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The goal is to have an understandable proof of DLMF 15.5.E8 that can be generalized, or not, to Generalized HyperGeometric functions. This result is also in Wolfram http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/20/02/05/
and Slater “Generalized Hypergeometric Functions” 1.4.1.8

Rewriting DLMF http://dlmf.nist.gov/15.5.E8:
$\left((1-z)\frac{\mathrm{d}}{\mathrm{d}z}(1-z)\right)^{n}(z^{c-1}(1-z)^{b-c}F(a,b;c;z))$
$=(c-1)_{n} z^{c-1-n}(1-z)^{b-c+n} F(a-1,b;c-1;z)$
with $n=1$ and, if you want, a an integer $<0$
$(\frac{\mathrm{d}}{\mathrm{d}z}\left(z^{c-1}(1-z)^{b-c+1}F\left(a,b;c;z\right)\right)$
$=(c-1)z^{c-2}(1-z)^{b-c} F(a-1,b;c-1;z)$
Now for a particular power k we have:
$(\Sigma_{k}\frac{d}{dz}(z^{c-1}(1-z)^{b-c+1}\frac{(a)_{k},(b)_{k}}{(c)_{k}} \frac{z^{k}}{k!})$
$=\Sigma_{k} \frac{z^{c + k-2} (1 - z )^{b-c} (a)_{k} (b)_{n} ( - b\cdot z + c - k\cdot z + k - 1)}{ k! (c)_{k} }$
$ = (c-n)_{n} z^{(c-n-1)}(1-z)^{(b-c+n)} F(a-n,b;c-n;z) $
I will settle for the case $n=1$
Maxima/wxMaxima code for examining terms
t:z^(c+k-1)(1-z)^(b-c+1)/k!;
tt:diff(t,z);
factor(tt);
s:z^(c-1)
(1-z)^(b-c+1)*hypergeometric([a,b],[c],z);
ss:diff(s,z);
factor(ss);

I will try to give bonus points for a proof using the results in (or similar to) DLMF http://dlmf.nist.gov/16

My underlying goal is the raising of a in F(a-1,b,c,x) by
$h(x,b,c)\int g(x,b,c)F(a-1,b,c,x)dx$
with h(),g() independent of a.
Perhaps it should be noted that when a is a negative integer, this a result on polynomials.

Progress: After thinking I realized that, using DLMF numbers, applying 15.8.1 and then 15.5.4 and then 15.8.1 in reverse will prove, at least "formally", 15.5.9. So I am presuming that a similar sequence might work for 15.5.8; although it gets a little confusing/complicated.
As far as Generalized HyperGeometric goes these seem to point to solving via. the Mellin transform, or not.

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1 Answer 1

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This is a lot simpler than it looks:

  1. Beat up an identity until it is in the form you want and is differentiable

  2. Differentiate

  3. Reverse the identity

Constructive comments welcomed.


Using http://dlmf.nist.gov/15.5.E4 and the chain rule
$\frac{d}{dz}\left(\left(\frac{-z}{1-z}\right)^{c-1}\cdot F\left(a,b;c;\frac{-z}{1-z}\right)\right) [diff]$
$=\left(\left(c-1\right)\cdot\left(\frac{-z}{1-z}\right)^{c-2}\cdot F\left(a,b;c-1;\frac{-z}{1-z}\right)\right)\cdot\left(\frac{d}{dz}\left(\frac{-z}{1-z}\right)\right)$
We have:
$\frac{d}{dz}\left(\left(\frac{-z}{1-z}\right)^{c-1}\cdot F\left(a,b;c;\frac{-z}{1-z}\right)\right) [xform]$ $=\left(\left(c-1\right)\cdot\left(\frac{-z}{1-z}\right)^{c-2}\cdot F\left(a,b;c-1;\frac{-z}{1-z}\right)\right)\cdot\frac{-1}{\left(1-z\right)^{2}}$


Constructing the LHS of http://dlmf.nist.gov/15.5.E8
$\left(z^{c-1}\cdot\left(1-z\right)^{b-\left(c-1\right)}\right)\cdot F\left(a,b;c;z\right)$ $=\left(z^{c-1}\cdot\left(1-z\right)^{b-\left(c-1\right)}\right)\cdot\left(1-z\right)^{-b}\cdot F\left(c-a,b;c;\frac{-z}{1-z}\right) $
$=\left(\frac{-z}{1-z}\right)^{c-1}\cdot\left(F\left(c-a,b;c;\frac{-z}{1-z}\right)\right)$

We have from [eq:diff]:
$\frac{d}{dz}\left(\left(z^{c-1}\cdot\left(1-z\right)^{b-\left(c-1\right)}\right)\cdot F\left(a,b;c;z\right)\right)$
$=\frac{d}{dz}\left(\left(\frac{-z}{1-z}\right)^{c-1}\cdot\left(F\left(c-a,b;c;\frac{-z}{1-z}\right)\right)\right)$
$=\left(\left(c-1\right)\cdot\left(\frac{-z}{1-z}\right)^{c-2}\cdot F\left(c-a,b;c-1;\frac{-z}{1-z}\right)\right)\cdot\frac{-1}{\left(1-z\right)^{2}} $


Reversing [eq:xform]
$\frac{d}{dz}\left(\left(z^{c-1}\cdot\left(1-z\right)^{b-\left(c-1\right)}\right)\cdot F\left(a,b;c;z\right)\right)$ $=\left(c-1\right)\cdot z^{c-2}\cdot\left(1-z\right)^{b-\left(c-2\right)}\cdot\left(1-z\right)^{-2}\cdot F\left(a-1,b;c-1;z\right)$
QED


And the indefinite integral
${\displaystyle \int}\left(c-1\right)\cdot z^{c-2}\cdot\left(1-z\right)^{b-c}\cdot F\left(a-1,b;c-1;z\right)$
$=\left(z^{c-1}\cdot\left(1-z\right)^{b-c+1}\right)\cdot F\left(a,b;c;z\right)$


Missing: detailed evaluation of regions of convergence

Generalization thoughts:

  1. http://dlmf.nist.gov/15.5.E4 is generalizable as http://dlmf.nist.gov/16.3.E4
  2. The process is dependent on the existence of an identity which, as far as my research goes, is specialized. I plan to work on that.
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