# Expressing $\zeta^k+\zeta^{-k}$ as a polynomial in $\zeta+\zeta^{-1}$.

Let $$\zeta$$ be an $$n$$-th root of unity and let $$\chi:=\zeta+\zeta^{-1}$$. Then $$\zeta^k+\zeta^{-k}=P_k(\chi)$$ where $$P_k\in\Bbb{Z}[X]$$ is a polynomial not depending on $$n$$. For example we have $$\begin{eqnarray*} \zeta^2+\zeta^{-2}&=&\chi^2-2, \qquad&\text{ so }&\qquad P_2&=&X^2-2,\\ \zeta^3+\zeta^{-3}&=&\chi^3-3\chi, \qquad&\text{ so }&\qquad P_3&=&X^3-3X,\\ \zeta^4+\zeta^{-4}&=&\chi^4-4\chi^2+2, \qquad&\text{ so }&\qquad P_4&=&X^4-4X^2+2,\\ \zeta^5+\zeta^{-5}&=&\chi^5-5\chi^3+5\chi, \qquad&\text{ so }&\qquad P_5&=&X^5-5X^3+5,\\ \zeta^6+\zeta^{-6}&=&\chi^6-6\chi^4+9\chi^2+18, \qquad&\text{ so }&\qquad P_6&=&X^6-6X^4+9X^2+18. \end{eqnarray*}$$ It isn't hard to see that $$P_{ab}=P_a\circ P_b=P_b\circ P_a$$ for all positive integers $$a$$ and $$b$$, and that we have a recurrence relation $$P_a=X^a-\sum_{i=1}\binom{a}{i}P_{a-2i},$$ where we take the convention that $$P_k=0$$ for all $$k<0$$, and $$P_0=1$$. My question is:

Is there a simple explicit expression for $$P_k$$?

• This may help you: en.wikipedia.org/wiki/Chebyshev_polynomials – Seewoo Lee Nov 15 '18 at 16:40
• @SeewooLee Can you turn that comment into an answer? – Pedro Tamaroff Nov 15 '18 at 16:43
• @SeewooLee Thank you for the link. So if I understand correctly we have $P_k(X)=2T_k(\frac{X}{2})$, where $T_k$ is the $k$-th Chebyshev polynomial of the first kind? – Servaes Nov 15 '18 at 16:46
• @SeewooLee Ok all is clear, thanks again. If you care to write up an answer, I'll gladly accept. – Servaes Nov 15 '18 at 17:09