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We'll do this at a few different precisions. First we show the result from the original input and also from machine doubles created from that input. sd = Table[ AstronomicalData["Sun", {"Declination", DateList[t]}], {t, 3155716800, 3155716800 + 86400*365.2425*10, 86400}]; sdvals = sd[[All, 1]]; Max[Abs[Fourier[sdvals]]] Max[Abs[Fourier[N[sdvals]]]] ...



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