If I try to code numerics I (sometimes) first do it in Maple and then let it generate the C (or VB) code. The advantage (besides avoiding some coding errors) is mainly testing, especially 'extreme' situations, where Maple gives me a precise answer. Since mostly I want some DLLs that interfacing can be done with almost no extra coding. Now it is not always a good idea just setting Maple to 16 Digits (as it works a different way), but to force it to use hardware floatings. So I want to apply 'evalhf' to procedures, to have similar error behaviour. Even if Maple can not generate C++ code (to have complex floats) it is often not to difficult to massage the result into a desired form (where an additional text editor is quite useful).

Georgios, you missed a bracket, it is ( -2.5 ) ^ (1.5), the sign is within the bracket ... and yes, the "small" part is a rounding error through translation The point is: i am just developing some code and want it to be executed by evalhf and wondered, why evalhf can not handle complex powers - but it should Axel

A little simplification, but no solution, may be it helps (hopefully not
typo): x^6+c*x^5+b*x^4+a*b*x^2+c*a^2*x+a^3, where degree=3 is missing and
a and x add to degree 3 (hope it is understandable), but can not see more,
it still looks quite general.

To get that I used
 
coeftayl(R(alpha), alpha=0,4) =  3*Delta; solve(%,c);
0=eval(R(alpha), c= %):
%/3;
 
subs(delta=-C/2,%):
subs(mu=sqrt(m),%): 
 
#indets(rhs(%), atomic): indets(%);
subs(alpha=x,%):
subs(m=a,%):
subs(Delta=b, %):
subs(C=c,%): 
sort(%,x):
 
S:=unapply(rhs(%), x); 

                     6    5      4          2        2    3
          S := x -> x  + x  c + x  b + a b x  + c x a  + a

thank you for correcting - and sorry for my sloppy posting

every polynomial has even number of real solutions (counted with multiplicity), thsi is just another formulation for the fundamental theorem of algebra i would try to set up a general solution in terms of roots and check, what the given coefficients mean for that

restart;
assume(k::nonnegint); getassumptions(k);
Sum(pochhammer(a,n)*pochhammer(b,n)/pochhammer(c,n)/n!*z^n,n = k .. infinity) 
value(%);
            {k::AndProp(integer, RealRange(0, infinity))}


            infinity
             -----                                      n
              \      pochhammer(a, n) pochhammer(b, n) z
               )     ------------------------------------
              /              pochhammer(c, n) n!
             -----
             n = k


                      hypergeom([a, b], [c], z)

This is only true for k=0 by the definiton of the according hypergeometric series.

restart;
assume(k::nonnegint); getassumptions(k);
Sum(pochhammer(a,n)*pochhammer(b,n)/pochhammer(c,n)/n!*z^n,n = k .. infinity) 
value(%);
            {k::AndProp(integer, RealRange(0, infinity))}


            infinity
             -----                                      n
              \      pochhammer(a, n) pochhammer(b, n) z
               )     ------------------------------------
              /              pochhammer(c, n) n!
             -----
             n = k


                      hypergeom([a, b], [c], z)

This is only true for k=0 by the definiton of the according hypergeometric series.

s0 := [[b = 9., c = 10.]]:      # let's say that was your solution (without := of course
theList:=op(s0);                # make it a list

                     theList := [b = 9., c = 10.]

[c-b-a = 0, c-(1+k)*b+k*a = 0]; # your problem, you can enter it this way for 'solve'
eval(%, theList);               # let Maple evaluate your problem, check the help on 'eval'

                  [1. - a = 0, 1. - 9. k + k a = 0]


Just a note: do not use C coding style like assigning within a command,
you may run into troubles ...

Is it towards what you where looking for?

                                 \   |     /
                           \\     \  ||   /    //
                            \\\   \\    //  ///
                             \\\ ####### ///
                         \\   ###       ###   //
                          \\ ##           ##//
                       --   ##             ##   --
                        --  ##  Platsch!!  ## --
                          // ##           ##\\
                         //   ###       ###   \\
                            ///  ####### \\\
                           ///   //    \\  \\\
                         //      /  | |  \    \\
                                /    |    \

                  Du wurdest gerade von einem Cyber Schneeball getroffen!
                  Der erste in diesem Jahr!
                  Damit startet die...
                  Schneeballschlacht '2006!!!!
                  Eine Regel hat das Spiel...
                  Du darfst nie jemanden treffen, der Dich bereits
                  getroffen hat.
                  Also...los gehts und treffe soviele Leute, wie
                  Du nur kannst, bevor sie Dich treffen!!

I stumbled around and: It is an exercise in Whittaker & Watson, p.123 (the location suggestes to attack it through residues calculus), also contained as special case in a paper "Some remarkable properties of sinc and related integrals" by Borwein & Borwein (W&W is p.9). I have not tried if the case n->infinity could be read off their paper. Also there are several references at Mathworld, especially a recursive way by Zimmermann (without prove of course) to be found under "sinc". Download 102_(Borwein et al)Sinc integrals analytically found.zip

I stumbled around and: It is an exercise in Whittaker & Watson, p.123 (the location suggestes to attack it through residues calculus), also contained as special case in a paper "Some remarkable properties of sinc and related integrals" by Borwein & Borwein (W&W is p.9). I have not tried if the case n->infinity could be read off their paper. Also there are several references at Mathworld, especially a recursive way by Zimmermann (without prove of course) to be found under "sinc". Download 102_(Borwein et al)Sinc integrals analytically found.zip

Thanks - i will again look into Vidunas paper, may be I get some idea (the solution I posted was some 'russian way', I am still looking for s.th. similar for the other formula I want to kill). The 2nd, Takayama, is nice: never thought that i will meet again dualizing complexes again and that in the context of intersection (co)homology combined with hypergeometric functions :-)

Thanks - i will again look into Vidunas paper, may be I get some idea (the solution I posted was some 'russian way', I am still looking for s.th. similar for the other formula I want to kill). The 2nd, Takayama, is nice: never thought that i will meet again dualizing complexes again and that in the context of intersection (co)homology combined with hypergeometric functions :-)

i could agree ... and understand the admin - here a blog is only named as such, but it actually is not personal place here, so you just can hope for some etiquette

Thanks for this nice formulated answer :-) Hm, in this special case I am looking for public infos, which would make it a bit difficult here ... More or less I am aware of the links, it is the specific, exceptional case I am looking for.