{2}}}{\sqrt{2 \pi } \sqrt[4]{z^2-1}}\\&Q_a^{\frac{1}{2}}(z)=\frac{i \sqrt{\pi } \left(\sqrt{z^2-1}+z\right)^{-a-\frac{1}{2}}}{\sqrt{2} \sqrt[4]{z^2-1}}\\\end{aligned}\)
WilsonToHyper \(\begin{aligned}&\text{wilsonPolynomial}(a,b,c,d,n,x)=\frac{\Gamma (a+b+n) \Gamma (a+c+n) \Gamma (a+d+n) \Fpq{4}{3}{-n,a+b+c+d+n-1,a-i \sqrt{x},a+i \sqrt{x}}{a+b,a+c,a+d}{1}}{\Gamma (a+b) \Gamma (a+c) \Gamma (a+d)}\\\end{aligned}\)
WilsonFromHyper \(\begin{aligned}&\boldsymbol{a+b+c-d-e-f+1-n=0} \qquad \Fpq{4}{3}{-n,a,b,c}{d,e,f}{1}=\frac{(\Gamma (d) \Gamma (e) \Gamma (f)) \text{wilsonPolynomial}\left(\frac{b+c}{2},-\frac{b}{2}-\frac{c}{2}+d,-\frac{b}{2}-\frac{c}{2}+e,-\frac{b}{2}-\frac{c}{2}+f,n,-\frac{1}{4} (b-c)^2\right)}{\Gamma (1+a+b+c-d-e) \Gamma (1+a+b+c-d-f) \Gamma (1+a+b+c-e-f)}\\\end{aligned}\)
JacobiPhiToHyper \(\begin{aligned}&\text{jacobiPhi}(a,b,c,z)=\Fpq{2}{1}{\frac{1}{2} (a+b-i c+1),\frac{1}{2} (a+b+i c+1)}{a+1}{-\sinh^2(z)}\\\end{aligned}\)
JacobiPhiFromHyper \(\begin{aligned}&\Fpq{2}{1}{a,b}{c}{z}=\text{jacobiPhi}\left(c-1,a+b-c,i (a-b),\sinh^{-1}\left(\sqrt{-z}\right)\right)\\\end{aligned}\)