Can the Difference of Two Hypergeometric Functions be Expressed as One Term?

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The discussion focuses on simplifying the difference between two hypergeometric functions, specifically _2F_1(1/2, b; 3/2; -x^2) and _2F_1(3/2, b+1; 5/2; -x^2) for a given range of b values. The user seeks to express this difference in a simpler form, ideally not involving hypergeometric functions. Using Maple, they derive a series representation that leads to a transformation involving a _3F_2 function. Additionally, the user questions the reliability of Maple's analytical solutions compared to Mathematica and explores the possibility of expressing the result in terms of exponential or trigonometric functions. The conversation emphasizes the complexities and potential errors in analytic solutions for hypergeometric functions.
BCox
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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?
 
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Using Maple, I get
<br /> {{}_2F_1(1/2,b;\,3/2;\,-{x}^{2})}-{{}_2F_1(3/2,b+1;\,5/2;\,-{x}^{2})} =<br /> \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) }}<br />
<br /> =\sum _{k=0}^{\infty }-{\frac { \left( -1 \right) ^{k}{x}^{2\,k}<br /> \left( 4\,b+6\,k+3 \right) \Gamma \left( b+k \right) }{ \left( 2\,k+<br /> 3 \right) \left( 2\,k+1 \right) \Gamma \left( b+1 \right) \Gamma <br /> \left( k \right) }}<br /> =1/15\,{x}^{2} \left( 4\,b+9 \right) <br /> {{}_3F_2(3/2,b+1,2/3\,b+5/2;\,7/2,2/3\,b+3/2;\,-{x}^{2})}<br />
 
g_edgar said:
Using Maple, I get
<br /> {{}_2F_1(1/2,b;\,3/2;\,-{x}^{2})}-{{}_2F_1(3/2,b+1;\,5/2;\,-{x}^{2})} =<br /> \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) }}<br />
<br /> =\sum _{k=0}^{\infty }-{\frac { \left( -1 \right) ^{k}{x}^{2\,k}<br /> \left( 4\,b+6\,k+3 \right) \Gamma \left( b+k \right) }{ \left( 2\,k+<br /> 3 \right) \left( 2\,k+1 \right) \Gamma \left( b+1 \right) \Gamma <br /> \left( k \right) }}<br /> =1/15\,{x}^{2} \left( 4\,b+9 \right) <br /> {{}_3F_2(3/2,b+1,2/3\,b+5/2;\,7/2,2/3\,b+3/2;\,-{x}^{2})}<br />


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?
 
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?
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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