On Thu, Nov 23, 2023 at 12:11:01PM +0800, Qian Yun wrote:
> >
> > Trouble here is with A(u,v) and B(u,v). Already when p is of degree
> > 4 with Galois group S(4) we have trouble: beside right solutions
> > there are spurious ones and it is hard to separate good ones from
> > spurious one.
> >
> > Consider
> >
> > p := 10*(a^4 + 1) + a
> >
> > and
> >
> > rootSum(a*log(x - a), p::SUP(EXPR(INT)))
> >
> > How you want to find right u, v? The approach I described
> > works around difficulty with u, v by working with coefficients
> > of p. We still have difficulty with final result, but it is
> > less problematic: we need to choose positive real root of
> > a polyniomial. Note that the polynomial is minimal polynomial
> > for the answer, so we need to choose this root directly or
> > indirectly. But we can avoid trouble with u and v.
> >
>
> In this particular case, B(u, v) = - v, when v is positive, B is
> negative. So for this case, limit when x is +infinity is
> rootSum(-%pi*v, V(v) where v is real and positive)
>
> V(v) can be achieved by resultant:
>
> p := 10*(a^4 + 1) + a
> puv := eval(p,a=u+%i*v)
> P := real puv
> Q := imag puv
> V := resultant(univariate(P, second kernels P), univariate(Q, second kernels
> P))::POLY INT
>
> 40960000000*v^16+20480000000*v^12+(-166400000)*v^10+(-17920000000)*v^8+(-86400000)*v^6+2559973000*v^4
>
> countRealRoots V returns 5, which is a 0 (ignore) and 2 positive v and
> 2 negative v (come in pairs), which is the correct number of v.
>
> This does not scale for more complex B(u, v) though.
The trouble is that AFAICS we have _no_ reasonable expression for
real roots. When V factors we can try to use countRealRoots to
find factors responsible for real roots. However, above V has
irreducible factor of degree 12 and need to sum over its real roots.
If wee look at u we get polynomial of degree 6, which may be easier
to handle (but then we need square roots to get v). In approach
that I sketched we get polynomial of degree 6 for final result.
BTW: Q is always divisible by v and v=0 can not give complex root,
so we can divide by it in "interesting" cases (that is when there
is no real root).
--
Waldek Hebisch
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