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Simplified regular tetration - andydude - 05/02/2009
Regular iteration of exponentials gives a power series of \( \exp_a^{y}(z) \) about z. In order to obtain regular tetration from this we must evaluate this at (z=1), but in doing so, it is no longer a power series. In order to make this a power series again we have to re-expand it about 'y' or 'a', which makes really messy power series. This messy power series is of the form
\( {}^{y}a = \sum_{i \ge 0} \sum_{j \ge 0} \sum_{k \ge 0} C_{ijk} (a - 1)^i \ln(a - 1)^j y^k \)
This is not a power series in 'a', because it involves both (a - 1) and \( \ln(a - 1) \). If we are to compare this with other methods, then it would be beneficial to have a true power series in 'a'. To this end, we can define the following function
\( X(u, v) = (e^{ue^{-u}}) \uparrow\uparrow \log_u(v) \)
and define an an inverse function such that \( Z(u, X(u, v)) = v \). There are numerous benefits to defining the function this way. Since all of the functions used are invertible, we can express tetration in terms of X: \( {}^{y}a = X(\ln({}^{\infty}a), \ln({}^{\infty}a)^y) \). We can also express the superlog in terms of Z: \( \text{slog}_a(z) = \log_{\ln({}^{\infty}a)}(Z(\ln({}^{\infty}a), z)) \). Superroots cannot be expressed with these functions, however.
To show that all of the logarithms are gone, here is the resulting power series of X:
\(
\begin{tabular}{rl}
-X(u, v+1)
& = -1 + uv + \frac{u^2v}{2} \\
& + {u^3}\left(\frac{2}{3}v + \frac{1}{2}v^2\right) \\
& + {u^4}\left(\frac{13}{24}v + \frac{1}{2}v^2\right) \\
& + {u^5}\left(\frac{43}{60}v + \frac{25}{24}v^2 + \frac{1}{3}v^3\right) \\
& + {u^6}\left(\frac{331}{720}v + \frac{19}{24}v^2 + \frac{1}{3}v^3\right) \\
& + {u^7}\left(\frac{1999}{2520}v + \frac{1231}{720}v^2 + \frac{7}{6}v^3 + \frac{1}{4}v^4\right) \\
& + {u^8}\left(\frac{17977}{40320}v + \frac{811}{720}v^2 + \frac{67}{72}v^3 + \frac{1}{4}v^4\right) \\
& + {u^9}\left(\frac{25705}{36288}v
+ \frac{85681}{40320}v^2 + \frac{281}{120}v^3 + \frac{9}{8}v^4 + \frac{1}{5}v^5\right) \\
& + {u^{10}}\left(\frac{296113}{518400}v
+ \frac{14143}{8064}v^2 + \frac{583}{288}v^3 + \frac{25}{24}v^4 + \frac{1}{5}v^5\right) \\
& + \cdots
\end{tabular}
\)
as you can see, some of the coefficients display a pattern, like the \( \frac{1}{n}v^n \) terms, but I don't know if this pattern continues. Even so, I think this form is much easier to analyze than picking a base and sticking with it for all calculations. For example, the Julia function of exponentials (evaluated at 1) can be expressed with this function as well.\begin{tabular}{rl}
-X(u, v+1)
& = -1 + uv + \frac{u^2v}{2} \\
& + {u^3}\left(\frac{2}{3}v + \frac{1}{2}v^2\right) \\
& + {u^4}\left(\frac{13}{24}v + \frac{1}{2}v^2\right) \\
& + {u^5}\left(\frac{43}{60}v + \frac{25}{24}v^2 + \frac{1}{3}v^3\right) \\
& + {u^6}\left(\frac{331}{720}v + \frac{19}{24}v^2 + \frac{1}{3}v^3\right) \\
& + {u^7}\left(\frac{1999}{2520}v + \frac{1231}{720}v^2 + \frac{7}{6}v^3 + \frac{1}{4}v^4\right) \\
& + {u^8}\left(\frac{17977}{40320}v + \frac{811}{720}v^2 + \frac{67}{72}v^3 + \frac{1}{4}v^4\right) \\
& + {u^9}\left(\frac{25705}{36288}v
+ \frac{85681}{40320}v^2 + \frac{281}{120}v^3 + \frac{9}{8}v^4 + \frac{1}{5}v^5\right) \\
& + {u^{10}}\left(\frac{296113}{518400}v
+ \frac{14143}{8064}v^2 + \frac{583}{288}v^3 + \frac{25}{24}v^4 + \frac{1}{5}v^5\right) \\
& + \cdots
\end{tabular}
\)
\( \mathcal{J}[\exp_a](1) = \ln(\ln({}^{\infty}a))\left[\frac{\partial}{\partial v} X(\ln({}^{\infty}a), v)\right]_{v=1} \)
so to summarize, these are the benefits of this simplified view of regular tetration:
- The coefficients of (-X) are always positive (except for the -1 in the first term), which means X itself has negative coefficients (except for the 1 in the first term).
- All of the logarithms in the direct expansion are gone, meaning they must have come from the sub-expansions of \( {}^{\infty}a \) in the power series.
- We can express tetration in terms of X, without any loss of generality.
- We can express superlogs in terms of the inverse function Z.
- We can express the Julia function of exponentials in terms of \( \frac{\partial X}{\partial v} \)
- \( X(u, 0) = e^{u} \)
- \( X(u, 1/u) = 0 \)
- \( X(u, 1) = 1 \)
- \( X(u, u) = e^{ue^{-u}} \)
- \( X(u, \infty) = e^{u} \)
- \( X(u, uv) = e^{ue^{-u}} \uparrow X(u, v) \)
Andrew Robbins
RE: Simplified regular tetration - bo198214 - 05/02/2009
We know that we can express the Abel function \( \alpha \) of \( f \) at \( a \) by
\( \alpha = \log_c \circ \chi \) where \( \chi \) is the Schröder function of \( f \) at \( a \), where \( c=f'(a) \).
This means that the inverse of the Abel function which is actually the superexponential can be expressed by
\( F(x)=\chi^{-1}(c^x) \) with appropriate translation along the x-axis choosen such that \( F(0)=1 \). Here \( c=\exp_b'(a)=\ln(b) \exp_b(a)=\ln(b)a=\ln(b^a)=\ln(a) \)
The coefficients of \( \eta+a=\chi^{-1} \), \( \eta_0=0 \), \( \eta_1=1 \), can be recursively computed from the equation
(*) \( \eta(cx)=f(\eta(x)) \), \( f(x)=b^{x+a}-a = a (b^x-1) \)
by the composition formula
\( (f\circ g)_n = \sum_{k=1}^n f_k (g^k)_n \) where \( (g^k)_n = \sum_{n_1+\dots+n_k = n} f_{n_1} \dots f_{n_k} \).
Now we put formula (*) in:
\( c^n \eta_n = \sum_{k=1}^n f_k (\eta^k)_n \)
On the right side \( \eta_n \) only occurs for \( k=1 \) in \( \eta^k \), namely in the summand \( f_1 (\eta^1)_n = c \eta_n \). Thatswhy we have the recursive formula:
\( \eta_n = \frac{1}{c^n - c} \sum_{k=2}^n f_k (\eta^k)_n \)
So each coefficient of \( \eta \) is a polynomial in \( \ln(b) \) (\( b \) base) and a rational function in \( c=\ln(a) \) (\( a \) fixed point).
and upto translation along the x-Axis we have \( {^xb}=\eta(c^x)+a \).
RE: Simplified regular tetration - andydude - 05/03/2009
So is X the inverse of the Schroeder function?
RE: Simplified regular tetration - bo198214 - 05/03/2009
(05/03/2009, 08:49 AM)andydude Wrote: So is X the inverse of the Schroeder function?Actually I am also somewhat puzzled.
I would suggest to always use variable \( b \) for the base.
In this terminology you wrote:
\( {}^{y}b = X(\ln({}^{\infty}b), \ln({}^{\infty}b)^y) \)
Now \( {^\infty b} \) is the fixed point which I denoted with \( a \).
And \( \ln(a) \) is the derivative at the fixed point which I denoted with \( c \).
So your formula is:
\( {}^{x}b = X(c, c^x) \)
while my formula is:
\( {^xb}=\eta(c^x)+a \)
Your coefficients are polynomials in \( c \), while my coefficients are rational functions in \( c \) and also use \( \ln(b) \). Though both must be equal up to translation. *headscratch*