10/21/2022, 08:29 AM

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?

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?