# Hypergeometric Properties

1. Sep 25, 2009

### BCox

Hello:

I need to simplify the following if possible

_2F_1(a,b;c;-x^2) - _2F_1(a+1,b+1;c+1;-x^2)

In fact, a= 1/2 and c=3/2 and b>=1. In other words, the difference above that I am interested in is more specifically

_2F_1(.5, b; 1.5; -x^2) - _2F_1(.5+1, b+1; 1.5+1; -x^2)

I know that
Arctan x = x* _2F_1(1/2, 1 ; 3/2; -x^2)
which is a special case of the first term when b=1.

But I am more interested in reducing the difference at the top for any b>=1. Can I express the difference above as one term (and hopefully not as hypergeometric fct)? And how?

2. Sep 26, 2009

### g_edgar

Using Maple, I get
${{}_2F_1(1/2,b;\,3/2;\,-{x}^{2})}-{{}_2F_1(3/2,b+1;\,5/2;\,-{x}^{2})} = \sum _{k=0}^{\infty }{\frac { \left( -1 \right) ^{k}\Gamma \left( b+k \right) {x}^{2\,k}}{\Gamma \left( b \right) \Gamma \left( k+1 \right) \left( 2\,k+1 \right) }}-\sum _{k=0}^{\infty }3\,{\frac { \left( -1 \right) ^{k}\Gamma \left( 1+b+k \right) {x}^{2\,k}}{\Gamma \left( b+1 \right) \Gamma \left( k+1 \right) \left( 2\,k+3 \right) }}$
$=\sum _{k=0}^{\infty }-{\frac { \left( -1 \right) ^{k}{x}^{2\,k} \left( 4\,b+6\,k+3 \right) \Gamma \left( b+k \right) }{ \left( 2\,k+ 3 \right) \left( 2\,k+1 \right) \Gamma \left( b+1 \right) \Gamma \left( k \right) }} =1/15\,{x}^{2} \left( 4\,b+9 \right) {{}_3F_2(3/2,b+1,2/3\,b+5/2;\,7/2,2/3\,b+3/2;\,-{x}^{2})}$

3. Sep 26, 2009

### BCox

Thank you for checking in Maple. Hmmm... two things
1. Mathematica software sometimes gives erroneous analytic solutions for integration. Do we fall into that kind of error w. Maple sometimes?
2. If the above is analytically correct, can we represent the solution as exponential or trig functions?

4. Nov 30, 2009

### BCox

We have the property such as this

Hypergeometric2F1[a,b,c,z] = (1-z)^(c-b-a)*Hypergeometric2F1[c-a,c-b,c,z]

If we wanted to keep the 2nd term of the hypergeometric function constant, what would the r.h.s. be?

Hypergeometric2F1[a,b,c,z] = something * Hypergeometric2F1[something,b,something,z]

What would the somethings be?