[VA] SRC#001 - Spiky Integral

[Valentin Albillo]

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Hi, Gerson:

(07-18-2018 03:07 AM)Gerson W. Barbosa Wrote:  Not meaning to abuse your good will, would you please check how many significant digits I get right for N=20000?

No, sorry, I can't verify your alleged numeric result for N=20,000, because:

• f(x) = Cos(x)*Cos(2*x)*...*Cos(20000*x) is not just 'spiky', it's "solid-area" spiky, with tens of thousands of extremely thin spikes crowding the [0,2*Pi] interval so much that very few samples, if any, fall on each individual spike even using a million samples. Thus, I'd have to use tens of millions of samples to compute the integral in [0,2*Pi] to any useful accuracy.
        
• f(x) is the product of 20,000 cosines, each of which is a number with absolute value <= 1 (it's exactly 1 only at x=0 and various fractions of Pi, none of which are ever sampled) and with average absolute value 2/Pi, so f(x) is typically ~ (2/Pi)^20000 ~ 10^(-3922) for almost every x in [0,2*Pi].
  
  This means I'd need to compute each sample using at least 4,000 decimal digits of precision, i.e.: evaluate the product of 20,000 cosines, each of them computed to 4,000 decimal digits, for *each* and every one of the many million samples. Trying to use less decimal digits, say 1,000 or 2,000, would result in f(x) evaluating to 0 for every x sampled, as would the integral itself.

Needless to say, computing I(20000) this way would require a truly humongous amount of time, certainly it would for my POPS (Pretty Old Pretty Slow) system, and regrettably I can't allocate that much running time to this task.

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)

Regards.
V.
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