HP50g simplifing a root

[Albert Chan]

(10-09-2020 05:21 PM)Albert Chan Wrote:  \(\sqrt[3]{A ± \sqrt{R}} = a ± \sqrt{r} \quad ⇒ \quad a = \large\frac{\sqrt[3]{A+\sqrt{R}}
\;+\; \sqrt[3]{A-\sqrt{R}}}{2} \)

Slightly off topics. I was curious how to calculate a accurately (assumed R > 0)
Since cube root is odd function, we pull out the sign.

a = sign(A)/2 * (³√(|A|+√R) + ³√(|A|-√R))

Last term can be rewritten as ³√(A²-R) / ³√(|A|+√R)

Let c = ³√(A²-R), d = ³√(|A|+√R)², we have:

a = sign(A)/2 * (d + c) / √d

Since both c, d are cube roots, we try identity c³ + d³ = (c + d) (c² - c d + d²)

c³ + d³ = (A² - R) + (A² + R + 2|A|√R) = |2A| * (|A|+√R) = |2A| * d√d

Substitute back in, many terms get cancelled, a = A / (c²/d - c + d)

If ³√(A ± √R) can be simplified, c turns to integer (we had shown this earlier).
In fact, the whole denominator turns to integer = a² + 3r = 4a² - 3c

Code:

function calc_a(A,R)
    local c = cbrt(A*A-R)
    local d = cbrt(A*A+R + 2*abs(A)*sqrt(R))
    return A / (c*c/d - c + d)
end

lua> function test_a(A,R)
:           local q = sqrt(R)
:           return calc_a(A,R), (cbrt(A+q) + cbrt(A-q))/2
:      end
lua>
lua> test_a(26, 15^2 * 3)
2      1.9999999999999982
lua> test_a(9416, 4256^2 * 5)
11    11.000000000000005
lua> test_a(300940299,103940300^2 * 101)
99    99.00000000000006