# Prove that $\sin(x+\frac{\pi}{n})$ converges uniformly to $\sin(x)$.

I've just starting learning uniform convergence and understand the formal definition. What I've got so far is:

$|\sin(x+ \frac{\pi}{n}) - \sin(x)| < \epsilon \ \ \ \ \forall x \in \mathbb{R} \ \ \ \$ for $n \geq N, \epsilon>0$

LHS = $|2\cos(x+\frac{\pi}{2n})\cdot \sin(\frac{\pi}{2n})| < \epsilon$

Am I going down the right route here? I've done some examples fine, but when trig is involved on all space, I get confused as to what I should be doing...

Any help at all would be VERY much appreciated, I have an analysis exam tomorrow and need to be able to practice this.

Thanks.

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Try working with the definition, you need the same $\delta$ to 'work' for all $x,y$. Maybe it is woth noting that since $sin(x)=sin(x+2\pi)$ you only need to show that the $\delta$ you chose works for all $x,y\in [0,2\pi]$. Edit: Also be carefull writing something like "$for n\in\mathbb{N}, \epsilon>0$ since you probably meant that for every $epsilon$ exist an $N$ s.t ... – Belgi May 31 '12 at 13:57
Your trig identity approach will work, though there are easier ways. The $\cos$ term that involves $x$ cannot get big, and of course $\sin(\pi/2n)$ can be made small. So your expression is $\le \pi/n$. – André Nicolas May 31 '12 at 14:11
@Belgi: where in the definition of uniform convergence does $\delta$ appear? – T. Eskin May 31 '12 at 14:12
Thankyou all for your responses. This is the first time I've posted on here and it took 10 minutes for me to have a few different approaches to the question - amazing! – Sarah_24 May 31 '12 at 14:29

Use the fact that the sine function's derivative has absolute value of at most one to see that $$|\sin(x) - \sin(y)| \le |x - y|.$$

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Thankyou very much - I had forgotten this and it makes the inequality very simple to work with. I think this is probably how our lecturer wants us to approach this question. – Sarah_24 May 31 '12 at 14:18
+1 Much thriftier and more elegant than what I did. – DonAntonio May 31 '12 at 14:25

What you did so far is fine. If you want to continue in the direction you started, it remains to notice that $$\left|2\cos\left(x+\frac{\pi}{2n}\right)\cdot \sin\frac{\pi}{2n}\right| \le 2\left|\sin\frac{\pi}{2n}\right|$$ and that for any given $\varepsilon>0$ you can choose $N$ such that you have inequality $$2\left|\sin\frac{\pi}{2n}\right|<\varepsilon$$ for $n>N$.

Can you see why this is true?

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Thanks - this is also very helpful and along the same lines as @ncmathsadist's response. Presumably, the LHS $\leq \frac{\pi}{n} < \frac{\epsilon}{2}$ and hence you can solve for an $N \leq n$. – Sarah_24 May 31 '12 at 14:25
Yes. Other possibility would be to use that $\lim\limits_{n\to\infty} \frac\pi{2n}=0$ and $\lim\limits_{x\to0} \sin x=0$. – Martin Sleziak May 31 '12 at 14:26

Can you prove that $\,\,\sin \pi/n\to 0\,\,,\,\,\cos \pi/n\to 1\,\,$? Then do some $\epsilon$-stuff, and use trigonometric identites of double angle to get Cauchy condition for uniform convergence:$$\left|\sin\left(x+\frac{\pi}{n}\right)-\sin\left(x+\frac{\pi}{m}\right)\right|\leq \left|\sin x\left(\cos\frac{\pi}{n}-\cos\frac{\pi}{m}\right)+\cos x\left(\sin\frac{\pi}{n}-\sin\frac{\pi}{m}\right)\right|$$ and remember that $\,\,\forall x\in\mathbb{R}\,,\,\,|\sin x+\cos x|<2$

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Thanks for your response - this looks like where I headed with it originally, but a lot messier than it needed to be! – Sarah_24 May 31 '12 at 14:27

First find the pointwise limit, that is where $\displaystyle \lim_{n \to \infty} f_{n}(x)=\displaystyle \lim_{n \to \infty} \sin\Bigl(x + \frac{\pi}{n}\Bigr)$ goes to. You can easily see that this limit is same as $\displaystyle \lim_{y \to 0} \sin\bigl(x +\pi \cdot y\bigr)= \sin{x}$. So your function converges to $f(x)=\sin(x)$. Now to prove uniform convergence we have to look at the sequence $$d_{n} = \sup_{x \in \mathbb{R}} \: \Bigl\{\:\Bigl|\sin\Bigl(x + \frac{\pi}{n}\Bigr) - \sin(x)\:\Bigr| : x \in \mathbb{R}\Bigr\}$$ and prove that $d_{n} \to 0$ for $n \in \mathbb{N}$.

We do the derivative test. Look at $f(x) = \sin\Bigl(x + \frac{\pi}{n}\Bigr) - \sin(x)$, $f'(x) = \cos\Bigl(x+\frac{\pi}{n}\Bigr) - \cos(x)$. Equating this to $0$ we see that \begin{align*} \cos\Bigl(x + \frac{\pi}{n}\Bigr) - \cos(x) &= 2 \cdot \sin\Bigl(x + \frac{\pi}{2n}\Bigr) \cdot \sin\Bigl(-\frac{\pi}{n}\Bigr) = 0 \end{align*} and so $x = -\frac{\pi}{2n}$. From this we see that $d_{n} \to 0$ and hence the convergence is uniform.

This question might be worth looking:

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You're using the derivative test to find $\max |f(x)|$. It's probably worth explaining why local extrema are also global extrema. And, of course, the extremal values are obtained at all points of the form $x=-\frac\pi{2n}+k\pi$, $k\in\mathbb Z$, but it's clear that the value of $|f(x)|$ at each of these points is the same. – Martin Sleziak May 31 '12 at 16:47
@Martin: Thanks Martin for the comment. – user9413 May 31 '12 at 16:56