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Patterns That Eventually Fail
2018-09-22T20:34:29Z - / Hacker News
We now know the two functions cross somewhere near but we don’t know if this is the first crossing! In the original pulse, the point (0,1/2) lies on a plateau, a perfectly constant segment with a half-width of 1. The process of repeatedly taking the moving average will nibble away at this plateau, shrinking its half-width by the half-width of the averaging window. Because this is how Hanspeter Schmid explained the infamous Borwein integrals: ∫sin(t)/t dt = π/2∫sin(t/3)/(t/3) × sin(t)/t dt = π/2∫sin(t/5)/(t/5) × sin(t/3)/(t/3) × sin(t)/t dt = π/2 … ∫sin(t/13)/(t/13) × … × sin(t/3)/(t/3) × sin(t)/t dt = π/2 But then the pattern is broken: ∫sin(t/15)/(t/15) × … × sin(t/3)/(t/3) × sin(t)/t dt < π/2 Here these integrals are from t=0 to t=∞. And Schmid came up with an even more persistent pattern of his own: ∫2 cos(t) sin(t)/t dt = π/2∫2 cos(t) sin(t/3)/(t/3) × sin(t)/t dt = π/2∫2 cos(t) sin(t/5)/(t/5) × sin(t/3)/(t/3) × sin(t)/t dt = π/2…∫2 cos(t) sin(t/111)/(t/111) × … × sin(t/3)/(t/3) × sin(t)/t dt = π/2 But: ∫2 cos(t) sin(t/113)/(t/113) × … × sin(t/3)/(t/3) × sin(t)/t dt < π/2 The first set of integrals, due to Borwein, correspond to taking the Fourier transforms of our sequence of ever-smoother pulses and then evaluating F(0).

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