08/30/2010, 03:09 AM
(This post was last modified: 08/30/2010, 03:33 AM by sheldonison.)
(08/28/2010, 11:21 PM)tommy1729 Wrote: that seems efficient and intresting.Thanks Tommy! I assume it has been considered, and probably calculated before. I think Kneser developed the complex periodic regular tetration for base e, and probably would've generated the coefficients. But I haven't seen them before. Perhaps Henryk (or someone else) could comment???
in fact i doubt it hasnt been considered before ?
I figured out the closed form equation for a couple more terms, and I have an equation that should generate the other terms, but I'm still working it, literally as I write this post!
\( a_2 = (1/2)/(L - 1) \)
\( a_3 = (1/6 + a_2)/(L*L - 1) \)
\( a_4 = (1/24 + (1/2)*a_2*a_2 + (1/2)*a_2 + a_3)/(L*L*L-1) \)
What I did is start with the equation:
\( \text{RegularSuperf}(z) = \sum_{n=0}^{\infty}a_nL^{nz} \)
and set it equal to the equation
\( \text{RegularSuperf}(z) = \exp{(\text{RegularSuperf}(z-1))} \)
Continuing, there is a bit of trickery in this step to keep the equations in terms of \( L^{nz} \), instead of in terms of \( L^{n(z-1)} \). Notice that \( L^{n(z-1)}=L^{(nz-n)}=L^{-n}L^{nz} \).
\( \text{RegularSuperf}(z) = \exp{(\text{RegularSuperf}(z-1))} =
\exp{( \sum_{n=0}^{\infty}\exp^{(L^{-n}a_nL^{nz})})} \)
This becomes a product, with \( a_0=L \) and \( a_1=1 \)
\( \text{RegularSuperf}(z) = \prod_{n=0}^{\infty}
\exp{(L^{-n}a_nL^{nz})} \)
The goal is to get an equation in terms of \( L^{nz} \) on both sides of the equation. Then I had a breakthrough, while I was typing this post!!!! The breakthrough is to set \( y=L^z \), and rewrite all of the equations in terms of y! This wraps the 2Pi*I/L cyclic Fourier series around the unit circle, as an analytic function in terms of y, which greatly simplifies the equations, and also helps to justify the equations.
\( \text{RegularSuperf}(z) = \sum_{n=0}^{\infty}a_ny^n
= \prod_{n=0}^{\infty}
\exp{(L^{-n}a_ny^n)} \)
The next step is to expand the individual Tayler series for the \( \exp
{(L^{-n}a_ny^n)} \), and multiply them all together (which gets a little messy, but remember a0=L and a1=1), and finally equate the terms in \( y^n \) on the left hand side equation with those on the right hand side equation, and solve for the individual \( a_n \) coefficients. Anyway, the equations match the numerical results.
I'll fill in the Tayler series substitution next time; this post is already much more detailed then I thought it was going to be! I figured a lot of this out as I typed this post!
- Sheldon