A function $f:\mathbb{Z}\mapsto\mathbb{R}$ is $$f_L(n)=\frac{1}{1+b\frac{n}{L}-\frac{1}{2}\big(\frac{n}{L}\big)^2}$$ where integer $n\in [0,L]$ s.t., $\rho=n/L\in [0,1]$. We want to evaluate the product $$P_L(j)=\prod\limits_{k<j}\bigg(1-\frac{f_L(n^*)}{f_L(k)}\bigg)$$ where $f_L(n^*)\le f_L(m)\forall m\in [0,L]$. We are interested in large $L$ limit.
An approach
My approach to evaluate $P_L(j)$: we can take $\log$ on both sides. The sum becomes integral in continuum limit (Remember large $L$). And we evaluate $P_L(j)$ as $$P_L(j)=\exp\bigg(c\int\limits_0^{\rho_j}\log\big(1-\frac{f_L(n^*)}{f_L(k)}\big) \bigg)\Theta(n^*-j)$$ As $f(n^*)$ is constant inside the integral, so solving the integral is not big deal. I hope problem is well stated uptill here. If not please tell in comments below. Jumping to the question:
Question:
- What should be $c$?
- Or, I am completely wrong with my approach so we there is no $c$ in the first place!
Consider instead
$$f(x)=\frac1{1+bx-\frac12x^2}$$
so that $f_L(n)=f(n/L)$ and $f_L(n^\star)\simeq\min\limits_{0\le x\le1}f(x)=M$ is the minimum. Note that if the denominator has no roots in $[0,1]$, then the minimum is either $f(0)$ or $f(1)$ depending on the sign of $b$.
Let $x_k=k/L$. We are essentially interested in
$$\prod_{k<j}\left(1-\frac M{f(x_k)}\right)=\exp\sum_{k<j}\ln\left(1-\frac M{f(x_k)}\right)$$
which is not a Riemann sum. Consider instead
$$\frac1L\sum_{k<j}\ln\left(1-\frac M{f(x_k)}\right)\underset{L\to\infty}\longrightarrow\int_0^{j/L}\ln\left(1-\frac M{f(x)}\right)~\mathrm dx$$
so that it converges to the integral $I_j$. We then have the asymptotic behavior given by
$$\exp\sum_{k<j}\ln\left(1-\frac M{f(x_k)}\right)\simeq\exp(I_jL)$$
as $L\to\infty$.