 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 ab=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 But now the major advantage, the extension to the real numbers. We can easily see that for for example , , . There and so , and . Now the good thing about each is that it has the fixed point 1 () and we can do regular iteration there. For k>2, it seems . Back to the operation we have or in other words we define . I didnt explicate it yet, but this yields quite sure and also on the positive reals. I will see to provide some graphs of xy 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)(a+1)=a+2 which changes to aa=a+1 by substituting a+1=a. However this seems to contradict (a+2)(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)(a+1)=a+2 which changes to aa=a+1 by substituting a+1=a. However this seems to contradict (a+2)(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. as you mentioned, so because it is the identity function, but by definition of addition! therefore, cannot exist. Andrew Robbins RE: The balanced hyperop sequence - bo198214 - 04/18/2008 andydude Wrote: as you mentioned, so because it is the identity function, but by definition of addition! therefore, 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 in tetration use the formula , starting with , 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 ) is quite simple, the principal Abel function is: for an attracting fixed point at 0 and an arbitrary starting point in the attracting domain of the fixed point. The formula remains still valid for an arbitrary attracting fixed point. The regular iteration is then . I am not completely sure about the convergence but this should be equivalent (if we substitute with its approximations) to In our case however the fixed point at 1 is repelling, so we take advantage of having an attracting fixed point together with : here and . However the convergence is so fucking slow, already for the Abel function of 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. xx where  is the balanced tetration. This was possible by a mixture of recurrent and power series formula. And this is the result: [attachment=658] blue: xx = x*x = x^2 green: xx = x^x red: xx 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]