Minimal Polynomial – Minimal Polynomial of cos(?/n)

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I know that $\cos(\pi/n)$ is a root of the Chebyshev polynomial $(T_n + 1)$, in fact it is the largest root of that polynomial, but often that polynomial factors. For example, if $n = 2 k$ then $\cos(\pi/n)$ is the largest root of $T_k$, which is a polynomial of lower degree, and if $n = 3$ then $\cos(\pi/n)$ is a root of $2 x – 1$, again lower degree than $T_3 + 1$.

How can I compute, for a given $n$, a polynomial in $\mathbb{Q}[x]$ of minimal degree that $\cos(\pi/n)$ is a root of?

Best Answer

The minimal polynomial of $\cos(2\pi/n)$ (by William Watkins and Joel Zeitlin, The American Mathematical Monthly Vol. 100, No. 5 (May, 1993), pp. 471-474) has full clarity on this matter (just take their result for even $n$ to resolve your case).

Related Solutions

Chebyshev Polynomials – Intuitive Explanation of Extremal Property

Well, let's try to avoid the hat.

Consider the dual (and obviously equivalent) problem: find the polynomial $p(x):[-1,1]\rightarrow [-1,1]$ of degree $n$ with the greatest possible leading coefficient. We have some information on values of $p$, and need something about its coefficient. Let's try Lagrange's interpolation. Take some $n+1$ values $t_1 < t_2 < \dots < t_{n+1}$ from $[-1,1]$ and write down (for $u(x)=(x-t_1)\dots(x-t_{n+1})$) the formula $$ p(x)=\sum p(t_i) \frac{u(x)/(x-t_i)}{u'(t_i)}. $$ Then take a look on coefficient of $x^n$. It equals $$ \sum \frac{p(t_i)}{u'(t_i)}. $$ We know that $|p(t_i)|\leq 1$, so the leading coefficient does not exceed $ \sum 1/|u'(t_i)|. $ Ok, when does equality occur? The answer is: $p$ should take values $(-1)^{n-i+1}$ in $t_i$. That is, we have to find a polynomial of degree $n$ with $n+1$ extremal values $\pm 1$ on $[-1,1]$. This may hold only if $t_1=-1$, $t_{n+1}=1$, and $t_2$, $\dots$, $t_n$ are roots of $p'$. So, $1-p^2(x)$ should be divisible by $(1-x^2)p'(x)$. Hereon the trigonometic substitution $x=\cos t$, $p=\cos f$ is very natural, as we know that $1-f^2$ is divisible by $f'$ for $f=\cos$. So we invent Chebyshev's polynomials.

Also, it is seen from Lagrange formula that they are extremal in many other problems with restrictions $|p(x)|\leq 1$ on $[-1,1]$. For example, the value in each specific point $x_0>1$ is maximized also for Chebyshev polynomial, it is proved by exactly the same way.

[Math] Computation of a minimal polynomial

To compute the minimal polynomial of integer multiple of an algebraic integer is easy, so the only thing you need for linear combinations is the minimal polynomials of sums. Now, note that if $A$ is the companion matrix of $\alpha$ and $B$ is the companion matrix of $\beta,$ then $A\otimes I + I\otimes B$ a companion matrix of $\alpha + \beta.$ (Of course, $A\otimes B$ is a companion matrix of $\alpha \beta.$)

Mathematica code:

frobeniusCompanion[poly_, x_] /; PolynomialQ[poly, x] := 
 Module[{n = Exponent[poly, x], coef}, 
  coef = CoefficientList[poly, x]; coef = -Most[coef]/Last[coef];
  SparseArray[{{1, j_} :> coef[[-j]], Band[{2, 1}] -> 1}, {n, n}]]


AA= frobeniusCompanion[x^5 -x -1, x]
CC = KroneckerProduct[AA, IdentityMatrix[5]] + KroneckerProduct[IdentityMatrix[5], AA]
 cp = CharacteristicPolynomial[Normal[CC], x]
 Factor[cp]

gives $$-\left(x^5-16 x-32\right) \left(x^{10}+3 x^6+11 x^5-4 x^2+4 x-1\right)^2$$

Best Answer

Related Solutions

Chebyshev Polynomials – Intuitive Explanation of Extremal Property

[Math] Computation of a minimal polynomial

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