[VA] SRC#001 - Spiky Integral

[Valentin Albillo]

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Hi, Gerson:

(07-19-2018 01:31 AM)Gerson W. Barbosa Wrote:  

(07-18-2018 11:39 PM)Valentin Albillo Wrote:  Anyway, on a more feasible scale and in case it still might be useful to you, this is what a sufficiently accurate algorithm should return for N=1,000:
      I(1000) = 0.0002742581536       (all digits shown are correct)

It has been useful!
On Free42 I get 0.000274258153298, which is good to 9 significant digits. Since that's based on an asymptotic formula, the result for N=20,000 should be even more accurate.

Seems likely. As I said in some previous post, I don't use any theoretical way to compute the numerators S(n) which, when divided by the respective powers of 2 (and times Pi), directly give the value of the integral. I simply compute the integral itself numerically using a quadrature algorithm, which for the case N=1,000 goes as follows (9 iterations):

      0    0.000319732675251708806710904860429290813827540315
      1    0.000293664662073707698824357734758191408742496989
      2    0.000272327835887749444493527986927062286291843477
      3    0.000274034638824020953737856637483929708504259670
      4    0.000274259426644445675375095867663504951685809235
      5    0.000274258154414552357469096400753858778167613322
      6    0.000274258153608375557632839196888728683604929730
      7    0.000274258153608378926807741119028301299681276600
      8    0.000274258153608378926807734432669808007752908528
      9    0.000274258153608378926807734432669808007979394750

So we get I(1000) = 0.000274258153608378926807734432669808007979394750 (all 48 decimal digits shown are correct)

Applying a multilevel extrapolation scheme to those iterations quickly gives in excess of 100 correct decimal digits. These quadrature-provided results are useful to check that the S(n)-provided ones are correct.

Regards.
V.