Lecture 16

More on Newton-Puiseux algorithm

A lemma

• There exists epsilon>0 such that for every point (u,v) with the property ||(u,v)|| < epsilon there exist no two values w₁ and w₂ such that for j=1,2 we have: u=w_j^nu v=g(w_j)
• GCD({j: g_j # 0} = 1

Various algebraic examples

• Macsyma's taylor_solve fails on examples that require a recursive application of the Newton-Puiseux algorithm. The following example f(x,y)=(x+y)²+(x+y)³ results in [false,false]. See the output We clearly have vertices(NP(f))={ (0,2), (1,1), (2,0) } Thus, there is only one characteristic power: 1. The substitution x=u, y=xv leads to u²(1+v)²(2+v)=0 After dividing by u² we end up with the equation (v+1)²(1+u(1+v))=0 Plugging in u=0 produces (v+1)²=0 with a root v=-1 of multiplicity 2. The shifted equation is v²(1+uv)=0 We feed it back into the algorithm and immediately factor out v², which leads to the solution y=-x (multiplicity 2) of the original equation upon back-substitution. The remaining factored equation 1+uv=0 has a constant term (Newton polygon contains (0,0)) and thus it should be discarded. Thus, the final answer is that our initial equation has a solution y=-x of multiplicity 2.
• Let f(x,y)=(x³+3xy²+y⁵)² The question arises, what happens to the Puiseux series algorithm when the series f(x,y) is not square-free. Macsyma produces the following output, which shows the same solution as for the un-squared f(x,y), except various constants will have equations with multiple roots.

An implementation of the Newton-Puiseux algorithm

• This implementation is limited to polynomials.
• It will apply Newton-Puiseux algorithm recursively, so it is more powerful then the built-in taylor_solve.