Reputation
2,243
Top tag
Next privilege 2,500 Rep.
Create tag synonyms
Badges
4 18
Newest
 Constituent
Impact
~13k people reached

Mar
31
comment two fixed points, same fractional iteration
Thanks! I need to lookup that paper.
Mar
12
comment What number tetrated by itself equals a googol?
$$b_1 \approx 3.22192863\; ; \;\;\; b_2 \approx 3.96367752$$ $$b_1 \uparrow \uparrow b_1 \approx 10^{100}$$ $$b_2 \uparrow \uparrow b_2 \approx 10^{10^{100}}$$ Analytic extension of tetration results using math.eretrandre.org/tetrationforum/showthread.php?tid=486
Jan
25
comment Which one is greater?
See en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation. It would be $$2 \uparrow \uparrow 3 = 2^{2^2} = ^{3}2 = 16$$
Jan
15
comment How to prove that $F(x) = \lim_{n \to \infty} F^{n}( f ' (0) \cdot F^{-n}(x)) $?
Another correction; although perhaps Mick wanted the form he posted; I'm not sure. I guess usually a similar equation is used for the Schroder equation; This is the form for the kth fractional iterate. $$g = F \circ g \circ F^{-1}$$
Jan
15
comment How to prove that $F(x) = \lim_{n \to \infty} F^{n}( f ' (0) \cdot F^{-n}(x)) $?
Mick copied from an equation with some details missing. $$g(z) = \lim_{n \to \infty} F^{[n]} (g'(0) \cdot F^{[-n]}(z))$$ here, $g'(0)$ refers to the derivative of the $\frac{1}{k}$ fractional iterate of interest, so that $(g'(0))^k = F'(0)$. In the example, F'(0)=4, and g'(0)=2, and we were calculating $g(z)= F^{0.5}(z)$
Jan
14
comment Fixed point and fractional iteration: if $F(k)=k$ then $F^{1\over n}(k)$ is another fixed point of $F$
Thanks for the bounty. I think if you restrict yourself to starting with $$F^{[1/n]}=\Psi(x)$$ and $\Psi(x)$ is single valued, then the resulting $F(x)$ is single valued. But this seems to me like working with $f(x)$ and $f^{[n]}(x)$ ... when one speaks of a fractional iterate, normally I would think of a general analytic function. So that's why I say your definition of a fractional iterate is flawed. It sounds like you are really interested in integer iterates ...
Jan
14
comment Fixed point and fractional iteration: if $F(k)=k$ then $F^{1\over n}(k)$ is another fixed point of $F$
I fixed up the counter example, by showing that it is the fractional iterate, $h(x)$ that takes on multiple values at the fixed point, depending on how many times we have iterated the function. And that leads to the contradiction.
Jan
14
revised Fixed point and fractional iteration: if $F(k)=k$ then $F^{1\over n}(k)$ is another fixed point of $F$
added 30 characters in body
Jan
14
revised Fixed point and fractional iteration: if $F(k)=k$ then $F^{1\over n}(k)$ is another fixed point of $F$
added 111 characters in body
Jan
14
revised Fixed point and fractional iteration: if $F(k)=k$ then $F^{1\over n}(k)$ is another fixed point of $F$
I fixed the explanation for the contradiction
Jan
14
revised Fixed point and fractional iteration: if $F(k)=k$ then $F^{1\over n}(k)$ is another fixed point of $F$
I fixed the explanation for the contradiction
Jan
13
revised Fixed point and fractional iteration: if $F(k)=k$ then $F^{1\over n}(k)$ is another fixed point of $F$
added 426 characters in body
Jan
13
comment Fixed point and fractional iteration: if $F(k)=k$ then $F^{1\over n}(k)$ is another fixed point of $F$
I'm not qualified to address generic X domain, but I do enjoy working with half iterates and analytic functions, and entire functions... But for non-entire $F$, how would you know what to set the domain to??? For the limit equation, all that is required is the fixed point at the origin, and the absolute value of the first derivative is at the origin is >1. I wanted to avoid Abel functions and Superfunctions, which is why I started with the Formal Taylor series solution for the half iterate. But then I needed to know the closest singularity for h, so I also had to use the limit equations.
Jan
13
revised Fixed point and fractional iteration: if $F(k)=k$ then $F^{1\over n}(k)$ is another fixed point of $F$
added 496 characters in body
Jan
13
comment Fixed point and fractional iteration: if $F(k)=k$ then $F^{1\over n}(k)$ is another fixed point of $F$
@mick, MphLee; I really like your question, btw, it just took me awhile to realize that MphLee's proof was correct iff the fractional iterate was entire ... and then I got really interested. I wanted to post as simple and counterexample as I could. I probably should've skipped the Taylor series, and stuck with the limit definition of $h(x)$, but I generated the formal Taylor series solution of $h(x)$ first ...
Jan
13
revised Fixed point and fractional iteration: if $F(k)=k$ then $F^{1\over n}(k)$ is another fixed point of $F$
deleted 10 characters in body
Jan
13
revised Fixed point and fractional iteration: if $F(k)=k$ then $F^{1\over n}(k)$ is another fixed point of $F$
added alternative half iterate and radius of convergence equation
Jan
13
revised Fixed point and fractional iteration: if $F(k)=k$ then $F^{1\over n}(k)$ is another fixed point of $F$
added alternative definition of $H(x)$
Jan
13
revised Fixed point and fractional iteration: if $F(k)=k$ then $F^{1\over n}(k)$ is another fixed point of $F$
minor clarification
Jan
13
revised Fixed point and fractional iteration: if $F(k)=k$ then $F^{1\over n}(k)$ is another fixed point of $F$
added 20 characters in body