Prove that $f_n(x) = 1 - \cos(\frac{x}{n})$ is uniformly convergent

Question

Prove that $f_n(x) = 1 - \cos(\frac{x}{n})$ is uniformly convergent in $[0,2\pi]$.

So I said this, and my professor said it is wrong:

$|1-\cos(\frac{x}{n})| \leq 2, $ then we take some $\epsilon >2 $, and we have for every $n \in \mathbb{N}$ and $\forall x \in [0,2\pi]$ that $|1-\cos(\frac{x}{n})| < \epsilon$ and it is uniformly convergent.

What is wrong about what I did?

Some MathJax advice: Named math operators should appear upright, and the common ones have their own MathJax code for this purpose (e.g. \sin, \log - see entry 11 in the MathJax guide). — Zev Chonoles, Jul 08 '13 at 19:44
You only proved it for $\epsilon >2$, you need to prove it for ALL $\epsilon >0$. — N. S., Jul 08 '13 at 19:44
@MhenniBenghorbal I get that it is maximal when $x = n\pi k$. Then what? — TheNotMe, Jul 09 '13 at 07:48

N. S. · Accepted Answer · 2013-07-08T19:55:10.297

4

As I said in my comment, your proof doesn't cover the $\epsilon \in (0,2)$ case, you need to prove for all $\epsilon$ which means also for these.

Hint For $n >4$ you can prove that

$$0< 1 - \cos(\frac{x}{n}) \leq 1 - \cos(\frac{2 \pi}{n}) \forall x \in [0,2 \pi] \,.$$

$n>4$ is needed to make sure we are in the first quadrant. This follows immediately from the monotony of $\cos(x)$ in the first quadrant.

[$n>2$ would actually suffice, but who cares :) ].

Now, use the the definition of $\lim_n \cos(\frac{2 \pi}{n})=1$ and the fact that there is no $x$ in this limit.

Edit Alternately, by the double angle formulas and $\sin(y) <y$:

$$0 \leq 1 - \cos(\frac{x}{n}) = 2 \sin^2(\frac{x}{2n}) \leq 2 (\frac{x}{2n})^2 \leq 2 (\frac{2 \pi}{2n})^2$$

edited Jul 08 '13 at 19:55

answered Jul 08 '13 at 19:48

N. S.

132,525

How do we prove the first inequality though? – TheNotMe Jul 08 '13 at 19:52
@TheNotMe If $n >4$ then $\frac{x}{n} \leq \frac{2\pi}{n}$ are in the first quadrant. Since $\cos(x)$ is decreasing in the first quadrant, you get $\cos(\frac{x}{n}) \geq \cos(\frac{2\pi}{n})$. – N. S. Jul 08 '13 at 19:56

score 1 · Answer 2 · answered Jul 08 '13 at 19:52

Note that $(f_n)$ is pointwise convergent to the zero function in $[0,2\pi]$ moreover we have $$\forall n\geq 4\quad |f_n(x)|\leq 1-\cos\left(\frac{2\pi}{n}\right),\quad\forall x\in[0,2\pi]$$ so $$||f_n||_\infty\to0$$ and then we have the uniform convergence of $(f_n)$ to the zero function.

Robert Lewis · Answer 3 · 2013-07-08T21:05:00.223

Proposition: Let $f_n(x) = 1 - \cos(\frac{x}{n})$. Then $f_n(x) \to 0$ uniformly in $x$ on the interval $[0, 2 \pi]$ as $n \to \infty$.

Proof: Pick any real $\epsilon > 0$. Then there exists a real $\delta > 0$ such that $|1 - \cos y| < \epsilon$ if $|y| < \delta$; this assertion follows immediately from the continuity of $\cos y$ and the fact that $\cos 0 = 1$. Now choose $N > \frac{2 \pi}{\delta}$ and integer $n > N$. Then for $x \in [0, 2 \pi]$, we have $\frac{x}{n} < \frac{x}{N} \le \frac{2 \pi}{N} < \delta$, whence $f_n(x) < \epsilon$. Note this holds for any $\epsilon$ and $N$ for all $ x \in [0, 2 \pi]$. Thus is the uniform convergence of $f_n(x)$ to $0$. established. QED.

Prove that $f_n(x) = 1 - \cos(\frac{x}{n})$ is uniformly convergent

3 Answers3