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 The balanced hyperop sequence bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 04/14/2008, 08:44 AM (This post was last modified: 04/14/2008, 08:49 AM by bo198214.) 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. « Next Oldest | Next Newest »

 Messages In This Thread The balanced hyperop sequence - by bo198214 - 04/14/2008, 08:44 AM RE: The balanced hyperop sequence - by andydude - 04/18/2008, 05:23 PM RE: The balanced hyperop sequence - by bo198214 - 04/18/2008, 05:58 PM RE: The balanced hyperop sequence - by bo198214 - 04/18/2008, 06:20 PM RE: The balanced hyperop sequence - by andydude - 04/20/2008, 02:28 AM RE: The balanced hyperop sequence - by bo198214 - 04/26/2008, 07:20 PM RE: The balanced hyperop sequence - by bo198214 - 11/30/2009, 11:37 PM

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