In this section we consider the expression $\displaystyle w'(z)=\frac{f_z}{f_w}$ at a singular point $(z,w)$ in which $f_z=0$. In this case we have the indeterminant form $\frac{0}{0}$. Consider first
$$ \begin{equation}f(z,w)=(13 z+3 z^2z^3)+(4+8 z4 z^2)w+(66 z)w^2+(4)w^3+(1)w^4=0 \end{equation} $$ with $$ \begin{equation} \displaystyle \frac{dw}{dz}=\frac{\frac{df}{dz}}{\frac{df}{dw}}=\frac{6 w^2+8 w z8 w+3 z^26 z+3}{4 w^312 w^212 w z+12 w4 z^2+8 z4}. \end{equation} $$ Solving for the singular points we obtain $\{s\}=(0,1)$, $(1,0)$. Simply plugging these values into the expression for the derivative we obtain: $$ \begin{aligned} \frac{dw}{dz}(0,1)&=\frac{1}{0} \\ \frac{dw}{dz}(1,0)&=\frac{0}{0}. \end{aligned} $$ In the second case we have the indeterminant form $\frac{0}{0}$ and the limiting case may or may not exist. It is the purpose of this section to describe how to determine this limit.
First, we need to understand that the variable $w$ in the expressions is not an independent variable but rather dependent upon $z$. A more accurate expression for the derivative would then be $$ \displaystyle \frac{dw}{dz}=\frac{\frac{df}{dz}}{\frac{df}{dw}}=\frac{6 w(z)^2w(z) (88 z)+3 z^26 z+3}{4 w(z)^312 w(z)^2+2 w(z) (66 z)4 z^2+8 z4}, $$ and we could then take the limit: $$ \begin{equation} \displaystyle \frac{dw}{dz}=\mathop\lim_{z\to 1}\bigg\{\frac{6 w(z)^2+8 w(z) z8 w(z)+3 z^26 z+3}{4 w(z)^312 w(z)^212 w(z) z+12 w(z)4 z^2+8 z4}\bigg\}. \end{equation} $$ However, we would need the expression for $w(z)$ or at least a numerical approximation for $w(z)$ and evaluate the limit over an analyticallycontinuous path over each branch sheet to $z=1$. We can do both in this case.

Evaluating the indeterminant form numerically:
From previous sections, we know how to solve the azimuthal equation numerically. This gives us, for a starting initial value, $w(z)$ over a branch sheet. We can then plug this numerical approximation into the derivative expression for a path which spirals into the singular point and in this way obtain an approximation to the limit if it exists and if the function is wellbehaved. If it doesn't exist, the value will not tend to converge and we can check for this situation. For example, we let $\displaystyle z(t)=1+\frac{e^{it}}{t}$ as $t$ varies from $\pi$ to $1000$ for example. Solving for f(z(tstart),w)=0 gives us the four starting values. We can then solve for $w(z)$ over $z(t)$ with the following Mathematica code:
Mathematica code
theFunction = (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
myz[t_]:=1+(Cos[t]+I Sin[t])/t;
startVals=w/.Solve[theFunction==0/.z>myz[\[Pi]],w]//Flatten
wstart=startVals[[1]];
tstart=\[Pi];
tend=100;
wDeriv=w'[t]==(((D[theFunction,z]/D[theFunction,w]) D[myz[t],t])/.{w>w[t],z>myz[t]})
myazsol=First[NDSolve[{wDeriv,w[tstart]==wstart},w,{t,tstart,tend},MaxSteps>1500000,MaxStepSize>0.0001]]; myDerivTrace[t_]=Evaluate[Flatten[w[t]/.myazsol]];
theDeriv=(D[theFunction,z]/D[theFunction,w])/.{z>myz[t],w>myDerivTrace[t]};
Where in the code above we select the particular starting value with the command wstart=startVals[[k]] for $k=1,2,3,4$. For example, letting wstart=startVals[[1]] and then plotting the real and imaginary component of theDeriv gives us the following plot:
In the plot, we can see the value of the derivative spiraling into the point around $0.5+0.5i$. The value of $z$ after $t=200$ is $1.00244  0.00436649$. Doing this for each starting value, we obtain the approximate derivatives:: $$\begin{array}{c} 0.503+0.498i \\ 0.5000.505i \\ 0.5020.002i \\ 1.499+0.003i \end{array} $$
We would now like to compare these numericallydetermined values to the actual limiting cases.

Evaluating the indeterminant form algebraically from the origin:
The reader is asked to review Section 7: Puiseux series, and Section 9: Polynomial solutions for background on the following.
Equation $(1)$ was created to obtain a polynomial solution as per Section 9. The solution is $$ w(z)=1+z^{1/4}+z^{1/2}+z^{3/4} $$ or in conjugate form as per section $(7)$: $$\begin{array}{cc} w_1(z)&=1+z^{1/4}+z^{1/2}+z^{3/4} \\ w_2(z)&=1+iz^{1/4}z^{1/2}iz^{3/4} \\ w_3(z)&=1iz^{1/4}z^{1/2}+iz^{3/4} \\ w_4(z)&=1z^{1/4}+z^{1/2}z^{3/4} \end{array} $$ where the principal value of $z^{1/4}$ is used. And note these solutions are analytic at $z=1$. Taking derivatives: $$\begin{aligned} w_1'(z)&=1/4z^{3/4}+1/2 z^{1/2}+3/4z^{1/4} \\ w_2'(z)&=i/4z^{3/4}1/2z^{1/2}3i/4z^{1/4} \\ w_3'(z)&=i/4z^{3/4}1/2z^{1/2}+3i/4z^{1/4} \\ w_4'(z)&=1/4z^{3/4}+1/2z^{1/2}3/4z^{1/4} \end{aligned} $$ so that: $$ \begin{equation} \begin{aligned} \frac{dw_1}{dz}(1)&=1.5 \\ \frac{dw_2}{dz}(1)&=1/21/2 i \\ \frac{dw_3}{dz}(1)&=1/2+1/2 i \\ \frac{dw_4}{dz}(1)&=1/2 \end{aligned} \end{equation} $$ and we could further show that we have the correspondence between the analytic values and numeric values: $$ \begin{equation} \begin{array}{c} \text{Numeric} & \text{Analytic} \\ 0.503+0.498i & 1/2+1/2i \\ 0.5010.505i & 1/21/2i \\ 0.5020.002 i & 1/2 \\ 1.498+0.003i & 1.5 \end{array} \end{equation} $$
We can see now how the functional dependence of $w$ on $z$ affects the limit. In fact, we can substitute the explicit values of $w_i(z)$ into $(3)$ above and take the limit as $z\to 0$ and obtain the results in $(4)$. For example, we obtain for $w_1(z)$, the following limit: $$ \mathop{\lim}_{z\to 1}\frac{6 \left(z^{3/4}+\sqrt[4]{z}+\sqrt{z}+1\right)^2\left(z^{3/4}+\sqrt[4]{z}+\sqrt{z}+1\right) (88 z)+3 z^26 z+3}{4 \left(z^{3/4}+\sqrt[4]{z}+\sqrt{z}+1\right)^312 \left(z^{3/4}+\sqrt[4]{z}+\sqrt{z}+1\right)^2+2 (66 z) \left(z^{3/4}+\sqrt[4]{z}+\sqrt{z}+1\right)4 z^2+8 z4}=3/2 $$ and do likewise for the other branch functions $w_i(z)$ to obtain the remaining values in table $(5)$.

Evaluating the indeterminant form algebraically from offset Puiseux series:
Finally, readers familiar with the sections on Puiseux series know we can translate the center of power expansion to $z=1$. We then compute the Puiseux series for the translated function: $$ F(z,w)=w^44 w^3+w^2 (66 (z+1))+w \left(4 (z+1)^2+8 (z+1)4\right)(z+1)^3+3 (z+1)^23 (z+1)+1 $$ now centered at the origin and obtain the following (nonpolynomial) expansions: $$ \begin{aligned} w_1(z)&=0.5 z + 0.0625 z^2  0.03125 z^3 + 0.0205078 z^4  0.0151367 z^5+\cdots \\ w_2(z)&=(0.50.5 I) z+0.125 z^2(0.0625 0.015625 I) z^3+(0.0390625 0.015625 I) z^4+\cdots \\ w_3(z)&=(0.5+0.5 I) z+0.125 z^2(0.0625 +0.015625 I) z^3+(0.0390625 +0.015625 I) z^4+\cdots\\ w_4(z)&=4. +1.5 z0.3125 z^2+0.15625 z^30.0986328 z^4+0.0698242 z^5+\cdots \end{aligned} $$ and then taking the derivatives of these expressions now at $z=0$, obtain the limits by inspection.
One important consequence of these results is shown in the ramifications about each singular point: $$ \begin{equation} \begin{array}{c} \text{Singular Point} & \text{Ramification} \\ 0 & \{\{1,2,3,4\}\} \\ 1 & \{\{1\},\{2\},\{3\},\{4\}\}\\ \infty & \{\{1,2,3,4\}\} \end{array} \end{equation} $$ And note the function does not ramify about the singular point at $z=1$. This is expected since the derivatives exists for all branch sheets.
Now consider $$ \begin{equation} (2 z^2+7 z^3)+(2 z)w+(6 z)w^2+(35 z^2)w^3 \end{equation} $$ with singular point $(0,0)$ and $\frac{dw}{dz}(0,0)=\frac{0}{0}$. And this time we will go directly to the Puiseux expansion around the origin. We obtain: $$ \begin{aligned} w_1(z)&=1. z  1. z^2  10.5 z^3  115.25 z^4  1361.63 z^5+\cdots \\ w_2(z)&=0.816497 \sqrt{z} + 1.5 z + 0.459279 z^{3/2} + 0.5 z^2 + 0.97969 z^{5/2}+\cdots \\ w_3(z)&=0.816497 \sqrt{z} + 1.5 z  0.459279 z^{3/2} + 0.5 z^2  0.97969 z^{5/2}+\cdots \end{aligned} $$ And we see immediately the derivative exists for the $1$cycle branch but does not exists for the $2$cycle branch. Let's take a third example: $$ \begin{equation} f(z,w)=(z^2+4 z^3)+(4 z^23 z^3)w+(3 z6 z^2+7 z^3)w^2+(39 z)w^3 \end{equation} $$ with the Puiseux expansion about the origin $$ \begin{aligned} w_1(z)&=(0.3466810.600468 i)z^{2/3}+0.333333 z+(0.240375\, 0.416342 i) z^{4/3}+(0.24396\, +0.422552 i) z^{5/3}+1.66667 z^2+\cdots \\ w_2(z)&=0.693361 z^{2/3}0.48075 z^{4/3}0.487921 z^{5/3}+1.66667 z^2+0.333333 z+\cdots \\ w_3(z)&=(0.346681+0.600468 I) z^{2/3}+0.333333 z+(0.240375 +0.416342 I) z^{4/3}+(0.24396 0.422552 I) z^{5/3}+1.66667 z^2+\cdots \end{aligned} $$ And in this third example we see the derivative does not exist at the singular point $(0,0)$.
From these observations we can then make the following assertion: The indeterminant form $\frac{dw}{dz}=\frac{0}{0}$ has a limit if and only if the singular point is not a pole and the Puiseux expansion does not have a term $z^{k/n}$ with $k\lt n$. It is a simple matter to identify poles. Those are the solutions to $a_n(z)=0$. And in principle, through the Newton Polygon algorithm, we have an algebraic method to determine if the Puiseux expansion has a $z^{k/n}$ term with $k \lt n$.
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