One approach to proving the expansion
[eq 9.1]
f(T_x(x,y,\alpha),T_y(x,y,\alpha) = f(x,y) + U f(x,y)\ \alpha + \frac{1}{2!} U^2\ \alpha^2 + ....
using "just calculus" is to define U so that U^n f(T_x(x,y,\alpha),T_y(x,y,\alpha) is (exactly) equal to D^n_\alpha f( T(_x(x,y,\alpha), T_y(x,y,\alpha)).
That will ensure
[eq. 9.2]
(U^n f(T_x(x,y,\alpha),T_y(x,y,\alpha)))_{|\ \alpha = 0 } = (D^n f(T_x(x,y,\alpha),T_y(x,y,\alpha)))_{|\ \alpha = 0}.
The only question left open will be whether
[eq. 9.3]
(U^n f(T_x(x,y,\alpha),T_y(x,y,\alpha)))_{|\ \alpha = 0 } = U^n f(x,y)
Settling that question may or may not require more advanced mathematics.
I'll summarize this approach (and ignore criticism that I use too many letters, in view of the fact that alternatives feel free to employ \phi,\psi,\eta,\xi, U, u_x, u_y,f,f_1, x, y, x_0,y_0,x_1,y_1,\epsilon, \delta x, \delta y, \delta \alpha.)The elements of the 1-parameter group are the mappings T(C): (A,B) \rightarrow ( T_x(A,B,C), T_y(A,B,C)).
We assume the parameterization with C is done so that
[Eq. 9.4]
T(0) = (A,B) (i.e. T(0) is the identiy map.)
[Eq. 9.5]
T( T(\beta), \alpha) = T(\beta + \alpha).
Assume all derivatives mentioned in this post exist.
The following two results were proved in post #49:
Theorem 9-1:
[Eq. 9.6]
\frac{\partial T_x}{\partial C} =\frac{\partial T_x}{\partial C}_{|\ C = 0}\frac{\partial T_x}{\partial A} + \frac{\partial T_y}{\partial C}_{|\ C = 0}\frac{\partial T_x}{\partial B}
[Eq. 9.7]
\frac{\partial T_y}{\partial C} =\frac{\partial T_x}{\partial C}_{|\ C = 0}\frac{\partial T_y}{\partial A} + \frac{\partial T_y}{\partial C}_{|\ C = 0}\frac{\partial T_y}{\partial B}
To prove theorem 9-1, let C = \alpha + \beta, differentiate both sides of the coordinate equations implied by T(\alpha + \beta) = T( T(\alpha),\beta) with respect to \beta. Set \beta = 0. Then deeply contemplate the verbal interpretation of the notation in the result!
Develop condensed notation for Theorem 9-1 by making the definitions:
[Eq. 9.8]
u_x(A,B) = \frac{\partial T_x}{\partial C}_{|\ C = 0}
[Eq. 9.9]
u_y(A,B) = \frac{\partial T_y}{\partial C}_{|\ C = 0}
[Eq. 9.10]
U_x(A,B,C) = u_x \frac{\partial T_x}{\partial A} + u_y \frac{\partial T_x}{\partial B}
[Eq 9.11]
U_y(A,B,C) = u_x \frac{\partial T_y}{\partial A} + u_y \frac{\partial T_y}{\partial B}
With that notation Theorem 9-1 amounts to:
[Eq. 9.12]
\frac{\partial T_x}{\partial C} = U_x
[Eq 9.13]
\frac{\partial T_y}{\partial C}= U_y
Define the differential operator U acting on a real valued function f of two variables (S,W) by:
[Eq. 9.14]
U f = U_x \frac{\partial}{\partial S} + U_y \frac{\partial}{\partial W}
An equivalent definition of U for a matrix group might be written in a simpler manner. In a matrix group, T_x and T_y are linear in A and B. Operations such as \frac{\partial T_x}{\partial A} or \frac{\partial T_y}{\partial B} will "pick-off" the functions of C that are the coefficients of A and B. (For example, T_x(x,y,\alpha) = x \cos(\alpha) - y \sin(\alpha),\ \frac{\partial T_x}{\partial x} = cos(\alpha).) For matrix groups, it may be simpler to write a definition of U that specifies the functions that are "picked off" directly by stating them as a matrix rather than doing this implicitly via the above definitions of U_x, U_y.
For U to be well defined, we must have specified a particular 1-parameter group since its definition depends on the definitions of U_x, U_y. The definition of U does not enforce any relationship between the variables S,W and the variables A,B,C involved with the group.
The most important application of U will be in the particular case when S = T_x(A,B,C), W=T_y(A,B,C).
Theorem 9-2: Let f(S,W) be a real valued function with S = T_x(A,B,C), W=T_y(A,B,C). Then
[Eq. 9.15]
\frac{\partial f}{\partial C} = U f
This proven by using the chain rule, theorem 9-1 and the definitions relating to U above.
If f(S,W) is understood to be evaluated at S = T_x(A,B,C), W=T_y(A,B,C) then functions derived from it such as \frac{\partial f}{\partial S} or \frac{\partial f}{\partial C} are understood to be evaluated at the same values. Hence Theorem 9-2 applies to them also and so we have results such as:
[Eq. 9.16]
\frac{\partial}{\partial C} \ (\frac{\partial f}{\partial S}) = U \ \frac{\partial f}{\partial S}
[Eq. 9.17]
\frac{\partial^2 f}{\partial C^2} = \frac{\partial}{\partial C} \ (\frac{\partial f}{\partial C}) = U \ \frac{\partial f}{\partial C} = U (U f)
Using the values A = x, B= y, C =\alpha the above reasoning shows (at least informally) that for any given integer n > 0
[Eq 9.18]
D^n_\alpha f(T_x(x,y,\alpha),T_y(x,y,\alpha)) = U^n f(T_x(x,y,\alpha),T_y(x,y,\alpha))
We now come to the question of how to prove eq 9.3, which said
(U^n f(T_x(x,y,\alpha),T_y(x,y,\alpha)))_{|\ \alpha = 0} = U^n f(x,y)
(and also, whether it is actually true for all 1-parameter groups!)
I can't decide whether this result requires some sophisticated continuity argument about differential operators or whether result follows just from verbally interpreting the notation and definitions involved with U.
We can contrast, the meanings of:
[expression 9-A]
U^n\ f(x,y)
[expression 9-B]
(U^n\ ( f(T_x(x,y,\alpha),T_y),T_y(x,y,\alpha)))_{|\ \alpha = 0}
Using eq, 9.4, we can rewrite f(x,y) so expression 9-A becomes:
[expression 9-C]
U^n\ f( T_x(x,y,0), T_y(x,y,0))
This establishes that the variables in the definition of U^n f(x,y) are S = T_x(A,B,C), W = T_x(A,B,C), A = x, B = y, C = 0 As I interpret 9-C, it means that the relationships among the variables are enforced (including the fact that C = 0) and then the operator U^n is applied to f(S,W).
My interpretation of expression 9-B is that the relationships A = x, B = y , C =\alpha, S = T_x(A,B,C), W= T_y(A,B,C) are enforced. The operation of U^n is applied to f(S,W). Then C is set to zero after all that is done.
The question of whether expression 9-C equals expression 9-B hinges on whether the operation of setting C = 0 commutes with the operation of applying U^n.
I don't know much about differential operators, but I'll speculate that proving 9.3 by a continuity property of operators would involve something like showing:
(U^n(f(T_x(x,y,\alpha),T_y(x,y,\alpha)))_{|\ \alpha = 0} = \lim_{\alpha \rightarrow 0} U^n f(T_x(x,y,\alpha),T_y(x,y,\alpha))
= U^n (lim_{\alpha \rightarrow 0} f(T_x(x,y,\alpha),T_y(x,y,\alpha)) = U^n f(T_x,(x,y,0),T_y(x,y,0)) = U^n f(x,y)
The significant feature of how U has been defined is that it does not involve any differentiations with respect to C. If there is a simple proof of 9.3 by parsing definitions, it would involve the claim that all functions involved in U^n, such as U_x(A,B,C), U_y(A,B,C), f(T_x(A,B,C),T_y(A,B,C)) and their various partial derivatives with respect to A, B give the same results, whether you first substitute C = 0 and perform the differentiations with respect to A, B or whether you do the differentiations first and then substitute C = 0.