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Let $P(x)$ be a 3rd-degree polynomial with integer coefficients, one of whose roots is $\cos(\pi/7)$. Compute $\frac{P(1)}{P(-1)}$

I saw this question in a contest math problem, and I know that it has something to do with Chebeskev(? spelling) Polynomials, but I have no idea what those are.

Teoc
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    Chebyshev polynomials. For example, the seventh Chebyshev polynomial $T_7$ satisfies $T_7(\cos x) = \cos 7x$. Whoever mentioned Chebyshev polynomials was probably suggesting that you calculate $\cos (7 \cdot \pi/7)$ in two different ways. – Keith Jun 30 '15 at 02:11
  • @Keith: I think your comment should be an answer. – user21820 Jun 30 '15 at 02:13
  • @Keith So $T_7(\cos \frac{\pi}{7})=\cos(\pi)=-1$, but how does that help me(It is not a root)? (Also $T_7$ is not 3rd degree?) – Teoc Jun 30 '15 at 02:15
  • @VladimirLenin You can factor $T_7(x) + 1$. I'm not saying this is the best way. Do you know about complex numbers? – Keith Jun 30 '15 at 02:19
  • @Keith Yes, but how do you find $T_7$? – Teoc Jun 30 '15 at 02:21
  • Use either trigonometric formulas to find $\cos 7x$ in terms of $\cos x$ or use De Moivre's formula. – Keith Jun 30 '15 at 02:22
  • @keith So $\Re{e^{7ix}}=\Re({e^{ix}}^7)$, which is tedious to compute. Is there a better way? – Teoc Jun 30 '15 at 02:24
  • There are recurrence relations to find the Chebyshev polynomials, but I don't know a faster way. – Keith Jun 30 '15 at 02:46
  • @VladimirLenin,http://math.stackexchange.com/questions/638874/factor-z7-1-into-linear-and-quadratic-factors-and-prove-that-cos-pi-7-c/638896#638896 shows that $$P(x)=8x^3-4x^2-4x+1$$ – lab bhattacharjee Jun 30 '15 at 08:16

3 Answers3

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I will use a method without Chebyshev polynomials.

Let $a = e^{\pi i/7}$. Let $b = \cos \pi/7 = (a + 1/a)/2$.

We have $0 = a^7 + 1 = (a + 1)(a^6 - a^5 + a^4 - a^3 + a^2 - a + 1)$. Since $a, a + 1 \ne 0$, we can divide by $(a + 1)a^3$ to get $$a^3 + \frac{1}{a^3} - a^2 - \frac{1}{a^2} + a + \frac{1}{a} - 1 = 0,$$ which can be rewritten as $$8b^3 - 4b^2 - 4b + 1 = 0.$$

The polynomial $P(b)$ on the left can easily be checked to have no rational roots, by the rational root theorem. Since it is of degree three, it cannot be factored any further using rational coefficients. Therefore any polynomial $P$ satisfying the conditions of the problem must be a constant multiple of this polynomial, hence the ratio $P(1)/P(-1) = -1/7$ is independent of the polynomial chosen.

Keith
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The quickest way is to find the polynomial $P$.

Let $\theta = \frac{\pi}{7}$. We know that $4\theta + 3\theta = \pi$ so $$ \cos(4\theta) = - \cos(3\theta)$$ .

Let $x = \cos(\theta)$, then the above equation turns to $$ 2(2x^2-1) -1 = -(4x^3 - 3x)$$ $$ 8x^4+4x^3-8x^2-3x+1=0$$ Consider the equation $\cos(4\alpha) = - \cos(3\alpha)$. We know that $\alpha = 0$ is a solution to this equation. So $x = \cos(0) = 1$ must be a solution to $$ 8x^4+4x^3-8x^2-3x+1=0$$ Thus we know that $X - 1$ divides $ 8X^4+4X^3-8X^2-3X+1$. And since $x = \cos(\theta)$ is not the root of $X - 1$, it must be the root of the quotient polynomial.

$8X^4+4X^3-8X^2—3X+1 = (X-1)(8X^3-4X^2-4X+1)$

$x = \cos(\theta)$ is the root of the polynomial $8X^3-4X^2-4X+1$ which is of 3rd degree.

corindo
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Hint: Chebyshev polynomials is right, but you don't need to know what they are to figure this one out. All you really need to know is that $p(\cos \theta) = \cos(7 \theta) + 1$ is a $7$th degree polynomial on $\cos \theta$, and one of the roots will be attained at $\cos \theta = \cos (\pi/7)$.

This polynomial has other roots. In fact, the roots of this polynomial are $$ \cos (\pi/7), \cos(3 \pi/7), \cos(5 \pi/7), \cos \pi, \cos(9\pi/7), \cos(11 \pi/7), \cos(13 \pi/7) $$ In particular, if we take $q$ to be the monic polynomial that's a multiple of $p$, we can write $$ q(x) = (x + 1)\left[(x - \cos (\pi/7))(x - \cos (3\pi/7))(x - \cos (5\pi/7))\right]^2 $$ From this polynomial, we can factor the polynomial $$ P(x) = (x - \cos (\pi/7))(x - \cos (3\pi/7))(x - \cos (5\pi/7)) $$ From here, we simply need to "plug in" $x = \pm 1$.

If we use complex numbers, then letting $z = e^{\pi i/7}$, we can evaluate $$ P(1) = \Re [(1 - z)(1 - z^2)(1 - z^3)]=\\ \Re[1 - z - z^2 - z^3 + z^3 + z^4 + z^5 - z^6]\\ P(-1) = -\Re [(1 + z)(1 + z^2)(1 + z^3)]=\\ \Re[1 + z + z^2 + 2z^3 + z^4 + z^5 + z^6]=\\ \Re[z^3] $$

Ben Grossmann
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