Need to find if a sequence of functions has uniform convergence

[natasha d]

Homework Statement

f$_{n}$ is is a sequence of functions in R, x$\in$ [0,1]
is f$_{n}$ uniformly convergent?
f = nx/1+n$^{2}$x$^{2}$

Homework Equations

uniform convergence $\Leftrightarrow$
|f$_{n}$(x) - f(x)| < $\epsilon$ $\forall$ n>= n$_{o}$ $\in$N

The Attempt at a Solution

lim f$_{n}$ = lim nx/1+n$^{2}$x$^{2}$ = 0 if
n→∞ n→∞ x$\in$ [0,1)



= lim (1/n)(1/(n$^{2}$x$^{2}$)=1) if x=1
n→∞

=0 as we sub x=1 and then sub the limit as 1/n = 0
hence seq converges to f(x) = 0, proving pointwise convergence

for uniform convergence we'll need a n$_{o}$ $\in$N independant of x
|f$_{n}$(x) - f(x)|<ε
ie |nx/(1+n$^{2}$x$^{2}$) - 0|<ε


cant seem to manipulate the inequalities for an n

 

[Dick]

I would suggest you try to find the maximum of $f_n(x)=\frac{nx}{1+n^2 x^2}$ on the interval [0,1]. Use calculus.

 

[natasha d]

would that be at x = 1?
$\Rightarrow$ n/$\epsilon$(1+n$^{2}$)< the required n
n is variable.. i don't get it

 

[Dick]

would that be at x = 1?
$\Rightarrow$ n/$\epsilon$(1+n$^{2}$)< the required n
n is variable.. i don't get it

No, no. Use calculus. We aren't to the epsilon-delta part yet. Take the derivative of f_n and set it equal to zero and try to figure out where the extreme points of the function are. The result will be informative. You want to figure out how your functions are behaving before you start thinking about the proof.

 

[natasha d]

tried differentiation got x=$\sqrt{1/n}$ by equating it to 0
what on Earth does that mean...
anyway how do we know the extremes of each function in the sequence of
functions is defined by the same extreme. Sorry if I am being an idiot.. but i seem to be not understanding something
wait what did you mean by the way the function is behaving?

 

[Dick]

tried differentiation got x=$\sqrt{1/n}$ by equating it to 0
what on Earth does that mean...
anyway how do we know the extremes of each function in the sequence of
functions is defined by the same extreme. Sorry if I am being an idiot..

You aren't being an idiot. My only problem with that is when I differentiated I got x=1/n. How did you get x=1/sqrt(n)? Don't give up on this. The answer is actually pretty simple once you get it right. What did you get for a derivative?

 

[Dick]

wait what did you mean by the way the function is behaving?

I mean in the sense of being about to sketch a graph of y=f_n(x). f_n doesn't approach 0.

 

[natasha d]

that was a stupid error actually
it came to something like this 1-n$^{2}$x$^{2}$=0
do we overlook the -ve x because of the interval [0,1]?

 

[Dick]

that was a stupid error actually
it came to something like this 1-n$^{2}$x$^{2}$=0
do we overlook the -ve x because of the interval [0,1]?

Yes, ignore the negative root. So what's the max of f_n and where does it occur?

 

[natasha d]

now I am lost completely
do you mean f is maximum at some x? for every n?

 

[Dick]

now I am lost completely
do you mean f is maximum at some x? for every n?

No, a different x for every n. The max is at x=1/n. What's f_n(1/n)?

 

[natasha d]

is it .. half?
your probably like this right nowhttp://fastfoodies.org/wp-content/uploads/Head-desk-1.jpg

 

[Dick]

is it .. half?
your probably like this right nowhttp://fastfoodies.org/wp-content/uploads/Head-desk-1.jpg

Yes, f_n(1/n)=1/2. I actually hadn't gotten to that point, but thanks. I knew you'd see it. So you see why that makes a problem for uniform continuity?

 

[natasha d]

does this have something to do with the for every ε part?
ie the diff between f$_{n}$(x) = 1/2 and f(1/n) = 0 is greater than some ε (ie ε < 1/2)
i think I've got it
Thank you Dick