The balanced hyperop sequence - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: The balanced hyperop sequence (/showthread.php?tid=145) The balanced hyperop sequence - bo198214 - 04/14/2008 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. RE: The balanced hyperop sequence - andydude - 04/18/2008 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 RE: The balanced hyperop sequence - bo198214 - 04/18/2008 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? RE: The balanced hyperop sequence - bo198214 - 04/18/2008 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 RE: The balanced hyperop sequence - andydude - 04/20/2008 bo198214 Wrote:An e-mail to me? I dont remember, did I reply? No you didn't. Perhaps it never got through. RE: The balanced hyperop sequence - bo198214 - 04/26/2008 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. Please resend it, have also a look at your private messages. RE: The balanced hyperop sequence - bo198214 - 11/30/2009 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]