andydude Wrote:This means that which relates back to the super-logarithm as follows:

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])

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

Gottfried Helms, Kassel