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I was reading a paper $[1]$ in which authors claimed that we can simplify below Gauss function to finite series if $m $ and $v$ are positive integers.

$$ _2F_{1}(v,m+v;m+1;x)=\psi\sum_{c=0}^{v-1} {v+m-1\choose c}{2v-2-c\choose v-1} \gamma^c$$

for values of $\psi$ and the$\gamma$ include x values as beow:

$$\gamma = \frac{1-x}{x}$$

However, I know $[2]$ the Hyper geometric, $_2F_1(\alpha,\beta;z;x)$,function terminates if $\alpha$ and $\beta$ is non-positive integers. This bothers me how such positive arguments can lead to a finite series representation.

Thanks


Reference:

[1] http://ieeexplore.ieee.org/xpl/login.jsp?tp=&arnumber=1247815&url=http%3A%2F%2Fieeexplore.ieee.org%2Fiel5%2F25%2F27947%2F01247815.pdf%3Farnumber%3D1247815

[2] Table of Integrals, Series and Products, I.S, Gradshteyn, et. al., 2007.

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    $\begingroup$ Exploit the hypergeometric differential equation and prove that the LHS is some polynomial multiplied by $(1-z)^{2\nu+1}$. After that, compute the coefficients of such polynomial. $\endgroup$ – Jack D'Aurizio Jul 20 '15 at 13:02
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    $\begingroup$ Moreover, the LHS has no singularity in the origin, and a pole of order $2\nu+1$ in $z=1$, so $\gamma$ has to be $\frac{x}{1-x}$. $\endgroup$ – Jack D'Aurizio Jul 20 '15 at 13:15
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    $\begingroup$ Note: $_2F_1(\alpha,\beta;z;x)$ terminates (i.e. is a polynomial) if $\alpha$ or $\beta$ are non-positive integers. $\endgroup$ – gammatester Jul 20 '15 at 13:24
  • $\begingroup$ (+1) Are you suggesting that start with $z(1-z){\ddot{u}} + (\lambda-(\alpha+\beta+1)z){\dot{u}}-\alpha\beta u=0$ and show this equation can wirtten as sum of elements similar to below: $$ \frac{k_1}{z^2} + \frac{k_2}{z(z-1)}+ \frac{k_3}{(z-1)^2}$$ $\endgroup$ – Cardinal Jul 20 '15 at 13:29
  • $\begingroup$ @gammatester it was typo thanks $\endgroup$ – Cardinal Jul 20 '15 at 13:32
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If you look at DLMF 15.8.1 you will see that:
$F\left(\begin{array}{c} a,b\\ c \end{array};z\right)=\frac{1}{\left(1-x\right)^{a}}F\left(\begin{array}{c} a,c-b\\ c \end{array};\frac{z}{z-1}\right)$
Now in your case $b=m+v,c=m+1$
So $c-b=m+1-m-v=1-v$
and if $v\in\mathcal{N}$
$-v$ is a negative integer.

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