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!

Reducing logarithms of factorials

  1. Sep 5, 2009 #1
    1. The problem statement, all variables and given/known data

    Part of a much bigger problem, but I am hung up on solving the following:

    [tex]ln\left [ \left(\frac{N+n}{2}\right ) ! \right ] = \left ( \frac{N+n+1}{2}\right) \frac{ln(N+n)}{2}\right )[/tex]

    I am trying to follow a proof in http://books.google.com/books?id=CD...on random walk&pg=PA205#v=onepage&q=&f=false" My confusion comes from eqn. 10.34. Clearly this is following the Stirling Approximation, where the associated substituted approximation is eqn. 10.33.

    2. Relevant equations

    The stirling approximation (Eqn. 10.33), as well as Eqns. 10.31 and 10.32.

    3. The attempt at a solution

    In trying to reduce the following:

    [tex]ln\left [ \left(\frac{N+n}{2}\right ) ! \right ] [/tex]

    I do not understand how I am to use the stirling approximation since it specifies the definition of [tex]ln(n!)[/tex], and not what I have above. Is there some identity that I am not remembering? I know that:

    [tex]ln\left [ \left(\frac{N+n}{2}\right ) ! \right ] [/tex]

    is NOT the same as

    [tex]ln\left [ \frac{(N+n)!}{2!} \right ] [/tex]

    But other than that I cannot remember a relevant identity. Thoughts?

    IHateMayonnaise

    EDIT: To solve use the Stirling's approximation twice. Nevermind :)
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?



Similar Discussions: Reducing logarithms of factorials
Loading...