Statistical Mechanics - Physics at Oregon State University

236 APPENDIX A. SOLUTIONS TO SELECTED PROBLEMS.
and using Stirling
we arrive at
log g(M, N) = log(N − 1 + M)! − log(N − 1)! − log M!
log N! = N log(N) − N + 1
2 log(2πN)
log g(M, N) = (N − 1 + M) log(N − 1 + M) − (N − 1) log(N − 1) − M log(M)
− 1 1 − 1 + M
log(2π) + log(N
2 2 (N − 1)M )
Insert M = xN and replace N − 1 by N:
log g(M, N) = N(1 + x) log(N(1 + x)) − N log N − xN log(xN)
The maximum occurs if ∂
∂x
Consider the function:
and we need to solve
− 1 1 + x
log(2π) + log(1
2 2 xN )
log g(M, N) = N(1 + x) log(1 + x) − xN log(x)
− 1 1 + x
log(2π) + log(1
2 2 xN )
g = 0 which gives:
N log(1 + x) − N log(x) + 1
2
f(x) = N log(x) + 1 1
2 x
f(x + 1) = f(x)
The slope of the function f(x) is given by:
df N
=
dx x
1 1
−
2 x2
1 1
−
1 + x x
= 0
which is positive for x > 1
2N . Because g = f(x + 1) − f(x) this shows that g
does not have a maximum. The largest value is at x = ∞, which makes sense
if you think about the physics. But my apologies for making the last part of the
problem misleading.