Section 7: Puiseux Series (examples)

$$ \newcommand{\bint}{\displaystyle{\int\hspace{-10.4pt}\Large\mathit{8}}} \newcommand{\res}{\displaystyle{\text{Res}}} \newcommand{\wvalx}{\underbrace{z^{\lambda_4}(c_4+w_5)}_{w_4}} \newcommand{wvalxx}{\underbrace{z^{\lambda_3}(c_3+\wvalx)}_{w_3}} \newcommand{wvalxxx}{\underbrace{z^{\lambda_2}\{c_2+\wvalxx\}}_{w_2}} \newcommand{wvalxxxx}{z^{\lambda_1}\big(c_1+\wvalxxx\big)} $$

Revision in progress . . .


Examples of power expansions about singular points using the Newton Polygon Method:


The reader is advised to read Section $6$ for background information pertaining to the method. In all cases studied in these pages, an expansion $P_i(z)$ about a point $z_c$ is first transformed to an expansion about the origin via the substitution $z\to z+z_c$. In the simple examples below, the power expansions are given in exact (rational) terms for easier comprehension. However, in some cases the results cannot be computed exactly restricting all subsequent computations to a finite precision. In these cases, the singular points are computed to very high precision, usually between $100$ to $1000$ digits and all remaining calculations limited to this precision which will in general decrease as the number of terms of the series is increased.

Consider the algebraic function $w(z)$ of degree $n$ given by the implicit expression \begin{equation} f(z,w)=a_0(z)+a_1(z)w+a_2(z)w^2+\cdots a_n(z)w^n=0 \end{equation} where $a_i(z)$ are polynomials with rational coefficients. We wish to compute the $n$ power expansions $\{P_i(z)\}$ of $w(z)$ centered at a singular point of $w(z)$.

We begin by first normalizing, if necessary, $f(z,w)$ and compute the normal exponent $E$ (see Section $6$). This consists of transforming $f(z,w)$ to eliminate any poles at the origin. We then represent a Puiseux expansion of $w(z)$ at $z_c$ as $$ P_i(z)=c_1z^{\lambda_1}+c_2 z^{\lambda_1+\lambda_2}+c_3 z^{\lambda_1+\lambda_2+\lambda_3}+\cdots $$ in which $\lambda_1\geq 0$ and $\lambda_i>0$ for $i>1$ and $c_i \in \mathbb{C}$. And we find the exponents $\lambda_i$ and coefficients $c_i$ by both constructing Newton polygons for the (normalized) function as well as by Newton Iteration. The points for the polygons are obtained as follows: We plot a set of points $\left(i,\textbf{ord}(a_i)\right)$ and then extract from this set of points, the lower Newton Leg described in Section $6$ with one additional requirement: The lower leg is the lower part of the convex hull with negative or zero slope for the first iteration and then only negative slopes for successive iterations. This process is iterated until simple segments are reached. Then a version of Netwon iteration due to Kung and Traub 2 is used to generate a desired number of terms for the series.

Example Functions

Click on the buttons inside the tabbed menu:

Function 1:

\begin{equation} f(z,w)= (z+z^2)+(1+z)w+w^2=0 \end{equation} This function has two singular points $\{1/3,-1\}$ and since $a_2(z)=1$, the normal exponent is zero. So that in order to compute the expansion about $1/3$, we first make the transformation $f(z,w)=f(z+1/3,w)=f_1(z,w)$ giving us: $$ f_1(z,w)=(4/9+5/3 z+z^2)+(4/3+z)w+w^2=0. $$ with the notation $f_i$ representing the $i$'th polygonal function used to generate the $i$'th Newton polygon. The singular points for $f_1(z,w)$ are $\{0,-4/3\}$ so that an expansion about the origin of $w(z)$ given by $f_1$ is an expansion of the original $w(z)$ at $z=1/3$. Next, determine the orders as per Section $6$. These are: $\text{ord}(a_0)=0$, $\text{ord}(a_1)=0$ and $\text{ord}(a_2)=0$. Therefore the first Newton Polygon consists of the points $(0,0)$, $(1,0)$ and $(2,0)$ and is shown in Figure $1$.

The lower Newton leg in this case is the degenerate case of a straight line with $\lambda_1=0$ and $\beta_1=0$. The characteristic equation is $K_1(x)=4/9+4/3 x+x^2$ with roots $\{-2/3,-2/3\}$ or $c_1=-2/3$. This means there are two series expansions at this singular point with first terms equal to $-2/3 z^{\lambda_1}=-2/3$. That is, we have initially $$ \begin{align*} P_1(z)&=-2/3+\cdots\\ P_2(z)&=-2/3+\cdots \end{align*} $$ and since the roots have multiplicity greater than one (i.e., are not simple), we generate $f_2(z,w)$: $$ \begin{align*} f_2(z,w)&=z^{-\beta_1}f_1\left(z,z^{\lambda_1}(-2/3+w)\right)\\ &=z^0 f_1(z,(-2/3+w))\\ &=(z+z^2)+zw+w^2 \end{align*} $$ and create a second Newton polygon. In the case of $f_2(z,w)$ we have $\text{ord}(a_0)=1$, $\text{ord}(a_1)=1$ and $\text{ord}(a_2)=0$ so the second Newton polygon is given by the points $(0,1)$, $(1,1)$ and $(2,0)$ producing the next Newton polygon shown in Figure $2$. The lower Newton leg of Figure $2$ is the red segment spanning the points $(0,1),(2,0)$. For this lower Newton leg, $\lambda_2=1/2$ and $\beta_2=1$. The characteristic equation $K_2(x)=1+x^2=0$ has simple roots $c_2=\{i,-i\}$ giving the next terms of the series as $iz^{\lambda_1+\lambda_2}=iz^{1/2}$, and $-iz^{\lambda_1+\lambda_2}=-iz^{1/2}$: $$ \begin{align*} P_1(z)&=-2/3+i z^{1/2}+\cdots\\ P_2(z)&=-2/3-iz^{1/2}+\cdots \end{align*} $$

Once simple roots are obtained for the characteristic equation, we can use Newton iteration to compute additional terms for $P_1(z)$ and $P_2(z)$. In order to use this method, we need a polynomial with integer powers, however, the procedure above will often produce an expression with fractional powers. But we can convert an expression $f_k(z,w)$ with fractional powers into one with integer powers using the substitutions given by $(3)$. $$ \begin{equation} \begin{aligned} \widehat{f}(z,w)&=\frac{1}{z^{\beta_k}} f_k(z,z^{\lambda_k} w) \\ \overline{f}(z,w)&=\widehat{f}(z^d,w) \end{aligned} \end{equation} $$ where $k$ is the index of the last polygonal function which did not produce a characteristic equation with multiple roots. Let $d$ be the lowest common denominator of the exponents $\{\lambda_j\}$ for $j=1,2,\cdots,k$. In this case, $k=2$ and $d=2$ with $\lambda_2=1/2$ and $\beta_2=1$. Therefore we have: $$ \begin{align*} \widehat{f}(z,w)&=\frac{1}{z^{\beta_k}} f_k(z,z^{\lambda_k} w) \\ &=z^{-1}\left[(z+z^2)+z(z^{1/2}w)+zw^2\right] \\ &=1+z+z^{1/2}w+w^2=0 \end{align*} $$ and $$ \begin{align*} \overline{f}(z,w)&=\widehat{f}(z^d,w)\\ &=1+z^2+zw+w^2 \end{align*} $$ In order to generate more terms of the series, we implement Kung and Traub's version of Newton Iteration on $\overline{f}$: $$ w_{j+1}=w_j-\text{mod}\left(\frac{\overline{f}(z,w_j)}{\overline{f}_w(z,w_j)},z^{2^{(j+1)}}\right);\quad w_0=\{i,-i\} $$ where the $\text{mod}$ function extracts all terms of the Taylor expanson of the quotient with power less than $2^{(j+1)}$. We review the process for $w_0=i$ (The case of $w_0=-i$ is done similarly). The first three Newton Iterations are: $$ \begin{align*} w_1(z)&=w_0-\text{mod}\left(\frac{\overline{f}(z,w_0)}{\overline{f}_w(z,w_0)},z^{2}\right)\\ &=i-\text{mod}\left(\frac{1}{2}z-\frac{i z^2}{4}+\frac{1}{8}z^3+\frac{i z^4}{16}-\frac{1}{32}z^5+\cdots,z^2\right)\\ &=i-\frac{1}{2}z \\ \\ w_2(z)&=w_1-\text{mod}\left(\frac{\overline{f}(z,w_1)}{\overline{f}_w(z,w_1)},z^{4}\right)\\ &=w_1-\text{mod}\left(-\frac{3i}{8}z^2,z^4\right)\\ &=i-\frac{1}{2} z+\frac{3i}{8} z^2 \\ \\ w_3(z)&=w_2-\text{mod}\left(\frac{\overline{f}(z,w_2)}{\overline{f}_w(z,w_2)},z^{8}\right)\\ &=i-\frac{1}{2} z+\frac{3i}{8} z^2 -\text{mod}\left(\frac{9 i}{128} z^4-\frac{27}{1024}iz^6+\frac{81 i}{8192}z^8+\cdots,z^8\right) \\ &=\left(i-\frac{1}{2} z+\frac{3i}{8} z^2 \right)-\left(\frac{9i}{128} z^4-\frac{27 i}{1024} z^6\right)\\ &=i-\frac{1}{2} z+\frac{3i}{8} z^2-\frac{9i}{128}z^4+\frac{27i}{1024} z^6 \end{align*} $$

We iterate the Newton iteration $p$ times to $w_p(z)$ and then compute the final form of this power expansion via $(4)$: \begin{equation} \begin{aligned} e_i&=\sum_{j=1}^i \lambda_j, \\ e_k&=\sum_{j=1}^k \lambda_j, \\ P(z)&=\frac{1}{z^{E}}\bigg\{\sum_{i=1}^{k-1} c_i z^{e_i}+z^{e_k} w_p(z^{1/d})\bigg\}, \end{aligned} \end{equation} where $E$ is the normal exponent (see Section 6 for a definition of $E$) of the function which is zero in this case, and $P_i(z)$ the series representation for a single branch sheet of $w(z)$. Recall we needed two polygon iterations to reach a simple segment so $k=2$. And $\displaystyle e_2=\lambda_1+\lambda_2=1/2$. Using the example above with three iterations, we have \begin{equation} \begin{aligned} P(z)& =-2/3+z^{1/2} w_p(z^{1/2})\\ &=-\frac{2}{3}+z^{1/2}\left(i-\frac{\sqrt{z}}{2}+\frac{3 i z}{8}-\frac{9 i z^2}{128}+\frac{27 i z^3}{1024}\right)\\ &=-\frac{2}{3}+i \sqrt{z}-\frac{z}{2}+\frac{3}{8} i z^{3/2}-\frac{9}{128} i z^{5/2}+\frac{27 i z^{7/2}}{1024}+\cdots \end{aligned} \end{equation}

Mathematica code

    
    fBar[z_,w_]:=1+z^2+z w+w^2;
    j=0;
    currentw=I;
    currentSize=1;
    maxSize=5;
    While[currentSize<=maxSize,
      numerator=fBar[z,currentw];
      deriv=D[fBar[z,w],w]/.w->currentw;
      seriesTerms=Normal[Series[numerator/deriv,{z,0,2^(j+1)-1}]];
      currentw-=seriesTerms;
      currentSize=Length@currentw;
      j++;
    ];
     
    

Function 2:

$$ f_1(z,w)=(z^4)+(2 z^2+z^4)w+(1+z^2+z^3)w^2+(z)w^3+(1/4-z/2)w^4+(-(1/2))w^5=0 $$

Figure 3 shows the first Newton polygon for this function. The first thing to note about the lower Newton leg is now there are two lower segments with intersecting points so has the potential for roots with multiplicities for their associated characteristic equations.

  1. Power series for $S_1$


  2. For the first segment $S_1$ we have $\lambda_1=2$, $\beta_1=4$. And $K_1(x)=1+2x+x^2$ with roots $\{-1,-1\}$. Therefore, the first terms of two series (the multiplicity of the root) are: $$ \begin{align*} P_1(z)=-z^2 \\ P_2(z)=-z^2 \\ \end{align*} $$ And since we have non-zero root of multiplicity greater than one, we iterate the recursive equation for this segment again letting $\beta_1=4$, $\lambda_1=2$ and $c_1=-1$ in the computation of $f_2$: $$ \begin{align*} f_2(z,w)&=z^{-\beta_1}f_1\left(z,z^{\lambda_2}(-1+w_2)\right)\\ &=z^{-4} f_1(z,z^2(-1+w_2))\\ &=\left(\frac{1}{4}z^4-\frac{1}{2}z^5+\frac{1}{2}z^6\right)\\ &+\left(-z^2+z^3-z^4+2z^5-\frac{5}{2}z^6\right)w\\ &+\left(1+z^2-2z^3+\frac{3}{2}z^4-3z^5+5z^6\right)w^2\\ &+\left(z^3-z^4+2z^5-5z^6\right)w^3\\ &+\left(\frac{1}{4}z^4-\frac{1}{2}z^5+\frac{5}{2}z^6\right)w^4\\ &+\left(-\frac{1}{2}z^6\right)w^5 \end{align*} $$ The next Newton Polygon for this iteration is in Figure 4. The lower Newton leg consists of a single segment. Had it consisted of multiple (sub) segments, we would have to iterate the Newton polygons for each sub-segment as part of the analysis for segment $S_1$.

    From Figure 4, we have $\beta_2=4$, $\lambda_2=2$ and the $K_2(x)=1/4-x+x^2$ with roots $(1/2,1/2)$. We can then add $1/2 z^{\lambda_1+\lambda_2}=1/2 z^4$ to the growing series: $$ \begin{align*} P_1(z)=-z^2+1/2 z^4 \\ P_2(z)=-z^2+1/2 z^4 \\ \end{align*} $$

    And since the roots are again multiple, we compute the third Newton Polygon shown in Figure 5 with segment data $\lambda_3=1$, $\beta_3=2$ with $K_3(x)=1/4+x+x^2$ having multiple roots $\{-1/2,-1/2\}$.: $$ \begin{align*}\\f_3(z,w)&=\left(\frac{1}{4}z^2+\frac{1}{2}z^3-\frac{7}{8}z^4-\frac{5}{8}z^5+\frac{9}{8}z^6+\frac{1}{4}z^7-\frac{39}{64}z^8-\frac{1}{32}z^9+\frac{5}{32}z^{10}-\frac{1}{64}z^{12}\right)\\&+\left(z-z^4-\frac{9}{4}z^5+\frac{17}{4}z^6+\frac{3}{2}z^7-\frac{29}{8}z^8-\frac{1}{4}z^9+\frac{5}{4}z^{10}-\frac{5}{32}z^{12}\right)w\\&+\left(1+z^2-2z^3+\frac{3}{2}z^4-\frac{3}{2}z^5+\frac{7}{2}z^6+3z^7-\frac{57}{8}z^8-\frac{3}{4}z^9+\frac{15}{4}z^{10}-\frac{5}{8}z^{12}\right)w^2\\&+\left(z^5-z^6+2z^7-\frac{9}{2}z^8-z^9+5z^{10}-\frac{5}{4}z^{12}\right)w^3\\&+\left(\frac{1}{4}z^8-\frac{1}{2}z^9+\frac{5}{2}z^{10}-\frac{5}{4}z^{12}\right)w^4\\&+\left(-\frac{1}{2}z^{12}\right)w^5\\\end{align*} $$

    We next add $-1/2 z^{\lambda_1+\lambda_2+\lambda_3}=-1/2 z^5$ to the series: $$ \begin{align*} P_1(z)=-z^2+1/2 z^4-1/2 z^5 \\ P_2(z)=-z^2+1/2 z^2-1/2 z^5 \\ \end{align*} $$ and iterate the Newton polygons a fourth time: $$ \begin{align*}\\f_4(z,w)&=\left(\frac{1}{2}z-\frac{5}{8}z^2-\frac{5}{8}z^3+\frac{21}{8}z^4-\frac{9}{4}z^5-\frac{39}{64}z^6+\frac{85}{32}z^7-\frac{7}{4}z^8-\frac{1}{4}z^9+\frac{17}{16}z^{10}-\frac{37}{64}z^{11}+\frac{5}{32}z^{13}-\frac{5}{64}z^{14}+\frac{1}{64}z^{15}\right)\\&+\left(-z^2+z^3-\frac{15}{4}z^4+\frac{23}{4}z^5-\frac{5}{4}z^6-\frac{59}{8}z^7+\frac{67}{8}z^8-\frac{11}{8}z^9-\frac{37}{8}z^{10}+\frac{123}{32}z^{11}-\frac{5}{8}z^{12}-\frac{15}{16}z^{13}+\frac{5}{8}z^{14}-\frac{5}{32}z^{15}\right)w\\&+\left(1+z^2-2z^3+\frac{3}{2}z^4-\frac{3}{2}z^5+2z^6+\frac{9}{2}z^7-\frac{81}{8}z^8+6z^9+\frac{45}{8}z^{10}-\frac{33}{4}z^{11}+\frac{25}{8}z^{12}+\frac{15}{8}z^{13}-\frac{15}{8}z^{14}+\frac{5}{8}z^{15}\right)w^2\\&+\left(z^6-z^7+2z^8-\frac{9}{2}z^9-\frac{3}{2}z^{10}+6z^{11}-5z^{12}-\frac{5}{4}z^{13}+\frac{5}{2}z^{14}-\frac{5}{4}z^{15}\right)w^3\\&+\left(\frac{1}{4}z^{10}-\frac{1}{2}z^{11}+\frac{5}{2}z^{12}-\frac{5}{4}z^{14}+\frac{5}{4}z^{15}\right)w^4\\&+\left(-\frac{1}{2}z^{15}\right)w^5\\\end{align*} $$ The fourth iteration is shown in Figure $6$ with segment data $\lambda_4=1/2$, $\beta_4=1$ and $K_4(x)=1/2+x^2$ giving roots of $\{\frac{i}{\sqrt{2}},-\frac{i}{\sqrt{2}}\}$.

    We have after three iterations of Newton polygons, reached a simple segment and can now use Newton iteration with seeds $w_0=\{\frac{i}{\sqrt{2}},-\frac{i}{\sqrt{2}}\}$. We first compute: $$ \small \begin{align*}\\ \widehat{f}(z,w)&=\left(\frac{1}{2}-\frac{5}{8}z-\frac{5}{8}z^2+\frac{21}{8}z^3-\frac{9}{4}z^4-\frac{39}{64}z^5+\frac{85}{32}z^6-\frac{7}{4}z^7-\frac{1}{4}z^8+\frac{17}{16}z^9-\frac{37}{64}z^{10}+\frac{5}{32}z^{12}-\frac{5}{64}z^{13}+\frac{1}{64}z^{14}\right)\\&+\left(-z^{3/2}+z^{5/2}-\frac{15 z^{7/2}}{4}+\frac{23 z^{9/2}}{4}-\frac{5 z^{11/2}}{4}-\frac{59 z^{13/2}}{8}+\frac{67 z^{15/2}}{8}-\frac{11 z^{17/2}}{8}-\frac{37 z^{19/2}}{8}+\frac{123 z^{21/2}}{32}-\frac{5 z^{23/2}}{8}-\frac{15 z^{25/2}}{16}+\frac{5 z^{27/2}}{8}-\frac{5 z^{29/2}}{32}\right)w\\&+\left(1+z^2-2z^3+\frac{3}{2}z^4-\frac{3}{2}z^5+2z^6+\frac{9}{2}z^7-\frac{81}{8}z^8+6z^9+\frac{45}{8}z^{10}-\frac{33}{4}z^{11}+\frac{25}{8}z^{12}+\frac{15}{8}z^{13}-\frac{15}{8}z^{14}+\frac{5}{8}z^{15}\right)w^2\\&+\left(z^{13/2}-z^{15/2}+2 z^{17/2}-\frac{9 z^{19/2}}{2}-\frac{3 z^{21/2}}{2}+6 z^{23/2}-5 z^{25/2}-\frac{5 z^{27/2}}{4}+\frac{5 z^{29/2}}{2}-\frac{5 z^{31/2}}{4}\right)w^3\\&+\left(\frac{1}{4}z^{11}-\frac{1}{2}z^{12}+\frac{5}{2}z^{13}-\frac{5}{4}z^{15}+\frac{5}{4}z^{16}\right)w^4\\&+\left(-\frac{z^{33/2}}{2}\right)w^5\\ \\ \overline{f}(z,w)&=\left(\frac{1}{2}-\frac{5}{8}z^2-\frac{5}{8}z^4+\frac{21}{8}z^6-\frac{9}{4}z^8-\frac{39}{64}z^{10}+\frac{85}{32}z^{12}-\frac{7}{4}z^{14}-\frac{1}{4}z^{16}+\frac{17}{16}z^{18}-\frac{37}{64}z^{20}+\frac{5}{32}z^{24}-\frac{5}{64}z^{26}+\frac{1}{64}z^{28}\right)\\&+\left(-z^3+z^5-\frac{15}{4}z^7+\frac{23}{4}z^9-\frac{5}{4}z^{11}-\frac{59}{8}z^{13}+\frac{67}{8}z^{15}-\frac{11}{8}z^{17}-\frac{37}{8}z^{19}+\frac{123}{32}z^{21}-\frac{5}{8}z^{23}-\frac{15}{16}z^{25}+\frac{5}{8}z^{27}-\frac{5}{32}z^{29}\right)w\\&+\left(1+z^4-2z^6+\frac{3}{2}z^8-\frac{3}{2}z^{10}+2z^{12}+\frac{9}{2}z^{14}-\frac{81}{8}z^{16}+6z^{18}+\frac{45}{8}z^{20}-\frac{33}{4}z^{22}+\frac{25}{8}z^{24}+\frac{15}{8}z^{26}-\frac{15}{8}z^{28}+\frac{5}{8}z^{30}\right)w^2\\&+\left(z^{13}-z^{15}+2z^{17}-\frac{9}{2}z^{19}-\frac{3}{2}z^{21}+6z^{23}-5z^{25}-\frac{5}{4}z^{27}+\frac{5}{2}z^{29}-\frac{5}{4}z^{31}\right)w^3\\&+\left(\frac{1}{4}z^{22}-\frac{1}{2}z^{24}+\frac{5}{2}z^{26}-\frac{5}{4}z^{30}+\frac{5}{4}z^{32}\right)w^4\\&+\left(-\frac{1}{2}z^{33}\right)w^5 \end{align*} $$ We now use Newton Iteration using the first root of the characteristic equation $K_4$: $$ w_{j+1}=w_j-\text{mod}\left(\frac{\overline{f}(z,w_j)}{\overline{f}_w(z,w_j)},z^{2^{(j+1)}}\right);\quad w_0=\frac{i}{\sqrt{2}} $$ $$ \begin{align*} w_1(z)&=w_0-\text{mod}\left(\frac{5 i}{8 \sqrt{2}} z^2-\frac{1}{2} z^3+\frac{9}{8 \sqrt{2}} z^4+\frac{13}{16} z^5,z^2\right)\\ &=\frac{i}{\sqrt{2}}-0=\frac{i}{\sqrt{2}}\\ \\ w_2(z)&=w_1-\text{mod}\left(\frac{\overline{f}(z,w_1)}{\overline{f}_w(z,w_1)},z^{4}\right)\\ &=w_1-\text{mod}\left(\frac{5 i}{8 \sqrt{2}} z^2-\frac{1}{2} z^3+\frac{9}{8 \sqrt{2}} z^4+\frac{13}{16} z^5,z^4\right)\\ &=\left(\frac{i}{\sqrt{2}}-\frac{5}{8 \sqrt{2}} z^2+\frac{1}{2} z^3\right)\\ \\ w_3(z)&=w_2-\text{mod}\left(\frac{\overline{f}(z,w_2)}{\overline{f}_w(z,w_2)},z^{8}\right)\\ &=\left(\frac{i}{\sqrt{2}}-\frac{5i}{8 \sqrt{2}} z^2+\frac{1}{2} z^3\right) -\text{mod}\left(-\frac{169 i z^4}{128 \sqrt{2}}+\frac{z^5}{2}+\frac{3251 i z^6}{1024 \sqrt{2}}-\frac{11 z^7}{8}-\frac{5249 i z^8}{8192\sqrt{2}}+\frac{183 z^9}{256}-\frac{504453 i z^{10}}{65536 \sqrt{2}},z^8\right)\\ \\ &=\frac{i}{\sqrt{2}}-\frac{5i}{8 \sqrt{2}} z^2+\frac{1}{2} z^3+\frac{169}{128 \sqrt{2}} z^4-\frac{1}{2}z^5+\frac{3251 i}{1024 \sqrt{2}} z^6+\frac{11}{8} z^7 \end{align*} $$ Iterating $p$ times we then compute: \begin{equation} \begin{aligned} e_i&=\sum_{j=1}^i \lambda_j, \\ e_k&=\sum_{j=1}^k \lambda_j, \\ w(z)&=\frac{1}{z^{E}}\bigg\{\sum_{i=1}^{k-1} c_i z^{e_i}+z^{e_k} w_p(z^{1/d})\bigg\}, \end{aligned} \end{equation} where $E$ is the normal exponent which is zero in this case. Recall we needed four polygon iterations to reach a simple segment so $k=4$. And $\displaystyle e_4=\lambda_1+\lambda_2+\lambda_3+\lambda_4=2+2+1+1/2=11/2$. Using the three iterates above we have $p=3$ and: $$ \begin{align*} P(z)&=\frac{1}{z^0}\bigg\{-z^2+1/2 z^4-1/2 z^5+z^{11/2}w_3(z^{1/2})\bigg\}\\ &=-z^2+\frac{z^4}{2}-\frac{z^5}{2}+\frac{i z^{11/2}}{\sqrt{2}}-\frac{5 i z^{13/2}}{8 \sqrt{2}}+\frac{z^7}{2}-\frac{169 i z^{15/2}}{128 \sqrt{2}}-\frac{z^8}{2}+\frac{3251 i z^{17/2}}{1024 \sqrt{2}}+\frac{11 z^9}{8}+\cdots \end{align*} $$

  3. Power series for $S_2$


  4. In the case of segment $S_2$, we have $\lambda_1=0$ and $\beta_1=0$ with $K(x)=x^2+1/4 x^4-1/2 x^5=0$ or $K(x)=1+1/4 x^2-1/2 x^3=0$ and although the roots can be determined exactly, we use the numeric values $\{-0.47527 - 1.07374 I, -0.47527 + 1.07374 I, 1.45054\}$ although the actual values are computed to say $100$ digits of precision. Since the roots are simple with $\lambda_1=0$, three series begin with these constants: $$ \begin{align*} P_3(z)&=-0.47527-1.07374i+\cdots\\ P_4(z)&=-0.47527+1.07374i+\cdots \\ P_5(z)&=1.45054+\cdots \end{align*} $$ And at this point we can immediately go into Newton iteration: $$ \begin{align*} \widehat{f}(z,w)&=\frac{1}{z^{\beta_k}} f_k(z,z^{\lambda_k}w)\\ \overline{f}(z,w)&=\tilde{f}(z^d,w) \end{align*} $$ lettin $k$=1: $$ \begin{align*} \widehat{f}(z,w)&=f_1(z,w) \\ \overline{f}(z,w)&=\widehat{f}(z,w)=f_1(z,w) \end{align*} $$ And the Newton iteration phase for segment $S_2$ becomes: $$ w_{i+1}=w_i-\text{mod}\left(\frac{\overline{f}(z,w_i)}{\overline{f}_w(z,w_i)},z^{2^{(i+1)}}\right);\quad w_0=\{-0.47527 - 1.07374 I, -0.47527 + 1.07374 I, 1.45054\} $$ The first terms of these series are: $$ \begin{align*} P_3(z)&-(1.0737 i+0.4753)-(0.5820-0.3302 i) z+(0.5365-0.2690 i) z^2+(0.2017 i+0.0359) z^3+(-0.3600 i-0.1986) z^4+\cdots \\ P_4(z)&=-(0.4753-1.0737 i)-(0.3302 i+0.5820) z+(0.2690 i+0.5365) z^2+(0.0359-0.2017 i) z^3+(0.3600 i-0.1986) z^4+\cdots \\ P_5(z)&=1.451+0.1639 z+0.9269 z^2-0.07173 z^3-0.6027 z^4+\cdots \end{align*} $$

Third Function: Polynomials in $w$

Now consider a polynomial strictly in terms of $w$: $$ \begin{align*} f(z,w)&=(w-1)(w-2)(w-3)^3\\ &=(-54)+(135)w+(-126)w^2+(56)w^3+(-12)w^4+(1)w^5 \end{align*} $$ The orders are $\{0,0\}$,$\{1,0\}$,$\{2,0\}$,$\{3,0\}$,$\{4,0\}$,$\{5,0\}$ which results in a degenerate polygon with the characteristic equation being identical to the polynomial, $-54.+135. x-126. x^2+56. x^3-12. x^4+x^5$ with the roots being $\{1,2,3\}$. When we substitute these roots into the first recursive equation, we obtain a second degenerate polygon with zero slope. For example, when we use the first root, we obtain $f_2(z,w)=(8)w+(-20)w^2+(18)w^3+(-7)w^4+(1)w^5$. Recall from the discussion, that we choose lower Newton legs with negative or zero slope only for the first polygon and then those only with negative slopes for successive iterations. In these cases, a finite series has been reached and in this particular example, results in solutions $P_1(z)=1$, $P_2(z)=2$ and $P_3(z)=3$.

Fourth Function: Fractional Polynomials

We now consider a function in which the Newton iteration of $w(z)$ terminates because the quotient $ \displaystyle\frac{\overline{f}(z,w_i)}{\overline{f}_w(z,w_i)}$ becomes zero. This is a case when $w(z)$ is a polynomial. Consider: $$ f_1(z,w)=(1 - 3 z + 3 z^2 - z^3) + (-4 + 8 z - 4 z^2) w + (6 - 6 z) w^2 + (-4) w^3 + (1) w^4 $$ with singular points $\{0,1\}$. We begin the expansion about the origin by computing the orders: $\{0,0\}$,$\{1,0\}$,$\{2,0\}$,$\{3,0\}$, $\{4,0\}$ which results in a degenerate Newton polygon with $\lambda_1=0$, $\beta_1=0$ and characteristic equation $K(x)=1-4 x+6x^2-4 x^3+x^4$ with roots $\{1,1,1,1\}$ and therefore the four series expansions about the origin take the form: $$\begin{align*} P_1(z)&=1+\cdots\\ P_2(z)&=1+\cdots \\ P_3(z)&=1+\cdots \\ P_4(z)&=1+\cdots \end{align*} $$ We next iterate the Newton Polygon again via $f_2$: $$ \begin{align} f_2(z,w)&=z^{-\beta_1}f_1\left(z,z^{\lambda_1}(1+w)\right)=0\\ &=z^0 f_1(z,(1+w))=0\\ &=\left(-z-z^2-z^3\right)+\left(-4z-4 z^2\right)w+\left(-6z\right)w^2+\left(1\right) w^4 \end{align} $$ with the associated Newton Polygon shown in Figure $7$.

From Figure $7$ we have $\lambda_2=1/4$, $\beta_2=1$ and $K_2(x)=-1+x^4$ with simple roots $\{-1,1,i,-i\}$. And with the appearance of simple roots, begin the Newton iteration phase of expansion by computing the iterated polynomial $\overline{F}$. In this case $k=2$, $d=4$, $\lambda_2=1/4$ and $\beta_2=1$. $$ \begin{equation} \begin{aligned} \widehat{f}(z,w)&=\frac{1}{z^{\beta_2}} f_k(z,z^{\lambda_2} w)=0 \\ &=z^{-1}f_2(z,z^{1/4}w)\\ &=\left(-1-1 z-1 z^2\right)+\left(-4 z^{1/4}-4 z^{5/4}\right) w+\left(-6 \sqrt{z}\right) w^2+(1) w^4 \end{aligned} \end{equation} $$ and $$ \begin{equation} \begin{aligned} \overline{f}(z,w)&=\widehat{f}(z^4,w)=0\\ &=\left(-1-1 z^4-z^8\right)+\left(-4 z-4 z^5\right) w+\left(-6 z^2\right) w^2+(1) w^4 \end{aligned} \end{equation} $$ And then iterate: $$ w_{i+1}=w_i-\text{mod}\left(\frac{\overline{f}(z,w_i)}{\overline{f}_w(z,w_i)},z^{2^{i+2}}\right);\quad w_0=\{1,-1,i,-i\} $$ $$ \begin{align*} w_1(z)&=w_0-\text{mod}\left(-z-\frac{5}{2} z^2-\frac{11}{2} z^3-\frac{53}{4} z^4-\frac{123}{4} z^5,z^2\right))\\ &=1+z\\ \\ w_2(z)&=w_1-\text{mod}\left(-z^2-\frac{3}{2} z^4-1 z^6-2 z^7+\frac{15}{4} z^8+\cdots,z^4\right)\\ &=1+z+z^2 \\ \\ w_3(z)&=w_2-\text{mod}\left(\frac{\overline{f_2}(z,w_2)}{\overline{f_2}_w(z,w_2)},z^{8}\right)\\ &=w_2-\text{mod}\left(0,z^8\right)\\ &=w_2+0 \end{align*} $$ and note $\displaystyle \frac{\overline{f}(z,w_2)}{\overline{f}_w(z,w_2)}=0$ in this case which means the iteration has terminated and we are left with $$ w_2(z)=1+z+z^2 $$ We then have: \begin{equation} \begin{aligned} e_i&=\sum_{j=1}^i \lambda_j, \\ e_k&=\sum_{j=1}^k \lambda_j, \\ w(z)&=\frac{1}{z^{E}}\bigg\{\sum_{j=1}^{k-1} c_i z^{e_i}+z^{e_k} w_p(z^{1/d})\bigg\}, \end{aligned} \end{equation} where $E$ is the normal exponent which is zero in this case. Recall we needed two polygon iterations to reach a simple segment so $k=2$. And $\displaystyle e_2=\lambda_1+\lambda_2=1/4$. Using the two iterates above we have $p=2$ and: $$ \begin{align*} P_1(z)&=\frac{1}{z^0}\bigg\{1+z^{1/4}w_2(z^{1/4})\bigg\}\\ &=1+z^{1/4}+z^{1/2}+z^{3/4} \end{align*} $$ The other three are done similarly or $P_1(z)$ can be conjugated to obtain these.


References:
1 Algebraic Curves, Walker, R.J.
2 All algebraic functions can be computed fast, H.T. Kung and J.F. Traub

No comments:

Post a Comment

Blog Archive