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Derivative of Gauss hypergeometric function ${}_{2}F_{1} (a,b,b+1,z)$


Given the Gauss hypergeometric function ${}_{2}F_{1} (a,b,c,z)$, its $n$th derivative can be written as

$\frac{\mathrm{d}^{n}}{\mathrm{d} z^{n}} {}_{2}F_{1} (a,b,c,z) = \frac{(a)_{n} (b)_{n}}{(c)_{n}} {}_{2}F_{1} (a+n,b+n,c+n,z) $

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

$\frac{\mathrm{d}}{\mathrm{d} z} {}_{2}F_{1} (a,b,b+1,z) = \frac{b \big( (1-z)^{-a} - {}_{2}F_{1} (a,b,b+1,z) \big)}{z}$

$\frac{\mathrm{d}^{2}}{\mathrm{d} z^{2}} {}_{2}F_{1} (a,b,b+1,z) = \frac{b \left(a (1-z)^{-a-1}-\frac{b \left((1-z)^{-a}-\, _2F_1(a,b;b+1;z)\right)}{z}\right)}{z}-\frac{b \left((1-z)^{-a}-\, _2F_1(a,b;b+1;z)\right)}{z^2}$

$\frac{\mathrm{d}^{3}}{\mathrm{d} z^{3}} {}_{2}F_{1} (a,b,b+1,z) = \ldots$

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. What is the property behind this result? I searched all the properties of the Gauss hypergeometric function in and in the book "Table of integrals, series and products" but with no luck.

To see this on Mathematica, you can compute the following to see the different results:

D[Hypergeometric2F1[a, b, c, z], {z, 1}] /. c -> b + 1

D[Hypergeometric2F1[a, b, b + 1, z], {z, 1}]

1 year ago

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