Cauchy Integral Experiment

[mike3]

The explicit requirement, to be "guaranteed" of convergence, appears to be that the node number be an odd number like 211, 215, etc. i.e. be odd and also for (N-1)/2+1 to be even (or equivalently (N-1)/2 to be odd). One may be able to get away with other numbers if they are sufficiently small (I think 209 works, but 409 does not). Not sure where it starts to fail at precisely.

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
tet(-1.8*I): 0.5439302090049257 - 1.233797070129863*I
tet(-1.7*I): 0.5689857178432989 - 1.210863129463098*I
tet(-1.6*I): 0.5958616741755516 - 1.184048130413531*I
tet(-1.5*I): 0.6244695115404595 - 1.152891336259040*I
tet(-1.4*I): 0.6546596527447695 - 1.116925217149086*I
tet(-1.3*I): 0.6862130782596795 - 1.075690135485824*I
tet(-1.2*I): 0.7188344751404906 - 1.028752709130253*I
tet(-1.1*I): 0.7521479533413454 - 0.9757276681381615*I
tet(-1.0*I): 0.7856963882146419 - 0.9163026217258810*I
tet(-0.9*I): 0.8189454132907991 - 0.8502646768632284*I
tet(-0.8*I): 0.8512929111838792 - 0.7775273409233825*I
tet(-0.7*I): 0.8820845103121792 - 0.6981556650513622*I
tet(-0.6*I): 0.9106350914594263 - 0.6123872292055688*I
tet(-0.5*I): 0.9362556726991349 - 0.5206464296113079*I
tet(-0.4*I): 0.9582843404237904 - 0.4235496900871865*I
tet(-0.3*I): 0.9761192253768883 - 0.3218997327286443*I
tet(-0.2*I): 0.9892510000207400 - 0.2166679077342638*I
tet(-0.1*I): 0.9972921071940174 - 0.1089647240256238*I
tet(0.0*I): 1.000000000000000 - 4.743384504624082 E-20*I
tet(0.1*I): 0.9972921071940174 + 0.1089647240256238*I
tet(0.2*I): 0.9892510000207400 + 0.2166679077342638*I
tet(0.3*I): 0.9761192253768883 + 0.3218997327286443*I
tet(0.4*I): 0.9582843404237904 + 0.4235496900871865*I
tet(0.5*I): 0.9362556726991349 + 0.5206464296113080*I
tet(0.6*I): 0.9106350914594263 + 0.6123872292055688*I
tet(0.7*I): 0.8820845103121792 + 0.6981556650513622*I
tet(0.8*I): 0.8512929111838792 + 0.7775273409233825*I
tet(0.9*I): 0.8189454132907991 + 0.8502646768632284*I
tet(1.0*I): 0.7856963882146419 + 0.9163026217258810*I
tet(1.1*I): 0.7521479533413454 + 0.9757276681381616*I
tet(1.2*I): 0.7188344751404906 + 1.028752709130253*I
tet(1.3*I): 0.6862130782596795 + 1.075690135485824*I
tet(1.4*I): 0.6546596527447695 + 1.116925217149086*I
tet(1.5*I): 0.6244695115404595 + 1.152891336259040*I
tet(1.6*I): 0.5958616741755516 + 1.184048130413531*I
tet(1.7*I): 0.5689857178432989 + 1.210863129463098*I
tet(1.8*I): 0.5439302090049257 + 1.233797070129863*I
tet(1.9*I): 0.5207318688313386 + 1.253292769813972*I
tet(2.0*I): 0.4993847992607353 + 1.269767235743538*I
tet(2.1*I): 0.4798492754148191 + 1.283606567671135*I
tet(2.2*I): 0.4620597753654270 + 1.295163164718386*I

And why does one need to bother with the residual, when one can plainly see the failure as the graph veers way off what it's supposed to be?

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