Section 0: Preliminaries

$$ \newcommand{\bint}{\displaystyle{\int\hspace{-10.4pt}\Large\mathit{8}}} \newcommand{\res}{\displaystyle{\text{Res}}} $$

We are interested in studying the function $w(z)$ given implicitly by $$ \begin{equation} f(z,w)=a_0(z)+a_1(z)w+a_2(z)w^2+\cdots+a_n(z)w^n \end{equation} $$ with $z,w\in \mathbb{C}$ and $a_i(z)$ polynomials with rational coefficients. $w(z)$ is $n$-valued except at singular points. In this discussion, we use the following definition of a singular point: We know for a function $w(z)$ given implicitly by $(1)$, we have $$w'(z)=-\frac{f_z}{f_w}$$ and so we find the points $(z,w)$ where $f(z,w)=f_w(z,w)=0$. We can find these points by first determining where the resultant $R(f,f_w)=0$. The resultant is a polynomial in $z$ the zeros of which are the $z$-values of the singular points. Finding $z$, we then solve for $w$ by plugging into $f(z,w)=0$. This is easily done in Mathematica with the code:

thezvals = z /. NSolve[Resultant[theFunction, D[theFunction, w], w] == 0, z]
thewvals = NSolve[{theFunction == 0, D[theFunction, w] == 0} /. z -> #] & /@ thezvals
Consider for example $$ \begin{equation} f(z,w)=-8z^2+6zw+(9z-1)w^2-9zw^3 \end{equation} $$ which yields the following singular points: \begin{array}{cc} \text{Singular number} & \text{(z,w)} \\ s_1 & \{-0.0134098,-0.0356913\} \\ s_2 & \{0.,0.\} \\ s_3 & \{0.0328578\, -0.0488795 i,-0.0577313-0.748737 i\} \\ s_4 & \{0.0328578\, +0.0488795 i,-0.0577313+0.748737 i\} \\ s_5 & \{0.614361,0.817821\} \\ \end{array}

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