Superlog with exact coefficients

[Gottfried]

This means that \( \beta(x) = \text{slog}_{(b^{1/b})}(x) \) which relates back to the super-logarithm as follows:
\( \text{slog}_{(e^{a(e^{-a})})}(x) = C + \frac{1}{\ln(a)}(
(x(e^{-a})-1) + \frac{a(x(e^{-a})-1)^2}{4(1-a)} + \frac{a^2(1+5a)(x(e^{-a})-1)^3}{36(a-1)^2(a+1)}
- \frac{a^4(2+a+3a^2)(x(e^{-a})-1)^4}{32(a-1)^3(a+1)(1+a+a^2)} + \dots )
\)



Andrew Robbins

One year late... and only partial progress.
I express this with my toolbox of matrices.
First we agree, that we can use for exp_b(x) the dxp_t(x/t-1) (while, however the full equivalence is then (dxp_t(x/t-1) + 1)*t )

For exp_b(x) I use the transposed Carleman-matrix Bb
V(x) ~ * Bb = V(b^x)~
For dxp_t(x) I use t^(1/t) = b ,u = log(t)

V(x/t-1) ~ *Ut = V( b^x/t- 1) ~

where Ut is the matrix-operator for dxp_t having the coefficients in its second column
Create it as "Utsym" symbolically with the parameter a (from your notation = log(t) = u) to dimension, say 5:

Utsym = dV(a,5) * VE(fS2F,5);

Now the iterative logarithm can be found by finding the matrix-log of Utsym. Since the diagonal is not the Id-matrix, I take the eigen-decomposition (efficiently configured for triangular matrices)

UtsymKenn = triEigSys(Utsym) ;

and this has in the three components UtsymKenn[2] =W, UtsymKenn[3]=D ,UtsymKenn[4] = W^-1
where D is diagonal and contains the eigenvalues [1,a,a^2,a^3,a^4], sucht that

Utsym = W * D * W^-1

The log of Utsym is then the re-composition with the log of the diagonal, (Log(D)) and this is the diagonalmatrix

Log(D) = diag([0,1,2,3,4])*lna

where I use the symbol lna for the ln(a) (since we want to keep it symbolically)

UtsymLog = W * lna* diag(0,1,2,3,4) * W^-1
... = lna* W * diag(0,1,2,3,4) * W^-1

The result is not a matrix-operator because the eigenvalues are not vandermonde; so we have to construct the matrixopertator from its second column:

UtJulia = matfromser(UtsymLog[,2])

\(
\begin{matrix} {rrrr}
& lna* (x & -\frac{a}{2(1-a)}*x^2 & + \frac{a^2(1-a)}{3!(1-a)(1-a^2)}*x^3 & - \frac{((a^3(2a+1)(1-a)}{4!(1-a^2)(1-a^3)}*x^4 & \dots )
\end{matrix} \)



This agrees with your representation except of a bit straightening of denominators, so far.

Now to proceed I needed the integral-representation, but unfortunately I don't have the appropriate "toolbox-matrix" for this at the moment...

So I'll try the remaining part another day

Gottfried