Approximations for the Logarithm Function

[Albert Chan]

We should compare sum with trapezoids vs squares, with same function evaluations (not intervals)

To simplify, we let t=s*x+1, to get integration limit from 0 to 1.
RHS integral value is simply averaged height of its integrand.

\(\displaystyle \ln(1+x) = \int_1^{1+x} \frac{1}{t}\,dt= \int_0^1 \frac{x}{x\,s + 1}\,ds \)

For midpoint rule with 3 function evaluation, we trisect intervals.

lua> function f(s) return x/(x*s+1) end
lua> x = 0.2
lua> log1p(x)
0.18232155679395465

Trapezoidal Rule, and squares equivalent:

lua> t1 = (f(0) + f(1))/2
lua> t2 = (t1 + f(1/2))/2
lua> t1, t2
0.18333333333333335      0.18257575757575759
lua> s1 = f(1/2)
lua> s2 = (s1 + f(1/6) + f(5/6))/3
lua> s1, s2
0.18181818181818182      0.1822650467811758

Simpson 1/3 rule, and squares equivalent:

lua> t2 + (t2-t1)/(4-1) -- weight = {1,4,1} / 6
0.18232323232323233
lua> s2 + (s2-s1)/(9-1) -- weight = {3,2,3} / 8
0.18232090490155006

Simpson 3/8 rule, and squares equivalent:

lua> require'fun'()
lua> function acc(s,x,y) return s+x*y end
lua> function dot(x,y) return foldl(acc,0,zip(x,y)) end

lua> dot({f(0/3),f(1/3),f(2/3),f(3/3)}, {1,3,3,1}) / 8
0.18232230392156865
lua> dot({f(1/8),f(3/8),f(5/8),f(7/8)}, {13,11,11,13}) / 48
0.18232121889089584

For ∫(1/x), sum with squares avoided edge bias, converge slightly better.