Abstract
Given a bivariate polynomial f(x, y), let ☎i(y) be a power series root of f(x, y) = 0 with respect tox, i.e., ☎i(y) is a function ofy such thatf(☎i(y),,y) = 0. If ☎i(y) is analytic aty = 0, then we have its power series expansion (1) $$\phi (y) = \alpha _0 + \alpha _1 y + \alpha _2 y^2 + \cdots + \alpha _r y^r + \cdots .$$ Let ☎i(k)(y) denote ☎i(y) truncated aty k , i.e., (2) $$\phi ^{(k)} (y) = \alpha _0 + \alpha _1 y + \alpha _2 y^2 + \cdots + \alpha _k y^k .$$ Then, it is well known that, given initial value ☎i(0)(y) = α0 ∈ C, the symbolic Newton’s method with the formula (3) $$\phi ^{(2^m - 1)} (y) \leftarrow \phi ^{(2^{m - 1} - 1)} (y) - \frac{{f(\phi ^{(2^{m - 1} - 1)} (y),y)}}{{\frac{{\partial f}}{{\partial x}}(\phi ^{(2^{m - 1} - 1)} (y),y)}} (mod y^{2m} )$$ computes $$\phi ^{(2^m - 1)} (y) (1 \le m)$$ in (2) with quadratic convergence (the roots are computed in the order $$\phi ^{(0)} (y) \to \phi ^{(2^1 - 1)} (y) \to \phi ^{(2^2 - 1)} (y) \to \cdots \to \phi ^{(2^m - 1)} (y))$$ .
Submitted Version (Free)
Published Version
Talk to us
Join us for a 30 min session where you can share your feedback and ask us any queries you have