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 Our tetration concept is a wrong way? Ansus Junior Fellow Posts: 20 Threads: 5 Joined: Aug 2008 11/06/2010, 12:08 AM (This post was last modified: 11/06/2010, 01:33 AM by Ansus.) By analogy with discrete multiplicative derivative of a function $f(x+1)\frac1{f(x)}$ we can define a discrete iterative derivative $f(x+1)\circ f^{[-1]}(x)$. The inverse operator to it (discrete iterative integral) is known as superfunction or flow. It exhibits many properties common to integral such as that it is more difficult to express in closed form than the derivative and that a constant parameter appears in the process. But what if to consider the local iterative integrals and derivatives? Lets define the local iterative derivative as follows: $f^{\circ}(x)=\lim_{h\to\infty} \left( f \left(x+\frac xh\right)\circ f^{[-1]}(x)\right)^{[h]}$ where in the square brackets is the number of iterations. And also by analogy we can define the local iterative integral as an inverse operator. As we can see, the local iterative integral of x+1 is addition, local iterative integral of addition is multiplication and local iterative integral of multiplication is exponentiation (just as in the case of discrete iterative integral). There is no reason to define tetration as discrete iterative integral of exponentiation as we usually do. Rather we could better define tetration as the local iterative integral of exponentiation, this new definition will rid us of any problems of interpolation over discrete points, problems of uniqueness etc. Although it will give us completely different function as "local tetration" The local tetration thus satisfies the following equation: $a^x=\lim_{h\to\infty} \left( f_a \left(f_a^{[-1]}(x)+\frac xh\right)\right)^{[h]}$ Any thoughts? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 11/06/2010, 01:30 AM (This post was last modified: 11/06/2010, 01:32 AM by mike3.) Well this approach looks to be of no use. Note that $f_a(f_a^{-1}(x) + \frac{1}{h})$ is $\frac{1}{h}$ iterate of the "subfunction" of $f_a$, which is uniquely given by $f_a(f_a^{-1}(x) + 1)$. Thus h iterates of it yields that subfunction, regardless of the specific superfunction used (i.e. a 1-cyclic transform of it will give the same subfunction, hence the same "iterative derivative".). Essentially, there is apparently no difference between this and the "discrete iterative derivative". Ansus Junior Fellow Posts: 20 Threads: 5 Joined: Aug 2008 11/06/2010, 04:57 AM (11/06/2010, 01:30 AM)mike3 Wrote: Well this approach looks to be of no use. Note that $f_a(f_a^{-1}(x) + \frac{1}{h})$ is $\frac{1}{h}$ iterate of the "subfunction" of $f_a$, which is uniquely given by $f_a(f_a^{-1}(x) + 1)$. Thus h iterates of it yields that subfunction, regardless of the specific superfunction used (i.e. a 1-cyclic transform of it will give the same subfunction, hence the same "iterative derivative".). Essentially, there is apparently no difference between this and the "discrete iterative derivative". Yes. I have noticed that. What about using the x instead of 1 as in the current post? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 11/06/2010, 06:37 AM (11/06/2010, 04:57 AM)Ansus Wrote: Yes. I have noticed that. What about using the x instead of 1 as in the current post? That would just seem to be saying that "x-iterated exp = x-iterated exp", because the right hand side ends up as iteration by x times of the subfunction, in this case, the exponential. « Next Oldest | Next Newest »

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