# uniform convergence of $f_n(x)=\left(\cos\frac{x}{\sqrt{n}}\right)^n$

Let a sequence of functions $$(f_n)_{n\in \mathbb{N}^*}$$ such that $$f_n(x)=\left(\cos\dfrac{x}{\sqrt{n}}\right)^n$$ defined on $$\mathbb{R}$$

1. Show the pointwise convergence on $$\mathbb{R}$$

2. Show that the sequence converges uniformly on $$[-A,A]$$ with $$A>0$$

Pointwise convergence

I fund that $$f_n(x)\longrightarrow e^{-x^2/2}$$

Uniform convergence

I must show that $$||f_n(x)-e^{-x^2/2}||_{\infty}\underset{n\to +\infty}{\longrightarrow }0$$

My first idea (*to be honest I've got no overview) is to write :

$$\bigg|f_n(x)-e^{-x^2/2}\bigg|$$, and I tried to find a bound.

As $$1-\dfrac{u^2}{2}\le\cos u\le 1-\dfrac{u^2}{2}+\dfrac{u^4}{24}\iff 1-\dfrac{x^2}{2n}\le \cos\dfrac{x}{\sqrt{n}}\le 1-\dfrac{x^2}{2n}+\dfrac{x^4}{24n^2}$$

Let $$g_1(x)=1-\dfrac{x^2}{2n}$$ and $$g_2(x)=1-\dfrac{x^2}{2n}+\dfrac{x^4}{24n^2}$$ and $$M_1, M_2$$ such that :

For $$M_1$$ that is but for $$M_2$$ I can find a finite bound $$|g_1(x)|

I repeat again I haven't got any overview about how to attack this problem

• It's not a duplicate, because the uniform convergence hasn't been proven – Stu Dec 1 '18 at 21:09
• I reopened after closing as a duplicate of this. The answer there does prove non-uniform convergence on $\mathbb{R}$, but fails to address uniform convergence on a compact interval. There probably is another duplicate somewhere. Note that this can be proved by squeezing with your estimates and using the inequality $0 \leqslant e^{-y} - \left(1 - \dfrac{y}{n}\right)^n \leqslant \dfrac{e^{-y}y^2}{n}$, – RRL Dec 2 '18 at 0:06

$$\cos(x/\sqrt{n})^n=(1-x^2/(2n)+a(x) x^4/(24n^2))^n$$
where $$-1 \leq a(x) \leq 1$$. This is the Lagrange remainder form of Taylor expansion.
Using the assumption $$x \in [-A,A]$$ gives that $$\cos(x/\sqrt{n})^n$$ is always between $$(1-x^2/(2n)+A^4/(24n^2))^n$$ and $$(1-x^2/(2n)-A^4/(24n^2))^n$$. Now you are set up to use a squeezing argument and just need to be careful to select a $$N(\varepsilon)$$ that doesn't depend on $$x$$ (but will depend on $$A$$).
One way to do this squeezing argument is to note that $$(1-x^2/(2n)+A^4/(24n^2))^n \\ =\exp(n \log(1-x^2/(2n)+A^4/(24n^2)) \\ <\exp(-x^2/2)$$
using the Taylor expansion of $$\ln(1+x)$$. Then you do something similar but slightly more complicated for the lower bound.