I just realized that it is quite easy in maple to compute the matrix power via diagonalization (the function is called "MatrixPower" and you can put float values as exponents), I just compare it with the natural tetration.
To get the dsexp (diagonalization super exponential) I compute the Carlemann matrix \( E \) of \( e^x \) then just take the \( t \)-th matrix power \( E^t \) via "MatrixPower" and get the value of row 1 and column 0, which is then \( \exp^{\circ t}(0) \) so the diagonalization tetration is \( e[4]t=\text{dsexp}(t)=\exp^{\circ t+1}(0)=\exp^{\circ t}(1) \).
For the comparison I compute \( \delta(t)=\text{dsexp}(\text{nslog}(t))-t \) which is always a periodic function with period 1.
And this is the resulting \( \delta(t) \) for matrix size of dsexp and nslog being 9 and precision 90 digits:
I think even in this low precision its recognizable that they are not equal. However I am currently preparing a plot in doubled precision which though takes some time, so I will add the graph later to this post.
edit: and here it is now:
hm, the amplitude decreased a lot, so I am again unsure ...
To get the dsexp (diagonalization super exponential) I compute the Carlemann matrix \( E \) of \( e^x \) then just take the \( t \)-th matrix power \( E^t \) via "MatrixPower" and get the value of row 1 and column 0, which is then \( \exp^{\circ t}(0) \) so the diagonalization tetration is \( e[4]t=\text{dsexp}(t)=\exp^{\circ t+1}(0)=\exp^{\circ t}(1) \).
For the comparison I compute \( \delta(t)=\text{dsexp}(\text{nslog}(t))-t \) which is always a periodic function with period 1.
And this is the resulting \( \delta(t) \) for matrix size of dsexp and nslog being 9 and precision 90 digits:
I think even in this low precision its recognizable that they are not equal. However I am currently preparing a plot in doubled precision which though takes some time, so I will add the graph later to this post.
edit: and here it is now:
hm, the amplitude decreased a lot, so I am again unsure ...