1. Limited time only! Sign up for a free 30min personal 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!

Continuous function

  1. May 9, 2007 #1
    How do I show that f(x)=sum (from n=0 to infinity) cos(nx)e^-nx is a continuous function? x is from (0, infinity)

    So, I need to show that the series converges uniformly. I'm trying to say that |cos nx e^-nx| <= |e^-nx| and use Weierstrass comparison, but I can't find a function M_n to use for Weierstrass.

    I feel like I'm missing something really simple...
    Last edited: May 9, 2007
  2. jcsd
  3. May 10, 2007 #2
    For positive [itex]n[/itex] what is [itex]e^{-nx}[/itex] always less than?
  4. May 10, 2007 #3
    1, but the sum of 1s is obviously not convergent.
  5. May 15, 2007 #4
    Okay, lets see:

    [tex]f(x) = \sum_{n = 0}^{\infty} e^{-nx}\cos(nx)[/tex]


    [tex]|e^{-nx}\cos(nx)| \leq e^{-nx}[/tex]

    Now, as you said, [itex]e^{-nx} < 1[/itex] for all [itex]x \in (0,\infty)[/itex] and nonnegative n. So, clearly, your "[itex]M_{n}[/itex]" is simply the (convergent) geometric sum

    [tex]\sum_{n = 0}^{\infty} e^{-nx}[/tex]
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Continuous function
  1. Continuous Functions (Replies: 2)

  2. Continuity function (Replies: 8)