Estimate logarithm quickly

[Albert Chan]

(08-21-2021 03:58 PM)Albert Chan Wrote:  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

My attempt to proof this, by showing there is no local extremum for 0 < w < 1
To get the extrema of above is messy. Instead, we solve extrema of ln(expression)

XCAS> f := (2-w)*ln(1-w) + (2+w)*ln(1+w)
XCAS> diff(f,w)

-ln(1-w) - (2-w)/(1-w) + ln(1+w) + (2+w)/(1+w)

XCAS> numer(ans()) // local extrema if df/dw = 0

-ln(w+1) + ln(-w+1) + w^2*ln(w+1) - w^2*ln(-w+1) + 2*w

Divide all by 2, then replace with identity atanh(x) = ln((x+1)/(1-x))/2

(w^2-1) * atanh(w) + w = (w^2-1) * (w + w^3/3 + w^5/5 + ...) + w

All coefficients of w's is positive, 1/(2k-1) - 1/(2k+1) = 2/(4k^2-1) > 0

XCAS> series((w^2-1)*atanh(w) + w, w,0,10)

2*w^3/3 + 2*w^5/15 + 2*w^7/35 + 2*w^9/63 + w^11*order_size(w)

Within 0 < w < 1, there is no local extremum, f is bounded by edges.

XCAS> limit(f, w=0, +1)      → 0
XCAS> limif(f, w=1, -1)       → -infinity

f = ln((1-w)^(2-w) * (1+w)^(2+w)) < 0       → (1-w)^(2-w) * (1+w)^(2+w) < 1