Why is this Taylor Expansion, Leading to the Boltzmann Distribution, Acceptable? 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.
 A: 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.
A: I think you are just getting a little lost in notation; there is nothing particularly deep happening.
The way the Taylor expansion is often defined by mathematicians (eg see wikipedia) is to relate the value of a function at some point $x$, with the value of the function (and derivatives) at some other point $a$, like so
\begin{equation}
f(x) = f(a) + \frac{df}{dx}\Big|_{x=a} (x-a) + \frac{1}{2} \frac{d^2f}{dx^2}\Big|_{x=a}  (x-a)^2 + \cdots
\end{equation}
This point of view makes sense if you want to ask questions like "what is the radius of convergence of the Taylor series done at a point $a$" that a mathematician is interested in.
In physics, we are almost always interested in using the Taylor series to approximate some function. Therefore, defining $|x - a| = \epsilon$, we are almost always interested in the case where $\epsilon$ is small (in some units).
We therefore tend to use notation that makes it easy to truncate the series when $\epsilon$ is small. To do this, we replace $x\rightarrow x-\epsilon$ and $a\rightarrow x$ in the above expression. Note this requires a change in how we think about what $x$ means. Previously, we thought of $a$ as a fixed reference point for the expansion, and $x$ as a test point that we can vary; we want to know the value of $f$ at $x$ given knowledge of $f$ and its derivatives at $a$. Now, $x$ is the reference point, and we vary $\epsilon$; we want to know the value of $f$ at $x-\epsilon$ given the value of $f$ and its derivatives at $x$.
It's also important to know that the variable "in the denominator of the derivative" is a dummy variable. We can use any symbol for it, since ultimately we are going to evaluate the derivative at the reference point $x$. For clarity, I will use $y$ as the variable in the derivative.
With all this in mind, we make the substitutions described in words above:

*

*$x \rightarrow x-\epsilon$

*$a \rightarrow x$

*$\frac{d^n f}{d x^n}\Big|_{x=a} \rightarrow \frac{d^n f}{d y^n}\Big|_{y=x}$
and we find
\begin{equation}
f(x-\epsilon) = f(x) + \frac{df}{dy}\Big|_{y=x} \left(-\epsilon\right) + \frac{1}{2}\frac{d^2 f}{dy^2}\Big|_{y=x} \left(-\epsilon\right)^2 + \cdots
\end{equation}
As you can see, this is an exactly equivalent way of representing the Taylor expansion as defined on wikipedia -- all we have done is change variables. This form is the same as the one in your question -- all we need to do to fully make the connection is rename $x$ as $E$, and $f(x)$ as $\ln \Omega(E)$.
A: Technically, you CAN take derivative with respect to $E^{\prime} = E-\varepsilon$. Expansion around $E^{\prime} = E$  should look like that then:
$$
\ln{\Omega\left( E^{\prime} \right)} \approx \ln{\Omega\left( E \right)} + \left. \frac{d \ln{\Omega\left( E^{\prime} \right)}}{ d E^{\prime} } \right|_{E^{\prime} = E} (E^{\prime} - E) = \ln{\Omega\left( E \right)} - \left. \frac{d \ln{\Omega\left( E^{\prime} \right)}}{ d E^{\prime} } \right|_{E^{\prime} = E} \varepsilon
$$
Compare to the traditional expansion around $x_0$:
$$
f(x) \approx f(x_0) + \left. \frac{d f(x)}{d x} \right|_{x=x_0}(x-x_0)
$$
Basically the same. The trick is that you can differentiate function over its argument, and if the argument is constant then set it to the constant value. Even if $E$ is constant, you still can take derivative over that. It's like saying "IF the energy changed, what would be the derivative?"
Or, in other words, you want to describe the behavior of a function, not its variables. You can rename the variables as you wish:
$$
f(x) \approx f(x_0) + \left. \frac{d f()}{d } \right|_{=x_0}(x-x_0),
$$
and that is still correct.
Sometimes, for convenience, to avoid drawing that ugly stick on the right, people write something like in your textbook. In conventional functions it can be written like that:
$$
f(x) \approx f(x_0) + \frac{d f(x_0)}{d x_0}(x-x_0)
$$
That is a little bit less traditional form, but still true. It's like you are saying at the very beginning at which point you take the derivative.
UPD: I was using partial derivatives initially. Sorry, got used to it.
