Cauchy Integral Experiment

[mike3]

.. So yeah, it fills the strip |Re(z)| <= 0.6 and thus |Re(z)| < 0.6. It gets fairly good accuracy (estimated by residuals where both points z and z-1 lie in that strip, 7-8 decimals past the point I'd guess) even out to z = +/- 0.6 +/- 20i, which is pretty good.

Congratulations, Mike!
You have covered all the complex plane!
Now you may fit your approximation with function
F_1(z)=L+exp(Lz+R)
at Im(z)>>1, and evaluate the fundamental mathematical constant R;
let us compare your value to that in my preprint.

I get this:

Code:

0*I yields 0.4061632794102940 - 1.099252992615197*I
0.5000000000000000*I yields 0.6924867156551292 - 1.081922402845001*I
1.000000000000000*I yields 0.8738145712104024 - 1.051093918320808*I
1.500000000000000*I yields 0.9773101871029590 - 1.018915638151093*I
2.000000000000000*I yields 1.031492620482252 - 0.9926056537509507*I
2.500000000000000*I yields 1.057916122757191 - 0.9740754861170568*I
3.000000000000000*I yields 1.070021771054289 - 0.9622395304732624*I
3.500000000000000*I yields 1.075221977003856 - 0.9551663982020895*I
4.000000000000000*I yields 1.077279549202432 - 0.9511366389234044*I
4.500000000000000*I yields 1.077991266194071 - 0.9489233675231232*I
5.000000000000000*I yields 1.078170621768531 - 0.9477440297511597*I
5.500000000000000*I yields 1.078165418055296 - 0.9471322674327784*I
6.000000000000000*I yields 1.078113478974864 - 0.9468227848420041*I
6.500000000000000*I yields 1.078063128309121 - 0.9466698992156713*I
7.000000000000000*I yields 1.078027585902012 - 0.9465959687249935*I
7.500000000000000*I yields 1.078008533900850 - 0.9465609164056555*I <----
8.000000000000000*I yields 1.078006096997444 - 0.9465453396600495*I
8.500000000000000*I yields 1.078023477849046 - 0.9465422813164446*I
9.000000000000000*I yields 1.078070187185964 - 0.9465552710178442*I
9.500000000000000*I yields 1.078165770855951 - 0.9466027419683657*I
10.00000000000000*I yields 1.078344685852032 - 0.9467320277490121*I
10.50000000000000*I yields 1.078661217763204 - 0.9470509993541728*I
11.00000000000000*I yields 1.079189117562023 - 0.9477938949578427*I
11.50000000000000*I yields 1.080000192286941 - 0.9494530867467299*I
12.00000000000000*I yields 1.081082058963924 - 0.9530355549982699*I
12.50000000000000*I yields 1.082104809256400 - 0.9605518099688068*I
13.00000000000000*I yields 1.081851640193614 - 7.259129558344381*I
13.50000000000000*I yields 1.076980538364589 - 7.290104554659523*I
14.00000000000000*I yields 1.059672426843578 - 7.352203210926850*I
14.50000000000000*I yields 1.014618322326413 - 7.479875821210950*I
15.00000000000000*I yields 0.9281072092851824 - 7.762132820922603*I
15.50000000000000*I yields 0.9300179598593906 - 2.115715210452333*I
16.00000000000000*I yields 1.485639670455890 - 2.924089684858254*I
16.50000000000000*I yields 2.329552728395440 - 3.385155818729130*I
17.00000000000000*I yields 3.143953114185848 - 3.648425512750720*I
17.50000000000000*I yields 3.914953676086398 - 3.844031274641113*I
18.00000000000000*I yields 4.661065077670659 - 4.015756778217886*I
18.50000000000000*I yields 5.394408693844471 - 4.178634130720164*I
19.00000000000000*I yields 6.121142024387684 - 4.337857556079375*I
19.50000000000000*I yields 6.843394731674929 - 4.494686993048395*I
20.00000000000000*I yields 7.559722817541824 - 4.648117828772436*I

The max accuracy appears to be where the "<----" is, giving 1.07800 - 0.94656i
or 1.07801 - 0.94656i (rounded). I think I'll need more decimals to get better convergence though, as I've only got like 8-9, so maybe we just have 1.0780 - 0.9465i / 1.0780 - 0.9466i? I think it takes at least 2x as many decimals in the accuracy of the tetration as the target amount of decimals we want for R due to cancellation, round off, etc.

Code used:

Code:

for(X=0,X=40,print(X/2.0,"*I yields ",log(Tetrate(v, 24.0, bias, X/2.0*I) - L) - (L*X/2.0*I)))

Also, is it time to plot the function on the z plane?