possible tetration extension part 1 Shanghai46 Junior Fellow Posts: 20 Threads: 6 Joined: Oct 2022 10/16/2022, 08:26 PM (This post was last modified: 10/17/2022, 10:15 AM by Shanghai46.) So, I was working with a friend of mine, and we came up with 2 formulas to try to extend tetration upon certain intervals. It actually uses a more general formula that I discovered (I might not be the first one) about recursive functions. I will first present the formula, then the tetration extensions.  1) THE FORMULA  the formula I'm talking about is about approximating non integer iteration of a function $$f(x)$$. This only works if the function converges to a finite values when iterated an infinite amount of times.  So here's the theorem :      "for any function $$f$$ where $$f^n(x)$$ equals the nth iteration of $$f(x)$$, where $$n$$ is a integer, and that the infinite iteration of $$f$$ equals a finite value $$\tau$$, $$f^k(x) = \lim_{n\rightarrow+\infty}(f^{-n}((f^n(x)-\tau)f'(\tau)^k +\tau))$$ for any complex number $$k$$ if $$f'(\tau)$$ doesn't equal 1 or 0." with this kind of iterated function, I've actually noticed that  $$\lim_{n\rightarrow+\infty}(\frac{f^{n+1}(x)-\tau}{f^{n}(x)-\tau})=f'(\tau)$$. It ALWAYS approaches $$f'(\tau)$$ when $$n$$ approaches $$+\infty$$ . This can be easely shown by rewritting the limit likewise :  $$\lim_{n\rightarrow+\infty}(\frac{f^{n+1}(x)-\tau}{f^n(x)-\tau})=\lim_{n\rightarrow+\infty}(\frac{f(f^n(x))-f(\tau)}{f^n(x)-\tau})$$, which actually is the lim definition of the derivative $$f'(\tau)$$, since $$f^{n+1}(x)$$ and $$f^n(x)$$ approaches $$\tau$$ . So in this case, we have a ratio (multiply by this to increase the rank of the iteration by one), so to iterate for example 0.5 times the function, we just multiply by the ratio 0.5 times. In other words : multiplying by the square root of the ratio. the only "problem" is that when the ratio is negative, non integer iteration of real functions will give complex results, but for example, with the function sqrt, it totally matches the real formula for non integer iterations of the sqrt (sqrt base $$2^a$$ for the ath iteration). But if we assume this to be the correct way to extand such iteration of functions, it's easy to see the use of that in tetration.   So this was the first part of my personnal tetration extension, I will probably do part 2 and 3 tomorow, anyway tell me what you think! bo198214 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 10/17/2022, 07:20 AM (This post was last modified: 10/17/2022, 07:23 AM by bo198214.) (10/16/2022, 08:26 PM)Shanghai46 Wrote: "for any function $f$ where $f^n(x)$ equals the nth iteration of $f(x)$, where $n$ is a integer, and that the infinite iteration of $f$ equals a finite value $\tau$, $$f^k(x) = \lim_{n\rightarrow+\infty}(f^{-n}((f(x)-\tau)f'(\tau)^k +\tau))$$ for any complex number $k$ if $f'(\tau)$ doesn't equal $1$ or $0$." Very well done! This formula is one way to obtain so called regular iteration (though I think you forgot an $$^n$$ in the formula!). (Btw. in this forum you need to use \$$and \$$ for inline formulas instead of $and$) One calls $$\tau$$ "fixed point" of the function ("attracting" when $$0<|f'(\tau)|<1$$ and "repelling" if $$|f'(\tau)|>1$$) and $$f'(\tau)$$ is called the "multiplier" of the fixed point). One important property of the so defined function $$f^k(x)$$ is: If the function is analytic at the fixed point $$\tau$$ then also $$f^k$$ is analytic at $$\tau$$, in older times "analytic" was called "regular" - hence "regular iteration". Another interesting thing is that if you have two fixed points, typically the regular iteration at the one fixed point does not coincide with the regular iteration at the other fixed point (for example for all functions $$f(x)=b^x$$ with \(1

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