Puiseux Series and the Newton Polygon algorithm
Part 4: Finite power series.
$$ \newcommand{\bint}{\displaystyle{\int\hspace{10.4pt}\Large\mathit{8}}} \newcommand{\res}{\displaystyle{\text{Res}}} $$The reader is asked to review the sections in sequential order as each succeeding section builds on a previous section.
In this part we design a method to compute $f(z,w)$ for a finite power series $w(z)=c_1+c_2 z^{1/d}+c_3 z^{2/d}+\cdots+c_k z^{k/d}$. We will need to determine a value of $n$ for the following expression: $$ \begin{aligned} f(z,w)&=(a_{1}+a_2 z+a_3 z^2+\cdots+a_n z^n)\\ &+(a_{10}+a_{11}z+a_{12} z^2+\cdots+a_{1n}z^n)w\\ &+ (a_{20}+a_{21}z+a_{22} z^2+\cdots+a_{2n}z^n)w^2\\ &\hspace{4pt}\vdots \\ &+(a_{n0}+a_{n1}z+a_{n2}z^2+\cdots+a_{nn} z^n)w^n=0 \end{aligned} $$ and then by substituting $w$ into $f(z,w)$, equate powers of $z$ to zero and then solve the resulting homogeneous system of equations $Ax=0$ for the constants $a_i$. We know such a system will have either the trivial solution or an infinite number of solutions as functions of a set of "free" variables, the number of free variables being equal to the difference between the number of variables and the rank of the coefficient matrix.
A homogeneous linear system of $M$ equations in $N$ unknowns with $M>N$, is overdetermined and will usually have only the trivial solution. However if the system of equations has enough linear dependencies so that the number of linearly independent equations is at most $N1$, then there will be an infinite number of solutions and it is this case we look for when determining $f(z,w)$ above: we simply set the free variables to say $1$ and then solve for the remaining ones. And we can determine the number of independent equations by computing the rank of the coefficient matrix. This is easily done in Mathematica using MatrixRank

First Example:
We wish to find a function $f(z,w)=0$ with $w(z)=1+z^{1/4}+z^{1/2}+z^{3/4}$. And we would like to determine a fourthdegree polynomial in $w$ with a minimum size of the polynomials $a_i(z)$. First consider $n=1$. That is, each $a_i(z)$ will have the form $a_i(z)=a+b z$. When we compute the resulting coefficient matrix we obtain a rank of $10$ with $10$ variables. This is not the minimum required $N1$ linearly independent equations. Next we choose $n=2$ or each polynomial $a_i(z)=a+bz+cz^2$. This gives us $15$ variables with a coefficient matrix of rank $15$. This is still an inadequate system for our purposes. Next we try $n=3$ with the form $$ \begin{aligned} f(z,w)&=(a_0+a_1 z+a_2 z^2+a_3 z^3)\\ &+ (a_{10}+a_{11} z+a_{12}z^2+a_{13} z^3) w \\ &+ (a_{20}+a_{21} z+a_{22}z^2+a_{23} z^3) w^2 \\ &+ (a_{30}+a_{31} z+a_{32} z^2+a_{33}z^3) w^3 \\ &+ (a_{40}+a_{41} z+a_{42} z^2+a_{43} z^3) w^4 \end{aligned} $$ When we compute $f(z,w)$ using $w$ above and equate (fractional) powers of $z$ to zero we obtain $20$ variables but this time the rank of the coefficient matrix is $19$ giving us the required minimum $N1$ linerarly independent equations. This means we can express $19$ of the variables in terms of one "free" variable which in this case is $a_0$. We simply let $a_0=1$ (or some other constant) and solve for the remaining constants. When $a_0=1$, we obtain for our answer $$ f(z,w)=(13 z+3 z^2z^3)+(4+8 z4 z^2)w+(66 z)w^2+(4)w^3+(1)w^4 $$
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