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Solving for x with ln and e

  1. Oct 22, 2005 #1
    The question is: for a Gaussian distribution what is the mathematical relationship between the FWHM and the standard deviation.

    The equations I'm using are:

    [TEX]N(x) =\frac {\ a}{2}[/TEX]
    [TEX]N(x) = Ae^- \frac {\ (x-x_2)^2}{2 sigma^2}[/tex]

    I equated the equations and started to solve for x. I know you take the ln of both sides to get rid of e. But a few of my friends have 2ln2 in their answer...how do you get ln2? isn't it ln 1/2?
     
  2. jcsd
  3. Oct 22, 2005 #2

    Tide

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    The "extra" factor of 2 arises from the fact that there are two values of x for which N = A/2 and the difference between them gives you the FWHM.
     
  4. Oct 22, 2005 #3
    Sorry, I mean the 2 in ln2, not the factor 2

    N(x) = A/2 ....the A cancels, giving you 1/2
    then taking the ln of both sides would give you ln 1/2, right? not ln2

    I'm getting [tex]sqrt(2*sigma*ln \frac {\1}{2}) = -(x-x_2)


    Thanks.
     
  5. Oct 22, 2005 #4

    Tide

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    You do understand that [itex]\ln \frac {1}{2} = - \ln 2[/itex]?
     
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