08/10/2007, 03:59 AM
For the rest of this post, assume \( x\ =\ e^{\frac {\tiny 1}{e}} \).
This is mainly a theoretical solution. It'll be slow to converge without using helper functions to speed up certain processes. I'll cover some of the helper functions I've come up with, but in a later post.
For an arbitrary large integer n, we can find the tetration, calling it z:
\( \begin{eqnarray}
z
& = & ^{\small n} x \ \\ \\[10pt]
\\
& = & e-{\normalsize \epsilon},\text{ where }{\normalsize \epsilon}\ >\ 0 \\
& = & e(1-{\normalsize \delta}),\ \text{where}\ {\normalsize \delta}\ =\ \frac{\epsilon}{e}\ >\ 0 \end{eqnarray} \)
It will be much easier to understand the following if we factor out the e.
Let's find the first logarithm. Remember the formula for logarithms in bases other than e:
\( \begin{eqnarray}
log_x(z)
& = & \frac{ln(z)}{ln(x)} \\ \\[10pt]
\\
& = & \frac{ln(z)}{\frac{\tiny 1}{e}} \\ \\[10pt]
\\
& = & e\ \times\ ln(z)\end{eqnarray} \)
Now, assuming z = e(1-d), let's see what we know about its logarithm and its second iterated logarithm:
\( \begin{eqnarray}
log_x(z)
& = & e\ \times\ ln\left(e(1\ -\ {\normalsize \delta})\right) \\
& = & e \left(1+ln(1\ -\ {\normalsize \delta})\right) \\
& = & e \left(1\ -\ {\normalsize \delta} -\ \frac{\delta^2}{2}\ -\ \frac{\delta^3}{3}\ -\ \dots\right)\ \text{using the series expansion of ln(1+t)}\end{eqnarray} \)
\(
\begin{eqnarray}
log_x\left(log_x(z)\right)
& = & e\ \times\ ln\left(e(1\ -\ {\normalsize \delta}\ -\ \frac{\delta^2}{2}\ -\ \frac{\delta^3}{3}\ -\ \dots)\right) \\
& = & e\left(1+ln(1\ -\ {\normalsize \delta}\ -\ \frac{\delta^2}{2}\ -\ \frac{\delta^3}{3}\ -\ \dots)\right) \\
& = & e\left(1\ -\ {\normalsize \delta}\ -\ 2\frac{\delta^2}{2}\ -\ 3.5\frac{\delta^3}{3}\ -\ \dots\right)\end{eqnarray}
\)
Note that as n grows arbitrarily large, we see that on the interval y in [n-2,n], x^^y is must be almost linear, because we have three points that are almost colinear:
\( \begin{eqnarray}
^{\small n-2} x & = & e\left(1\ -\ {\normalsize \delta}\ -\ {\Large (2)}\frac{\delta^2}{2}\ -\ {\Large (3.5)}\frac{\delta^3}{3}\ -\ \dots\right)\\
^{\small n-1} x & = & e\left(1\ -\ {\normalsize \delta}\ -\ {\Large (1)}\frac{\delta^2}{2}\ -\ {\Large (1)}\frac{\delta^3}{3}\ -\ \dots\right)\\
^{\small n-0} x & = & e\left(1\ -\ {\normalsize \delta}\ -\ {\Large (0)}\frac{\delta^2}{2}\ -\ {\Large (0)}\frac{\delta^3}{3}\ -\ \dots\right)
\end{eqnarray}
\)
If you can't tell at a glance that the interpolating function must be linear in the limit as epsilon goes to 0, let me know. However, I assume that at this point, I've made my case. The interpolating function is linear, and since we're using tetration to integer hyperpowers to find the endpoints of our interval, we can solve exactly. We then use iterative logarithms (integer iteration count, again, so we can solve exactly).
It's important that everybody can agree that this solution is both exact and unique. The uniqueness of this solution can then be used to solve other bases, with some hope that those solutions will be unique as well. In other words, other solutions might meet the iterated exponential property and be infinitely differentiable, but they would be off by some cyclic factor in the tetrational exponent.
Given that I only have a bachelor's degree in computer science, with no formal graduate math study, I can only assume this exact solution must be known.
However, I haven't found it in my cursory reading of what's available on the internet, so apparently this solution, assuming it's already known, is buried in the literature. Perhaps it's considered too technical for us mere mortals. Anyway, can someone please point me to where this has been independently derived? Thanks.
(There is, of course, a chance that my solution is not unique, which is to say, that I've overlooked some small detail. But I really don't see that I could have missed anything, unless my series expansion for the second logarithm is wrong. However, I've triple-checked it.)
For background reference to how I derived this solution, see my sci.math.reference posts here:
http://groups.google.com/group/sci.math....dcdd1fe33d
This is mainly a theoretical solution. It'll be slow to converge without using helper functions to speed up certain processes. I'll cover some of the helper functions I've come up with, but in a later post.
For an arbitrary large integer n, we can find the tetration, calling it z:
\( \begin{eqnarray}
z
& = & ^{\small n} x \ \\ \\[10pt]
\\
& = & e-{\normalsize \epsilon},\text{ where }{\normalsize \epsilon}\ >\ 0 \\
& = & e(1-{\normalsize \delta}),\ \text{where}\ {\normalsize \delta}\ =\ \frac{\epsilon}{e}\ >\ 0 \end{eqnarray} \)
It will be much easier to understand the following if we factor out the e.
Let's find the first logarithm. Remember the formula for logarithms in bases other than e:
\( \begin{eqnarray}
log_x(z)
& = & \frac{ln(z)}{ln(x)} \\ \\[10pt]
\\
& = & \frac{ln(z)}{\frac{\tiny 1}{e}} \\ \\[10pt]
\\
& = & e\ \times\ ln(z)\end{eqnarray} \)
Now, assuming z = e(1-d), let's see what we know about its logarithm and its second iterated logarithm:
\( \begin{eqnarray}
log_x(z)
& = & e\ \times\ ln\left(e(1\ -\ {\normalsize \delta})\right) \\
& = & e \left(1+ln(1\ -\ {\normalsize \delta})\right) \\
& = & e \left(1\ -\ {\normalsize \delta} -\ \frac{\delta^2}{2}\ -\ \frac{\delta^3}{3}\ -\ \dots\right)\ \text{using the series expansion of ln(1+t)}\end{eqnarray} \)
\(
\begin{eqnarray}
log_x\left(log_x(z)\right)
& = & e\ \times\ ln\left(e(1\ -\ {\normalsize \delta}\ -\ \frac{\delta^2}{2}\ -\ \frac{\delta^3}{3}\ -\ \dots)\right) \\
& = & e\left(1+ln(1\ -\ {\normalsize \delta}\ -\ \frac{\delta^2}{2}\ -\ \frac{\delta^3}{3}\ -\ \dots)\right) \\
& = & e\left(1\ -\ {\normalsize \delta}\ -\ 2\frac{\delta^2}{2}\ -\ 3.5\frac{\delta^3}{3}\ -\ \dots\right)\end{eqnarray}
\)
Note that as n grows arbitrarily large, we see that on the interval y in [n-2,n], x^^y is must be almost linear, because we have three points that are almost colinear:
\( \begin{eqnarray}
^{\small n-2} x & = & e\left(1\ -\ {\normalsize \delta}\ -\ {\Large (2)}\frac{\delta^2}{2}\ -\ {\Large (3.5)}\frac{\delta^3}{3}\ -\ \dots\right)\\
^{\small n-1} x & = & e\left(1\ -\ {\normalsize \delta}\ -\ {\Large (1)}\frac{\delta^2}{2}\ -\ {\Large (1)}\frac{\delta^3}{3}\ -\ \dots\right)\\
^{\small n-0} x & = & e\left(1\ -\ {\normalsize \delta}\ -\ {\Large (0)}\frac{\delta^2}{2}\ -\ {\Large (0)}\frac{\delta^3}{3}\ -\ \dots\right)
\end{eqnarray}
\)
If you can't tell at a glance that the interpolating function must be linear in the limit as epsilon goes to 0, let me know. However, I assume that at this point, I've made my case. The interpolating function is linear, and since we're using tetration to integer hyperpowers to find the endpoints of our interval, we can solve exactly. We then use iterative logarithms (integer iteration count, again, so we can solve exactly).
It's important that everybody can agree that this solution is both exact and unique. The uniqueness of this solution can then be used to solve other bases, with some hope that those solutions will be unique as well. In other words, other solutions might meet the iterated exponential property and be infinitely differentiable, but they would be off by some cyclic factor in the tetrational exponent.
Given that I only have a bachelor's degree in computer science, with no formal graduate math study, I can only assume this exact solution must be known.
However, I haven't found it in my cursory reading of what's available on the internet, so apparently this solution, assuming it's already known, is buried in the literature. Perhaps it's considered too technical for us mere mortals. Anyway, can someone please point me to where this has been independently derived? Thanks.
(There is, of course, a chance that my solution is not unique, which is to say, that I've overlooked some small detail. But I really don't see that I could have missed anything, unless my series expansion for the second logarithm is wrong. However, I've triple-checked it.)
For background reference to how I derived this solution, see my sci.math.reference posts here:
http://groups.google.com/group/sci.math....dcdd1fe33d
