Prove that $\prod\limits_{k=1}^{[n/2]} (3+2\cos\frac{2kπ}{n}) =F_n$

Question

$F_1=1; F_2=2; F_{n+1}=F_{n}+F_{n-1}$ (Fibonacci)

Prove that $\prod\limits_{k=1}^{[n/2]} \left(3+2\cos\dfrac{2kπ}{n}\right) =F_n$

please help me :(

Answer 1

Plugging in $n=2$ in the LHS gives $3+2\cos(\pi)=1$, so I will assume that the Fibonacci sequence is defined by $F_0=0,F_1=1,F_2=1\ldots$ and so on. Letting $\varphi=\frac{1+\sqrt5}{2}$ gives $\varphi^2-\varphi-1=0$ and $\varphi^4-3\varphi^2+1=0$. Moreover, that last equation $x^4-3x^2+1=0$ is also satisfied by $\varphi^{-1}$. We shall use the following well-known facts about Fibonacci numbers and Chebyshev polynomials $T_n(x)$, $n\geq 0$.

• Binet's Formula: $F_n=\frac{1}{\sqrt5}\left(\varphi^n-(-\varphi^{-1})^n\right)$.
• Recurrence Relation for $T_n(x)$: $T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$.

Plugging in $x=-\frac{3}{2}$ in the recurrence relation and rearranging some terms gives the formula: $\frac{2}{5}\left(T_{n+1}\left(-\frac{3}{2}\right)-1\right)=-3\cdot\frac{2}{5}\left(T_{n}\left(-\frac{3}{2}\right)-1\right)-\frac{2}{5}\left(T_{n-1}\left(-\frac{3}{2}\right)-1\right)-2$, denoted by (*).

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Now, let $P_n$ be the product $\prod\limits_{k=1}^{[n/2]} (3+2\cos\dfrac{2kπ}{n}) $ on the LHS of the question. Since $P_n>0$, it suffices to show that $P_n^2=F_n^2$ for all $n\geq 0$. We establish this in three steps.

1. For all $m\geq 0$, we have $F_{2m}^2=3F_{2m-1}^2-F_{2m-2}^2-2$ and $F_{2m+1}^2=3F_{2m}^2-F_{2m-1}^2+2$.

This can be easily proved by induction on $m$ and Binet's formula. Note that the two equations $\varphi^4-3\varphi^2+1=0$ and $(\varphi^{-1})^4-3(\varphi^{-1})^2+1=0$ are used in the inductive step.

2. For all $m\geq 0$, we have $F_{2m}^2=\frac{2}{5}\left(T_{2m}\left(-\frac{3}{2}\right)-1\right)$ and $F_{2m+1}^2=-\frac{2}{5}\left(T_{2m+1}\left(-\frac{3}{2}\right)-1\right)$.

This is again an easy induction on $m$. Just use step 1 and the formula (*).

3. For all $m\geq 0$, we have $P_{2m}^2=\frac{2}{5}\left(T_{2m}\left(-\frac{3}{2}\right)-1\right)$ and $P_{2m+1}^2=-\frac{2}{5}\left(T_{2m+1}\left(-\frac{3}{2}\right)-1\right)$.

This is done by analyzing the real roots of $T_n(x)-1$. The equation $T_{2m}(x)-1=0$ has double zeros at each $\cos\dfrac{2kπ}{2m}$, $1\leq k\leq m-1$, and two trivial zeros $x=\pm 1$. This gives $$T_{2m}(x)-1=2^{2m-1}(x-1)(x+1)\prod\limits_{k=1}^{m-1} (x-\cos\dfrac{2kπ}{2m})^2.$$ On the other hand, $T_{2m+1}(x)-1$ has double zeros at each $\cos\dfrac{2kπ}{2m+1}$, $1\leq k\leq m$, and one trivial zero $x=1$. This gives $$T_{2m+1}(x)-1=2^{2m}(x-1)\prod\limits_{k=1}^{m} (x-\cos\dfrac{2kπ}{2m+1})^2.$$ Plugging in $x=-\frac{3}{2}$ in the above equations easily yieds $P_{2m}^2=\frac{2}{5}\left(T_{2m}\left(-\frac{3}{2}\right)-1\right)$ and $P_{2m+1}^2=-\frac{2}{5}\left(T_{2m+1}\left(-\frac{3}{2}\right)-1\right)$.

We have shown that $P_{2m}^2=\frac{2}{5}\left(T_{2m}\left(-\frac{3}{2}\right)-1\right)=F_{2m}^2$ and $P_{2m+1}^2=-\frac{2}{5}\left(T_{2m+1}\left(-\frac{3}{2}\right)-1\right)=F_{2m+1}^2$. The proof is complete.