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!
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!
Regards
Shanghai46
Shanghai46