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.


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.


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;
Expand[Integrate[t^(b-1) (1-t)^(c-b-1)/(1-t z)^a,{t,0,1},
  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;
  (2 v Integrate[t^(b-1) (1-t)^(c1-b-1)/(1-t z)^a,{t,0,1},
      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},
      Gamma[c2]/(Gamma[b] Gamma[c2-b]))/(v+u-1)]//FullSimplify

And an example,

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

  • $\begingroup$ Thanks. I just put it into Wolfram Alpha and got a less concise form, will try to get a hold of Mathematica. Before now I had been finding it with values substituted for v. To give some background I'm looking at actually is 6.621 - 1. Gradshteyn and Rhyzik (page 699) which uses $$_2F_1\left(\frac{v+u}{2},\frac{v+u+1}{2};v+1;-\frac{\beta^2}{\alpha^2}\right)$$ for $u$ = 2. Might need to find an expression for a higher value of $u$ later. $\endgroup$ – geometrikal May 19 '12 at 12:58
  • $\begingroup$ @geometrikal, I have added a bit on calculating that expression for higher values of $u$. Hope it helps. $\endgroup$ – Antonio Vargas May 19 '12 at 17:31
  • $\begingroup$ Thanks for the detailed answer, and the code for Mathematica. That will help a lot. $\endgroup$ – geometrikal May 20 '12 at 3:45

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.


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.

  • $\begingroup$ Nice work on figuring out a proper derivation! $\endgroup$ – Antonio Vargas May 20 '12 at 6:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.