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Topic: An independent integration test suite
Replies: 42   Last Post: Jul 25, 2013 6:09 PM

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 daly@axiom-developer.org Posts: 100 Registered: 11/9/05
Re: An independent integration test suite
Posted: Jul 24, 2013 2:31 AM

On Wednesday, July 24, 2013 2:17:57 AM UTC-4, Nasser M. Abbasi wrote:
> On 7/24/2013 12:21 AM, Albert Rich wrote:
>

> > On Tuesday, July 23, 2013 2:32:20 PM UTC-10, da...@axiom-developer.org wrote:
>
> >
>
> >> We differ on some results because Rubi (or whatever program you are using
>
> >> to create the optimal results) assumes that square roots have only a single
>
> >> positive value. Axiom does not want to make this simplification so given
>
> >> sqrt(3)*sqrt(7)*sqrt(21) - 21
>
> >> Axiom will not simplify this to zero but the Rubi test suite does.
>
> >
>
>
>

> > Surly there must be a way to tell Axiom to choose the principal
>
> > branch so sqrt(4)-2 will simplify to zero?
>
> >
>
> > Albert
>
> >
>
>
>
> I remember reading that some math people, 100 or 200 years ago,
>
> decided that sqrt (of non-negative, non-complex values) was single valued
>
> function and its value is the non-negative root (ie. principal
>
> square root).
>
>
>
> Does any one have a reference if this is true and when this
>
> change was decided on? Or may be it always was like this?
>
>
>
> I looked at wiki now
>
> http://en.wikipedia.org/wiki/Square_root
>
> and do not spot a date on this.
>
>
>
> --Nasser

Axiom implements a function called zero? from AlgebraicNumber
which can claim that the simple numeric example I posted is zero.
I tried to give a trivial example to illustrate the idea rather
than the full equation causing the problem. I post the full
equation below.

The zero? function works on Algebraic Numbers. If two algebraic
numbers have the same norm (after deleting repeated roots) then
they are certainly conjugates. However, in the more complicated
case posted below, the zero? function cannot decide that the
numerator is identically zero. Is there an algorithm to show
that it is or is not?

Tim