Exercise 2.3.1

Examples 2.3.1 and 2.3.2 showed that the roots of y 3 + 41 y 2 + 138 y + 125 are the cubes of the roots of y 3 + 2 y 2 3 y + 5 . Verify this numerically.

Answers

Proof. We repeat Examples 2.3.1 and 2.3.2 with Sage :

We build the Groebner basis of the ideal e 1 y 1 , e 2 y 2 , e 3 y 3 , where e 1 , e 2 , e 3 are the elementary symmetric polynomials in x 0 , x 1 , x 2 :

e = SymmetricFunctions(QQ).e()
e1, e2, e3 = e([1]).expand(3),e([2]).expand(3),e([3]).expand(3)
R.<x0,x1,x2,y1,y2,y3> = PolynomialRing(QQ, order = ’degrevlex’)
J = R.ideal(e1-y1, e2-y2, e3-y3)
G = J.groebner_basis()

We compute the coefficients of f = ( x x 0 3 ) ( x x 1 3 ) ( x x 2 3 ) as polynomials in x 1 , x 2 , x 3 :

f = (x-x0^3) * (x-x1^3) * (x-x2^3)
coeffs = f.coefficients(x, sparse = False)
coeffs = map(lambda c : R(c), coeffs)
coeffs

[ x 0 3 x 1 3 x 2 3 , x 0 3 x 1 3 + x 0 3 x 2 3 + x 1 3 x 2 3 , x 0 3 x 1 3 x 2 3 , 1 ]

The same coefficients as polynomials in σ 1 , σ 2 , σ 3 :

var(’sigma_1,sigma_2,sigma_3’)
ncoeffs = [c.reduce(G) for c in coeffs]
nncoeffs = [c.subs(y1 = sigma_1,y2 = sigma_2,y3 = sigma_3) for c in ncoeffs]
nncoeffs

[ σ 3 3 , σ 2 3 3 σ 1 σ 2 σ 3 + 3 σ 3 2 , σ 1 3 + 3 σ 1 σ 2 3 σ 3 , 1 ]

We apply the substitution σ 1 2 , σ 2 3 , σ 3 5 and compute the polynomial p whose roots are α 1 3 , α 2 3 , α 3 3 , where α 1 , α 2 , α 3 are the roots of y 3 + 2 y 2 3 y + 5 .

nncoeffs = [c.subs(sigma_1 = -2, sigma_2 = -3, sigma_3 = -5) for c in nncoeffs]
p = sum(nncoeffs[i]*y^i for i in range(1+f.degree(x)))
p

y 3 + 41 y 2 + 138 y + 125

Numerical verification:

S.<y>  = PolynomialRing(ComplexField(prec = 40))
[c[0] for c in S(p).roots()]

[ 37 . 399476110 , 1.8002619448 0.31835473525 i , 1.8002619448 + 0.31835473525 i ]

q = y^3+2*y^2-3*y+5
l = [c[0]^3 for c in q.roots()]
l

[ 37 . 399476110 , 1.8002619448 0.31835473525 i , 1.8002619448 + 0.31835473525 i ]

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2022-07-19 00:00
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