Lecture 15

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Newton-Puiseux algorithm

It is possible to "solve" the equation: f(x,y)=0 where f(x,y) is analytic at (0,0) and f(0,0)=0
The solution takes the form of the Puiseux series
The Macsyma function taylor_solve will perform an algorithm (Newton-Puiseux) in order to solve this equation.
The following sample output illustrates the solution obtained for the equation x³+3xy²+y⁵
Every power series obtained in the solution has the form y=g(x^1/q) where g(u) is analytic at 0 and g(0)#0 (not equal to 0, Macsyma style!) , and q is a certain integer. Moreover, we may assume that GCD({i: f_i#0})=1, so that the multi-valued mapping x ----> g(x^1/q) where |x| < epsilon, and epsilon is sufficiently small, parameterizes a connected component of the set V(epsilon) = { (x,y)#(0,0): f(x,y)=0, |x| < epsilon ^1/q, |y| < delta } for some delta > 0.
A Puiseux series is simply a shorthand for the pair of parametric equations x=u^q y=g(u) which is supposed to parameterize the solutions of f(x,y)=0 near 0.

Newton-Puiseux algorithm and the resolution of singularities

A definition

The Newton polygon NP(f) of the power series f(x,y) in C[[x,y]] is the convex hull of the set of the points of the integer lattice: {(m,n): there is k<=m and l<=n such that f_k,l # 0}

An example

If f(x,y)=x³+3xy²+y⁵ then the Newton polygon NP(f) is

Every solid or transparent circle is in NP(f)
Every solid circle is an extreme point (vertex) of NP(f)
The point (2,1) is on the boundary of NP(f) but not a vertex of NP(f); pay attention to such points!
The slopes of the boundary segments of the Newton polygon (except for those lying on the axis) are related to the leading powers of the power series solutions to the equation f(x,y)=0; the slopes are -1 and -3; the powers are of the form -1/slope, i.e. 1 and 1/3.

A theorem

Let f(x,y) be a power series in C[[x,y]] such that f(0,y)# 0 (this property is called "regularity" in y)

there exist excatly N power series solutions of the equation f(x,y)=0 where N is the order of 0 as a zero of f(0,y), i.e. f(0,y)=cy^N+O(y^N+1), c#0.
Each power series is of the form y=g(x^1/q) where g(u) is a power series in u and q is an integer. The power series of g(u) depends on a finite number parameters k₀,k₁,... which satisfy algebraic equations of the form h_l(k_l)=0
In general, each power series has to be counted with a certain multiplicity to add up to N
The multiplicity is 1 if the power series is square free i.e. when factored in the ring C[[x,y]] into a product of powers of prime elements, it will have each prime factor in power 1. However, the algorithm in general will produce a correct multiplicity of each power series.
If f(x,y) is convergent in a neghborhood of (0,0) then each of the power series g(u) is convergent. Moreover, every sufficiently close to (0,0) solution of the equation (x,y) is obtained from exactly one power series for exactly one choice of constants k_l.

Proof

This is really a sketch of the proof, but it describes the algorithm for calculating power series solutions with enough detail to produce the solutions by hand. We are going to illustrate the steps with the equation x³+3xy²+y⁵=0.

First, we divide the series f(x,y) by the maximal monomial x^a y^b so that f(0,y) and f(x,0) are non-zero power series. This is equivalent to saying that there is an edge of NP(f) on each axis. The zero solution y=0+... is counted with multiplicity b. If after this step f(0,0)#0 then there are no solutions to f(x,y)=0 and we discard this case.
Let (m_j,n_j) be the vertices of the Newton polygon NP(f); let us assume that 0=m₀ < m₁ < ... < m_r n₀ > n₁ > ... > n_r=0
Let alpha_j be the negative slopes of the edges of the Newton polygon, i.e. alpha_j= -(m_j+1-m_j)/(n_j+1-n_j)= p_j/q_j
We consider the substitution, where p=p_j and q=q_j: x=u^q y=u^pv The equation f(x,y)=0 can be written as follows: sum_m,n f_m,nu^mq+np vⁿ = 0
Let us study the minimum of the linear functional (m,n) ---> mq + np on the Newton polygon: L=min{ mq+np: (m,n) in NP(f) }; we note that (p,q) is a vector normal to the edge of NP(f) containing the edge connecting the vertex P_j=(m_j,n_j) to the next vertex P_j+1=(m_j+1,n_j+1) thus the minimum L is assumed on both vertices P_j and P_j+1.
We re-write the equation f(x,y)=0 as follows sum_{{(m,n):mq+np=L}} f_m,n vⁿ = - sum_{{(m,n):mq+np>L}} f_m,n u^mq+np-L vⁿ This equation if of the form F(v)=G(u,v)=uG₁(u,v) i.e. the right-hand side is small when u is near 0.
If the polynomial F(v) is linear in v, say, F(v)=cv, c#0 then we solve the equation F(v)=G(u,v) using the method of undetermined coefficients. The Implicit Function Theorem guarantees that the solution v=g(u) is analytic, or it is a formal power series with no fractional powers. Let us tentatively assign multiplicity 1 to every such power series. Later on we will propagate the multiplicity up the computation tree.
If deg(F)>1 then let us factor the polynomial F(v): F(v)=c product_{j:1<=j<=t} (v-k_j)^r_j Subsequently, for every j we translate the variable v by k_j. Thus, for every j we study the equation F(v+k_j)=G(u,v+k_j) We note that this equation explicitly involves k_j which is a solution of the equation F(k_j)=0. This is how new constants algebraic over the initial coefficients appear in the final power series g. We call our algorithm recursively on the equation h_j(u,v)=0 where h_j(u,v)=F(v+k_j)-G(u,v+k_j)
Let us assume that every power series v=g(u^1/nu) obtained in the previous step as a solution of h_j(u,v)=0 has some tentative multiplicity s. This equation could be re-written as a pair u=w^nu v=g(w) This series will be back-substituted into the equations x=u^q, y=u^pv to obtain a solution to the equation f(x,y)=0. The resulting series y=x^p/qg(x^{1/(q nu)}) will be assigned multiplicity r_js. This could be re-written of course as a pair of equations: x=w^{nu q} y=w^{nu p} (k_j+g(w)) (Remember that v was shifted!) It is not difficult to see that if p and q are co-prime, this parameterization yields no point twice, provided that GCD({j: g_j #0 })=1.
A priori the above algorithm could never terminate. However, as long as we can split the equation into two with a lower degree of y, we may proceed. Moreover, the powers of y split between the resulting equations, and it drops along any branch of the tree of polynomials created by the algorithm. Thus the resulting process must terminate.

A remark on algebraic constants

The algorithm requires keeping track of a number of constants k₀, k₁, k₂,... which are algebraic over the initial coefficients of the equation f(x,y)=0. Also, in the recursive steps we need to be able to determine whether various polynomial expressions in this constants are actually equal to 0 in order to find the Newton polygon. The full generality requires that the information about the constant k_j be kept in the form of its minimal polynomial, i.e. an equation P_j(k_j)=0, where P is a monic polynomial with rational coefficients and has minimal possible degree. The theory of algebraic extensions lets us reduce every rational expression in constants k_j to a polynomial sum_J a_J k^J where J=(j₀,j₁,...,) is a multi-index and the degree of this polynomial in k_j is strictly less than the degree of P_j. This expression is zero iff all coefficients are zero. All algorithms involved are constructive and available in computer algebra systems. They are

division of polynomials in one variable (long division)
computation of GCD over rationals (Euclid algorithm)
factorization of polynomials over rationals
square-free part of a polynomial, which is the same as f(x)/ GCD(f(x),f'(x)) when the field of coefficients has characteristic 0.

A practical consideration when F(v) factors

In the above algorithm, when F(v) factors, each factor should be considered separately. Factorization over complex numbers cannot be constructively performed and thus in the above Newton-Puiseux algorithm we should find all irreducible factors of F(v) over the rational numbers and proceed separately with each factor. Thus, let P_j(v), be all rational irreducible factors of F(v). For each factor we introduce an algebraic "constant" k_j (in practice, it is just another variable) satisfying the equation P_j(k_j)=0. We branch the algorithm for each j. This modification to the algorithm yields the previous version if we assume that factorization over the complex numbers can be performed constructively, i.e. every P_j(v) is linear.

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