@Gui314 so what is the actual question? Why not post the actual task?

Example: if you have 1/((1-x)^(1/2)) given as DE then it will be real in some domain and not real in other ranges

And now I am loogged in with my usual account and can not delete or mark the above reply. Funny.

The second summand behaves smoothly and seems to positive always while the first one jumps with sign changes (poles or something like that)? So what does Matlab say, have you been able to check it?

for what is worth:

Yes, factor is a way to go - as already said. But the question is vague.

What is (possibly) meant: the radical generated by one (1) multivariate
polynomial. That heavily depends on the ring or field (x^2-2 shows that).

A generic source for that is a book like Atiyah & MacDonald.

From former life I would prefer to see it over the complex numbers (or
similar) instead of Z or Q (which is more towards number theory). Though
I become rusty on it, sigh.

Then this polynomial defines a hypersurface. As a topolopical space it
can be decomposed (common components = intersections have lower dimension)
and here.

Now these components are hypersurfaces again - they do not have lower
dimensions and generated by one polynomial each, if passing to their
reduced structure (IIRC, c.f. Kunz, Commutative Algebra -it is more or
less Krull's Hauptideal Theorem) (NB: over the Reals 'strange' things
may happen).

That way a Radical is nothing but to find the 'reduced' structure and
eliminates 'nilpotency'. Not more. But more interesting and a natural
extension is: can one re-discover the topological decomposition from an
algebraic view = polynomial ring?

The answer is "yes, in geometry over good fields". Roughly one has that
the factors are just the components and the 'powers' are the nilpotency.

If (!) "factor" works then one is fine with it.

What I miss from the Groebner approach in Maple that I can see it. For
me the help pages are not good. And if the system answer is empty it
does not help much, no?


Coming back to the initial question (in what sense ever): say you have
"the" radical (if your are able to define the ring of coefficients). Now
what do you want to do with it?

with(PolynomialIdeals):
> a*b^2 + sqrt(41)*b*c - a^3*c^2; %^7;
> p:=expand(%);
> A:=<%>;

Radical(A) gives not result, factor(p) gives (a*b^2+41^(1/2)*b*c-a^3*c^2)^7

It is a very unendearing habit to track users in person and in Germany the
company would get into serious troubles for that disgusting behaviour.

Anyway: I mostly do not log in for viewing and any credentials are deleted
after posting. Also all the tracking stuff is killed.

Auf Deutsch: es ist unglaublich, dass Marketing das auch noch stolz verkündet.

"factor" will do the same, allowing to eliminate multiplicity.

Besides that the poster does not say over which ring or field /s)he wants
to work (and that may be essential, say for units). It is a bit 'illposed'.

(x-sqrt(2))*(x+sqrt(2)); expand(%); a:=<%>;

gives <x^2 - 2> and

Radical(a);

gives a again, it does not decompose into 'geometric' components.

The link given in the initial post does not say much.

Years ago I used http://www.curveexpert.net/ , essentially it had a bundle of models and one would have selected either some specific or a group or let it run over all of them - having a final look which one to pick.

And I go with vv and ecterrab, until having seen something convincing.

Hm ... 0.^(-3.0) gives float(infinity) and 0^(-3) "Error, numeric exception: division by zero" in M 18.2 :-(

So it is the convention about 0^1 = what (or 1 / 0^1 = ...) ?

@acer  - I do not understand the reason for converting to globals - would you mind to give some insight?

@vv: Yes, sorry, I had an error.

Using Tom's suggestion I do:

f := proc(x)
            local theta:=5/1000, r_theta:=10000;
            sol(parameters = [x]);
            rhs(sol(theta)[2]);
       end proc;

plot(f - 10000, 0 .. 200, numpoints=5, smartview=false);

and then use Brent's solver to find a solution (for the above (!), so still
vv's remarks about dsolve remain)

ff:= x -> f(x) - 10000;

ff(2);
ff(200);
                -9997.99995958107590363339972359
                45607.4437934495406364817203467

x0:=zBrent(ff, 2, 200);
'ff(x0)': '%'= evalf(%);

                        steps used = 23

                199.961306129991433460330340342

                                        -22
                     ff(x0) = -1.6487 10   

mx0,ex0:=SFloatMantissa(x0), SFloatExponent(x0);

# Digits = 30
'ff(SFloat(mx0-1, ex0))': '%'= %;
'ff(SFloat(mx0, ex0))': '%'= %;
'ff(SFloat(mx0+1, ex0))': '%'= %;
                                                 -22
            ff(SFloat(mx0 - 1, ex0)) = -4.7356 10   
                                               -22
              ff(SFloat(mx0, ex0)) = -1.6487 10   
                                                -22
             ff(SFloat(mx0 + 1, ex0)) = 3.524 10   

Brent Root Finder:
http://www.mapleprimes.com/posts/38757-Brents-Method-For-Root-Finding

@vv in this case Maple seems to return a constant solution

Already a simple filter might help: only show threads if initiator has 5 points.

Might be unfair for new members (until being confirmed/upgraded), but meanwhile I do not read any 0-point-posts

Diff(f(x), x): %= convert(%, D);

                          d
                          -- f(x) = D(f)(x)
                          dx

eval(%, f=sin);
                          d
                          -- sin(x) = cos(x)
                          dx

You can write it as integral

hypergeom([2/3, 2/3, 2/3],[-1/3, 4/3],z) = 
  2/9*Pi*3^(1/2)/GAMMA(2/3)^3*Int(-(3*v*z-1)/(1-v)^(1/3)/v^(1/3)/(-v*z+1)^(5/3),v = 0 .. 1);


            /               1                                   \
            |      1/2     /                                    |
            |  Pi 3       |               3 v z - 1             |
        2/9 |-----------  |   - ----------------------------- dv|
            |          3  |            1/3  1/3           5/3   |
            |GAMMA(2/3)  /      (1 - v)    v    (-v z + 1)      |
            \              0                                    /

For z=99/100 a numerical evaluation is fast (despite the singularity) and
symbolically it gives two expressions of 2F1, good enough for numerics,
even for high precision:

hypergeom([2/3, 2/3],[4/3],99/100) - 99/100*hypergeom([5/3, 5/3],[7/3],99/100);