$\bullet~$Problem: Show that $\cos\bigg(\dfrac{2\pi}{n}\bigg)$ is an algebraic number [where $n$ $\in$ $\mathbb{Z}/\{0\}$].
$\bullet~$ My approach:
Let's consider the following polynomial in $\mathbb{Z}[x]$ in recursive terms. \begin{align*} &T_{0}(x) = 1\\ &T_{1}(x) = x\\ &T_{n + 1}(x) = 2x T_{n}(x) - T_{n-1}(x) \end{align*} $\bullet~$ $\textbf{Claim:}$ The polynomial $T_{n}(x)$ for any $n$ $\in$ $\mathbb{N}$ satisfies the following \begin{align*} T_{n}(\cos(\theta)) = \cos(n\theta) \end{align*} $\bullet~$Proof: We'll use induction on $n$ for this proof.
At first, we easily obtain that for $n = 0$ the given is true.
Now for some $n = k$, we assume that \begin{align*} T_{k}(\cos(\theta)) = \cos(k\theta) \end{align*} Therefore we need to prove for $n = (k + 1)$.
Now from the recursion relation of $T_{n}(x)$ we have \begin{align*} T_{k + 1}(\cos(\theta)) & = 2 \cos(\theta)T_{k}(\cos(\theta)) - T_{k -1}(\cos(\theta))\\ & = 2 \cos(\theta) \cos(k\theta) - \cos((k -1)\theta)\\ & = 2 \cos(\theta) \cos(k\theta) - \cos(k\theta) \cos(\theta) - \sin(k\theta)\sin(\theta)\\ & = \cos((k + 1)\theta) \end{align*} Hence by induction hypothesis, we obtain that our claim is true.
Therefore we have \begin{align*} T_{n}\Bigg(\cos\bigg(\frac{2\pi}{n}\bigg)\Bigg) = \cos(2\pi) = 1 \end{align*} Therefore we just need to consider a polynomial $P(x) = T_{n}(x) - 1.~$ As $T_{n}(x) \in \mathbb{Z}[x]$ it implies $P(x) \in \mathbb{Z}[x]$
Therefore we have $\cos\big(\frac{2\pi}{n}\big)$ is an algebraic number.
Please check the solution and point out the glitches :)
from Hot Weekly Questions - Mathematics Stack Exchange
Ralph Clausen
Post a Comment