closed form for regular superfunction expressed as a periodic function

[sheldonison]

On to the generic base implementation, for the periodic regular superfunction. Right now, I don't have a working routine for either Mike's suggestion of Bell's polynomial, or Gottfried's matrix routine. So, I'm still doing this "longhand". B is the base. L is the fixed point for base B. I started with my last post, and edited for a generic base.

\( \text{superf}_{B}(z) = \lim_{n \to \infty} B^{[n](L + L\ln(B)^{z-n})} =
\lim_{n \to \infty} \exp^{[n]} ((L+(L\ln(B))^{z-n})/\ln(B)) \)

\( \text{period}=2\pi i/\ln(L*\ln(B)) \)
\( \text{period}=2\pi i/(L*\ln(B)+\ln(\ln(B))) \)
:: verified if B=sqrt(2), L=4, period = 19.236*I

\( \text{RegularSuperf_B}= \sum_{n=0}^{\infty}a_n(L\ln(B))^{nz} \)
a_0=L, a_1=1,
substitute: \( y=(L\ln(B))^z \)

\( \text{RegularSuperf_B}= \sum_{n=0}^{\infty}a_ny^n \)

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

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

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

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

6) equate terms with the same y coefficient
\( \ln(B)^2a_2L^2y^2=\ln(B)^2Ly^2/2! + \ln(B)La_2y^2 \)
\( \ln(B)^3a_3L^3y^3=\ln(B)^3Ly^3/3! + \ln(B)La_3y^3 + \ln(B)^2La_2y^3 \)
\( \ln(B)^4a_4L^4y^4=\ln(B)^4Ly^4/4! + \ln(B)La_4y^4 + \ln(B)^3La_2y^4/2 + \ln(B)^2La_3y^4 + \ln(B)^2La_2^2y^4/2 \)
\( \ln(B)^5a_5L^5y^5=\ln(B)^5Ly^5/5! + \ln(B)La_5y^5 + \ln(B)^4La_2y^5/6 + \ln(B)^2La_4y^5 + \ln(B)^3La_2^2y^5/2 + \ln(B)^3La_3y^5/2 + \ln(B)^2La_3a_2y^5 \)

cancelling, and rearranging terms
\( \ln(B)a_2L - a_2=\ln(B)/2! \)
\( \ln(B)^2a_3L^2 - a_3=\ln(B)^2/3! + \ln(B)a_2 \)
\( \ln(B)^3a_4L^3 - a_4=\ln(B)^3/4! + \ln(B)^2a_2/2 + \ln(B)a_3 + \ln(B)a_2^2/2 \)
\( \ln(B)^4a_5L^4 - a_5=\ln(B)^4/5! + \ln(B)^3a_2/6 + \ln(B)a_4 + \ln(B)^2a_2^2/2 + \ln(B)^2a_3/2 + \ln(B)a_3a_2 \)

\( a_2 = (\ln(B)/2) / (\ln(B)L-1) \)
\( a_3 = (\ln(B)^2/3! + \ln(B)a_2) / (\ln(B)^2L^2 - 1) \)
\( a_4 = (\ln(B)^3/4! + \ln(B)^2a_2/2 + \ln(B)a_3 + \ln(B)a_2^2/2) / (\ln(B)^3L^3 - 1) \)
\( a_5 = (\ln(B)^4/5! + \ln(B)^3a_2/6 + \ln(B)a_4 + \ln(B)^2a_2^2/2 + \ln(B)^2a_3/2 + \ln(B)a_3a_2) / (\ln(B)^4L^4 - 1) \)

I haven't verified these equations, though the last set matched the discreet Fourier analysis for terms a_0 through a_5. In theory, using Mike's Bell expansion, or perhaps Gottfried's matrix, one could write a program that would generate the coefficients for a known interesting base, like sqrt(2). The equations should match the upper super exponential for sqrt(2).
- Sheldon