arcsinc( 1-y ), for small y

[Albert Chan]

Inspired by Happy Coincidence for the approximate solution of x tan(x) = k.

I tried the same trick for sin(x)/x = k, using Mathematica.

> <<Calculus`Pade`
> sinc[x_] := Sin[x] / x
> Pade[sinc[x], {x,0,4,4}] // Simplify
⇒ (166320 - 22260 x^2 + 551 x^4) / (166320 + 5460 x^2 + 75 x^4)

> Solve[166320 - 22260 x^2 + 551 x^4 == 0, x] // N
⇒ {{x → -3.14572}, {x → 3.14572}, {x → -5.52302}, {x → 5.52302}}

Pade approximant numerator had 2 roots very close to ±Pi, so solve this instead:

\(\Large { x^2-\pi^2 \over sinc(x) } = { x^2-\pi^2 \over k}\)

We don't want expression too complicated, and limit it to solving Quadratics:

> lhs = Pade[(x^2-Pi^2) / sinc[x], {x,0,4,2}];
> soln = Solve[ Numerator[lhs] * k - Denominator[lhs] * (x^2-Pi^2) == 0, x]

Although it look like a quartic, it is really a quadratic, with variables x^2.

Instead of solving asinc(k) for x, I choose solving asinc(1-y) for x.
This is a rule I learned from Acton's book, "Numerical Methods that works", p71

Quote:Store the small quantites; Compute the larger ones; Never subtract nearly equal quantities.

\(\large asinc(1-y) ≈ \Large \sqrt{\frac{12y}{1-0.20409y \;+ \sqrt{1 - 0.792y - 0.0318y^2}}} \)

The estimate formula is good for full y range.
y = 1.00 → worst over-estimated: abs error = 0.00023, rel error = 0.0075%
y = 0.85 → worst under-estimated: abs error = 0.00021, rel error = 0.0075%

Previous post example is close to worst case, y = 1-1/6.5 ≈ 0.846154
Using above formula, we get asinc(1/6.5) ≈ 2.71111, under-estimated 0.00020