# how to show sinx uni converges

• October 26th 2008, 05:12 AM
szpengchao
how to show sinx uni converges
$y_{n}(x)=\sin(x), 0\leq |x| \leq 2\pi n$ and

$y_{n}(x)=0, 2\pi n< |x| <\infty$
• October 27th 2008, 02:42 AM
CaptainBlack
Quote:

Originally Posted by szpengchao
$y_{n}(x)=\sin(x), 0\leq |x| \leq 2\pi n$ and

$y_{n}(x)=0, 2\pi n< |x| <\infty$

$y_n(x)$ converges pointwise to $\sin(x)$ ,but for all $n$ there exists $x_n$ such that $|y_n(x_n)-\sin(x_n)|=1$.

CB