Estimate logarithm quickly

[Albert Chan]

(11-20-2021 02:02 PM)Albert Chan Wrote:  |ln(n)| ≥ |n-1| / ((1 + 4*√n + n)/6) // Simpson's Rule

A simpler proof of inequality is convert it to atanh(y)
Assume x>1, then y = (x-1)/(x+1) > 0

To avoid square mess, we apply Doerfler's formula with squared argument.
ln(x) = atanh(y = (x-1)/(x+1))*2       → atanh(y) = ln(x = (1+y)/(1-y))/2

XCAS> D2(x) := 3*(x*x-1)/(1 + 4*x + x*x)
XCas> factor(D2((1+y)/(1-y)) /2)       → -3*y/(-3+y^2)

This is just pade(atanh(y),y,4,2), which expands to:

y/(1-y^2/3) = y + y^3/3 + y^5/3² + y^7/3³ + ...

For y>0, atanh(y) = y + y^3/3 + y^5/5 + y^7/7 + ... is bigger.
For x>1, ln(x), which atanh(y) were derived from, is biggger than D2(x)

Because of symmetry, For 0<x<1, D2(x) = -D2(1/x), same as ln(x).
Thus the proof can be extended from x > 1, to x > 0

XCAS> D2(2), D2(1/2)       → (9/13 , -9/13)