How are $x$ and $a$ defined in the Taylor Series Expansion of the Boltzmann Distribution?

Question

I am reading a book on statistical mechanics in which the author uses a Taylor Expansion to derive the Boltzmann distribution. He starts with the following equation for the probability of any state $\epsilon$:

$$P(\epsilon) \propto \Omega(E- \epsilon) \times 1$$

The author takes the natural logarithm of both sides and expands the right hand side in a taylor expansion about $\epsilon = 0$, to get this:

$$ln\left(\Omega(E- \epsilon)\right) = ln(\Omega(E)) - \frac{d \ ln(\Omega(E))}{dE}\epsilon + ...$$

I don't understand how the author has reached this, the corresponding standard taylor series expansion about $a$ is:

$$f(x) = f(a) +\frac{d \ f(x)}{dx}(x-a)+...$$

I don't see which part of $ln\left(\Omega(E- \epsilon)\right)$ should be considered $x$, and which part should be considered $a$. Should $(E- \epsilon)$ all be considered $x$, should it be only $E$ or $\epsilon$?

Answer 1

In your post it is clear that $x=\epsilon$ and $a=0$. As such let $$ f(x)=\log\Omega(E-x). $$ Expanding about $x=0$ we subsequently have for the first two terms of the Taylor seires $$ \tag{1} f(x)=f(0)+\left(\frac{\mathrm d}{\mathrm dx}\log\Omega(E-x)\Big|_{x=0}\right)x+\cdots. $$ Upon inspection $f(0)=\log\Omega(E)$. Now consider some function $g$ and observe that $$ \frac{\mathrm d}{\mathrm dx}g(E-x)\Big|_{x=0}=-g^\prime(E-x)\Big|_{x=0}=-g^\prime(E). $$ But we can get the same result by writing $$ -\frac{\mathrm d}{\mathrm dE}g(E)=-g^\prime(E). $$ Thus, $$ -\frac{\mathrm d}{\mathrm dE}g(E)=\frac{\mathrm d}{\mathrm dx}g(E-x)\Big|_{x=0}. $$ Bringing both results together yields $$ \tag{2} f(x)=\log\Omega(E)-\frac{\mathrm d}{\mathrm dE}\log\Omega(E)x+\cdots, $$ which is the result in your textbook.