(12C) Square Root

[Albert Chan]

It might be better not asking the user to supply a possibly bad guess.

Instead, user is limited to enter value between 0.1 to 10
For example, N=12345, user entered n=1.2345, √N = 100 √n

A naive guess = 0.5*(1+n) might create big relative errors.
Worst case at the edge, n=0.1 or 10

max_rel_err = |1 - 0.5*11/√10| ≈ 0.7393 ≈ 74%

Each iteration reduce max_rel_err to around ½(prev_rel_err)², we got:
Ref: http://www.azillionmonkeys.com/qed/sqroot.html

0.7393 → 0.2732 → 0.03733 → 0.0006968 → 2.428e-7 → 2.947e-14

Thus, a maximum of 5 iterations to reach 10 digits accuracy.

A better guess = k * (1+n), such that rel_err(n=1) = - rel_err(n=10).
In other words, |rel_error| for n = 0.1 to 10 have 3 peaks, with W shape.

XCas> simplify(solve(1-k*2/1 = k*11/sqrt(10) - 1, k))     → k = (22√10 - 40)/81 ≈ 0.365

guess = 0.365 * (1 + n) reduced max_rel_err to 27%, thus required at most 4 iterations.

Ref: Numerical Analysis by Ridgway Scott, section 1.3.2, the best start