Dmitrii Kouznetsov's Tetration Extension

[Kouznetsov]

Thats always the difficulty with those numerical comparison. If the difference is small does that mean that they are equal?

Henryk asked me to calcilate my tetration with displaced basic interval.
In the paper http://www.ils.uec.ac.jp/~dima/PAPERS/2008analuxp99.pdf ,
the basic algorithm evaluates tetration(\( z \)) in the range
\( |\Re(z)|<1 \); precisely (14 digits) in the range \( |\Re(z)|<0.8 \);
I use it for \( |\Re(z)| \le 0.5 \), and bring to this interval other values using the recurrent relation.
Then, I shifted the interval; the new implementation evaluates the tetration for
\( -1.5\le \Re(z) \le 0.5 \).
I run the test; it is precise for \( -1.25\le \Re(z) \le 0.25 \).
I use it for \( -1\le \Re(z) \le 0 \), bringing other cases to this intergal.
The plots for old algorithm and for the new one look identical.
Then I plot the difference \( f=f(z)=\mathrm{(old-new)}*10^{14} \).
   
The grid covers the range
\( -10\le\Re(z)\le 10 \)
\( -4\le\Im(z)\le 4 \)
with unity step.
Levels \( \Re(f)=-2 \) are shown with thick pink lines
Levels \( \Re(f)=-1 \) are shown with thin pink lines
Levels \( \Re(f)= 1 \) are shown with thin green lines
Levels \( \Re(f)= 2 \) are shown with thck green lines

Levels \( \Im(f)=-2 \) are shown with thick red lines
Levels \( \Im(f)=-1 \) are shown with thin red lines
Levels \( \Im(f)= 1 \) are shown with thin blue lines
Levels \( \Im(f)= 2 \) are shown with thck blue lines

If variation of function esceed 10 per step of the mesh, then the line is not drawn; so, the right hand side of the figure in vicinity of the real axix left blank; but there, the deviation between these two functions is larger than 1.e-14 and, perhaps, even larger than unity; the only the plotter cannot identify the position of lines, tey are too dence.

I see the jumps of the difference at the integer and at half-integer values of the real part; the jumps are at the level of \( 10^{-14} \). Each of the implementations has its own errors, and they can be revealed comparing two finctions. I attribute the humps at the half-integer values to the old algorithm, and jumps at the integer values to the new algorithm. Comparing these two algorithms, I cannot say that one of them is somehow better than another one.

Conclusion: In order to see from numerical evaluation, that two functions are not the same, you should plot them in the complex plane. It is better than to compare the Tailor series.