10/18/2022, 06:13 AM
(This post was last modified: 10/18/2022, 06:21 AM by Shanghai46.)
So, we saw in the previous part a formula which can be used to calculate half (or even complex) iteration of any function which converge. (it doesn't work if it doesn't)
Now it's easy to see how it could be used in tetration. For any x, \({^x}r\) converges to a finite value when r belongs to \([e^{-e}, e^{1/e}]\), and tetration can be defined as an iteration of the function \(r^x\), repeating it x times.
So, for example, with \(sqrt2\), infinite tetration converges to 2.
So the value \(\tau\) is equal to the value in which it converges, like in this example, and here \(\lambda\) equals the derivative of the function \(r^x\) at \(\tau\). So it equals \(\ln(\tau) \). Since we, with the formula, technically calculate half iteration of the function for big values (like 20.5),to get to lower values, we just need to apply the log a certain amount of times.
So we can drive our final formula :
$${^r}x =(\lim_{n\rightarrow{+}\infty}(\log_{x} ^{n-k}((({^n}x-\tau)*\lambda^p) +\tau))$$
Where k+p=r, k being an integer and |p|<1, and \(\log_x^n\) means taking n times the log (or iterating n times the log base x).
With this formula, we can define \({^r}x\) for values of r lower than -2, BUT NOT whole negatives below or equal to minus 2. For example, r=-1 is ok, same for r=-3.64, r=-15.9, but r=-2,r=-3,r=-4, ... is forbidden.
The formula looks hard but is really easy to understand, for example, to calculate ${^{0.5}}(\sqrt2)$, we take a big value of ${^n}(\sqrt2)$ (like $n=23$), we substract $2$, multiply by $\ln(2)^{0.5}$, we add $2$, and take the $\log_{\sqrt2}$ $n$ times ($23$ times here). And we get a good approximation.
With this formula, I have the following results :
\({^{-0.5}}({\sqrt2})\)≈0.6290566121
\({^{1.3}}({\sqrt2})\)≈1.49334127
\({^{-{\sqrt2}}}({\sqrt2})\)≈-1
\({^{-{2.5}}}({\sqrt2})\) ≈0.8390270267+9.064720284i
etc...
Here is the graph of \({^{x}}({\sqrt2})\) from -2 to 4.
For example, here are some values of \({^x}0.5\) :
\({^{0.5}}0.5\) ≈ 0.6297281585+0.2178921319i
\({^{-1.9}}0.5\) ≈ 2.487456488+1.718168223i
etc...
Here is the graph of \({^x}0.5\) from -2 to 5. (red is the real part, blue is the imaginary part)
Anyway, tell me what you think!
Now it's easy to see how it could be used in tetration. For any x, \({^x}r\) converges to a finite value when r belongs to \([e^{-e}, e^{1/e}]\), and tetration can be defined as an iteration of the function \(r^x\), repeating it x times.
So, for example, with \(sqrt2\), infinite tetration converges to 2.
So the value \(\tau\) is equal to the value in which it converges, like in this example, and here \(\lambda\) equals the derivative of the function \(r^x\) at \(\tau\). So it equals \(\ln(\tau) \). Since we, with the formula, technically calculate half iteration of the function for big values (like 20.5),to get to lower values, we just need to apply the log a certain amount of times.
So we can drive our final formula :
$${^r}x =(\lim_{n\rightarrow{+}\infty}(\log_{x} ^{n-k}((({^n}x-\tau)*\lambda^p) +\tau))$$
Where k+p=r, k being an integer and |p|<1, and \(\log_x^n\) means taking n times the log (or iterating n times the log base x).
With this formula, we can define \({^r}x\) for values of r lower than -2, BUT NOT whole negatives below or equal to minus 2. For example, r=-1 is ok, same for r=-3.64, r=-15.9, but r=-2,r=-3,r=-4, ... is forbidden.
The formula looks hard but is really easy to understand, for example, to calculate ${^{0.5}}(\sqrt2)$, we take a big value of ${^n}(\sqrt2)$ (like $n=23$), we substract $2$, multiply by $\ln(2)^{0.5}$, we add $2$, and take the $\log_{\sqrt2}$ $n$ times ($23$ times here). And we get a good approximation.
With this formula, I have the following results :
\({^{-0.5}}({\sqrt2})\)≈0.6290566121
\({^{1.3}}({\sqrt2})\)≈1.49334127
\({^{-{\sqrt2}}}({\sqrt2})\)≈-1
\({^{-{2.5}}}({\sqrt2})\) ≈0.8390270267+9.064720284i
etc...
Here is the graph of \({^{x}}({\sqrt2})\) from -2 to 4.
For example, here are some values of \({^x}0.5\) :
\({^{0.5}}0.5\) ≈ 0.6297281585+0.2178921319i
\({^{-1.9}}0.5\) ≈ 2.487456488+1.718168223i
etc...
Here is the graph of \({^x}0.5\) from -2 to 5. (red is the real part, blue is the imaginary part)
Anyway, tell me what you think!
