Ln(x) using repeated square root extraction

[Albert Chan]

(03-03-2022 11:20 PM)Thomas Klemm Wrote:  We can compare this to your solution:

\(\sqrt{2(1 + \varepsilon) - 1} - 1\)

The Taylor series agrees for the first two terms:

\(\varepsilon - \frac{\varepsilon^2}{2} + \frac{\varepsilon^3}{2} - \frac{5 \varepsilon^4}{8} + \frac{7 \varepsilon^5}{8} + \mathcal{O}(\varepsilon^6)\)

log1p(ε) = 2*atanh(y) = 2*(y+y^3/3+y^5/5+...) ≈ 2y, where y=ε/(2+ε)

2ε/(2+ε) is simpler than √(1+2ε)-1, and twice as accurate, when ε is tiny.

CAS> series(2*ε/(2+ε),ε) // estimate for log1p(ε)

ε - 1/2*ε^2 + 1/4*ε^3 - 1/8*ε^4 + 1/16*ε^5 + ε^6*order_size(ε)

CAS> (1/3-1/2)/(1/3-1/4)

-2.

Is is funny sqrt approximation formula give better estimate for log1p, than sqrt itself.

CAS> pade(√(1+2*x)-1,x,2,2)

2*x/(x+2)