On Fri, Dec 01, 2023 at 07:55:45AM +0800, Qian Yun wrote:
>
>
> On 12/1/23 05:51, Waldek Hebisch wrote:
> > On Tue, Nov 28, 2023 at 07:02:16PM +0800, Qian Yun wrote:
> > > I have examined integrals where rootSum is over degree 4, from a
> > > subset of Nasser's test.
> >
> > Martin Weltz (clicliclic) points out to example from Timofeev:
> >
> > integrate(tan(x)/(sqrt(tan(x)) - 1)^2, x)
> >
> > where root sum is over roots of
> >
> > a^5 - (1/2)*a^3 - (5/8)*a^2 + (9/64)*a - 1/64
>
> Is this integral example wrong? "internalIntegrate" shows a different
> rootSum.
This is essentially what ')trace INTEF )math' shows me. The
only difference is that I use 'a' as variable name istead of
generated symbol name.
> So it looks like that this "Riboo real functions" method is effective
> for rootSum over polynomial with FRAC(INT) coefficients.
>
> As for POLY(INT) coefficients, 'countRealRoots' fails and we have to
> do case by case treatment, and for some cases, we have to assume
> different signs (branches), giving a list of possible forms as result.
Yes, parameters in coefficients make things more tricky. We can
still do something if our sign deternination works. For example
'sign' treats 'a^2' as nonnegative and 'exp(a)' as positive.
> If you are working on the code, please continue, so our efforts will
> not be duplicated.
I have put almost all that I did into messages. As I wrote,
I think we should implement a few easy cases:
- degree 3 with one real root
- a^4 + c with c of know sign
- factorizable resolvent for degree 4
- maybe few other
AFAICS a^4 + c is most important as it appears in many examples.
I would limit this to cases when 'sign' gives us answer or
when split between cases is simple and easy.
I am not sure if you want to do them. I could do first two and
third for rational coefficients. Or maybe you want more general
code.
--
Waldek Hebisch
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