Cox derivation of the laws of probability I have read Jaynes' Probability Theory: The Logic of Science a while ago, but mostly skimmed over parts of his derivations that I didn't immediately understand. Now I'm trying to really understand it, but it appears he mostly skips over steps he sees as obvious or trivial.
Now, I've been able to construct most of the missing steps, but there's one that's stumping me (page 32). He goes from
$\frac{S(x)}{y}=1-e^{-q}$
to
$S(y) = S[S(x)] + e^{-q}S(x)S'[S(x)]+O(e^{-2q})$
And I have no idea how. Can anyone explain it to me?
And it'd be even better if someone could point me to a more "friendly" derivation of the axioms of probability. He seems to make things up as he goes along (such as the variable q(x, y) or the function J(q)) without justifying why he decided that was the best step to take or where it came from, and it'd be great to have something that explained those steps to me.
--Edit:
S(x) is the function such that $S[w(A|B)] = w(\bar A|B)$ (so eventually one finds that $S(x) = (1-x^m)^{1/m}$), where w(A|B) is the function that relates the plausibility that A is true given that B is true to a real number between 0 and 1 (and that must obey the relation $w(AB|C) = w(A|C)w(B|AC) = w(B|C)w(A|BC)$).
Using x = w(A|C), y = w(B|C) and Jaynes is analysing the case where $\bar B = AD$ which yields the functional equation $xS(\frac{S(y)}{x}) = yS(\frac{S(x)}{y})$ and then he tries to analyse the behaviour of $S(1-\delta)$ for $\delta\rightarrow 0^+$ and to do so he defines $\delta = e^{-q}$ with q(x, y) defined as above and J(q) defined by $e^{-J(q)} = S(1-e^{-q})$.
 A: This isn't at all obvious, but here goes:
$$\frac{S(x)}{y}=1-\exp\{-q\}$$
Because $\exp\{-q\}$ is small we can take reciprocals to get
$$\frac{y}{S(x)}=1+\exp\{-q\}+O(\exp\{-2q\})$$
So
$$y=S(x)\left[1+\exp\{-q\}+O(\exp\{-2q\})\right]$$
Apply $S$ to both sides
\begin{align*}
S(y)&=S\left[S(x)\left(1+\exp\{-q\}+O(\exp\{-2q\})\right)\right]\\
&=S\left[S(x)+\exp\{-q\}S(x)+O(\exp\{-2q\})\right]
\end{align*}
And now Jaynes does a Taylor expansion! In particular we expand $S$ about $S(x)$ treating $exp\{-q\}S(x)$ as small.
$$S(y)=S\left[S(x)\right]+\exp\{-q\}S(x)S'\left[S(x)\right]+O(\exp\{-2q\})$$
We win.
EDIT: as for a better guide to Cox's theorem, I seem to recall that K. S. Van Horn's paper "Constructing a logic of plausible inference: a guide to Cox's Theorem" is quite readable. It's here.
EDIT2: Here's how the expansion goes: If we have an expression of the form $S[a+d]$ where $d$ is small then we can write
$$S[a+d]\simeq S[a]+S'[a]d$$
where the error in the approximation is $O(d^2)$. This is what we've done here but with $a=S(x)$ and $d=\exp\{-q\}S(x)$
