I was reading Intro to Modern Statistical Mechanics by David Chandler, on page 63. He states the following:

we can expand $\ln\Omega(E-E_v)$ in the Taylor series $$\ln\Omega(E-E_v) = \ln\Omega(E) - E_v\frac{d\ln\Omega}{dE} + \cdots $$

Here $E_v$ denotes the energy for a particular system state $v$.

I was trying to compare this to the Taylor series formula: $$ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!}+ \cdots$$

But I am not sure I am following what the author said.

  • $\begingroup$ Note that the Taylor serie for logarithm has very slow convergence. With advantage is used e.g the Padé approximation instead. See www.nezumi.demon.co.uk/consult/logx.htm $\endgroup$
    – Poutnik
    Mar 25, 2020 at 0:18

1 Answer 1


Remember that $\Omega$ is a function of $E$ so you have to use the chain rule. The equivalent mathematical formulation is - $$\ln(f(a-x))\approx\ln(f(a))-\underbrace{\frac{f'(a)}{f(a)}}_{\frac{\partial\ln(f(x))}{\partial x}\large|_{x=a}}x+\dots$$

and so on... Use the following command in mathematica/WolfraAlpha to get expansion to order $n$ (replace $n$ with integer)

Series[Log[f[a - x]], {x, 0, n}]
  • 2
    $\begingroup$ Good answer, although the Mathematica command seems like an unnecessary addition. $\endgroup$
    – David Z
    Mar 25, 2020 at 0:05

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