2
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.