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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 ...2012-05-31
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    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 $\le \pi/n$.2012-05-31
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    @Belgi: where in the definition of uniform convergence does $\delta$ appear?2012-05-31
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    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!2012-05-31

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