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Need to find if a sequence of functions has uniform convergence

  1. Mar 12, 2012 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations
    uniform convergence [itex]\Leftrightarrow[/itex]
    |f[itex]_{n}[/itex](x) - f(x)| < [itex]\epsilon[/itex] [itex]\forall[/itex] n>= n[itex]_{o}[/itex] [itex]\in[/itex]N

    3. The attempt at a solution

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



    = lim (1/n)(1/(n[itex]^{2}[/itex]x[itex]^{2}[/itex])=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[itex]_{o}[/itex] [itex]\in[/itex]N independant of x
    |f[itex]_{n}[/itex](x) - f(x)|<ε
    ie |nx/(1+n[itex]^{2}[/itex]x[itex]^{2}[/itex]) - 0|<ε


    cant seem to manipulate the inequalities for an n
     
  2. jcsd
  3. Mar 12, 2012 #2

    Dick

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    I would suggest you try to find the maximum of [itex]f_n(x)=\frac{nx}{1+n^2 x^2}[/itex] on the interval [0,1]. Use calculus.
     
  4. Mar 12, 2012 #3
    would that be at x = 1?
    [itex]\Rightarrow[/itex] n/[itex]\epsilon[/itex](1+n[itex]^{2}[/itex])< the required n
    n is variable.. i dont get it
     
  5. Mar 12, 2012 #4

    Dick

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    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.
     
    Last edited: Mar 12, 2012
  6. Mar 12, 2012 #5
    tried differentiation got x=[itex]\sqrt{1/n}[/itex] 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 im being an idiot.. but i seem to be not understanding something
    wait what did you mean by the way the function is behaving?
     
    Last edited: Mar 12, 2012
  7. Mar 12, 2012 #6

    Dick

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    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?
     
  8. Mar 12, 2012 #7

    Dick

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    I mean in the sense of being about to sketch a graph of y=f_n(x). f_n doesn't approach 0.
     
  9. Mar 12, 2012 #8
    that was a stupid error actually
    it came to something like this 1-n[itex]^{2}[/itex]x[itex]^{2}[/itex]=0
    do we overlook the -ve x because of the interval [0,1]?
     
  10. Mar 12, 2012 #9

    Dick

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    Yes, ignore the negative root. So what's the max of f_n and where does it occur?
     
  11. Mar 12, 2012 #10
    now im lost completely
    do you mean f is maximum at some x? for every n?
     
  12. Mar 12, 2012 #11

    Dick

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    No, a different x for every n. The max is at x=1/n. What's f_n(1/n)?
     
  13. Mar 13, 2012 #12
    is it .. half?
    your probably like this right nowhttp://fastfoodies.org/wp-content/uploads/Head-desk-1.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  14. Mar 13, 2012 #13

    Dick

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    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?
     
    Last edited by a moderator: May 5, 2017
  15. Mar 13, 2012 #14
    does this have something to do with the for every ε part?
    ie the diff between f[itex]_{n}[/itex](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
     
    Last edited: Mar 13, 2012
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