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Multiplying Roots of a Polynomial

asked 2019-01-17 05:29:29 -0500

LukeC93 gravatar image

Hello - I'm new to SAGE, so trying to get to grips with the basics!

I have a polynomial $f = x^{2}((x+\frac{1}{x})^{2} - 1) $. I would like to be able to do to things:

1) See $f$ in the form $\beta(x - \alpha_1)(x-\alpha_2)(x-\alpha_3)(x-\alpha_4)$

2) Know the value of $\beta (\prod_i \alpha_i)$

Using

f.factor

f.roots

Does not give me the linear factorisation, nor can I see a simple way of getting the product of the roots.

In the long term I'm hoping to do this for polynomials of larger degree, so any advice would be great!

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answered 2019-01-23 12:24:03 -0500

BrentBaccala gravatar image

updated 2019-01-23 12:24:52 -0500

You can also get exact results by using the Algebraic Field QQbar, which avoids the need of manually constructing a splitting field (like rburing shows in his answer).

In Sage 8.4, I'd also suggest setting QQbar's display option to radical instead of decimal (the default). Try both and see the difference; it's the simplest way I know how to explain it!

sage: QQbar.options.display_format = 'radical'
sage: R.<x> = PolynomialRing(QQbar)
sage: f = x^2*((x+1/x)^2-1)
sage: f.factor()
(x - 1/2*I*sqrt(3) - 1/2) * (x + 1/2*I*sqrt(3) - 1/2) * (x - 1/2*I*sqrt(3) + 1/2) * (x + 1/2*I*sqrt(3) + 1/2)
sage: mul([p.constant_coefficient() for p,m in f.factor()])
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answered 2019-01-17 15:18:49 -0500

Emmanuel Charpentier gravatar image

Let's see...

sage: f=x^2*((x+1/x)^2-1)

The roots are :

sage: f.roots(multiplicities=False)
[-sqrt(1/2*I*sqrt(3) - 1/2),
 sqrt(1/2*I*sqrt(3) - 1/2),
 -sqrt(-1/2*I*sqrt(3) - 1/2),
 sqrt(-1/2*I*sqrt(3) - 1/2)]

Therefore, the factorized polynom is :

sage: prod([x-u for u in f.roots(multiplicities=False)])
(x + sqrt(1/2*I*sqrt(3) - 1/2))*(x - sqrt(1/2*I*sqrt(3) - 1/2))*(x + sqrt(-1/2*I*sqrt(3) - 1/2))*(x - sqrt(-1/2*I*sqrt(3) - 1/2))

And the roots' products is:

sage: prod(f.roots(multiplicities=False)).expand()
1

Whereas moving to a polynomial ring, more specialized, is often useful, it is not necessary here...

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Comments

I would argue polynomial rings are simpler than the symbolic ring. It depends on your point of view. In any case, it is good to see both approaches.

rburing gravatar imagerburing ( 2019-01-17 16:39:49 -0500 )edit

Thank you - this is perfect for what I'm after!

LukeC93 gravatar imageLukeC93 ( 2019-01-18 03:11:45 -0500 )edit

For fun, you can check what happens for f = x^5 - x - 1 in both answers.

rburing gravatar imagerburing ( 2019-01-18 03:36:23 -0500 )edit
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answered 2019-01-17 11:03:43 -0500

rburing gravatar image

updated 2019-01-17 11:11:42 -0500

Do you want exact results or approximations? For exact results, you can define

R.<x> = PolynomialRing(QQ)
f = R(x^2*((x+1/x)^2-1))
K.<a> = f.splitting_field()

so you can do:

sage: f.change_ring(K).factor()
(x - a) * (x + a - 1) * (x - a + 1) * (x + a)
sage: f.change_ring(K).roots()
[(a, 1), (-a + 1, 1), (a - 1, 1), (-a, 1)]
sage: beta = f.leading_coefficient()
sage: myprod = beta*prod(r[0]^r[1] for r in f.change_ring(K).roots()); myprod
1

By Vieta's formulas what you get is just $(-1)^n f_0$ where $n = \deg(f)$ and $f_0$ is the constant coefficient of $f$.

sage: myprod == (-1)^f.degree()*f.constant_coefficient()
True

If you are wondering what $a$ is in the discussion above: it is a root of

sage: a.minpoly()
x^2 - x + 1

As for numerics, you can do e.g.:

sage: f.change_ring(QQbar).roots()
[(-0.500000000000000? - 0.866025403784439?*I, 1),
(-0.500000000000000? + 0.866025403784439?*I, 1),
 (0.500000000000000? - 0.866025403784439?*I, 1),
 (0.500000000000000? + 0.866025403784439?*I, 1)]
sage: beta*prod(r[0]^r[1] for r in f.change_ring(QQbar).roots())
1.000000000000000? + 0.?e-18*I
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Asked: 2019-01-17 05:29:29 -0500

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Last updated: Jan 23