Re: A day of a CAS super-hero

From: G. A. Edgar (edgar_at_math.ohio-state.edu.invalid)
Date: 03/23/05 

Date: Wed, 23 Mar 2005 13:53:58 -0500

In article <iU20e.1411$hg.368@news01.roc.ny>, Alec Mihailovs
<alec@mihailovs.com> wrote:

> "G. A. Edgar" <edgar@math.ohio-state.edu.invalid> wrote in message
> news:220320051033188835%edgar@math.ohio-state.edu.invalid...
> > In article <nJM_d.972$up.628@news02.roc.ny>, Alec Mihailovs
> > <alec@mihailovs.com> wrote:
> >
> >> "Vladimir Bondarenko" <vb@cybertester.com> wrote in message
> >> news:1111143620.659214.39110@z14g2000cwz.googlegroups.com...
> >> >
> >> > 10. Input evalf(hypergeom([1/2, -3/2], [], 1));
> >>
> >> That can be evaluated in Maple 9.52 as
> >>
> >> exp(-1/2)/2*(BesselI(0,1/2)*sqrt(Pi)-BesselK(0,1/2)/sqrt(Pi)*I);
> >>
> >> evalf(%,20);
> >>
> >> 0.57164762453873852155 - 0.15816730823096183197 I
> >>
> >> Alec Mihailovs
> >> http://math.tntech.edu/alec/
> >>
> >>
> >>
> >>
> >
> > OK, using Borel summability, I got
> >
> > 1+Int(((t-1)*EllipticK(sqrt(t))+(1-2*t)*EllipticE(sqrt(t)))*exp(-t)/t/Pi,
> > t=0..infinity);
> >
> > which agrees numerically with the above. Is there a way to evaluate
> > this integral as those Bessel functions?
>
> I got slightly different formula using Borel summation. The Borel transform
> of hypergeom([1/2,-3/2],[],x) is hypergeom([1/2,-3/2],[1],x) and Borel
> summation formula gives
>
> int(hypergeom([1/2,-3/2],[1],x)*exp(-x),x=0..infinity);
> infinity
> /
> |
> | hypergeom([-3/2, 1/2], [1], x) exp(-x) dx
> |
> /
> 0
> convert(%,StandardFunctions);
>
> infinity
> /
> |
> |
> |
> /
> 0
>
> //2 x \ 1/2 / 4 x \ 1/2 \
> ||--- - 2/3| EllipticK(x ) |- --- + 8/3| EllipticE(x )|
> |\ 3 / \ 3 / |
> |--------------------------- + -----------------------------|
> \ Pi Pi /
>
> exp(-x) dx
>
> that numerically evaluates to the same number. Maple seems to be not able to
> convert that to Bessel functions expression.
>
> Alec Mihailovs
> http://math.tntech.edu/alec/
>
>

It's interesting to see two different ways of doing this as Borel
summation. Since we need a series of the form z*f(z),
and ours isn't, you did z*F(z) and when you put in z=1 we get F(1),
while I wrote it as F(z) = 1+z*G(z) and did the transform on the
part without the 1. Mine comes out as:

> 1-3/4*int(hypergeom([-1/2,3/2],[2],t)*exp(-t),t=0..infinity);

    3 / /[-1 3] \ \
1 - - int|hypergeom|[--, -], [2], t| exp(-t), t = 0 .. infinity|
    4 \ \[2 2] / /

> convert(%,StandardFunctions);

          // / (1/2) \ / (1/2)\ / (1/2)\
     3 || 4 \t - 1/ \1 + t / EllipticK\t /
 1 - - int||- ---------------------------------------------
     4 \\ 3 t Pi

                          / (1/2)\\ \
      (-4 + 8 t) EllipticE\t /| |
    + ----------------------------| exp(-t), t = 0 .. infinity|
                 3 t Pi / / 

-- 
G. A. Edgar                               http://www.math.ohio-state.edu/~edgar/