Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Statistical Physics Basics

  1. Sep 21, 2008 #1
    1. The problem statement, all variables and given/known data

    Flip N fair coins. The distribution for different numbers of heads and tails should be peaked at N/2. When N is very large, the peak will be very high. Let x = N(head)-N/2, required to find an expression for this distribution near the peak, i.e. x<<N.

    2. Relevant equations

    Strling Approx
    ln(1+x)=x for small x

    3. The attempt at a solution

    30bjvyr.jpg

    clearly incorrect because square of (N/2)! must be smaller than the product of (N/2+x)! and (N/2-x)!, but I can't find where goes wrong.
     
  2. jcsd
  3. Sep 21, 2008 #2
    1. Going over to the 5th line, you omitted [itex]-x \ln{(\frac{N}{2}+x)}[/itex] and [itex]x \ln{(\frac{N}{2}-x)}[/itex]. These give rise to additional corrections.
    [itex]-x \ln{(\frac{N}{2}+x)}+x\ln{(\frac{N}{2}-x)}=-x\ln{\left(\frac{1+2x/N}{1-2x/N}\right)}\approx -x\ln{(1+4x/N)}\approx -4x^2 /N[/itex]

    2. In the 7th line, you need to be more careful when expanding to the 2nd order. You should proceed as below.
    [itex]\ln \left(\frac{N/2}{N/2+x}\right) = \ln \left(\frac{1}{1+2x/N}\right) \approx \ln \left( 1-\frac{2x}{N}+\frac{4x^2}{N^2} \right) \approx -\frac{2x}{N}+\frac{4x^2}{N^2}-\frac{1}{2} {\left(-\frac{2x}{N}+\frac{4x^2}{N^2} \right)}^{2} [/itex]

    [itex] \approx -\frac{2x}{N}+\frac{4x^2}{N^2}-\frac{1}{2} \frac{4x^2}{N^2} = -\frac{2x}{N}+\frac{2x^2}{N^2}[/itex]

    and similarly

    [itex]\ln \left(\frac{N/2}{N/2-x}\right) \approx \frac{2x}{N}+\frac{2x^2}{N^2} [/itex]


    Then, the final answer becomes [itex]-2x^2 /N[/itex] rather than [itex]4x^2 /N [/itex].
     
  4. Sep 21, 2008 #3
    Thanks, but it doesn't make sense... when N is very large, the peak should be very narrow, but from the equation, N becomes larger then the exponential decreases much slower.
     
  5. Sep 22, 2008 #4
    You are right. The width of the peak actually increases, as sqrt(N).
    Yet, if you consider (# of heads)/(# of coins), the width decreases as 1/sqrt(N).
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook