# Homework Help: Pointwise & Unifrom Convergence

1. Apr 13, 2010

### kingwinner

1. The problem statement, all variables and given/known data
Find all intervals on which the sequence of functions
fn(x) = x2n / [n + x2n], n≥1, converges uniformly.

2. Relevant equations
3. The attempt at a solution
I think we'll need to compute the pointwise limit first.
For x=0 or x=1, the pointwise limit is 0.
But what is its pointwise limit for x>0?

Any help is appreciated!

Last edited: Apr 13, 2010
2. Apr 13, 2010

### Gib Z

Thing thing to do would be to consider the cases (0,1), x=1 and x>1 separately.

3. Apr 13, 2010

### kingwinner

OK, I think for all of the cases x=0,x=1,0<x<1, the pointwise limit would be 0.

But how can we compute the pointwise limit for x>1??

4. Apr 13, 2010

### Gib Z

Write $$\frac{x^{2n}}{x^{2n}+n} =\frac{x^{2n}+n - n}{x^{2n}+n} = 1 - \frac{n}{n+ x^{2n}}$$

On the right hand side, the second term is certainly always less than $$\frac{n}{x^{2n}}$$. You have a linear term divided by an exponential term, so what is the limit of that term?

5. Apr 13, 2010

### kingwinner

n/x2n ->0 for x>1

x2n / [n + x2n] ≥ 1 - n/x2n

1 - n/x2n->1, but still how can we compute the limit of x2n / [n + x2n]?

Last edited: Apr 13, 2010
6. Apr 13, 2010

### Gib Z

I'm not sure you know where all the inequalities are pointing at the moment. Can you see how

$$0 < \frac{n}{n+x^{2n} } < \frac{n}{x^{2n}}$$

You know what the term on the far right approaches, so what must the middle term approach? And so then what is the limit of the function if |x|> 1 ?

Last edited: Apr 13, 2010
7. Apr 13, 2010

### kingwinner

n/(n+x2n < n/x2n
=> -n/(n+x2n > -n/x2n
=> 1 -n/(n+x2n > 1 -n/x2n

Thus, x2n / [n + x2n] > 1 - n/x2n

1 - n/x2n -> 1

But still this doesn't say what x2n / [n + x2n] would converge to...

8. Apr 13, 2010

### Gib Z

Okay, just read post 6 again and tell me what the term on the far right converges to. And so, what do you conclude that the middle term goes to?

9. Apr 13, 2010

### kingwinner

That middle term goes to 0, so fn(x) -> 1 for |x|>1 and fn(x)->0 otherwise. Now how can we determine uniform convergence. I know that for any interval [a,b] containing 1 or -1 since it's discontinuous there, we won't have uniform convergence, but how can we prove uniform convergence on [a,b] where a>1?

Last edited: Apr 13, 2010
10. Apr 13, 2010

### Gib Z

Come on, try something! and Show me what you have done, and where you get stuck.

Of course you have to apply a definition of uniform convergence and see if the function satisfies it on some intervals.

11. Apr 13, 2010

### kingwinner

fn(x) = x2n / [n + x2n]

For the pointwise limit, I got fn(x) -> 1 for |x|>1 and fn(x)->0 otherwise.

So I know that for any interval [a,b] containing 1 or -1, we won't have uniform convergence, but how can we prove uniform convergence on [a,b] where a>1??
This is what I got, but I'm stuck...
fn(x) <= b2n / [n + a2n] < b2n / [n + 1]
How can we prove that the RHS ->0 ?