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Thread: Function of a function with ln(x)

  1. #1
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    Function of a function with ln(x)

    Consider the equation: $f(x) = \ln (f(x))$

    Does there exist a function that satisfies this equation? It is obvious that $f(x)$, if it exists, is not differentiable. Taking the derivative of both sides would yield $f'(x) = \dfrac{f'(x)}{f(x)}$. But, then you could cancel out the $f'(x)$ terms and you find that $f(x)=1$, which is a contradiction. Therefore, $f(x)$ could not be differentiable (if it exists at all).

    It could also be written as the limit to the sequence of functions $f_0(x) = x$ and $f_n(x) = \ln \left( f_{n-1}(x) \right)$ should the limit function exist.

    Never mind. I figured it out. No, the function does not exist. Proof:

    The domain of $f_n(x)$ would be $^ne$ where the superscript before the number refers to tetration. So, $^2e = e^e$, $^3e = e^{e^e}$. But then the limit domain would not exist.
    Last edited by SlipEternal; Feb 22nd 2018 at 06:58 PM.
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    Re: Function of a function with ln(x)

    Quote Originally Posted by SlipEternal View Post
    Proof:

    The domain of $f_n(x)$ would be $^ne$ where the superscript before the number refers to tetration. So, $^2e = e^e$, $^3e = e^{e^e}$. But then the limit domain would not exist.
    The editing timed out before I could finish updating it. I meant the domain for $f_n(x)$ with $n\ge 2$ would be $x \in (^{n-2}e,\infty)$.
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    Re: Function of a function with ln(x)

    What about the complex logarithm instead of the real logarithm? It has two fixed points. I think the limit function would tend to converge solutions of the form $x=e^{e^{\cdot^{\cdot^{\cdot^x}}}} $
    Last edited by SlipEternal; Feb 23rd 2018 at 03:14 AM.
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    Re: Function of a function with ln(x)

    Quote Originally Posted by SlipEternal View Post
    What about the complex logarithm instead of the real logarithm? It has two fixed points. I think the limit function would tend to converge solutions of the form $x=e^{e^{\cdot^{\cdot^{\cdot^x}}}} $
    $x = -W(-1) \approx 0.318132 -1.33724 i$

    where $W(x)$ is the Lambert W function is one of these fixed points.

    It is a very unstable fixed point. The slightest variation sends $e^{e^{e^{\vdots^x}}}$ off to infinity.
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    Re: Function of a function with ln(x)

    Quote Originally Posted by romsek View Post
    $x = -W(-1) \approx 0.318132 -1.33724 i$

    where $W(x)$ is the Lambert W function is one of these fixed points.

    It is a very unstable fixed point. The slightest variation sends $e^{e^{e^{\vdots^x}}}$ off to infinity.
    Very true. But, check out what Wolframalpha gives for $x=e^{e^{e^x}}$: Wolfram|Alpha: Computational Knowledge Engine
    That has four solutions.

    Then $x=e^{e^{e^{e^x}}}$ also has four solutions according to wolframalpha, but obviously not the same as the solutions for $e^{e^{e^x}}$.

    So, it is unclear how many solutions you might find as you take increasing heights to the tower of $e$'s. I don't have any intuition to determine how many solutions there might be at each step (other than multiples would preserve roots). So, if I had six levels of $e$, it would include all of the solutions to three levels. I'm not sure the correct terminology here, but I hope it is clear what I mean by "levels".

    Perhaps after a finite number of steps, there are an infinite number of solutions, and perhaps those would become more stable. But, I really have no intuition to know why the number of solutions changes as we add levels to the tower. Perhaps it could be an interesting topic to consider one day, but not right now.
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