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

asked 2022-07-23 16:52:53 +0200

georgeafr
13 ●2

updated 2022-07-23 18:54:41 +0200

slelievre
17654 ●22 ●160 ●348 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

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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 ( 2022-07-23 21:27:40 +0200 )edit

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 ( 2022-07-23 21:31:45 +0200 )edit

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 ( 2022-07-23 22:19:08 +0200 )edit

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

Emmanuel Charpentier ( 2022-07-23 22:47:00 +0200 )edit

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answered 2022-07-23 20:47:10 +0200

rburing

10644 ●4 ●77 ●211 https://www.rburing.nl/

updated 2022-07-23 20:49:00 +0200

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.

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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 ( 2022-07-23 21:59:57 +0200 )edit

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

georgeafr ( 2022-07-24 00:38:49 +0200 )edit

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

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