#2481 hgfred([-1/2,a+1],[a+2],x);

closed
nobody
None
5
2012-11-10
2012-10-19
No

(%i1) hgfred([-1/2,a+1],[a+2],x);
expt: undefined: 0 to a negative exponent.

Discussion

  • Raymond Toy

    Raymond Toy - 2012-10-31

    Maxima is using the A&S 15.2.6 to derive the value from F(-1/2, a+1; a+1; x). But that's equal to sqrt(1-x) and 15.2.6 is not valid in that case. Perhaps 15.2.16 should have been used instead?

     
  • Raymond Toy

    Raymond Toy - 2012-11-02

    A&S 15.2.16 doesn't help, but A&S 15.2.14 gives an answer because hgfred([a+1,b],[c],z) and hgfred([a,b+1],[c],z) can be expressed in terms of associated Legendre function and an elementary function.

    Assuming I did things correctly,

    hgfred([-1/2,a+1],[a+2],x) =
    (2^(a+1)*assoc_legendre_p(a,-a-1,sqrt(1-z))*gamma(a+2)*z^(-a/2-1/2)
    +(2*a+2)*sqrt(1-z))
    /(2*a+3)

    I don't have any algorithm to determine that unfortunately.

    Perhaps it's best that maxima just returns the 2F1 function itself instead of generating an error.

     
  • Raymond Toy

    Raymond Toy - 2012-11-10
    • status: open --> closed
     
  • Raymond Toy

    Raymond Toy - 2012-11-10

    Fixed in git. The formula is not applied if a+1/2 or b+1/2 is zero which would cause a division by zero.

     

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