Why, in any system of equations, does Sage give repeated solutions only once?

georgeafr
13 ●2

slelievre
17789 ●22 ●163 ●352 http://carva.org/samue...

For example, the following system of equations has two roots equal to one:

var('a,b,c,d,e,f,g,x')
sols = solve([(a - g) + 2,  - (a*g - b) - 1, - (b*g - c), - (c*g - d) + 10, - (d*g - e) - 22, - f*g - 2, - (e*g - f) + 14], a,b,c,d,e,f,g, solution_dict=true)
[{g: sol[Eg]} for sol in sols]

Six, instead of seven, solutions are provided by Sage, because repeated root 1 is presented only once:

[{g: -1.825113562621674},
 {g: -0.04961250951921104 - 1.781741795058668*I},
 {g: -0.04961250951921104 + 1.781741795058668*I},
 {g: 0.2000320307495195},
 {g: 1.724306472919419},
 {g: 1}]

If you don't know which is the repeated root and you want to know which is it, how to do it, because with the use of multiplicities, or f.root, you cannot know

Comments

Solving via SR's solve isn't exact : given :

var('a,b,c,d,e,f,g,x')
Sys=[(a - g) + 2,  - (a*g - b) - 1, - (b*g - c), - (c*g - d) + 10, - (d*g - e) - 22, - f*g - 2, - (e*g - f) + 14]
vars=list(set(flatten([u.variables() for u in Sys])))
sols=solve(Sys, vars, solution_dict=True)

The solutions do not check :

sage: [[u.subs(s).is_zero() for u in Sys] for s in sols]
[[False, False, False, False, False, False, False],
 [False, False, False, False, False, False, False],
 [False, False, False, False, False, False, False],
 [False, False, False, False, False, False, False],
 [False, False, False, False, False, False, False],
 [True, True, True, True, True, True, True]]

Emmanuel Charpentier ( 2 years ago )

A better (checkable) solution is to use the ring of polynomials of QQbar :

Vars=[str(u).upper() for u in vars]   #  My damn laziness...
R1=PolynomialRing(QQbar, Vars)
R1.inject_variables()
DD=dict(zip(vars, R1.gens()))
PSys=[R1(u.subs(DD)) for u in Sys]  # Even worse laziness...
J1=R1.ideal(*PSys)
V1=J1.variety()

Then the results can be checked :

sage: [[u.subs(s).is_zero() for u in PSys] for s in V1]
[[True, True, True, True, True, True, True],
 [True, True, True, True, True, True, True],
 [True, True, True, True, True, True, True],
 [True, True, True, True, True, True, True],
 [True, True, True, True, True, True, True],
 [True, True, True, True, True, True, True]]

But I have no idea about your multiplicity problem for now...

Emmanuel Charpentier ( 2 years ago )

Thank you Emmanuel, this is a good response to my question, but really what I am looking for is a solution like that of the answer below

georgeafr ( 2 years ago )

@georgeafr : @rburing is right. He was faster|smarter than me... My remark was not (even tentatively) an answer to your question...

Emmanuel Charpentier ( 2 years ago )

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answered 2 years ago

rburing

10964 ●4 ●80 ●219 https://www.rburing.nl/

updated 2 years ago

Here are the roots with multiplicities, using a different solution method:

sage: R.<a,b,c,d,e,f,g,x> = QQ[]
sage: I = R.ideal([(a - g) + 2,  - (a*g - b) - 1, - (b*g - c), - (c*g - d) + 10, - (d*g - e) - 22, - f*g - 2, - (e*g - f) + 14])
sage: F_g = I.elimination_ideal([a,b,c,d,e,f,x]).gen(0).polynomial(g); F_g
g^7 - 2*g^6 + g^5 - 10*g^3 + 22*g^2 - 14*g + 2
sage: F_g.roots(QQbar)
[(-1.825113480833768?, 1),
 (0.2000320256287094?, 1),
 (1, 2),
 (1.724306474243468?, 1),
 (-0.04961250951920454? - 1.781741795058673?*I, 1),
 (-0.04961250951920454? + 1.781741795058673?*I, 1)]

Indeed, the root 1 has multiplicity two.

link

Comments

With the same question: is it possible obtain a solution something like previous one if I starts directly from g^7 - 2g^6 + g^5 - 10g^3 + 22g^2 - 14g + 2?

georgeafr ( 2 years ago )

Thanks a lot @rburing, good response. It satisfies my search.

georgeafr ( 2 years ago )

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Why, in any system of equations, does Sage give repeated solutions only once?

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