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

AI Thread Summary
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
Messages
16
Reaction score
0
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?
 
Mathematics news on Phys.org
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?
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top