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

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SUMMARY

The discussion centers 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), where a=1/2 and c=3/2. The user seeks to express this difference as a single term, ideally not involving hypergeometric functions. Using Maple, the user derives a series representation and explores the possibility of expressing the solution in terms of exponential or trigonometric functions. The conversation also touches on the reliability of Maple versus Mathematica for obtaining analytic solutions.

PREREQUISITES
  • Understanding of hypergeometric functions, specifically _2F_1.
  • Familiarity with series expansions and convergence.
  • Knowledge of special functions and their properties.
  • Experience with mathematical software such as Maple or Mathematica.
NEXT STEPS
  • Research the properties of hypergeometric functions, particularly transformations and identities.
  • Learn about series convergence and how it applies to hypergeometric functions.
  • Explore the use of Maple for symbolic computation and its capabilities in handling hypergeometric functions.
  • Investigate the relationship between hypergeometric functions and trigonometric or exponential functions.
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Mathematicians, physicists, and engineers working with special functions, particularly those involved in analytical simplifications and transformations of 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?
 

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