Graeffe's root squaring method

[Albert Chan]

(11-23-2022 07:10 PM)Albert Chan Wrote:  P(x) = 2 + 3x + 5x^2 + 7x^3 + 11x^4 + ...

All P roots inside complex unit circle.

We can assume min root real part closer to -1 than +1, to keep imag part small.
Root squaring process gives min abs root ≈ 0.8, so I would first try x = 0.8*cis(3/4*pi ≈ 2.36)

Taylor series:

P(x+y*i) = P(x) + P'(x)*(y*i) + P''(x)*(y*i)^2/2! + P'''(x)*(y*i)^3/3! + ...

P with positive coefficients, if x>0, we have |P(x)| > |P(-x)|

My intution is that if x>0, it take bigger imaginery part "correction" to get P(x+y*i) = 0
This implied min abs root is in Q2 (or Q3 for conjugate root)

This is an educated guess.