I'm going through "Probability Theory  The logic of science" written by E.T. Jaynes and I have a problem with one step on page 27/28 in the proof of the product rule. The idea here is that we have a function $F$ for which: $$(ABC)=F[(BC), (ABC)]$$ We want to proof its associativity: $$F[F(x, y), z] = F[x, F(y,z)]$$ We assume that F is differentiable. We denote: $$u = F(x,y)$$ $$v = F(y,z)$$ $$F_1(x,y) = \frac{\partial F}{\partial x}$$ $$F_2(x,y) = \frac{\partial F}{\partial y}$$ With this assumptions, we want to prove: $$F(x, v) = F(u, z)$$ We differentiate w.r.t. x and y: $$F_1(x,v)=F_1(u,z)F_1(x,y)$$ $$F_2(x,v)F_1(y,z)=F_1(u,z)F_2(x,y)$$ We then use: $$G(x,y)=\frac{F_2(x,y)}{F_1(x,y)}$$ And get equation E1: $$G(x,v)F_1(y,z)=G(x,y)$$ Which can be rewritten as equation E2: $$G(x,v)F_2(y,z)=G(x,y)G(y,z)$$ Now there's the part is hard for me to understand. We differentiate E1 w.r.t. z and E2 w.r.t. y. The proof states that left hand sides of both derivatives are the same. How is that? When I'm using product rule of calculus and chain rule, I get summation in E1 left side: $$G_2(x,v)F_2(y,z)F_1(y,z)+G(x,v)F_{12}(y,z)$$ and for equation E2, left side equals: $$G_2(x,v)F_1(y,z)F_2(y,z)+G(x,v)F_{21}(y,z)$$ So there's the quation  where's my mistake? I can see that first product of both equations derivative are the same but how this equations are the same? I can see two options: first, I shouldn't use product rule here (why?), second, for some reason $$F_{12}(y,z)=F_{21}(y,z)$$ but I can't see reason for any of these. I've checked original Cox proof in his book "The algebra of probable inference" but it's quite the same as in Jaynes book.

Did you ever end up finding an answer to this? – Moo Oct 29 '17 at 05:48

The order that you do the two partial differentiations does not matter \begin{eqnarray*} \frac{ \partial^2 F(y,z) }{\partial y \partial z} = \frac{ \partial^2 F(y,z) }{\partial z \partial y} \end{eqnarray*} and thus $F_{12}(y,z)=F_{21}(y,z)$. – Donald Splutterwit Oct 30 '17 at 00:57

1As the OP mentioned in a comment, that would require that the second partial derivatives are continuous as per Schwarz's theorem. As far as I'm aware there weren't any assumptions made that would guarantee this. – Moo Oct 30 '17 at 03:34
2 Answers
Upon reading this chapter again, I believe the symmetry of the second derivatives was part of the assumptions Cox made. Jaynes suggests looking at Aczél's Lectures on Functional Equations and Their Applications for a derivation of the general solution without assumptions. He also revisited it here. Unfortunately, it's paywalled but his lack of assumptions is mentioned in the abstract:
Abstract Associativity is regarded as functional equation. Two ways of obtaining its general continuous strictly increasing solutions on real intervals are described. No differentiability or existence of neutral element or commutativity is pre‐assumed. Comments are made, among others on applications, and a generalization is also presented.
 73
 7
Yes, the double derivatives are the same whatever the performance order.
$$\begin{align}F_{12}(s,t) ~=~& \frac{\mathrm d ~}{\mathrm d s}\frac{\mathrm d ~}{\mathrm d t}F(s,t) \\[1ex] =~& \frac{\mathrm d ~}{\mathrm d t}\frac{\mathrm d ~}{\mathrm d s}F(s,t) \\[1ex]=~& F_{21}(s,t)\end{align}$$
 119,730
 6
 49
 108

5Thanks for answer. Unfortunately, I'm still not sure, why we can use Schwarz's theorem. We assume that F is continous and differentiable, is it enough to assure the symmetry of second derivative? – Marek G. Aug 05 '16 at 08:52