# Thread: how to show if they r uni. convergent

1. ## how to show if they r uni. convergent

$f_{n}(x)=log(x+\frac{1}{n})
f_{n}(x)=cos(\frac{x}{n})$

on (0,1)

2. Originally Posted by silversand
$f_{n}(x)=log(x+\frac{1}{n})
f_{n}(x)=cos(\frac{x}{n})$

on (0,1)
I'm a little confused about what you have written.

I assume those are two different functions?

$f_n(x)=\log(x+\frac{1}{n})$ and $f_n(x)=\cos(\frac{x}{n})$

Also, are you using $\log(x)$ to mean the same thing as $\ln(x)$?

3. ## yes

yes. they are two cases. and log means ln

4. Originally Posted by silversand
$f_{n}(x)=log(x+\frac{1}{n})
f_{n}(x)=cos(\frac{x}{n})$

on (0,1)
Second function: We know that this function converges to $\cos(0)=1$, the question is whether it does so uniformly. Using the difference-to-product identity, we have:

$\left|\cos(0)-\cos\left(\frac{x}{n}\right)\right| = \left|-2\sin\left(\frac{x}{2n}\right)\sin\left(-\frac{x}{2n}\right)\right|\leq 2\left|\sin\left(\frac{x}{2n}\right)\right|$ because $|\sin(y)|\leq 1$ for all $y$.

Since $2\sin\left(\frac{x}{2n}\right)$ is increasing and positive on $(0,1)$, we know that $2\left|\sin\left(\frac{x}{2n}\right)\right|\leq 2\sin\left(\frac{1}{2n}\right)$.

It is now clear that for all $\epsilon>0$, we can choose $n$ large enough so that $\left|1-\cos\left(\frac{x}{n}\right)\right| \leq 2\sin\left(\frac{1}{2n}\right)<\epsilon$

If you need an explicit formula, I think choosing $n>\frac{1}{2\sin^{-1}(\frac{\epsilon}{2})}$ will do the trick.