Reducing logarithms of factorials

[IHateMayonnaise]

Homework Statement

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

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

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.

Homework Equations

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

The Attempt at a Solution

In trying to reduce the following:

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

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

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

is NOT the same as

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

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

IHateMayonnaise

EDIT: To solve use the Stirling's approximation twice. Nevermind :)

 

[mighty2000]

Hello IHateMayonnaise,

To solve this problem, you can use Stirling's approximation twice. First, use it to approximate the factorial term in the left side of the equation:

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

Then, use it again to approximate the term in the right side of the equation:

\left ( \frac{N+n+1}{2}\right) \frac{ln(N+n)}{2}\right ) \approx \left ( \frac{N+n+1}{2}\right) \left ( \frac{N+n}{2}\right) - \frac{N+n}{2}

Now, you can simplify the equation by cancelling out the common terms (i.e. -\frac{N+n}{2}) and rearranging the remaining terms:

\left ( \frac{N+n}{2}\right)ln\left ( \frac{N+n}{2}\right ) = \left ( \frac{N+n+1}{2}\right) \left ( \frac{N+n}{2}\right)

\Rightarrow ln\left ( \frac{N+n}{2}\right ) = \frac{N+n+1}{2}

Finally, you can solve for N by exponentiating both sides and simplifying:

\frac{N+n}{2} = e^{\frac{N+n+1}{2}}

\Rightarrow N = 2e^{\frac{N+n+1}{2}} - n

I hope this helps! Let me know if you have any other questions.