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Testing for Uniform Continutity with Weierstrass MTest

  1. Oct 2, 2013 #1
    1. The problem statement, all variables and given/known data

    I am trying to make sure I am using the Weierstrass test correctly. So, on the following expression: $$f_n(x) = \frac{x}{1+nx^2}, -1 \le x \le 1$$

    I looked at it this way. First I checked to see what happens at -1 and 1. The "plug in" test gets me a limit of 1/2 at x=1 and n=1, and a limit of 1/2 at x=-1 and n=1. All-righty then, we do a little L'Hôpital's rule: $$\lim_{n \rightarrow \infty} f_n(x) = \frac{\frac{d}{dn}(x)}{\frac{d}{dn}(1+ nx^2)}= \frac {0}{x^2}=0 $$

    So I know this converges to zero. But does it do so uniformly? Well, I can see that this function has a bound at x=1/2. So I will use [itex]\frac{1}{1+n}[/itex] as my M-function. Giving us:
    [itex]M_k = \frac{1}{1+n},[/itex] and [itex]\sum_{n=1}^{\infty}\frac{1}{1+n}[/itex].

    [itex]\frac{1}{1+n}[/itex] is always going to be smaller than [itex]\frac{x}{1+nx^2}[/itex] for any n with a particular x. We can check his out: since x is between -1 and 1, and n≥1, the [itex]1+nx^2[/itex] is always going to be less than 1+n.

    So we can say $$ \left| \frac{x}{1+nx^2} \right| \ge \frac{1}{1+n}$$ for all n. In addition to that, [itex]\frac{1}{1+n} \lt \infty[/itex] for all n, but the other condition doesn't hold, so we don't have uniform convergence.

    Did I get it right?

    Thanks!
     
  2. jcsd
  3. Oct 2, 2013 #2

    Office_Shredder

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    L'Hopital's rule isn't applicable since only the denominator goes to infinity as n goes to infinity (incidentally the observation of this gives you that the limit is zero)

    When x = 0 your claimed inequality is 0 > 1/(1+n) which is false (and in general I think for all x it points the wrong way)

    The Weierstrass M-test is a statement about a sum of functions, not a sequence of functions, so you shouldn't need it here. Are you trying to determine whether
    [tex] \lim_{n\to \infty} f_n(x) = 0 [/tex]
    is a uniformly convergent limit, or are you trying to decide if
    [tex] \sum_{n=0}^{\infty} f_n(x) [/tex]
    is uniformly convergent?
     
  4. Oct 2, 2013 #3
    I am trying to find the limit and decide if it is uniformly convergent on the interval. What steps should I have taken?
     
  5. Oct 2, 2013 #4

    Office_Shredder

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    You should start by writing the definition of uniform convergence down... once you have that we can work through how to show it holds (or observe that it doesn't hold)
     
  6. Oct 2, 2013 #5
    Sorry, should edit this: Well as I understand it if [itex]\sum_{n=0}^{\infty}f_n \rightarrow f[/itex] on a given interval then it is uniformly converging. Is that correct? The M-test I thought also helped to establish that.
     
  7. Oct 2, 2013 #6
    by the way the convergence is for the sequence, (the book says) so I assume they mean the series is uniformly converging.
     
  8. Oct 2, 2013 #7

    Office_Shredder

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    No, the sum might converge pointwise but not uniformly, the M-test is a way of checking whether it converges uniformly.

    If the book says the sequence is uniformly converging, then they are asking whether
    [tex] \lim_{n\to \infty} \frac{x}{1+nx^2} [/tex]
    converges uniformly to its limit, which is zero. It is NOT asking if
    [tex] \sum_{n=1}^{\infty} \frac{x}{1+nx^2} [/tex]
    is converging uniformly, unless it at some point says "Is the series whose terms are:" or writes out the sum explicitly. In fact you can check that when x=1 that sum doesn't even converge.
     
  9. Oct 2, 2013 #8
    so then, even though the inequality I had doesn't hold all the time, isn't that sort of the point? We don't have M greater than the orginal limit, not all the time, so it isn't uniformly converging. No?
     
  10. Oct 2, 2013 #9

    Office_Shredder

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    What is M? I'm going to re-iterate a previous suggestion, which is write down the definition of uniform convergence, and then plug in your function and see what it is asking you to prove/disprove.
     
  11. Oct 3, 2013 #10
    M is supposed to be a sequence of numbers, yes? Anyhow the book says: we say that [itex]f_n[/itex] converges uniformly on a set E (an interval in this case, -1 to 1) if given ε > 0 we can find a positive integer N s.t. for all n≥N [itex]| f_n(x) - f(x)| < \epsilon[/itex] [itex] \forall x[/itex] in the set E.
     
  12. Oct 3, 2013 #11
    But given that definition of uniform convergence, I honestly don't find it that helpful. I am trying to follow the instructions on the exercise, you know? It would help if you could tell me where the mistakes are. When you say "what is M?" I would answer, a sequence of nonnegative numbers, and in the M-test the point is that it is always greater than the function in the sequence -- that is, [itex]M_k \ge |u_k|[/itex] for any given k. Did I not do that?

    (Try not to be too mysterious here. I am not a math major, I am a physics guy in the context of this class and I am NOT familiar with much of the vocabulary of higher math - pretend I am the dumbest ever student you have had).
     
  13. Oct 3, 2013 #12

    Office_Shredder

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    The M-test is for deciding on the uniform convergence of a series, not a sequence. You answered the question in the OP incorrectly because you were attempting to talk about the uniform convergence of the series whose terms are fn(x), not the sequence whose terms are fn(x) - these are two very different things and it is important to keep straight which one you are dealing with. If this distinction is confusing to you, then you should go back to your notes on sequences and series and study on them, because precise terminology is required to answer mathematics questions, and this terminology is definitely a prerequisite to understanding the topic you are studying.

    OK, so let's plug in fn(x) and f(x) into your definition of uniform convergence:
    [tex] \left| f_n(x) - f(x) \right| = \left| \frac{x}{1+nx^2} - 0 \right| = \left| \frac{x}{1+nx^2} \right|[/tex]

    The definition of convergence says that if you fix x, as n goes to infinity this thing goes to zero, which is something that you have proven. The definition of uniform convergence says that if you let x be arbitrary in [-1, 1], you can still find a small upper bound on this expression as n gets large. A good starting point for this problem is to take the expression
    [tex] \left| \frac{x}{1+nx^2} \right| [/tex]
    and find out, for fixed n, what is the largest value it can take on [-1,1].
     
  14. Oct 3, 2013 #13
    OK, the largest value it can take is going to be 1/2, assuming n>=1. If n=0 it is 1. So I would posit that the upper bound is 1.
     
  15. Oct 3, 2013 #14
    and i the upper bound is 1, and as n--> infinity the expression goes to zero no matter what, for any fixed X, does it not fit the definition of uniform convergence?
     
  16. Oct 3, 2013 #15

    jbunniii

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    That's not the answer I get. The upper bound decreases as ##n## increases. How did you find the maximum value of this function?

    By the way, Office_Shredder already pointed out that this has nothing to do with the Weierstrass M test. I just thought I'd mention that it also has nothing to do with uniform continuity. So the thread title is doubly misleading. You might want to change it to something like "testing sequence for uniform convergence."
     
    Last edited: Oct 3, 2013
  17. Oct 3, 2013 #16
    I plugged in the values at each end of the interval. Then I checked what happens when n and x are both zero, which would maximize the function it seemed to me since the denominator would be 1 and it can't be less than 1. I even graphed it to be sure. I tried taking a derivative to see whe it was zero as well, but that didnt seem to work well.
     
  18. Oct 3, 2013 #17

    jbunniii

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    What happened when you took the derivative? What is
    $$\frac{d}{dx}\left(\frac{x}{1+nx^2}\right)$$
     
  19. Oct 3, 2013 #18
    Taking a derivative I did the following:

    $$\frac{d}{dx}\left(\frac{x}{1+nx^2}\right)= \frac{d}{dx} (x(1+nx^2)^{-1})= -x(1+nx^2)^{-2}(2nx)+(1+nx^2)^{-1}=\left(\frac{-2nx^2}{(1+nx^2)^2}+\frac{x}{1+nx^2}\right)$$ which can be "simplified" to $$\left(\frac{-2nx+x(1+nx^2)}{(1+nx^2)^2}\right)$$

    and I notice that the numerator goes to zero at x=0, which would say to me that it reaches a max or minimum there, yes?
     
  20. Oct 3, 2013 #19

    jbunniii

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    How did you get the ##x## in the numerator in the second term of:
    $$\left(\frac{-2nx^2}{(1+nx^2)^2}+\frac{x}{1+nx^2}\right)$$
    Your simplification doesn't look consistent with the previous line, either. Here is what I get for the derivative:
    $$\frac{d}{dx}\left(\frac{x}{1+nx^2}\right) = \frac{(1+nx^2)(1) - (x)(2nx)}{(1+nx^2)^2} = \frac{1-nx^2}{(1+nx^2)^2}$$
    For what value(s) of ##x## is this expression equal to zero?
     
  21. Oct 3, 2013 #20
    You're right; that extra x was the mistake that was making the whole thing difficult for me earlier. I see it never goes to zero for either n or x at zero (though we'd be starting with n=1, no?) but it does go to 1 when x=0 and n=0, and at n=1 for any x it goes between zero and 1/2, and at n=2 it goes to -1/3, and at n=3 it goes to -2/4 et cetera. Am I on to something here if I say that it does seem to always be <1?
     
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