§15.5 Derivatives and Contiguous Functions

§15.5(i) Differentiation Formulas

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

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

 15.5.10 $\left(z\frac{\mathrm{d}}{\mathrm{d}z}z\right)^{n}=z^{n}\frac{{\mathrm{d}}^{n}}% {{\mathrm{d}z}^{n}}z^{n},$ $n=1,2,3,\dots$. ⓘ Symbols: $\frac{\mathrm{d}\NVar{f}}{\mathrm{d}\NVar{x}}$: derivative of $f$ with respect to $x$, $n$: integer and $z$: complex variable Referenced by: §15.5(i) Permalink: http://dlmf.nist.gov/15.5.E10 Encodings: TeX, pMML, png See also: Annotations for §15.5(i), §15.5 and Ch.15

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

§15.5(ii) Contiguous Functions

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

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

By repeated applications of (15.5.11)–(15.5.18) any function $F\left(a+k,b+\ell;c+m;z\right)$, in which $k,\ell,m$ are integers, can be expressed as a linear combination of $F\left(a,b;c;z\right)$ 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 ${z(1-z)(a+1)(b+1)}F\left(a+2,b+2;c+2;z\right)+{(c-(a+b+1)z)(c+1)}F\left(a+1,b+% 1;c+1;z\right)-{c(c+1)}F\left(a,b;c;z\right)=0.$

Further contiguous relations include:

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