In Stephen Blundell's "Concepts in Thermal Physics" chapter 4 he derives the Boltzmann distribution. The equation that leads to the Taylor expansion is the following:
$$P_s(\epsilon) \propto \Omega(E-\epsilon)$$
where $P_s(\epsilon)$ is the probability of a system in thermal equilibrium with a reservoir being a microstate $s$ of energy $\epsilon$, and $\Omega(E-\epsilon)$ is the number of microstates associated with a reservoir of energy $E-\epsilon$. In order to get the Boltzmann distribution he performs a Taylor expansion of the following function:
$$\ln(\Omega(E-\epsilon))$$
where $\Omega(E-\epsilon)$ is the number of microstates associated with a reservoir of energy $E-\epsilon$, $E$ is the total energy of the reservoir and an attached system and is therefore constant, and $\epsilon$ is the energy of the attached system. Here $E \gg \epsilon$, and we take the Taylor expansion about $\epsilon = 0$ to get:
$$\ln(\Omega(E-\epsilon)) = \ln(\Omega(E)) -\frac{d\ln(\Omega(E))}{dE}\epsilon + \dots$$
I don't understand how he has arrived at this equation. It seems to me setting $\epsilon = 0$ is the same as setting $a = E$ in a standard Taylor expansion, so that part is fine, but surely as we are representing $x$ from the normal taylor expansion as $x = E-\epsilon$, the second term should be:
$$-\frac{d\ln(\Omega(E))}{d(E-\epsilon)}\epsilon$$
Why doesn't the second term have this form as its denominator? Surely I can't set the $\epsilon$ in the denominator to zero without setting the multiplying $\epsilon$ to zero also and wiping out the whole term?
Edit:
As $E$ is the total energy of the reservoir plus the system it is my assumption it stays constant and cannot be a variable.
Best Answer
I'm not sure what your objection is. A generic Taylor expansion might look like $$f(x-\epsilon) = f(x) + f'(x) (-\epsilon)+ \frac{1}{2} f''(x) (-\epsilon)^2 + \ldots$$
Perhaps the confusion is related to using the Liebniz notation rather than the prime notation for the derivative? Try defining $f(E) = \ln(\Omega(E))$ and then using the prime notation.