@Henryk
Good point, (about the \( f^{\circ t} \) being a function-parameter, not the value-parameter) I didn't realize that because of the vague notation that we use. I'm going to use a very verbose notation, like the kind used for other transforms (like Fourier \( \mathcal{F}[f](x) \) and Laplace \( \mathcal{L}[f](x) \) transforms) so that we can get this right.
Starting with the Abel functional equation:
\( \mathcal{A}[f](f^{\circ t}(x)) = \mathcal{A}[f](x) + t \)
\( \frac{\partial}{\partial x} \mathcal{A}[f](f^{\circ t}(x)) (f^{\circ t})'(x) = \frac{\partial}{\partial x} \mathcal{A}[f](x) \)
\( \mathcal{J}[f](f^{\circ t}(x)) = (f^{\circ t})'(x) \mathcal{J}[f](x) \)
George Szekeres defines the Julia function as a function that satisfies the above functional equation and is also the reciprocal of the derivative of the Abel function. This is exactly what has just been said about "ilog" so these should be the same function.
The discussion above defines the same function as:
\( \mathcal{J}[f](x) = \left[\frac{\partial}{\partial t} f^{\circ t}(x)\right]_{t=0} \)
such that:
\( \mathcal{J}[f^{\circ t}](x) = t \mathcal{J}[f](x) \)
which according to you and Ecalle, is also related as:
\( \mathcal{J}[f](x) = \frac{1}{\frac{\partial}{\partial x}
\mathcal{A}[f](x)} \)
which is exactly the definition of a Julia function.
I think what we should take away from this is that the two equations:
[update]I fixed the t=1 problem.[/update]
Andrew Robbins
Good point, (about the \( f^{\circ t} \) being a function-parameter, not the value-parameter) I didn't realize that because of the vague notation that we use. I'm going to use a very verbose notation, like the kind used for other transforms (like Fourier \( \mathcal{F}[f](x) \) and Laplace \( \mathcal{L}[f](x) \) transforms) so that we can get this right.
Starting with the Abel functional equation:
\( \mathcal{A}[f](f^{\circ t}(x)) = \mathcal{A}[f](x) + t \)
\( \frac{\partial}{\partial x} \mathcal{A}[f](f^{\circ t}(x)) (f^{\circ t})'(x) = \frac{\partial}{\partial x} \mathcal{A}[f](x) \)
\( \mathcal{J}[f](f^{\circ t}(x)) = (f^{\circ t})'(x) \mathcal{J}[f](x) \)
George Szekeres defines the Julia function as a function that satisfies the above functional equation and is also the reciprocal of the derivative of the Abel function. This is exactly what has just been said about "ilog" so these should be the same function.
The discussion above defines the same function as:
\( \mathcal{J}[f](x) = \left[\frac{\partial}{\partial t} f^{\circ t}(x)\right]_{t=0} \)
such that:
\( \mathcal{J}[f^{\circ t}](x) = t \mathcal{J}[f](x) \)
which according to you and Ecalle, is also related as:
\( \mathcal{J}[f](x) = \frac{1}{\frac{\partial}{\partial x}
\mathcal{A}[f](x)} \)
which is exactly the definition of a Julia function.
I think what we should take away from this is that the two equations:
\( \mathcal{J}[f](f^{\circ t}(x)) = (f'(x))^t \mathcal{J}[f](x) \)
\( \mathcal{J}[f^{\circ t}](x) = t \mathcal{J}[f](x) \)
do not contradict each other in any way (although they would if we continued using the vague notation). And that now that we see that these are talking about the same function, now we know 2 things about 1 function, as opposed to 1 thing about 2 functions...\( \mathcal{J}[f^{\circ t}](x) = t \mathcal{J}[f](x) \)
[update]I fixed the t=1 problem.[/update]
Andrew Robbins
