jaydfox Wrote:Perhaps I'm missing the point of the tau function (other than to move the fixed point to 0 for building a power series).

I think so.

Quote:It seems like you are using tau to move points to the vicinity of the fixed point, then using the inverse of tau to move the fixed point to 0, iterating, moving the fixed point back to its proper location, then moving the transformed point back to its new location. If I'm reading this correctly, I don't think your tau function is the right way to move points towards the fixed point.

No, here the coefficients of the powerseries (or the Taylor development) is the interesting thing, not the value of the function.

We have a unique (up to \( 2\pi i k \)) regular (continuous) iteration (coefficients of the iterated power series) at a fixed point, which can be finitely computed by the coefficients at that fixed point.

So if we want to compute the coefficients of a Taylor expansion of the continously iterated function at a non-fixed point we first compute the coefficients at the fixed point (which is a non-finite operation in terms of the original coefficients, because it involves limits), then regularly iterating there, getting new coefficients and transforming these coefficients back to the original point. This is the sense of the formula \( \tau_a\circ (\tau_a^{-1}\circ f\circ \tau_a)^{\circ t}\circ \tau_a^{-1} \).

There are also formulas to directly compute a regular iterate which seems more in your interest.

This is the real number version: If you have an attracting fixed point at 0, i.e. \( f(0)=0 \) and \( 0<q<1 \) for \( q=f'(0) \), then the regular iterate is:

\( f^{\circ t}(x)=\lim_{n\to\infty} f^{\circ -n}(q^t f^{\circ n}(x)) \)

In our case (\( f=\exp \)) we have repelling fixed points so \( f^{-1}=\ln \) has attracting fixed points, then the formula is

\( f^{\circ t}(x)=\lim_{n\to\infty} f^{\circ n}(q^t f^{\circ -n}(x)) \).

So now our fixed point \( a \) is not at 0. So we consider

\( g:=\tau_a^{-1}\circ f\circ \tau_a \) which has a repelling fixed point at 0.

Then let \( g'(0)=f'(a)=:r \) and use our above formula:

\( g^{\circ t}(x)=\lim_{n\to\infty} g^{\circ n}(r^t g^{\circ -n}(x)) \).

\( (\tau_a^{-1}\circ f^{\circ t}\circ\tau_a)(x)=\lim_{n\to\infty} \tau_a^{-1}\circ f^{\circ n}\circ \tau_a(r^t (\tau_a^{-1} \circ f^{\circ -n}\circ \tau_a)(x)) \)

\( f^{\circ t}=\lim_{n\to\infty} f^{\circ n}\circ \tau_a(r^t (\tau^{-1}_a\circ f^{\circ -n})(x)) \)

\( f^{\circ t}(x)

=\lim_{n\to\infty} f^{\circ n}(a+(r^t(-a+(f^{\circ -n}(x)))) \)

\( f^{\circ t}(x)=\lim_{n\to\infty} f^{\circ n}(a(1-r^t) + r^t f^{\circ -n}(x)), r=f'(a) \).

For an analytic function \( f \) the limit formula yields again an analytic function and this function is exactly the function described by the coefficient formula, thatswhy they both are called regular iteration.

The regular iteration is the only one that has no singularity at the fixed point.

One also can immediatly see that the obove formula for \( f=\exp \) has singularities at \( \exp^{\circ k}(0) \).

For example Kneser developed the regular solution at the first fixed point \( a \). He also observed the singularities and of course that the solution had complex values for real arguments.

Thatswhy he modified his solution with a holomorphic transformation \( \alpha \): \( f_2^{\circ t}=\alpha^{-1}\circ f^{\circ t}\circ \alpha \) so that it yielded real values for real arguments and no singularities at the real line.

Of course his new solution was no more regular and hence had a singularity at the fixed point \( a \). And so will have every real solution at every fixed point, because no regular solution at some fixed point will be real (though strictly this was not verified yet).