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I was wondering if the following function exists: lets say you have

$$f^1(x) = \ln(x)$$ $$f^2(x) = \ln(\ln(x))$$ $$f^3(x) = \ln(\ln(\ln(x)))$$ $$f^4(x) = \ln(\ln(\ln(\ln(x))))$$ and so forth

is there a way to generalize $$f^n(x)$$ when $n$ can be a non-integer value?

I found this : Generalised logaritmic function but it deosn't look like there was any update

Also if such a function does exist, I was wonderin if it would be possible to have $$\frac{d}{dx} f^c(\Gamma(x)) = \text{constant}$$ for some $c$ because for $n = 1$ we get this: https://www.wolframalpha.com/input/?i=plot+derivative+of+ln%28gamma%28x%29%29+from+1+to+10000000000

and for $n = 2$ we get this: https://www.wolframalpha.com/input/?i=plot+derivative+of+ln%28ln%28gamma%28x%29%29%29+from+1+to+10000000000

which maybe implies that some $c$ in between would have the property above

Gottfried Helms
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Sasha1296
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    You can nest arbitary many logarithms, but the necessary magnitude of the input increases very rapidly. – Peter Oct 19 '20 at 18:31
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    This is equivalent to asking for iterates of $\exp(x)=e^x$, which is equivalent to asking for base $e$ tetration on arbitrary heights. – Simply Beautiful Art Oct 28 '20 at 20:51
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    It could exist (why not) but the domain of $f^n$ defined on $\mathbb R$ gets smaller as $n$ increases. I am not sure it has much practical purpose. For example, $f^6(10^{100})$ is not defined (as a real function). – PC1 Jun 21 '23 at 17:41

1 Answers1

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The usual way to calculate fractional iterations of functions is to use Newton's series as described here:

$$f^{[s]}(x)=\sum_{m=0}^{\infty} \binom {s}m \sum_{k=0}^m\binom mk(-1)^{m-k}f^{[k]}(x)$$

For the function $\ln x$ the series diverges, but if you take $\ln(x+1)$ it works fine, for instance, here is the half-iteration of $\ln(x+1)$ using this series:

enter image description here

Mathematica code:

n := 50
it := 0.5
func := Function[x, Log[x + 1]]
iter[k_Integer, x_] := Nest[func, x, k]
Iter[x_, i_] := 
 Sum[Binomial[i, m]*
   Sum[(-1)^(k - m)*Binomial[m, k]*iter[k, x], {k, 0, m}], 
     {m, 0, n}]
ReImPlot[{Iter[x, it]}, {x, 0, 2}, AspectRatio -> Automatic]
Anixx
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