Question

Derivative of gauss hypergeometric function

Answer 1

Given the Gauss hypergeometric function 2F1(a,b,c,z)2F1(a,b,c,z), its nnth derivative can be written as

dndzn2F1(a,b,c,z)=(a)n(b)n(c)n2F1(a+n,b+n,c+n,z)dndzn2F1(a,b,c,z)=(a)n(b)n(c)n2F1(a+n,b+n,c+n,z)

However, when c=b+1c=b+1, it appears that one can write the nnth derivative differently. For instance, with Mathematica, I obtain

ddz2F1(a,b,b+1,z)=b((1−z)−a−2F1(a,b,b+1,z))zddz2F1(a,b,b+1,z)=b((1−z)−a−2F1(a,b,b+1,z))z

d2dz22F1(a,b,b+1,z)=b(a(1−z)−a−1−b((1−z)−a−2F1(a,b;b+1;z))z)z−b((1−z)−a−2F1(a,b;b+1;z))z2d2dz22F1(a,b,b+1,z)=b(a(1−z)−a−1−b((1−z)−a−2F1(a,b;b+1;z))z)z−b((1−z)−a−2F1(a,b;b+1;z))z2

d3dz32F1(a,b,b+1,z)=…d3dz32F1(a,b,b+1,z)=…

Although less compact, the interesting thing of this form is that all Gauss hypergeometric functions appearing in the derivatives are the same as the original function