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Wierstrauss m test caucy criteria

  1. Nov 29, 2008 #1
    1. The problem statement, all variables and given/known data
    So, I missed class the day Weierstrass M-test and cauchy criterion. I have a theorem for Weierstrass M-test, but the book gives no examples on how to use it. I figure cauchy criterion has to do with series, and states that a series is convergent if the partial sums is cauchy. If this one is wrong please let me know. I made it up, just now. We should be in the sequence and series of functions chapter, and there are is a lot of things that look cauchy (fn(x) - f(x)) < epsilon, but none called "cauchy criterion for series" like the email I get.


    2. Relevant equations
    Suppose {fn} is a sequence of functions defined on E, and {Mn} is a sequenc of nonnegative real numbers such that | fn(x) | <= Mn for all x in E, all positive integers n. If
    the Sum Mn converges then the sum of fn converges uniformly.


    3. The attempt at a solution
    My guess is that how to use Weierstrass M test is sort of like a comparison test to derive uniform convergence. If you can get a series like 1/x^2 to converge and the sequence of functions is always less than 1/x^2, then the sequence of functions is uniformly convergent. Is this sounding right? There is no homework on this, so I don't have any examples to solve.

    I'm really confused, they did a chapter in a day, which means I probably am not expected to get the entire depth down, but I get sad if I can't get it all the way.
     
  2. jcsd
  3. Nov 29, 2008 #2

    Office_Shredder

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    You have the Cauchy criterion correct, but it's often written as the series converges iff for all epsilon > 0, there exists N such that n,m>N implies

    [tex]| \sum_{k=m}^{n}a_k | < \epsilon [/tex]

    (which is equivalent of course, but just written out in a more conventional form)

    The M-test you have the right idea... for example, I take the sequence of functions fn = 1/xn for 1/2<x<3/4

    Then

    [tex]\sum_{n=1}^{\infty}f_n(x)[/tex] converges uniformly to [tex]\frac{1}{1-x}[/tex]

    Proof:

    Clearly it converges pointwise (geometric series). It's uniform as for any x, |fn(x)|<= (3/4)n which we'll call Mn

    Clearly [tex]\sum_{n=1}^{\infty}M_n[/tex] converges, and hence the series
    [tex]\sum_{n=1}^{\infty}f_n(x)[/tex] converges uniformly.

    Notice the domain is important here... if we had picked 0<=x<1, this would have failed as we would have gotten the smallest possible value for each Mn would be 1, and [tex]\sum_{n=1}{\infty}M_n[/tex] no longer converges, and you can't conclude the series converges uniformly to 1/(1-x)

    You may have noticed that conveniently our point picked was one of the endpoints of the domain we were looking at.... this is often the case, since the M-test is usually applied to power series, and each monomial term is going to be monotonically increasing.

    Wikipedia has another example

    http://en.wikipedia.org/wiki/Uniform_convergence#Exponential_Function

    it's done over the complex numbers, but if instead of open disks you substitute open intervals, and instead of C you substitute R, it's exactly the same
     
  4. Nov 30, 2008 #3
    Should that be fn = x^n for 1/2 < x < 3/4. Then it should work on any interval [0,a] , where a < 1 correct?

    So, the real Idea, is that you find the M. You choose one that fn is going to be less than for all n, and then because you can get Mn less than epsilon, you can clearly get all the fn's less than that epsilon for free. If that's correct, this is another example of why I love analysis; it's really a playful subject.
     
  5. Nov 30, 2008 #4

    Office_Shredder

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    Uhh... yeah, you're right. My bad
     
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