# Tetration Forum

Full Version: The balanced hyperop sequence
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Till now we always discussed right-bracketed tetration, i.e. with the mother law:
a[n+1](b+1)=a[n](a[n+1]b)

here however I will introduce that balanced mother law:

a[n+1](2b) = (a[n+1]b) [n] (a[n+1]b)

a major difference to the right-bracketing hyperopsequence is that we can only derive values of the form 2^n for the right operand. Though this looks like a disadvantage, it has the major advantage being able to uniquely (or at least canonicly) being extended to the real/complex numbers.

First indeed we notice, that if we set a[1]b=a+b and if we set the starting condition a[n+1]1=a then the first three operations are indeed again addition multiplication and exponentiation:

by induction
$a[2](2^{n+1})=(a[2]2^n)+(a[2]2^n)=a2^n+a2^n=2a2^n=a2^{n+1}$
$a[3](2^{n+1})=(a[3]2^n)(a[3]2^n)=a^{2^n}a^{2^n}=a^{22^n}=a^{2^{n+1}}$

But now the major advantage, the extension to the real numbers. We can easily see that

$x[k+1]2^n=f_{k}^{\circ n}(x)$ for $f_k(x)=x[k]x$

for example $x[4]1=x$, $x[4]2=x[3]x=x^x$, $x[4]4=(x^x)[3](x^x)=(x^x)^{x^x}=x^{xx^x}$. There $f_3(x)=x^x$ and so $x[4]2^0=x=f^{\circ 0}(x)$, $x[4]2^1=x^x=f_3(x)$ and $x[4]2^2=x^{xx^x}=f_3(f_3(x))$.

Now the good thing about each $f_k$ is that it has the fixed point 1 ($k>1$) and we can do regular iteration there. For k>2, it seems ${f_k}'(1)=1$.

Back to the operation we have
$x[k+1]2^t=f_k^{\circ t}(x)$ or in other words we define

$x[k+1]y={f_k}^{\circ (\log_2 y)}(x)$.
I didnt explicate it yet, but this yields quite sure $x[2]y=xy$ and $x[3]y=x^y$ also on the positive reals.

I will see to provide some graphs of x[4]y in the future.
The increase rate of balanced tetration should be between the one left-bracketed tetration and right-bracketed/normal tetration.

I also didnt think about zeration in the context of the balanced mother law. We have (a+1)[0](a+1)=a+2 which changes to a[0]a=a+1 by substituting a+1=a. However this seems to contradict (a+2)[0](a+2)=a+4. So maybe there is no zeration here.
bo198214 Wrote:I also didnt think about zeration in the context of the balanced mother law. We have (a+1)[0](a+1)=a+2 which changes to a[0]a=a+1 by substituting a+1=a. However this seems to contradict (a+2)[0](a+2)=a+4. So maybe there is no zeration here.

Yes, I completely agree with you. But I would prove it differently, as I did in an email to you awhile back.

$x[n+1]2^t = f_n^t(x)$ as you mentioned, so
$x[1]1 = f_0^0(x) = x$ because it is the identity function, but
$x[1]1 = x + 1$ by definition of addition!
therefore, $f_0(x) = x[0]x$ cannot exist.

Andrew Robbins
andydude Wrote:$x[n+1]2^t = f_n^t(x)$ as you mentioned, so
$x[1]1 = f_0^0(x) = x$ because it is the identity function, but
$x[1]1 = x + 1$ by definition of addition!
therefore, $f_0(x) = x[0]x$ cannot exist.

This is a nice proof, thanks.

Quote:... prove it differently, as I did in an email to you awhile back.

An e-mail to me? I dont remember, did I reply?
If we instead of using the formula $z=c^z$ in tetration use the formula $z=z^z$, starting with $z=c$, we get the balanced tetration fractal. It is *much* simpler than the tetration fractal resembling the easy handling of the extension of this kind of tetration.
Here some pictures
[attachment=314]

[attachment=315]

[attachment=316]

[attachment=317]
The basic structure is similar to that of the tetration fractal, see
here, however it lacks its complexity. So it looks rather stupid
bo198214 Wrote:An e-mail to me? I dont remember, did I reply?

No you didn't. Perhaps it never got through.
The iterational formula for parabolic iteration (like for $f(x)=x^x$) is quite simple, the principal Abel function is:

$\alpha(x)=\lim_{n\to\infty} \frac{f^{\circ n}(x)-f^{\circ n}(x_0)}{f^{\circ n+1}(x_0)-f^{\circ n}(x_0)}$ for an attracting fixed point at 0 and an arbitrary starting point $x_0$ in the attracting domain of the fixed point.
The formula remains still valid for an arbitrary attracting fixed point. The regular iteration is then

$f^{\circ t}(x)=\alpha^{-1}(t+\alpha(x))$.
I am not completely sure about the convergence but this should be equivalent (if we substitute $\alpha$ with its approximations) to
$f^{\circ t}(x)=\lim_{n\to\infty}f^{\circ -n}(t(f^{\circ n+1}(x_0)-f^{\circ n}(x_0))+f^{\circ n}(x))$

In our case however the fixed point at 1 is repelling, so we take advantage of $f^{-1}$ having an attracting fixed point together with $f^{\circ t}=(f^{-1})^{\circ -t}$:

$f^{\circ t}(x)=\lim_{n\to\infty}f^{\circ n}(t(f^{\circ -n}(x_0)-f^{\circ -(n+1)}(x_0))+f^{\circ -n}(x))$

here $f(x)=x^x$ and $f^{-1}(x)=e^{W(\ln(x))}$.

However the convergence is so fucking slow, already for the Abel function of $f^{-1}$ that I am not able to post a graph yet!

andydude Wrote:No you didn't. Perhaps it never got through.

Oh guys, long time its ago since this thread was started.
But now I am finally able to post the premiere graph of balanced selftetration, i.e. x[4]x where [4] is the balanced tetration.
This was possible by a mixture of recurrent and power series formula.
And this is the result:
[attachment=658]

blue: x[2]x = x*x = x^2
green: x[3]x = x^x
red: x[4]x

Note that 2[n]2 = 4 in balanced hyperoperations of arbitrary rank n.

and here with aspect ratio 1 and the identity in black as comparison, including 0:
[attachment=659]