From: Vladimir Bondarenko on
Hello,

erfc(3/2*(-1)^(3/4))
+erfc((1-I)*(1+sqrt(2)/4))
+(1+I)*(I*FresnelC(3/sqrt(2*Pi))
-I*FresnelC(2/sqrt(Pi)+1/sqrt(2*Pi))
-FresnelS(3/sqrt(2*Pi))
+FresnelS(2/sqrt(Pi)+1/sqrt(2*Pi))
);

?
Cheers,

Vladimir Bondarenko

Co-founder, CEO, Mathematical Director

http://www.cybertester.com/ Cyber Tester Ltd.

----------------------------------------------------------------

"We must understand that technologies
like these are the way of the future."

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http://groups.google.com/group/sci.math/msg/9f429c3ea5649df5

"...... the challenges imply that a solution is built within the
framework of the existent CAS functions & built-in definitions."

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From: Mate on
On Jul 12, 10:43 am, Vladimir Bondarenko <v...(a)cybertester.com> wrote:
> Hello,
>
>  erfc(3/2*(-1)^(3/4))
> +erfc((1-I)*(1+sqrt(2)/4))
> +(1+I)*(I*FresnelC(3/sqrt(2*Pi))
>        -I*FresnelC(2/sqrt(Pi)+1/sqrt(2*Pi))
>        -FresnelS(3/sqrt(2*Pi))
>        +FresnelS(2/sqrt(Pi)+1/sqrt(2*Pi))
>        );
>
>                       ?

evalc(simplify(value(convert(%,compose,erf,Int,exp))));

2


Mate
From: Vladimir Bondarenko on
On Jul 12, 11:37 am, Mate <mmat...(a)personal.ro> wrote:
> On Jul 12, 10:43 am, Vladimir Bondarenko <v...(a)cybertester.com> wrote:
>
> > Hello,
>
> >  erfc(3/2*(-1)^(3/4))
> > +erfc((1-I)*(1+sqrt(2)/4))
> > +(1+I)*(I*FresnelC(3/sqrt(2*Pi))
> >        -I*FresnelC(2/sqrt(Pi)+1/sqrt(2*Pi))
> >        -FresnelS(3/sqrt(2*Pi))
> >        +FresnelS(2/sqrt(Pi)+1/sqrt(2*Pi))
> >        );
>
> >                       ?
>
> evalc(simplify(value(convert(%,compose,erf,Int,exp))));
>
>                                   2
>
> Mate

Perfect!

I wonder if you'd like to try to crack Mr. Py's challenge? :)

http://groups.google.com/group/sci.math.symbolic/browse_thread/thread/c056c0c768d26b56#

(I do remember your desire for a good motivation about the
challenges; in fact, almost all of them are not the random
ones; even better, once I promised you to disclose publicly
the reason why these challenges are the links of the same
chain; now I am closer to such a position; I hope to reach
it within 12-24 months...)

Best wishes,

Vladimir
From: Mate on
On Jul 12, 12:06 pm, Vladimir Bondarenko <v...(a)cybertester.com> wrote:
> On Jul 12, 11:37 am, Mate <mmat...(a)personal.ro> wrote:
>
>
>
> > On Jul 12, 10:43 am, Vladimir Bondarenko <v...(a)cybertester.com> wrote:
>
> > > Hello,
>
> > >  erfc(3/2*(-1)^(3/4))
> > > +erfc((1-I)*(1+sqrt(2)/4))
> > > +(1+I)*(I*FresnelC(3/sqrt(2*Pi))
> > >        -I*FresnelC(2/sqrt(Pi)+1/sqrt(2*Pi))
> > >        -FresnelS(3/sqrt(2*Pi))
> > >        +FresnelS(2/sqrt(Pi)+1/sqrt(2*Pi))
> > >        );
>
> > >                       ?
>
> > evalc(simplify(value(convert(%,compose,erf,Int,exp))));
>
> >                                   2
>
> > Mate
>
> Perfect!
>

Hello Vladimir,

actually the answer should have been even shorter (without the erf
converson) but there is
a _severe bug_ in convert/Int for erfc where erfc is treated as
erfi !!!
The wrong (integral) definition appears in FunctionAdvisor for erfc.
I mention this because I know you collect such material.

Mate

From: Vladimir Bondarenko on
On Jul 12, 12:50 pm, Mate <mmat...(a)personal.ro> wrote:
> On Jul 12, 12:06 pm, Vladimir Bondarenko <v...(a)cybertester.com> wrote:
>
>
>
>
>
> > On Jul 12, 11:37 am, Mate <mmat...(a)personal.ro> wrote:
>
> > > On Jul 12, 10:43 am, Vladimir Bondarenko <v...(a)cybertester.com> wrote:
>
> > > > Hello,
>
> > > >  erfc(3/2*(-1)^(3/4))
> > > > +erfc((1-I)*(1+sqrt(2)/4))
> > > > +(1+I)*(I*FresnelC(3/sqrt(2*Pi))
> > > >        -I*FresnelC(2/sqrt(Pi)+1/sqrt(2*Pi))
> > > >        -FresnelS(3/sqrt(2*Pi))
> > > >        +FresnelS(2/sqrt(Pi)+1/sqrt(2*Pi))
> > > >        );
>
> > > >                       ?
>
> > > evalc(simplify(value(convert(%,compose,erf,Int,exp))));
>
> > >                                   2
>
> > > Mate
>
> > Perfect!
>
> Hello Vladimir,
>
> actually the answer should have been even shorter (without the erf
> converson) but there is
> a  _severe bug_  in convert/Int for erfc where erfc is treated as
> erfi !!!
> The wrong (integral) definition appears in FunctionAdvisor for erfc.
> I mention this because I know you collect such material.
>
> Mate

Hello Mate,

Thank you very much for pointing out the bug!

In fact, one of the Cyber Tester goals is to
help improving the current commercial and
free CASs via providing their teams with the
bugs of all kind.

Right now our resources are not significant
enough to test all these 20,000 functions
in Mathematica, Maple, Axiom/FriCAS, Maxima
and hybrid systems like MATHLAB, MathCAD etc

Now we focused on Rubi, and again achieved
the performance no human being could boast.

The bug you shared with us is not discovered
yet by our machinery, - so thanks again!

On the positive note, I hope to announce our
next big steps within just several months
coming.

Best wishes from algorithmic Simferopol,

Vladimir