Context is I’m trying to compute the Feynman Kernel for the harmonic oscillator
Here introduced the matrix $B_{N}$ of size $N-1$ and its eigenvalues: $$ \lambda_{j} = 2\frac{1 - k(\frac{T}{N})^{2}}{1 + k(\frac{T}{N})^{2}} - 2\cos(\frac{\pi j}{N}) $$ for $j$ in $\{1,\dots,N-1\}$.
I’m trying to show the following convergence: $$\lim_{N \rightarrow \infty} \frac{\det(B_{N})}{N} = \frac{\sin(2\sqrt{k}T)}{2\sqrt{k}T}$$
But I don’t understand the argument used by my teacher in the picture I joined thereafter (Please tell me if you don’t see the picture in that case I will write everything here). In fact I feel like he said “these functions have the same zeros when $N \rightarrow \infty$ so they are equal”.
I tried another way by using the fact the $\det(B_N)$ is the product of eigenvalues, then I try to find a correspondence with Euler formula $\operatorname{sinc}(\pi x) = \Pi_{k} (1-\frac{x^{2}}{k^{2}})$ but I don’t succeed.
Please help me with this proof!
Greetings
A better approach (I think): the determinant has a closed form. Namely, we have $$ 2^{n-1}\prod_{j=1}^{n-1}\left(\cos\theta-\cos\frac{\pi j}n\right)=\frac{\sin n\theta}{\sin\theta} $$ (a.k.a. $U_{n-1}(\cos\theta)$ using the Chebyshev polynomials of the second kind). Thus $$ \frac{\det B_N}N=\frac{\sin N\theta_N}{N\sin\theta_N},\qquad \theta_N=\arccos\frac{1-k(T/N)^2}{1+k(T/N)^2}=\arcsin\frac{2\sqrt{k}\ T/N}{1+k(T/N)^2} $$ and clearly both $N\theta_N$ and $N\sin\theta_N$ tend to $2T\sqrt k$ as $N\to\infty$.