(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.