Cauchy Integral Experiment

[Kouznetsov]

I'm not sure how calculating the values on the imag axis for where it succeeds in converging is useful for debugging this, though.

But if you really want it, here's what I get using 811 nodes (as this meets the given requirement) and 32 normalized followed by 6 unnormalized iterations with A = 24:

Code:

residual mag at 0: 0.0000000007856242620156700
residual mag at 2.2*I: 0.0000000009481759175558504
bias: 0.000000009638303087908689 - 1.135168310191088 E-16*I
tet(-2.2*I): 0.4620597753654270 - 1.295163164718386*I
tet(-2.1*I): 0.4798492754148191 - 1.283606567671135*I
tet(-2.0*I): 0.4993847992607353 - 1.269767235743538*I
tet(-1.9*I): 0.5207318688313386 - 1.253292769813972*I
...

I attach the plot of the tetrational, performed with my function FSEXP,
along the imahinary axis: Black shows the real part; Red shows the imaginary part
   
In the same figure, I plot the deviation of the values by Mik from vallues by FSEXP,
scaled with factor 10^9; real part with green and imaginary part with blue.

I see, the deviation is at the level of 10^-9 and looks pretty smooth.
Mik already got 9 correct digits.
I expect, Mik can get 12 digits, if drops the step of integration with factor 0.1

Also, isn't Simpson's rule based on a quadratic, not a cubic? At least that's what my calc. text said.

Mik, the Simpson is cubic. if d is distance between nodes, the error is of order of d^3
The cetnered rectangles give the quadratic order.
The trapecies gives the quadratic order.
The off-centered rectangles give linear order; the error is roughly proportional to d.

Mik, how far from the imaginary axis can you go with your approximation?
Can you fill the strip |Re(z)|<0.6 ?