05/03/2009, 02:49 AM
(This post was last modified: 05/05/2009, 03:51 AM by Base-Acid Tetration.)
Could anyone tell me what "regular iteration" means? just generally, not for any specivic method.
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question about regular iteration, also about tetration properties in general
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05/03/2009, 02:49 AM
(This post was last modified: 05/05/2009, 03:51 AM by Base-Acid Tetration.)
Could anyone tell me what "regular iteration" means? just generally, not for any specivic method.
(05/03/2009, 02:49 AM)Tetratophile Wrote: Could anyone tell me what "regular iteration" means? just generally, not for any specivic method. Its regular if the iterates are differentiable (or at leas asymptotically differentiable) at the fixed point. This implies that \( (f^{[t]})'(x_0) = f'(x_0)^t \). Perhaps you have tried that already yourself: If you try graphically find the half iterate you start at a certain interval and then continue to the right and to the left. Depending how you choose the function on the initial segment, the function gets more wobbly or more smooth. Without a fixed point I dont know a criterion of the most unwobbly function, however at the fixed point you can express this unwobblyness by being differentiable. For the wobblyness of a function without fixed point see the pictures made by Gottfried in the thread another uniqueness musing.
05/04/2009, 04:23 AM
(This post was last modified: 05/04/2009, 04:26 AM by Base-Acid Tetration.)
(05/03/2009, 07:05 AM)bo198214 Wrote: Its regular if the iterates are differentiable (or at leas asymptotically differentiable) at the fixed point. This implies that \( (f^{[t]})'(x_0) = f'(x_0)^t \). do you mean that the derivative of the function iterated \( t \) times at the fixed point is the derivative at the fixed point to the \( t^{th} \) power (multiplied t times)? (05/04/2009, 04:23 AM)Tetratophile Wrote:(05/03/2009, 07:05 AM)bo198214 Wrote: Its regular if the iterates are differentiable (or at leas asymptotically differentiable) at the fixed point. This implies that \( (f^{[t]})'(x_0) = f'(x_0)^t \). exactly. You see that is satisfied for natural numbered iterations: \( (f^{[2]})'(x_0)=f'(f(x_0)) f'(x_0)= f'(x_0)^2 \) \( (f^{[3]})'(x_0)= ... \) \( (f^{[n+1]})'(x_0) = f'(f^{[n]}(x_0)) (f^{[n]})'(x_0) = f'(x_0) f'(x_0)^n = f'(x_0)^{n+1} \).
05/04/2009, 04:18 PM
05/05/2009, 03:27 AM
(This post was last modified: 05/12/2009, 03:55 AM by Base-Acid Tetration.)
So, Andysus bo198214, \( \frac{d}{dx}\exp^{\circ x} (\pm W(-1)) = \exp'(\pm W(-1))^x = \exp(\pm W(-1))^x \)? Where do you get from this to derive your tetration?
Does \( \lim_{y \to \infty} {}^{\pm iy} e = \mbox{fixed point of logarithm}= \pm W(-1) \), as in Kouznetsov's, in your tetration? Also, how does one get the differential equation \( \ln (a)\frac{({}^{x+1} a)'}{({}^x a)'}={}^x a \)? Edit: Oh, it comes from the definition of tetration. \( {}^{x} a = a^{{}^{x-1} a} \) \( [{}^{x} a]' = ({}^{x-1} a)' a^{{}^{x-1} a} \, \ln a \) \( \frac{({}^{x} a)'}{ \ln{a} \, ({}^{x-1} a)' } = {}^x a, \, {}^0 a=1 \)
05/10/2009, 09:32 AM
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