1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Weierstrass M test

  1. Mar 21, 2004 #1

    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

    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!"?

  4. Mar 22, 2004 #3

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    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]


    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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:


    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

    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

    User Avatar
    Science Advisor
    Homework Helper

    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).


    [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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Weierstrass M test
  1. Weierstrass Fourier (Replies: 5)