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 Tetration below 1 bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 08/31/2007, 02:10 PM (This post was last modified: 08/31/2007, 02:16 PM by bo198214.) Daniel Wrote:The other values of $b$ have hyperbolic fixed points and could be solved using Schroeder's functional equation, Bell matrices or other equivalent methods. For all values of $b$ in $(0,1)$ the function $\exp_b$ is strictly decreasing and has a fixed point $x_0$ with $\exp_b'(x_0)<0$. So Schroeder, Bell and equivalent methods do not fit. Strictly increasing and decrasing functions can be seen a bit in similarity with negative and positive numbers. For example $f^{\circ 2n}$ is strictly increasing and $f^{\circ 2n+1}$ is strictly decreasing for $f$ being strictly decreasing. Which is similar to $x^{2n} > 0$ and $x^{2n+1} < 0$ for $x<0$. Accordingly there are no strictly monotonic iterative square roots (or generally even roots) of strictly decreasing functions. (regardless whether the square root is increasing or decreasing, the iterative square must be increasing, but it is decreasing). Now you can say: Ok, then we allow non-monotonic functions (perhaps in analogy to the complex numbers, where we can extract square roots). But there is another thing: Proposition. The iterative root of a strictly monotonic continuous function (on an open interval X) is continuous if and only if it is strictly monotonic. So if we give up strict monotony of the root we also give up continuity, which then makes no more sense. Hence all the exponents $\frac{1}{2n}$ are lacking for tetration with $b<1$ and so we have no appropriate $\text{sexp}_b(\frac{1}{2n})$. If we use the analogy with real numbers then $b^x$, $b<0$, can be defined as $b^x=-|b|^x$ if $x=\frac{m}{2n+1}$ but it can not be defined for the cancelled fraction $x=\frac{m}{2n}$. (At this place I also want to correct some nonsense that I stated in my first post. It has to be: "because $x^{2n}$ is not surjective for $x<0$, $b^{1/2}$ can not be defined. This gives also trouble with $b^{n}=b^{2n/2}=?$.") So on the reals there is no way to continuously define $b^x$ for $b<0$. However if we extend our domain to the complex numbers there is the way, or rather the ways $b^x=\exp_{b,k}(x)=\exp(x(\log(b)+2\pi i k))$. Here for each $x=\frac{1}{2n+1}$ there is a $k$ such that indeed $b^x=-|b|^x$: $\log(b)=\log(-1|b|)=\log(-1)+\log|b|=\pi i +\log|b|$ $\exp_{b,k}(x)=\exp(x\cdot \log|b| + (2k+1)\pi i x)=|b|^x e^{(2k+1)\pi i x$ $\exp_{b,n}(\frac{1}{2n+1})=|b|^x e^{ \pi i }=-|b|^x$. However there is no function that $b^{1/(2n+1)}=-|b|^{1/(2n+1)}$ for all n. The suggestion of Daniel to simply iterate the strictly increasing $\exp_b^{\circ 2}$ does not really help about our problem as we still can not define the fractional iterates $\exp_b^{\circ \frac{1}{2n}}$. The only possibility I see is to extend the tetration for bases in $(0,1)$ to complex values. I think there are possibilities for (multiple) continuous iterative square roots there from which we can define a family of super exponentials $\text{sexp}_{b,k}(x)$ such that for each $x=\frac{1}{2n+1}$, we have $\text{sexp}_{b,n}(x)=1/\text{sexp}_{1/b}(x)$, where the reciprocal is surely wrong, but just there to give a rough idea. « Next Oldest | Next Newest »

 Messages In This Thread Tetration below 1 - by bo198214 - 08/29/2007, 06:14 PM RE: Tetration below 1 - by Daniel - 08/30/2007, 09:32 PM RE: Tetration below 1 - by jaydfox - 08/30/2007, 11:08 PM RE: Tetration below 1 - by jaydfox - 08/30/2007, 11:20 PM RE: Tetration below 1 - by Daniel - 08/31/2007, 12:47 AM RE: Tetration below 1 - by GFR - 09/02/2007, 01:30 PM RE: Tetration below 1 - by bo198214 - 09/02/2007, 01:40 PM RE: Tetration below 1 - by GFR - 09/02/2007, 05:38 PM RE: Tetration below 1 - by jaydfox - 09/03/2007, 03:58 PM RE: Tetration below 1 - by bo198214 - 09/03/2007, 04:07 PM RE: Tetration below 1 - by jaydfox - 09/03/2007, 04:36 PM RE: Tetration below 1 - by jaydfox - 09/05/2007, 11:24 PM RE: Tetration below 1 - by GFR - 09/06/2007, 12:01 AM RE: Tetration below 1 - by jaydfox - 09/06/2007, 03:28 AM RE: Tetration below 1 - by jaydfox - 09/06/2007, 07:21 AM RE: Tetration below 1 - by bo198214 - 03/26/2008, 04:51 PM

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