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.
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.
