Estimate logarithm quickly
08-21-2021, 03:58 PM
Post: #2
 Albert Chan Senior Member Posts: 2,611 Joined: Jul 2018
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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