possible tetration extension part 3
#1
So this is part 3. As we've seen, I've been using a formula which works with iterated functions that eventually converges. But for numbers bigger than \(e^{1/e}\),the infinite tetration diverges towards plus infinity. So it doesn't seem to be applicable here. Let's look at the inverse of tetration : logarithms

Well, the infinite iteration of the log base a of any number x do converge, but it does not if x is a tetration of b (like for log base 2,the numbers 0, 1, 2, 4, 16, 65536,... go to an undefined value, which is log(0)). We could still try to do by by approximating the values, like taking 2.001 instead of 2. Well for unknown reasons, the formula seems to break even if we do that, because it give incoherent results (I probably should add more restrictions but since I don't have any idea about why it doesn't work, well..)
So, let's look at the function log_b(x+1). If we take a huge number, log_b(x+1)≈log_b(x),and it eventually become equal if x tend towards infinity. So if we could do half times the log_x(x+1) for example, we could calculate the half tetration of x. In fact, we could do tetration backwards, taking a big tetration of x, applying a non integer number of times the log_x(the big number +1), then taking enough times the regular log_x(resulting number) to get to any tetration, and it actually works!

So let's consider 2 functions : \(f(a) = \log_x(a+1)\) and \(g(a) = x^a-1\). Here g(f(a)) = f(g(a)) = a


We can notice thanks to this definition that : 

$$\lim_{a\rightarrow~+\infty}(f(a)) = \log_x(a)$$ 

$$g^n(f^{n+k}(a)) = f^k(a)$$

$$\lim_{r\rightarrow~+\infty}(f^k({^r}x)) = {^{r-k}}x$$

So, to have a non integer tetration of x, we need to know how to iterate k times f(x) where k is a non whole number (like knowing how to iterate 0.5 time the function).

If we iterate an infinite number of times $f$, we can notice it actually converges to a finite value we will call \(\tau\). This value equals 0 for all x superior or equal to e, but is greater than 0 for all x inferior to e. 
Here are some examples :       
$$x=10,\tau = 0$$        
$$x=e,\tau = 0$$
$$x=2,\tau = 1$$  
$$x=1.8,\tau = 1.672225$$   
$$x=1.5,\tau ≈ 3.939176$$

The graph of tau in relation to x : 
     
So, now, we can also notice that the ratio between \(f^{n+1}-\tau\) and \(f^n-\tau\) converges to a value we will call \(\lambda\) (basically my formula.) 

$$\lim_{n\rightarrow~+\infty}(\frac{f^{n+1}-\tau}{f^{n}-\tau})=\lambda$$

What's interesting is that \(\lambda=\frac{1}{\ln(x)*(1+\tau)}\). Here we devide by \(\lambda\) because we want to do a negative amount of time the \(\log(x+1)\) function, which gets closer and closer to tetration when you approach infinity. Here's the graph of what \(\lambda\) is in relation to x :

Now, we can deduce that : $$\lim_{n\rightarrow~+\infty}(((f^n(a)-\tau)/\lambda^k)+\tau)=f^{k+n}(a)$$

So, we can also deduce : $$\lim_{n\rightarrow~+\infty}(g^n(((f^n(a)-\tau)/\lambda^k)+\tau) )=f^{k}(a)$$

And we can drive our final formula :

$${^r}x=\lim_{n\rightarrow~+\infty}({\log_x}^p( g^n(((f^n({^m}x)-\tau)/\lambda^k)+\tau)))={^{m-k-p}}x$$   

Where r=m-k-p, r is any real number (not equal to any whole negative numbers below -1), m and p are natural numbers and k any real integer number which |k|<1. 

This seem complicated, but let's take an example. To calculate \({^{-0.5}}10\), we take \({^2}10\) (m=2), we iterate f a lot of times (n times), wa substract \(\lambda\) which is 0, we multiply by \({\frac{1}{\ln(10)}^{0.5} }\) (k=0.5), we add \(\lambda\) (still 0) we iterate g the same number of times as f to cancel it out (n times), so here we have an approximation of \({^{1.5} }10\), so we iterate \(\log_{10}\) twice (p=2) to get an approximation of \({^{-0.5}}10\).

Here are some values I got :

$${^{-1.3}}10≈-0.227195279$$  
$${^{0.4}}10≈2.020355078$$    
$${^{1.5}}10≈294.61347005$$    
$${^{-1.1}}2≈-0.1345962639$$   
$${^{0.9}}2≈0.9109247189$$   
$${^{1.7}}2≈5.322181918$$

Here is the graph of 10 tetration r and 2 tetration r :


Voila, I hope it was clear and well explained.
Tell me what you think. Is it coherent? Is it logical to extend tetration by this definition?


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#2
(10/21/2022, 08:29 AM)Shanghai46 Wrote: $${^r}x=\lim_{n\rightarrow~+\infty}({\log_x}^p( g^n(((f^n({^m}x)-\tau)/\lambda^k)+\tau)))={^{m-k-p}}x$$   

Where r=m-k-p, r is any real number (not equal to any whole negative numbers below -1), m and p are natural numbers and k any real integer number which |k|<1. 

This seem complicated, but let's take an example. To calculate \({^{-0.5}}10\), we take \({^2}10\) (m=2), we iterate f a lot of times (n times), wa substract \(\lambda\) which is 0, we multiply by \({\frac{1}{\ln(10)}^{0.5} }\) (k=0.5), we add \(\lambda\) (still 0) we iterate g the same number of times as f to cancel it out (n times), so here we have an approximation of \({^{1.5} }10\), so we iterate \(\log_{10}\) twice (p=2) to get an approximation of \({^{-0.5}}10\).

It is not clear to me how you choose the values m and p. Above you just chose m=p=2, but you could have taken any other natural number (say m=p=5), and I guess the result would be different.
Or do you take another limit with \(m=p\to \infty\)?
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#3
(10/21/2022, 07:23 PM)bo198214 Wrote:
(10/21/2022, 08:29 AM)Shanghai46 Wrote: $${^r}x=\lim_{n\rightarrow~+\infty}({\log_x}^p( g^n(((f^n({^m}x)-\tau)/\lambda^k)+\tau)))={^{m-k-p}}x$$   

Where r=m-k-p, r is any real number (not equal to any whole negative numbers below -1), m and p are natural numbers and k any real integer number which |k|<1. 

This seem complicated, but let's take an example. To calculate \({^{-0.5}}10\), we take \({^2}10\) (m=2), we iterate f a lot of times (n times), wa substract \(\lambda\) which is 0, we multiply by \({\frac{1}{\ln(10)}^{0.5} }\) (k=0.5), we add \(\lambda\) (still 0) we iterate g the same number of times as f to cancel it out (n times), so here we have an approximation of \({^{1.5} }10\), so we iterate \(\log_{10}\) twice (p=2) to get an approximation of \({^{-0.5}}10\).

It is not clear to me how you choose the values m and p. Above you just chose m=p=2, but you could have taken any other natural number (say m=p=5), and I guess the result would be different.
Or do you take another limit with \(m=p\to \infty\)?
Technically it wouldn't give exactly the same number, but the limit would be the same one, so it's not really changing anything, but the simpler the better
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#4
(10/21/2022, 08:29 AM)Shanghai46 Wrote: $${^r}x=\lim_{n\rightarrow~+\infty}({\log_x}^p( g^n(((f^n({^m}x)-\tau)/\lambda^k)+\tau)))={^{m-k-p}}x$$   

Where r=m-k-p, r is any real number (not equal to any whole negative numbers below -1), m and p are natural numbers and k any real integer number which |k|<1. 

(10/21/2022, 08:16 PM)Shanghai46 Wrote: Technically it wouldn't give exactly the same number, but the limit would be the same one, so it's not really changing anything, but the simpler the better

You described letting \(^mx\) go to a big number and then taking the fractional iterate of \(f\), but it does not reflect in your formula. So I guess what you mean with "limit" in your comment is the limit over \(m\), so the formula would look like:
$${^r}x=\lim_{m\to\infty}\lim_{n\to\infty}({\log_x}^m( g^n(((f^n({^m}x)-\tau)/\lambda^k)+\tau)))$$  

Is that what you mean? (Computationally of course one would just look for a suitably big \(^mx\) ) Just in the clarification phase Smile
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#5
(10/22/2022, 08:10 AM)bo198214 Wrote:
(10/21/2022, 08:29 AM)Shanghai46 Wrote: $${^r}x=\lim_{n\rightarrow~+\infty}({\log_x}^p( g^n(((f^n({^m}x)-\tau)/\lambda^k)+\tau)))={^{m-k-p}}x$$   

Where r=m-k-p, r is any real number (not equal to any whole negative numbers below -1), m and p are natural numbers and k any real integer number which |k|<1. 

(10/21/2022, 08:16 PM)Shanghai46 Wrote: Technically it wouldn't give exactly the same number, but the limit would be the same one, so it's not really changing anything, but the simpler the better

You described letting \(^mx\) go to a big number and then taking the fractional iterate of \(f\), but it does not reflect in your formula. So I guess what you mean with "limit" in your comment is the limit over \(m\), so the formula would look like:
$${^r}x=\lim_{m\to\infty}\lim_{n\to\infty}({\log_x}^m( g^n(((f^n({^m}x)-\tau)/\lambda^k)+\tau)))$$  

Is that what you mean? (Computationally of course one would just look for a suitably big \(^mx\) ) Just in the clarification phase Smile

What I mean is that the larger an number is, the closer Log_a(x+1) will be to Log_a(x).
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#6
(10/22/2022, 09:50 AM)Shanghai46 Wrote:
(10/22/2022, 08:10 AM)bo198214 Wrote:
(10/21/2022, 08:29 AM)Shanghai46 Wrote: $${^r}x=\lim_{n\rightarrow~+\infty}({\log_x}^p( g^n(((f^n({^m}x)-\tau)/\lambda^k)+\tau)))={^{m-k-p}}x$$   

Where r=m-k-p, r is any real number (not equal to any whole negative numbers below -1), m and p are natural numbers and k any real integer number which |k|<1. 

(10/21/2022, 08:16 PM)Shanghai46 Wrote: Technically it wouldn't give exactly the same number, but the limit would be the same one, so it's not really changing anything, but the simpler the better


You described letting \(^mx\) go to a big number and then taking the fractional iterate of \(f\), but it does not reflect in your formula. So I guess what you mean with "limit" in your comment is the limit over \(m\), so the formula would look like:
$${^r}x=\lim_{m\to\infty}\lim_{n\to\infty}({\log_x}^m( g^n(((f^n({^m}x)-\tau)/\lambda^k)+\tau)))$$  

Is that what you mean? (Computationally of course one would just look for a suitably big \(^mx\) ) Just in the clarification phase Smile

What I mean is that the larger an number is, the closer Log_a(x+1) will be to Log_a(x).

Is that a "yes" to the double limit formula?
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#7
(10/22/2022, 10:09 AM)bo198214 Wrote:
(10/22/2022, 09:50 AM)Shanghai46 Wrote:
(10/22/2022, 08:10 AM)bo198214 Wrote:
(10/21/2022, 08:29 AM)Shanghai46 Wrote: $${^r}x=\lim_{n\rightarrow~+\infty}({\log_x}^p( g^n(((f^n({^m}x)-\tau)/\lambda^k)+\tau)))={^{m-k-p}}x$$   

Where r=m-k-p, r is any real number (not equal to any whole negative numbers below -1), m and p are natural numbers and k any real integer number which |k|<1. 

(10/21/2022, 08:16 PM)Shanghai46 Wrote: Technically it wouldn't give exactly the same number, but the limit would be the same one, so it's not really changing anything, but the simpler the better


You described letting \(^mx\) go to a big number and then taking the fractional iterate of \(f\), but it does not reflect in your formula. So I guess what you mean with "limit" in your comment is the limit over \(m\), so the formula would look like:
$${^r}x=\lim_{m\to\infty}\lim_{n\to\infty}({\log_x}^m( g^n(((f^n({^m}x)-\tau)/\lambda^k)+\tau)))$$  

Is that what you mean? (Computationally of course one would just look for a suitably big \(^mx\) ) Just in the clarification phase Smile

What I mean is that the larger an number is, the closer Log_a(x+1) will be to Log_a(x).

Is that a "yes" to the double limit formula?
Technically I don't think so
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#8
(10/22/2022, 04:42 PM)Shanghai46 Wrote: Technically I don't think so

Come on, be a bit more elaborative! It should be in your interest to make your finding more understandable!

Let me reiterate: you write
(10/21/2022, 08:29 AM)Shanghai46 Wrote: $${^r}x=\lim_{n\rightarrow~+\infty}({\log_x}^p( g^n(((f^n({^m}x)-\tau)/\lambda^k)+\tau)))={^{m-k-p}}x$$   

Where r=m-k-p, r is any real number (not equal to any whole negative numbers below -1), m and p are natural numbers and k any real integer number which |k|<1. 

This formula is not defining \({^r}x\) in a unique way (apart from the typo "k any real integer number" - what you mean is "k any real number"), because for different choices of m and p we get different results. Say x=2 and r=0.9; when we choose
m=p=2 then \({^{0.9}}2\approx\) 1.8482158
m=p=3 then \({^{0.9}}2\approx\) 1.8802187
m=p=4 then \({^{0.9}}2\approx\) 1.8802194
(which btw seems to be an error in your given value 0.9109247189 which is not \({^{0.9}}2\) but \({^{-0.1}}2\) because \(1={^0}2 < {^{0.9}}2 < {^1}2 =2 \) )

So to make it a proper definition (that does not depend on the choice of m and p), I suggested to take the limit \(m=p\to\infty\) which would be just the execution of what you described in text form in the beginning. So I am wondering a bit why you reject this formula.
For simplicity lets just assume that \(0<r<1\) if that helps to accept the formula.
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#9
seems like using the typical dexp(x) = exp(x) - 1 in the usual way.
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#10
(10/23/2022, 06:50 PM)tommy1729 Wrote: seems like using the typical dexp(x) = exp(x) - 1 in the usual way.

No, actually it is rather similar to your 2sinh method!

The general principle is the following: Say we have some flow \(f^{s+t}(x)=f^s(f^t(x))\)
And the limit 
$$ F^t(x)=\lim_{n\to\infty} \log_b^{\circ n}(f^t(\exp_b^{\circ n}(x))) $$ 
exists then \(F^t\) is again a flow, because roughly (this is not a proof):
\begin{align}
F^s(F^t(x)) &= \lim_{n\to\infty} \log_b^{\circ n}(f^s(\exp_b^{\circ n}(\log_b^{\circ n}(f^t(\exp_b^{\circ n}(x))))))\\
                  &= \lim_{n\to\infty} \log_b^{\circ n}(f^{s+t}(\exp_b^{\circ n}(x)))\\
                  &= F^{s+t}(x)
\end{align}

You use the regular iterates of \(f(x)=2\sinh(x)\) as flow, while Shanghai46 uses the regular iterates of \(f(x)=b^x-1\) as flow. Because both functions satisfy (if one extends sinh to other bases accordingly):
\begin{align}
\lim_{n\to\infty} \log_b^{\circ n}(f(\exp_b^{\circ n}(x))) &= \exp_b(x) \\
F^{t+1}(x) &= \exp_b(F^t(x))
\end{align}

And this is the main dilemma with those approaches they are just countless!

I think also Peter Walker used a similar approach.
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