Let
$$(f_n(x))_{n=1}^{\infty}=\left\{\arctan\left(\frac{\sqrt{n}x}{(8n^2+x^2)^{3/2}}\right\}\right)_{n=1}^{\infty}.$$ Does $(f_n(x))_{n=1}^{\infty}$ uniformly converges?
Solution:
I calculated $f(x)=\lim f_n(x)$ $$f(x)= \lim_{n\to\infty} f_n(x)=\lim_{n\to\infty}\arctan\left(\frac{\sqrt{n}x}{(8n^2+x^2)^{3/2}}\right)=0$$
According to the definition of uniform convergence below
"Suppose $S$ is a set and $f_n : S → R$ is a real-valued function for every natural number $n$. We say that the sequence $(f_n)_{n\in \mathbb{N}}$ is uniformly convergent with limit $f : S → R$ if for every $ε > 0$, there exists a natural number $N$ such that for all $x ∈ S$ and all $n ≥ N$ we have $|f_n(x) − f(x)| < ε$. Consider the sequence $α_n = sup_x |f_n(x) − f(x)|$ where the supremum is taken over all $x ∈ S$. Then $f_n$ converges to $f$ uniformly if and only if $α_n$ tends to 0."
I have to calculate $ α_n=\sup|f_n(x) − f(x)| =0$ $$α_n=\sup|f_n(x) − f(x)|=\sup \left|\arctan\left(\frac{\sqrt{n}x}{(8n^2+x^2)^{3/2}}\right)\right|$$
From this point I dont know how to continue but it seems that we need to solve $x$ for which holds
$$\sup \left|\arctan\left(\frac{\sqrt{n}x}{(8n^2+x^2)^{3/2}}\right)\right|=0$$
so I have to solve $x$ in this equation
$$\left(\frac{\sqrt{n}x}{(8n^2+x^2)^{3/2}}\right)=0$$
Is that right? Thx
Everything up to the bolded text is correct. But "solve $x$ for which holds..." is way off the mark. Finding $x$ where a function is small does not tell you much about the supremum of that function.
The key point here is that the argument of the arctangent, $$\frac{\sqrt{n}x}{(8n^2+x^2)^{3/2}} $$ converges to $0$ uniformly. One way to see this is to use the arithmetic-geometric inequality $a+b\ge 2\sqrt{ab}$:
$$(8n^2+x^2)^{3/2}\ge \left(2\sqrt{ 8n^2 x^2}\right)^{3/2} = 28^{3/2} n^{3/2} |x|^{3/2} $$ Hence, $$\frac{\sqrt{n}|x|}{(8n^2+x^2)^{3/2}}\le \frac{\sqrt{n}|x|}{28^{3/2} n^{3/2} |x|^{3/2}}$$ which gives uniform convergence to $0$ on the set $|x|\ge 1$.
When $|x|\le 1$, the simple estimate $$\frac{\sqrt{n}|x|}{(8n^2+x^2)^{3/2}}\le \frac{\sqrt{n} }{(8n^2 )^{3/2}}$$ suffices.
Since the arctangent is continuous at $0$ and $\arctan 0 = 0$, it follows that $f_n\to 0$ uniformly.