Estimate logarithm quickly
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08-21-2021, 03:58 PM
Post: #2
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RE: Estimate logarithm quickly
(08-21-2021 03:39 PM)Albert Chan Wrote: atanh(1/(2*c*c-1)) > atanh(1/c) / (2c) = (a-b)/2 / (2c) This is not yet proofed (n-x > 0, c = n/x > 1) To proof it, it required this inequality (also not proofed, suggestions welcome): (1-w)^(2-w) * (1+w)^(2+w) < 1, where 0 < w < 1 XCAS> series((1-w)^(2-w) * (1+w)^(2+w), w, 0, 10) 1 - w^4/3 - 4*w^6/15 - 10*w^8/63 - 4*w^10/45 + w^11*order_size(w) Taylor series suggested the expression peaked at w = 0, and plots support it. For now, assuming it is true ... (1-w)^(2-w) * (1+w)^(2+w) < 1 (1-w*w)^2 < ((1-w)/(1+w))^w (1/(1-w*w))^2 > ((1+w)/(1-w))^w Let c = 1/w, c > 1 (c*c/(c*c-1))^2 > ((1+1/c)/(1-1/c)) ^ (1/c) ln(c*c/(c*c-1)) * 2 > ln((1+1/c)/(1-1/c)) / c Simpilfy with identity ln(z) = 2*atanh((z-1)/(z+1)) (2*atanh(1/(2*c*c-1)) * 2 > (2*atanh(1/c)) / c atanh(1/(2*c*c-1) > atanh(1/c) / (2c) |
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