$u_n(x)= \prod_{k=1}^{n} \big(1 +\frac{x}{n} f(\frac{k}{n}) \big)$ : asymptotic

Question

• $f$ is continuous from $[0,1]$ to $\mathbb{R}$, $ f \ne 0$
• $x \in [-A,A]$ , $A$ is fixed
• $u_n(x)= \prod_{k=1}^{n} \big(1 +\frac{x}{n} f(\frac{k}{n}) \big)$
• $L=\int_{0}^{1} f(z) dz$

The goal of the exercise is to find an equivalent on $n$ of $ ~~~ u_n(x) - x \int_{0}^{1} f(z) dz$

I have proved the simple convergence : $u_n(x) \underset{n \to \infty}{\to} e^{Lx}$
I have proved the uniform converge $u_n(x)$ for all $x \in [-A,A]$

Here are the $4$ steps of the exercise :

1. $u_n(x)-e^{Lx} \sim e^{Lx} [ \ln (u_n(x)- Lx ] $. It is done.
2. $\left| \ln(1+u) - u + \frac{u^2}{2} \right|\leq |u| ^3 $ for $| u | \leq \frac{2}{3}$ done here
3. Show that there is a intercept $D$ such that : $$ \ln(u_n(x)) -Lx = x \lbrack \frac{1}{n} \sum_{k=1}^{n} f(\frac{k}{n}) - L \rbrack - \frac{x^2}{2 n^2} \sum_{k=1}^{n} f^2(\frac{k}{n}) + D \frac{ \mid x \mid ^3 }{n^2}$$
4. Find an equivalent of $\frac{1}{n} \sum_{k=1}^{n} f(\frac{k}{n}) - L$. I asked the question here
5. Deduce the equivalent of $ u_n(x) -Lx$

The step $4$ seems false, but can we prove the step 3 ? Thanks

Answer 1

Expriming the expression

$$ u_n(x)= \exp\left(\sum_{k=1}^n\ln\left(1+\dfrac{x}{n}f(\frac{k}{n})\right)\right)$$

Using Taylor-Lagrange on $g_n=\ln(1+\frac{x}{n}f(\frac{k}{n}))$ between $x$ and $0$

$$ \ln(u_n(x))=\sum_{k=1}^n\dfrac{x}{n}f(\frac{k}{n})-\dfrac{x^2}{2n^2}\sum_{k=1}^nf(\frac{k}{n})^2+O(\frac{1}{n^2})$$

Using Riemann sums :

$$\ln(u_n)=x \int_{0}^1f(t)dt-\dfrac{x^2}{2n}\int_{0}^1f(t)^2dt+O(\dfrac{1}{n^2})$$

Then taking the exponential and making asymptotic development of the exponential terms

$$u_n(x)-Lx=(Lx)(1-\dfrac{x^2}{2n}L'+O(\dfrac{1}{n^2}))-Lx$$

Hence

$$ \boxed{u_n(x)-Lx \sim -\dfrac{x^2}{2n}\int_{0}^1f^2(t)dt}$$