User:Michael assis

Testing:

$\displaystyle {}(c-a){}_{2}F_{1}(a-1,b;c;z)+(2a-c-az+bz){}_{2}F_{1}(a,b;c;z)+a(z-1){}_{2}F_{1}(a+1,b;c;z)=0$

$\displaystyle {}(c-b){}_{2}F_{1}(a,b-1;c;z)+(2b-c-bz+az){}_{2}F_{1}(a,b;c;z)+b(z-1){}_{2}F_{1}(a,b+1;c;z)=0$

$\displaystyle {}c(c-1)(z-1){}_{2}F_{1}(a,b;c-1;z)+c[c-1-(2c-a-b-1)z]{}_{2}F_{1}(a,b;c;z)+(c-a)(c-b)z{}_{2}F_{1}(a,b;c+1;z)=0$

$\displaystyle {}[c-2a-(b-a)z]{}_{2}F_{1}(a,b;c;z)+a(1-z){}_{2}F_{1}(a+1,b;c;z)-(c-a){}_{2}F_{1}(a-1,b;c;z)=0$

$\displaystyle {}(b-a){}_{2}F_{1}(a,b;c;z)+a{}_{2}F_{1}(a+1,b;c;z)-b{}_{2}F_{1}(a,b+1;c;z)=0$

$\displaystyle {}(c-a-b){}_{2}F_{1}(a,b;c;z)+a(1-z){}_{2}F_{1}(a+1,b;c;z)-(c-b){}_{2}F_{1}(a,b-1;c;z)=0$

$\displaystyle {}c[a-(c-b)z]{}_{2}F_{1}(a,b;c;z)-ac(1-z){}_{2}F_{1}(a+1,b;c;z)+(c-a)(c-b)z{}_{2}F_{1}(a,b;c+1;z)=0$

$\displaystyle {}(c-a-1){}_{2}F_{1}(a,b;c;z)+a{}_{2}F_{1}(a+1,b;c;z)-(c-1){}_{2}F_{1}(a,b;c-1;z)=0$

$\displaystyle {}(c-a-b){}_{2}F_{1}(a,b;c;z)-(c-a){}_{2}F_{1}(a-1,b;c;z)+b(1-z){}_{2}F_{1}(a,b+1;c;z)=0$

$\displaystyle {}(b-a)(1-z){}_{2}F_{1}(a,b;c;z)-(c-a){}_{2}F_{1}(a-1,b;c;z)+(c-b){}_{2}F_{1}(a,b-1;c;z)=0$

$\displaystyle {}c(1-z){}_{2}F_{1}(a,b;c;z)-c{}_{2}F_{1}(a-1,b;c;z)+(c-b)z{}_{2}F_{1}(a,b;c+1;z)=0$

$\displaystyle {}[a-1-(c-b-1)z]{}_{2}F_{1}(a,b;c;z)+(c-a){}_{2}F_{1}(a-1,b;c;z)-(c-1)(1-z){}_{2}F_{1}(a,b;c-1;z)=0$

$\displaystyle {}[c-2b+(b-a)z]{}_{2}F_{1}(a,b;c;z)+b(1-z){}_{2}F_{1}(a,b+1;c;z)-(c-b){}_{2}F_{1}(a,b-1;c;z)=0$

$\displaystyle {}c[b-(c-a)z]{}_{2}F_{1}(a,b;c;z)-bc(1-z){}_{2}F_{1}(a,b+1;c;z)+(c-a)(c-b)z{}_{2}F_{1}(a,b;c+1;z)=0$

$\displaystyle {}(c-b-1){}_{2}F_{1}(a,b;c;z)+b{}_{2}F_{1}(a,b+1;c;z)-(c-1){}_{2}F_{1}(a,b;c-1;z)=0$

$\displaystyle {}c(1-z){}_{2}F_{1}(a,b;c;z)-c{}_{2}F_{1}(a,b-1;c;z)+(c-a)z{}_{2}F_{1}(a,b;c+1;z)=0$

$\displaystyle {}[b-1-(c-a-1)z]{}_{2}F_{1}(a,b;c;z)+(c-b){}_{2}F_{1}(a,b-1;c;z)-(c-1)(1-z){}_{2}F_{1}(a,b;c-1;z)=0$

$\displaystyle {}c[c-1-(2c-a-b-1)z]{}_{2}F_{1}(a,b;c;z)+(c-a)(c-b)z{}_{2}F_{1}(a,b;c+1;z)-c(c-1)(1-z){}_{2}F_{1}(a,b;c-1;z)=0$

$\displaystyle {}[c-2a-(b-a)z]F+a(1-z)F^{a+}-(c-a)F^{a-}=0$

$\displaystyle {}(b-a)F+aF^{a+}-bF^{b+}=0$

$\displaystyle {}(c-a-b)F+a(1-z)F^{a+}-(c-b)F^{b-}=0$

$\displaystyle {}c[a-(c-b)z]F-ac(1-z)F^{a+}+(c-a)(c-b)zF^{c+}=0$

$\displaystyle {}(c-a-1)F+aF^{a+}-(c-1)F^{c-}=0$

$\displaystyle {}(c-a-b)F-(c-a)F^{a-}+b(1-z)F^{b+}=0$

$\displaystyle {}(b-a)(1-z)F-(c-a)F^{a-}+(c-b)F^{b-}=0$

$\displaystyle {}c(1-z)F-cF^{a-}+(c-b)zF^{c+}=0$

$\displaystyle {}[a-1-(c-b-1)z]F+(c-a)F^{a-}-(c-1)(1-z)F^{c-}=0$

$\displaystyle {}[c-2b+(b-a)z]F+b(1-z)F^{b+}-(c-b)F^{b-}=0$

$\displaystyle {}c[b-(c-a)z]F-bc(1-z)F^{b+}+(c-a)(c-b)zF^{c+}=0$

$\displaystyle {}(c-b-1)F+bF^{b+}-(c-1)F^{c-}=0$

$\displaystyle {}c(1-z)F-cF^{b-}+(c-a)zF^{c+}=0$

$\displaystyle {}[b-1-(c-a-1)z]F+(c-b)F^{b-}-(c-1)(1-z)F^{c-}=0$

$\displaystyle {}c[c-1-(2c-a-b-1)z]F+(c-a)(c-b)zF^{c+}-c(c-1)(1-z)F^{c-}=0$

• Abramowitz, Milton; Stegun, Irene Ann, eds. (1983) [June 1964]. "Chapter 15". Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. Applied Mathematics Series. Vol. 55 (Ninth reprint with additional corrections of tenth original printing with corrections (December 1972); first ed.). Washington D.C.; New York: United States Department of Commerce, National Bureau of Standards; Dover Publications. p. 558. ISBN 978-0-486-61272-0. LCCN 64-60036. MR 0167642. LCCN 65-12253.