Procedure for evaluating the hypergeometric series $_2F_1\left\{\frac{v+2}{2},\frac{v+3}{2};v+1;z\right\}$

Question

I'm trying to work out the procedure to get the following hypergeometric series into a simpler form, for all postive integer $v$: $$ _2F_1\left\{\frac{v+2}{2},\frac{v+3}{2};v+1;z\right\}$$

For example, plugging this into Wolfram Alpha gives for $v$ = 1, $$\frac{1}{(1-z)^{3/2}}$$ for $v$ = 2, $$\frac{4 (2 \sqrt{1-z} \,(z-1)-3 z+2)}{3 \sqrt{1-z}\, (z-1) z^2}$$ for $v$ = 3, $$-\frac{2 (3 z^2+4 (2 \sqrt{1-z}-3) z-8 \sqrt{1-z}+8)}{(1-z)^{3/2} z^3}$$ and so on.

I'm guessing a transformation is repeatedly applied until a terminating form of the hypergeometric series is obtained. For $v$ = 1, applying Euler's transformation, $$_2F_1 (a,b;c;z) = (1-z)^{c-a-b}{}_2F_1 (c-a, c-b;c ; z)$$ gives the correct form; however, I cant work out what is used for $v$ = 2 and higher.

Answer 1

According to Mathematica,

$$ _2F_1\left(\frac{v+2}{2},\frac{v+3}{2};v+1;z\right) = \frac{2^v \left(1+\sqrt{1-z}\right)^{-v} \left(1+v \sqrt{1-z}\right)}{(1+v) (1-z)^{3/2}}. $$

I would be surprised if this was not known, especially considering how close it is to the known cases $_2F_1(a,a+1/2;2a+1;z)$ and $_2F_1(a,a+1/2;2a;z)$. Indeed, if $b=(v+2)/2$ and $c = (v-1)/2$ then

$$\begin{align} _2F_1\left(\frac{v+2}{2},\frac{v+3}{2};v+1;z\right) &= {}_2F_1\left(b,b+\frac{1}{2};2b-1;z\right) \\ &= (1-z)^{-3/2} {}_2F_1\left(c,c+\frac{1}{2};2c+2;z\right). \end{align}$$

The DLMF gives Prudnikov et al. (1990, pp. 468–488) as a reference for elementary representations, which I'll check out from the library tomorrow.

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Edit.

With a little coaxing, Mathematica seems able to evaluate $$ _2F_1\left(\frac{v+u}{2},\frac{v+u+1}{2};v+1;z\right) $$

for all nonnegative integer $u$ through the use of the integral representation

$$ _2F_1(a,b;c;z) = \frac{\Gamma(c)}{\Gamma(b)\,\Gamma(c-b)} \int_0^1 \frac{t^{b-1} (1-t)^{c-b-1}}{(1-t z)^a}\,dt. $$

For even $u$ we can evaluate it using the code

u=2 n;
a=(v+u)/2;
b=(v+u+1)/2;
c=v+1;
Expand[Integrate[t^(b-1) (1-t)^(c-b-1)/(1-t z)^a,{t,0,1},
    Assumptions->Re[v]>u-1&&Re[z]<=1] 
  Gamma[c]/(Gamma[b] Gamma[c-b])]//FullSimplify

One such example is

$$\begin{align} &_2F_1\left(\frac{v+4}{2},\frac{v+5}{2};v+1;z\right) \\ &\qquad= \frac{2^v v^2 \left(1+\sqrt{1-z}\right)^{-v} \Gamma(v+1) \left(v (1-z)^{3/2}-6 (z-1)\right)}{(1-z)^{7/2} \Gamma(v+4)} \\ &\hspace{2cm} + \frac{2^v \left(1+\sqrt{1-z}\right)^{-v} \Gamma(v+1) \left(6 + 9 z+v \sqrt{1-z} (11+4 z)\right)}{(1-z)^{7/2} \Gamma(v+4)}. \end{align}$$

For odd $u$ we can reduce it back to the even case (sort of) by using the identity

$$\begin{align} &{}_2F_1\left(\frac{v+u}{2},\frac{v+u+1}{2};v+1;z\right) \\ &\qquad = \frac{2 v {}_2F_1\left(\frac{v+u-1}{2},\frac{v+u}{2},v,z\right)-(v-u+1) {}_2F_1\left(\frac{v+u-1}{2},\frac{v+u}{2},v+1,z\right)}{v+u-1}, \end{align}$$

which is courtesy of Mathematica and is probably some combination of Gauss' relations for the contiguous hypergeometric functions.

The code is thus

u=2 n+1;
a=(v+u-1)/2;
b=(v+u)/2;
c1=v;
c2=v+1;
Expand[
  (2 v Integrate[t^(b-1) (1-t)^(c1-b-1)/(1-t z)^a,{t,0,1},
        Assumptions->Re[v]>u+1&&Re[z]<=1] 
      Gamma[c1]/(Gamma[b] Gamma[c1-b])
   -(v-u+1) Integrate[t^(b-1) (1-t)^(c2-b-1)/(1-t z)^a,{t,0,1},
        Assumptions->Re[v]>u-1&&Re[z]<=1] 
      Gamma[c2]/(Gamma[b] Gamma[c2-b]))/(v+u-1)]//FullSimplify

And an example,

$$\begin{align} &{}_2F_1\left(\frac{v+3}{2},\frac{v+4}{2};v+1;z\right) \\ &\qquad = \frac{2^v v \left(1+\sqrt{1-z}\right)^{-v} \Gamma(v+1) \left(3\sqrt{1-z} + v(1-z)\right)}{(1-z)^{5/2} \Gamma(v+3)} \\ &\hspace{2cm} + \frac{2^v \left(1+\sqrt{1-z}\right)^{-v} \Gamma(v+1) \left(2+z\right)}{(1-z)^{5/2} \Gamma(v+3)}. \end{align}$$

Answer 2

Answering my own question with this post:

Tables of Hypergeometric Functions

There is a link to a paper that describes a method for performing the transformations. The method is quite complicated and not worth rewriting here. Apparently the python package sympy has a function "hyperexpand" that can work it out.

Update:

Thanks again to Antonio for the hint to start from a known form. I've worked out the process: Substituting $a = v/2$ we get $$ _2F_1\left\{\frac{v+2}{2},\frac{v+3}{2};v+1;z\right\} = \,_2F_1\left\{a+1,a+\frac{3}{2};2a+1;z\right\} $$ Since [1], $$ _2F_1\left\{a,a+\frac{1}{2};2a+1;z\right\} = 2^{2a}(1 + \sqrt{1-z})^{-2a} $$ and [2],

$$ _2F_1\{a_1 + 1,a_2;b;z\} = \left(\frac{z}{a_1}\frac{d}{dz} + 1\right) \,_2F_1\{a_1, a_2;b;z\}, $$ $$ _2F_1\{a_1,a_2+1;b;z\} = \left(\frac{z}{a_2}\frac{d}{dz} + 1\right) \,_2F_1\{a_1, a_2;b;z\}, $$

then, $$ _2F_1\left\{a+1,a+\frac{3}{2};2a+1;z\right\} = \left(\frac{z}{a_1}\frac{d}{dz} + 1\right)\left(\frac{z}{a_2}\frac{d}{dz} + 1\right)\left(2^{2a}(1 + \sqrt{1-z})^{-2a}\right).$$

The rest follows by doing the differentiation and substuting back $v/2 = a$, and gives the same result as described in Antonio's answer.