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Uniform convergence

  1. Dec 21, 2008 #1
    does {fn} converge uniformly? fn(x)=nx^2/1+nx

    I can see that fn converges pointwise to f(x)=x. I know, for epsilon>0, I need to find N st for n >or equal to N, |fn(x)-f(x)|<epsilon.

    |fn(x)-f(x)|=x/1+nx but then I am stuck.
  2. jcsd
  3. Dec 21, 2008 #2
    Why is that the case?
  4. Dec 21, 2008 #3


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    Homework Helper

    To be clear, are you considering convergence on all of [tex] \mathbb{R} [/tex] or
    only on [tex] [0,1] [/tex]? Also, your sequence of functions is

    f_n(x) = \frac{x}{1+nx}

    correct? If it really is (as you wrote)

    f_n(x) = nx^2 + nx

    then it should be clear that the only place it converges is at [tex] x = 0 [/tex]. Assuming the first version is correct, continue reading.

    One idea: note that

    |f_n(x) - x| = \frac{x}{1+nx} \le \frac{x}{nx} = \frac 1 n \quad \forall x

    Given an [tex] \epsilon > 0 [/tex], how would you choose an appropriate value of [tex] N [/tex]?
  5. Dec 21, 2008 #4
    for dirk_mec1,

    fn(x) converges pointwise to x because lim as n-->infinity of fn(x) equals x.
  6. Dec 22, 2008 #5
    On what domain are these functions fn defined (like statdad already asked you) ?
  7. Dec 23, 2008 #6
    I will prefer the domain to be [0,infinity]...for x is not defined at -1/n, which makes it a bit complicated.
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