I think I may have found something new because it's giving the correct results. I'm using the functional-roots of $logx$ to calculate super-logarithms. You can read this post of mine to get the idea : How to find a function $f(x)$ such that $f(f(x))=\log_ax$?

By functional square-root, I mean a function $f(x)$ such that $f(f(x))=logx$. I've not introduced this new term. By $n^{th}$ functional root of a given function, I mean the function which has to be applied $n$ times to get the required function.

$slog_ax$ is defined as the number of times logarithm with base $a$ is applied to $x$ to get to 1. But this definition fails when an integral number of logarithms can't be applied to $x$ to get to 1. The idea is that we don't necessarily have to apply logarithms continuously. If $k$ number of $n^{th}$ functional root of $log_ax$ can be applied to $x$ to get to 1, then $slog_ax=\frac{k}{n}$. This is in agreement with what I've calculated.

First start with, let's say $3^3=27$ or. $^23=27$ which gives $3=^{0.5}27$

or $slog_{27}3=0.5$

This is analogous to $log_{a^n}a=\frac{1}{n}$.

Now, we can't apply an integral number of logarithms with base 27 to 3 to get to 1 because applying the logarithm even once will get us smaller than 1. But if the functional square-root of $log_{27}x$ when evaluated at $x=3$ gives us 1 , then I'm correct. Here's how I'm calculating the approximate functional square-root of $log_{27}x$ when x is close to 3 because we need to apply that function to 3:

The approximation of $log_{27}x$ around $x=2.9$ is: $$log_{27}x\approx 0.0196+\frac{x}{9.557}$$. ......(1)

Now, if we assume the functional square-root to be $f(x)=ax+b$, then $f(f(x))=a^2x+ab+b$. By comparing like power of $x$ in this equation and equation (1), we get , $a=0.323$ and $b=0.0148$. So the functional square-root approximately is: $$f(x)=0.323x+0.0148$$ So, this function, when evaluated at $x=3$ should get us very close to 1. It's value at $x=3$ is 0.9838 which is very close to 1. So, I think I'm correct. I've assumed $f(x)=ax+b$. If we assume $f(x)$ of higher degree, we'll have more terms to compare and I think the functional square root will be more accurate. The actual functional square root will get us exactly to 1.

THIS is my question: Is there anything wrong with this extension of superlogarithms? The extensions on wikipedia don't sound as natural as mine.

UPDATE: I tried to do the same thing with $slog_{16}2$ which should be equal to $\frac{1}{3}$ because $^32=16$. This means the functional cube root of $log_{16}x$ when applied once to 2 should equal to 1. By assuming the functional cube root to be $f(x)=ax+b$, we get $f(f(f(x)))=a^3x+a^2b+ab+b$. After comparing this with the first two terms of taylor series of $log_{16}x$ around $x=2$, the approximate functional cube root of $log_{16}x$ is: $0.5649x-0.0587$ which evaluates to 1.07 for $x=2$ which is very close to 1 which means the super-logarithm is equal to $\frac{1}{3}$ as expected because $^32=16$.

If we try to do the same thing for calculating $slog_{65536}2$ which should be equal to $\frac{1}{4}$ because $^42=65536$, then after applying the functional 4th root of $log_{65536}x$ to 2, we get 0.91 which isn't very close to 1. But to evaluate the functional 4th root, I've assumed it to be $ax+b$. So, I think the functional 4th root isn't that accurate. I think if we assume the 4th root to be $ax^2+bx+c$ or of higher degree, then after getting the constants, evaluating that function at $x=2$ should get us closer to 1.

UPDATE: If we evaluate the 4th functional root by taking the taylor series of of $log_{65536}x$ around $x=1.5$, then then even by taking the 4th root as $ax+b$, the 4th root comes out to be $0.495x-0.0172$ which evaluates to $0.97$ at $x=2$ which is very close to 1. Last time, I had used the approximation at $x=2$ which gave me $0.9$.

  • $\begingroup$ What's the motivation for this? $\endgroup$ – user301988 Feb 10 '17 at 7:51
  • $\begingroup$ @selfawareuser: Logarithms can be defined in a similar way as the number of times x is divided by a to get to 1. I tried to extend this dedinition when x is not of the form $a^n$. That gave me motivation to extend the definition of superlogarithms in a similar way. Read this post of mine:math.stackexchange.com/questions/2134589/…. $\endgroup$ – Dove Feb 10 '17 at 7:55
  • $\begingroup$ @Yves Daoust: I've not given any new definitions. I've given a way to extend the definition. Is there anything wrong with this extension? $\endgroup$ – Dove Feb 10 '17 at 8:05
  • $\begingroup$ @Yves Daoust: I've given a way to calculate functional-roots using polynomilas which is in agreement with calculations and isn't nearly as stressful as calculating Kneser's functions. $\endgroup$ – Dove Feb 10 '17 at 8:14
  • $\begingroup$ @Yves Daoust: Try evaluating the approximate functional cube root of $\log_{3^{3^3}}x$ using my polynomial method. I bet applying that function once to 3 will get you to 1. $\endgroup$ – Dove Feb 10 '17 at 8:20

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