# §15.5 Derivatives and Contiguous Functions

## §15.5(i) Differentiation Formulas

 15.5.1 $\frac{d}{dz}\mathop{F\/}\nolimits\!\left(a,b;c;z\right)=\frac{ab}{c}\mathop{F% \/}\nolimits\!\left(a+1,b+1;c+1;z\right),$
 15.5.2 $\frac{{d}^{n}}{{dz}^{n}}\mathop{F\/}\nolimits\!\left(a,b;c;z\right)=\frac{% \left(a\right)_{n}\left(b\right)_{n}}{\left(c\right)_{n}}\*\mathop{F\/}% \nolimits\!\left(a+n,b+n;c+n;z\right).$
 15.5.3 $\left(z\frac{d}{dz}z\right)^{n}\left(z^{a-1}\mathop{F\/}\nolimits\!\left(a,b;c% ;z\right)\right)=\left(a\right)_{n}z^{a+n-1}\mathop{F\/}\nolimits\!\left(a+n,b% ;c;z\right).$
 15.5.4 $\frac{{d}^{n}}{{dz}^{n}}\left(z^{c-1}\mathop{F\/}\nolimits\!\left(a,b;c;z% \right)\right)=\left(c-n\right)_{n}z^{c-n-1}\mathop{F\/}\nolimits\!\left(a,b;c% -n;z\right).$
 15.5.5 $\left(z\frac{d}{dz}z\right)^{n}\left(z^{c-a-1}(1-z)^{a+b-c}\mathop{F\/}% \nolimits\!\left(a,b;c;z\right)\right)=\left(c-a\right)_{n}z^{c-a+n-1}(1-z)^{a% -n+b-c}\*\mathop{F\/}\nolimits\!\left(a-n,b;c;z\right).$
 15.5.6 $\frac{{d}^{n}}{{dz}^{n}}\left((1-z)^{a+b-c}\mathop{F\/}\nolimits\!\left(a,b;c;% z\right)\right)=\frac{\left(c-a\right)_{n}\left(c-b\right)_{n}}{\left(c\right)% _{n}}(1-z)^{a+b-c-n}\*\mathop{F\/}\nolimits\!\left(a,b;c+n;z\right).$
 15.5.7 $\left((1-z)\frac{d}{dz}(1-z)\right)^{n}\left((1-z)^{a-1}\mathop{F\/}\nolimits% \!\left(a,b;c;z\right)\right)=(-1)^{n}\frac{\left(a\right)_{n}\left(c-b\right)% _{n}}{\left(c\right)_{n}}(1-z)^{a+n-1}\*\mathop{F\/}\nolimits\!\left(a+n,b;c+n% ;z\right).$
 15.5.8 $\left((1-z)\frac{d}{dz}(1-z)\right)^{n}\left(z^{c-1}(1-z)^{b-c}\mathop{F\/}% \nolimits\!\left(a,b;c;z\right)\right)=\left(c-n\right)_{n}z^{c-n-1}(1-z)^{b-c% +n}\*\mathop{F\/}\nolimits\!\left(a-n,b;c-n;z\right).$
 15.5.9 $\frac{{d}^{n}}{{dz}^{n}}\left(z^{c-1}(1-z)^{a+b-c}\mathop{F\/}\nolimits\!\left% (a,b;c;z\right)\right)=\left(c-n\right)_{n}z^{c-n-1}(1-z)^{a+b-c-n}\*\mathop{F% \/}\nolimits\!\left(a-n,b-n;c-n;z\right).$

Other versions of several of the identities in this subsection can be constructed with the aid of the operator identity

 15.5.10 $\left(z\frac{d}{dz}z\right)^{n}=z^{n}\frac{{d}^{n}}{{dz}^{n}}z^{n},$ $n=1,2,3,\dots$.

See Erdélyi et al. (1953a, pp. 102–103).

## §15.5(ii) Contiguous Functions

The six functions $\mathop{F\/}\nolimits\!\left(a\pm 1,b;c;z\right)$, $\mathop{F\/}\nolimits\!\left(a,b\pm 1;c;z\right)$, $\mathop{F\/}\nolimits\!\left(a,b;c\pm 1;z\right)$ are said to be contiguous to $\mathop{F\/}\nolimits\!\left(a,b;c;z\right)$.

 15.5.11 $\displaystyle(c-a)\mathop{F\/}\nolimits\!\left(a-1,b;c;z\right)+\left(2a-c+(b-% a)z\right)\mathop{F\/}\nolimits\!\left(a,b;c;z\right)+a(z-1)\mathop{F\/}% \nolimits\!\left(a+1,b;c;z\right)$ $\displaystyle=0,$ 15.5.12 $\displaystyle(b-a)\mathop{F\/}\nolimits\!\left(a,b;c;z\right)+a\mathop{F\/}% \nolimits\!\left(a+1,b;c;z\right)-b\mathop{F\/}\nolimits\!\left(a,b+1;c;z\right)$ $\displaystyle=0,$ 15.5.13 $\displaystyle(c-a-b)\mathop{F\/}\nolimits\!\left(a,b;c;z\right)+a(1-z)\mathop{% F\/}\nolimits\!\left(a+1,b;c;z\right)-(c-b)\mathop{F\/}\nolimits\!\left(a,b-1;% c;z\right)$ $\displaystyle=0,$ 15.5.14 $\displaystyle c\left(a+(b-c)z\right)\mathop{F\/}\nolimits\!\left(a,b;c;z\right% )-ac(1-z)\mathop{F\/}\nolimits\!\left(a+1,b;c;z\right)+(c-a)(c-b)z\mathop{F\/}% \nolimits\!\left(a,b;c+1;z\right)$ $\displaystyle=0,$ 15.5.15 $\displaystyle(c-a-1)\mathop{F\/}\nolimits\!\left(a,b;c;z\right)+a\mathop{F\/}% \nolimits\!\left(a+1,b;c;z\right)-(c-1)\mathop{F\/}\nolimits\!\left(a,b;c-1;z\right)$ $\displaystyle=0,$ 15.5.16 $\displaystyle c(1-z)\mathop{F\/}\nolimits\!\left(a,b;c;z\right)-c\mathop{F\/}% \nolimits\!\left(a-1,b;c;z\right)+(c-b)z\mathop{F\/}\nolimits\!\left(a,b;c+1;z\right)$ $\displaystyle=0,$ 15.5.17 $\displaystyle\left(a-1+(b+1-c)z\right)\mathop{F\/}\nolimits\!\left(a,b;c;z% \right)+(c-a)\mathop{F\/}\nolimits\!\left(a-1,b;c;z\right)-(c-1)(1-z)\mathop{F% \/}\nolimits\!\left(a,b;c-1;z\right)$ $\displaystyle=0,$ 15.5.18 $\displaystyle c(c-1)(z-1)\mathop{F\/}\nolimits\!\left(a,b;c-1;z\right)+{c\left% (c-1-(2c-a-b-1)z\right)}\mathop{F\/}\nolimits\!\left(a,b;c;z\right)+(c-a)(c-b)% z\mathop{F\/}\nolimits\!\left(a,b;c+1;z\right)$ $\displaystyle=0.$

By repeated applications of (15.5.11)–(15.5.18) any function $\mathop{F\/}\nolimits\!\left(a+k,b+\ell;c+m;z\right)$, in which $k,\ell,m$ are integers, can be expressed as a linear combination of $\mathop{F\/}\nolimits\!\left(a,b;c;z\right)$ and any one of its contiguous functions, with coefficients that are rational functions of $a,b,c$, and $z$.

An equivalent equation to the hypergeometric differential equation (15.10.1) is

 15.5.19 ${z(1-z)(a+1)(b+1)}\mathop{F\/}\nolimits\!\left(a+2,b+2;c+2;z\right)+{(c-(a+b+1% )z)(c+1)}\mathop{F\/}\nolimits\!\left(a+1,b+1;c+1;z\right)-{c(c+1)}\mathop{F\/% }\nolimits\!\left(a,b;c;z\right)=0.$

Further contiguous relations include:

 15.5.20 $z(1-z)\left(\ifrac{d\mathop{F\/}\nolimits\!\left(a,b;c;z\right)}{dz}\right)=(c% -a)\mathop{F\/}\nolimits\!\left(a-1,b;c;z\right)+(a-c+bz)\mathop{F\/}\nolimits% \!\left(a,b;c;z\right)=(c-b)\mathop{F\/}\nolimits\!\left(a,b-1;c;z\right)+(b-c% +az)\mathop{F\/}\nolimits\!\left(a,b;c;z\right),$
 15.5.21 $c(1-z)\left(\ifrac{d\mathop{F\/}\nolimits\!\left(a,b;c;z\right)}{dz}\right)=(c% -a)(c-b)\mathop{F\/}\nolimits\!\left(a,b;c+1;z\right)+c(a+b-c)\mathop{F\/}% \nolimits\!\left(a,b;c;z\right).$