2

From a somewhat heuristic derivation in a physics problem, I found the identity

$$ \prod_{\ell=1}^{N-1}\left[1\pm x \cos\left(\frac{\ell \pi}{N}\right)\right] = \frac{\lambda_{+}^N - \lambda_{-}^N}{\lambda_+ - \lambda_-}; \qquad \lambda_{\pm} = \frac{1}{2} \left(1 \pm \sqrt{1-x^2}\right) $$

I assume this identity is already known (for example, the $x=1$ case reduces to a version of the well known formula discussed here, here, and many other times on math.stackexchange), but I am unfamiliar with the relevant literature.

Does anyone know/Can anyone construct an analytical proof for this result?

motherboard
  • 652

1 Answers1

1

The relevant litterature, as you call it, could be the ubiquitous Chebyshev polynomials of the second type $U_N(x)$. Here is a proof using them.

I will refer to your formula at the next step, i.e. with $N$ instead of $N-1$.

One finds here the following expression :

$$U_N(x)=\dfrac{(x+\sqrt{x^2-1})^{N+1}-(x-\sqrt{x^2-1})^{N+1}}{2 \sqrt{x^2-1}}\tag{1}$$

Remark : it can look strange that we get polynomials in this way, but a close look at the RHS show that all square roots are finally cancelled.

If we set $x=\dfrac{1}{X}$ in (1), we get :

$$U_N(\dfrac{1}{X})=\dfrac{1}{X^N}\dfrac{(1+\sqrt{1-X^2})^{n+1}-(1-\sqrt{1-X^2})^{n+1}}{2 \sqrt{1-X^2}} $$

which is your RHS expression (up to a factor of $2^N$ and $X^N$ that we will find back later on).

Besides (formula (18) here), we have the following formula :

$$U_N(x)=2^N \prod_{k=1}^N \left(x-\cos\left(\tfrac{k\pi}{N+1} \right) \right) \tag{2}$$

Setting $x=\dfrac{1}{X}$ in (2), we get the RHS of your expression, with factors $X^N$ and $2^N$ that are placed just where they have to be finally.

Jean Marie
  • 81,803