Thread Rating:
  • 0 Vote(s) - 0 Average
  • 1
  • 2
  • 3
  • 4
  • 5
The balanced hyperop sequence
#1
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



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 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.
Reply
#2
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.

as you mentioned, so
because it is the identity function, but
by definition of addition!
therefore, cannot exist.

Andrew Robbins
Reply
#3
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?
Reply
#4
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
   

   

   

   
The basic structure is similar to that of the tetration fractal, see
here, however it lacks its complexity. So it looks rather stupid Wink
Reply
#5
bo198214 Wrote:An e-mail to me? I dont remember, did I reply?

No you didn't. Perhaps it never got through.
Reply
#6
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.
Reply
#7
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:
   

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:
   
Reply


Possibly Related Threads...
Thread Author Replies Views Last Post
  applying continuum sum to interpolate any sequence. JmsNxn 1 2,056 08/18/2013, 08:55 PM
Last Post: tommy1729
  A random question for mathematicians regarding i and the Fibonacci sequence. robo37 0 1,819 08/07/2011, 11:17 PM
Last Post: robo37
  find an a_n sequence tommy1729 1 1,740 06/04/2011, 10:10 PM
Last Post: tommy1729
  Name for sequence y=0[0]0. y=1[1]1, y=2[2]2, y= n[n]n, y=x[x]x, y=i[i]i, y=z[z]z etc? Ivars 10 7,845 05/03/2008, 08:58 PM
Last Post: GFR



Users browsing this thread: 1 Guest(s)