Results 1 to 2 of 2

Math Help - Full Width Half Max question??

  1. #1
    Newbie
    Joined
    May 2010
    Posts
    13

    Full Width Half Max question??

    I have a plotted a graph of:
    \frac{I(t)}{|E_p|^2}=e^{(-2\alpha t^2)}

    Now I understand that the maximum is when t=0, however what about the full width half max?? Why is the half height 1/e??

    Can someone help me obtain the full width half max of this?

    Thanks
    Chris
    Last edited by chutsu; May 16th 2010 at 01:39 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,368
    Thanks
    1313
    I had to google "full width half max". Apparently, it is the distance between x_1 and x_2 for points such that f(x_1)= f(x-2) = 1/2 the maximum value of the function.

    You have correctly calculated that the maximum value of this function occurs at x= 0 and, so, is e. But the "half max" is not 1/e, it is, as you should guess, half of e: e/2.

    You need to solve f(x)= e^{-2\alpha x^2}= \frac{e}{2}. Since f is symmetric about x= 0, the "full width at half maximum" is twice that x value.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Full Binary Tree Question
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: May 12th 2010, 05:22 PM
  2. Replies: 1
    Last Post: April 29th 2010, 08:48 PM
  3. Replies: 0
    Last Post: March 5th 2010, 11:03 AM
  4. Half Width Half Maximum
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 27th 2008, 04:30 AM
  5. Half-full frustum
    Posted in the Geometry Forum
    Replies: 9
    Last Post: December 23rd 2006, 10:05 AM

Search Tags


/mathhelpforum @mathhelpforum