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Weierstrass M test

  1. Mar 21, 2004 #1
    Hi,


    I don't really grasp the M test for convergence and I have several Hmwk problems and figured that if someone could explain how to do one of them then I should be able to figure the rest out.

    ok - the problem is this:

    Use the M test to establish the uniform convergence of:

    (Sum from n - infinite) exp(in^2x) / n! for |z| < r where r < infinite.


    Any help would be really appreciated

    rachel.
     
    Last edited: Mar 22, 2004
  2. jcsd
  3. Mar 21, 2004 #2
    I think you need to clear up your notation first.

    What is "exp(in^2x / n!"?

    cookiemonster
     
  4. Mar 22, 2004 #3

    matt grime

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    The test states that

    [tex]\sum a_n(x)[/tex]

    converges absolutely and uniformly on some set of values of X if for all x in X |a_n(x)|<= M_n and

    [tex]\sum M_n[/tex]

    converges.


    So all you need to do is show the n'th term's absolute value satisfies this rule for some constant M_n, and the sum of these constants is convergent.
     
  5. Mar 22, 2004 #4
    Hi, the notation exp(in^2x) / n!
    means e to the power of i times n squared times x, all dividewd by n factorial where e is the exponential function, i is the aquare root of minus 1, n runs from 0 to infinity.


    I still dont understand how to do this problem. :(
     
    Last edited: Mar 22, 2004
  6. Mar 22, 2004 #5

    matt grime

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    Reread the statement of teh Weierstrass M test. It takes no understnading to actually use the test:

    a_n(x) = exp(ixn^2)/n!

    So what can you say about the absolute value of the n'th term in the series? So what is the sum of these values? Hence...?
     
  7. Mar 22, 2004 #6
    well - i probably wouldnt have posted a message if I understood the M test now would I?



    well - if the nth term is less than infinity, then exp(in^2x) is very big for some huge but not infinite value, and the denominator n! is also very big. Great, I still dont know if it converges or not!!
     
  8. Mar 22, 2004 #7

    matt grime

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    Firstly can you tell me if x is a real number or a complex number? your initial post half indicates complex by saying |z|<r some r, yet you use x inside the sum.

    If x is real this question is easy, though it appears from your last post that it might be complex.

    Understanding why the M test works and using it are two entirely different things.
     
  9. Mar 22, 2004 #8

    matt grime

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    Let us assume x is real because I think the question is wrong if we assume x is complex.

    the n'th term in the series is

    a(n) = exp{ixn^2}/n!


    |a(n)| = 1/n!, let this be M_n

    then sum M_n is convergent so the initial series is unifomrly convergent on the whole of the real line.

    I don't think it works for complex x since x = -it, for t some real number requires us to sum:

    e^{tn^2}/n!

    and that series deoesn't even converge as the ratio of consecutive terms is:

    e^{t(n+1)^2 -tn^2}/(n+1) = e^(2tn+t)/(n+1) which tends to infinity as n tends to infinity.
     
    Last edited: Mar 22, 2004
  10. Mar 22, 2004 #9
    Hi,

    Thank-you Matt. I appreciate the time you've spent looking at this.

    Yes - it is a complex question but with x real.

    So - essentially all you have to do is take the modulus of whatever you are summing over, and then see of the number gets closer to zero as n goes to infinity. Is it is straightforward as that?
     
  11. Mar 22, 2004 #10

    matt grime

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    Not quite - you must find for each n, a number M_n such that |a_n(x)|< M_n for all x in the domain, where a_n(x) is the n'th term in the series. It will be a different M_n for each n.

    Then the sum from 1 to infinity of the M_n must be finite (which is not the same as the M_n tending to 0).


    Example:

    [tex]\sum x^n/n![/tex]

    for x in the complex numbers and |x| < 10, say.

    a_n(x) = (x^n)/n!

    so |a_n(x)| < 10^n/n! := M_n

    now sum 1 to infinity of M_n converges hence the function is unifomrly continuous on the disc |x|<10
     
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