Tetration Forum
possible tetration extension part 1 - Printable Version

+- Tetration Forum (https://tetrationforum.org)
+-- Forum: Tetration and Related Topics (https://tetrationforum.org/forumdisplay.php?fid=1)
+--- Forum: Mathematical and General Discussion (https://tetrationforum.org/forumdisplay.php?fid=3)
+--- Thread: possible tetration extension part 1 (/showthread.php?tid=1656)



possible tetration extension part 1 - Shanghai46 - 10/16/2022

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!


RE: possible tetration extension part 1 - bo198214 - 10/17/2022

(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 \(1<b<e^{1/e}\)) - which is a real pitty for us tetration folks, because you don't know what the "right" tetration is.


RE: possible tetration extension part 1 - Shanghai46 - 10/17/2022

Yeah if it's the same case as for half dervivatives, well there would we no real tetration. 
But my method is mathematically correct, can be demonstrated, is simple and gives coherent results. I will show the results when I'll do part 2 and 3.


RE: possible tetration extension part 1 - bo198214 - 10/17/2022

(10/17/2022, 11:21 AM)Shanghai46 Wrote: I will show the results when I'll do part 2 and 3.

I am looking forward!


RE: possible tetration extension part 1 - Shanghai46 - 10/17/2022

(10/17/2022, 02:19 PM)bo198214 Wrote:
(10/17/2022, 11:21 AM)Shanghai46 Wrote: I will show the results when I'll do part 2 and 3.

I am looking forward!
Btw, I'm not gonna spoil but it seems that this formula can be used for bases between \(e^{-e} \) and \(e^{1/e}\). I actually use a trick to use this formula for all real bases higher than \(e^{1/e}\),like 10 tetration x or higher.


RE: possible tetration extension part 1 - bo198214 - 10/17/2022

(10/17/2022, 04:34 PM)Shanghai46 Wrote: Btw, I'm not gonna spoil but it seems that this formula can be used for bases between \(e^{-e} \) and \(e^{1/e}\). I actually use a trick to use this formula for all real bases higher than \(e^{1/e}\),like 10 tetration x or higher.

Now, it gets really interesting!!!


RE: possible tetration extension part 1 - Catullus - 10/31/2022

(10/17/2022, 07:20 AM)bo198214 Wrote: (Btw. in this forum you need to use \\( and \\) for inline formulas instead of \( and \))
On this forum, why are \\( and \\) used, instead of \( and \)? On the Hyperoperations Wiki, \( and \) are used.