Sharp EL-W506T vs. Sharp EL-W516T

[Albert Chan]

From the plot, it is hard to tell which converge area faster.
(assuming we do not know the simpson numbers ahead of time)

I tried a u-transformed t1 (blue curve), and compare against t2 (red curve).
u-transformed t1 is both flatter and wider then t2, yet simpsons converge slower.

To apply u-transformation to integral, keeping integral limit from 0 to 1, use this:

\(\int _0 ^ 1 f(x) dx = \int _0 ^1 6u(1-u) f(u^2(3-2u))du \)

My revised guess is bell-shaped curve accelerated convergence.

Think Trapezoids.
Note: Simpson's rule is trapezoids with corrections, S_2n = T_2n + (T_2n-T_n)/3

Bell-shaped curve have inflection point at µ ± σ
Within this region, trapezoids will under-estimate area
Outside this region, trapezoids will over-estimate area.
Summing all trapezoids will thus suppress overall errors.

If the guess is right, all we needed is half a bell curve.
Instead of shifting all the way to 1, say the shift is k

\(\large \int _a ^ b {c \over r^2} dr = \int _{\log(a-k)} ^{\log(b-k)} {c\;e^x \over (e^x + k)^2} dx\)

Solve for k so that integrand at x=log(a-k) is flat.

XCas> s := numer(diff(c*e^x/(e^x+k)^2))     → (-c)*exp(x)^2+c*k*exp(x)
XCas> x := log(a-k)
XCas> solve(s=0, k)                                       → [a/2, a]

Since x=log(a-a) = -∞, we have k=a/2

Here is the trapezoid and simpson numbers for k=a/2
It converge so fast that 64 intervals have both numbers converged !

Code:

n       trapezoids       simspons 
2     170601681322   113751147000
4      88198771695    60731135152
8      63864757960    55753420048
16     62564915758    62131635024
32     62563050657    62562428956
64     62563050656    62563050656

Due to error cancellations, trapezoids numbers actually converge slightly faster