closed form for regular superfunction expressed as a periodic function

[sheldonison]

i must say though that i dont completely get how " fourier methods" lead to the closed forms of a_n ... any fourier expert/sheldon want to explain ?

further i believe regular superfunction means expanded at upper fixpoint or at least an equivalent of that.

I agree with Tommy when he says "Fourier methods" (complex Fourier analysis of a discreet number of points), won't lead to a closed form, and I didn't use Fourier methods to generate the coefficients for a_4, and a_5. Also, so far, I've only used this method for the regular superfunction for base(e), although it would presumably work for any regular periodic superfunction (using the upper fixed point, for bases<eta).

I'm going to start over, and add a little more detail as well. What I'm doing here is a lot like what is done for intuitive tetration, but here the equations are exact. Here are the steps. Hopefully, these steps will be a little clearer.

1) realize that the regular superfunction is periodic, and express that superfunction as the infinite sum of a periodic/Fourier series. For base(e), the fixed point is "L", and the period is 2Pi*i/L.
\( \text{RegularSuperf}(z) = \sum_{n=0}^{\infty}a_nL^{nz} \)

2) do the substitution, \( y=L^z \), and rewrite the equations in terms of "y". This is because all of the periodic terms decay to the fixed point, at +I*infinity, and at -real infinity. The equation in y has the same coefficients a_0...a_n, as the equation in z.
\( \text{RegularSuperf}(z) = \sum_{n=0}^{\infty}a_ny^n \)

3) figure out the coefficients for z+1. \( y_{z+1}=L^{z+1}=L*L^z=L*y \)
\( \text{RegularSuperf}(z+1)=
\sum_{n=0}^{\infty}a_nL^ny^n \)

4) Apply the definition of the regular superfunction, f(z+1)=exp(f(z))
\( \text{RegularSuperf}(z+1)=\exp({\text{RegularSuperf}(z)}) \)
\( \text{RegularSuperf}(z+1) = \sum_{n=0}^{\infty}a_nL^ny^n =
\exp({\sum_{n=0}^{\infty}a_ny^n}) \)

5) express this as a product. Then expand the Taylor series for each term in the product, and substitute in \( a_0=L \) \( a_1=1 \).

\( \sum_{n=0}^{\infty}a_nL^ny^n=
\prod_{n=0}^{\infty}\exp({a_ny^n}) \)
\( \sum_{n=0}^{\infty}a_nL^ny^n=
\prod_{n=0}^{\infty}( {
\sum_{p=0}^{\infty}({(a_ny^n)^p/p!})} ) = \)
\(
\\(L)\times
\\(1 + y + y^2/2! + y^3/3! + y^4/4! + y^5/5! ....)\times
\\(1+a_2y^2+a_2^2y^4/2!+a_2^3y^6/3!+a_2^4y^8/4! ...)\times
\\(1+a_3y^3+a_3^2y^6/2!+a_3^3y^9/3!+a_3^4y^{12}/4! ...)\times
\\(1+a_4y^4+a_4^2y^8/2!+a_4^3y^{12}/3!+ ...)\times
\\(1+a_5y^5+a_5^2y^{10}/2!+a_5^3y^{15}/3! + ...)\times ... \)

6) equate terms with the same y coefficient, and figure out the pattern (I haven't yet)
\( a_2L^2y^2=Ly^2/2! + La_2y^2 \)
\( a_3L^3y^3=Ly^3/3! + La_3y^3 + La_2y^3 \)
\( a_4L^4y^4=Ly^4/4! + La_4y^4 + La_3y^4 + La_2y^4/2+ La_2^2y^4/2 \)
\( a_5L^5y^5=Ly^5/5! + La_5y^5 + La_4y^5 + La_3y^5/2 +
La_3a_2y^5+La_2^2y^5/2 + La_2y^5/6 \)
cancelling, and rearranging terms
\( a_2L-a_2=1/2 \)
\( a_3L^2-a_3=1/6 + a_2 \)
\( a_4L^3-a_4=1/24 + a_3 + a_2/2+ a_2^2/2 \)
\( a_5L^4-a_5=1/120 + a_4 + a_3/2 + a_3a_2+a_2^2/2 + a_2/6 \)

\( a_2=0.5/(L-1) \)
\( a_3=(1/6 + a_2)/(L^2-1) \)
\( a_4=(1/24 + a_3 + a_2/2+ a_2^2/2)/(L^3-1) \)
\( a_5= (1/120 + a_4 + a_3/2 + a_3a_2+a_2^2/2 + a_2/6)/(L^4-1) \)

And that would give you the four non-trivial terms I have closed form solutions for, although I haven't verified \( a_5 \), and wouldn't be surprised if it has a typo (edit: a_5 is also verified correct).
- Sheldon