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Show ##\{nx^{n}(1-x)\}_{n = 0}^{\infty}## converges uni

  1. Jun 5, 2015 #1
    1. The problem statement, all variables and given/known data
    show ##\{nx^{n}(1-x)\}## converges uni on [0,1]

    2. Relevant equations


    3. The attempt at a solution
    Note, that ##\sum nx^n(1-x)## converges uni on [0,1] by Abels' Theorem. Therefore the series follows the cauchy criterion,
    ##\forall \epsilon > 0##, ##\exists N > 0 ## such that if n,n > N then

    $$\left|(m+1)x^{m+1}(1 - x) + ... + nx^{n}(1- x)\right| < \epsilon\,\,\, \forall x \in [0,1]$$,

    Now since ##nx^n(1 - x) \geq 0## for all n we can say

    $$\left|nx^{n}(1 - x) - mx^{m}(1 - x)\right| < \epsilon\,\,\, \forall x \in [0,1].$$

    Which implies that ##\{nx^n(1 - x)\}## is uniformly cauchy ##\forall x \in [0,1]## and therefore uniformly convergent for ##x \in [0,1]##

    Dont know if this is right
     
  2. jcsd
  3. Jun 6, 2015 #2

    wabbit

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    Care to elaborate ?

    Note that once you've said that, you're done - if the series converges uniformly so does the sequence of course, there is no need for a Cauchy argument. But your first step is mysterious (and unnecessary).
     
  4. Jun 6, 2015 #3
    Well my version of Abel's theorem states
    Let ##g(x) = \sum a_{n}x^{n}## that converges at some point R > 0. Then the series converges uniformly on [0,R]. I used the cauchy argument since we never proved series uniformly convergent implies the sequence is uniformly convergent.
     
  5. Jun 6, 2015 #4

    micromass

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    That is:
    1) Incorrect, if ##g(x)## converges at some point ##R>0##, then you are only guaranteed convergence on ##[0,r]## for ##r<R##.
    2) Not what Abel's theorem says.

    See http://jlms.oxfordjournals.org/content/s1-39/1/81.extract [Broken]
     
    Last edited by a moderator: May 7, 2017
  6. Jun 6, 2015 #5
    i am quite literally picking this up verbatim from my text
    it is attached to this post
     

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    Last edited by a moderator: May 7, 2017
  7. Jun 6, 2015 #6

    micromass

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    Which book is this? The flaw in the proof seems to be assuming that ##\left( \frac{x}{R} \right)^n## is monotone decreasing, which is not the case for ##x=R##.
     
  8. Jun 6, 2015 #7
    understanding analysis, Abbot
     
  9. Jun 6, 2015 #8

    micromass

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    OK, never mind. The theorem and proof is correct. Your proof is correct too then.
     
  10. Jun 6, 2015 #9

    micromass

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    But, your proof is unnecessarily complicated by the way. It is much easier to find the maximum of the function on each interval, and to prove those maxima converge to ##0##. There is no real reason to transform this into a power series.
     
  11. Jun 6, 2015 #10
    Yeah we havent been given that theorem, this was me trying to tackle this thing with the tools available
     
  12. Jun 7, 2015 #11

    wabbit

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    Actually, your sequence ## f_n(x)=nx^n(1-x) ## does not converge uniformly on ## [0,1] ##.
    It converges pointwise to zero on that interval. However, if you set ## x_n=\frac{n}{n+1} ## which is the value at which ## f_n ## reaches its maximum over ## [0,1] ## , you find that ##f_n(x_n)=(\frac{n}{n+1})^{n+1}=(1-\frac{1}{n+1})^{n+1} \ _{\overrightarrow{n\rightarrow\infty}}\frac{1}{e}>0 ## and this proves that the convergence is not uniform.

    Edit : the reason your proof is incorrect, and the theorem you quote is not applicable here, is that ## \sum nx^n(1-x) ## is not a power series. You can of course write it as ## (1-x)\sum nx^n ## but then you will note that the power series ## \sum n x^n ## does not converge for ## x=1 ## so you cannot use that theorem.
     
    Last edited: Jun 7, 2015
  13. Jun 7, 2015 #12
    ok, so if it doesnt converge uni then isnt it weird that
    $$\lim_{n\rightarrow \infty}\int_{0}^{1} f_{n} \rightarrow 0?$$

    Isn't uniform convergence a necessary condition for that?
     

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  14. Jun 7, 2015 #13

    wabbit

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    No, uniform convergence is a sufficient condition for the convergence of the integral, but by no means a necessary one - as this example shows.
     
  15. Jun 7, 2015 #14

    Ray Vickson

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    This just implies that the series ##\sum_{n=0}^N x^n## converges uniformly on ##[0, 1 - \epsilon]## for any small ##\epsilon > 0##. Thus, ##\sum_{n=0}^N (1-x)x^n## converges to 1 uniformly on ##[0, 1-\epsilon]##. This does not say what happens right at ##x = 1##, because then you have a series of terms that all = 0 (and, somehow, you want that series to converge to 1?)
     
  16. Jun 7, 2015 #15
    hmmm, never said anything about converging to 1...I said that the theorem implies that it converges uni on [0,1]. It does say what happens at x = 1. The original series equals zero at x = 1. Therefore it converges at some point x = R = 1 > 0. Therefore it converges for [0,R] = [0,1]
     
  17. Jun 7, 2015 #16

    wabbit

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    As I mentionned in my previous post, the original series you wrote is not a power series so you cannot apply Abel's theorem to it.
     
  18. Jun 7, 2015 #17
    my only issue wabbit is that in my class we did not get the theorem you applied to show it is not uni conv. i do not know how to approach it without that theorem
     
  19. Jun 7, 2015 #18

    Ray Vickson

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    You did not mention convergence to 1, but I did.

    You have a series that converges uniformly to 1 for ##0 \leq x \leq 1 - \epsilon## (for any small ##\epsilon > 0##), but which converges to 0 when ##x = 1## exactly. You may call that uniform convergence on [0,1], but I---for one---would not.
     
  20. Jun 7, 2015 #19
    I do not see where you are getting that the series converges to 1. thats not even the point wise limit. Maybe i am misunderstanding you
     
  21. Jun 7, 2015 #20

    wabbit

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    I did not use a theorem in particular, could you be more specific as to what you find unclear in the proof I gave ?
     
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