The Zero Set of an Electrical Field

from a Finite Number of Point Charges:

One, Two, and Three Dimensions

Tamás Erdélyi, Joseph Rosenblatt, Rebecca Rosenblatt

###### Abstract

We study the structure of the zero set of a finite point charge electrical
field $F=(X,Y,Z)$ in $\mathbb{R}^{3}$. Indeed, mostly we focus on a finite
point charge electrical field $F=(X,Y)$ in $\mathbb{R}^{2}$. The well-known
conjecture is that the zero set of $F=(X,Y)$ is finite. We show that this is
true in a Special Case: when the point charges for $F=(X,Y)$ lie on a line. In
addition, we give fairly complete structural information about the zero sets
of $X$ and $Y$ for $F=(X,Y)$ in the Special Case. A highlight of the paper
states that in the Special Case the zero set of $F=(X,Y)$ contains at most
$9M^{2}4^{M}$ points, where $M$ is the number of point charges.

##  1 Introduction

This article gives the results that we know to hold for the zero sets of
finite point charge electrical fields $F=(X,Y,Z)$ in $\mathbb{R}^{3}$. While
some general facts in this setting are clear, in order to get complete proofs
we have to focus for now on the case where the charges are in a plane. That
is, $F=(X,Y)$ is a finite point charge field in $\mathbb{R}^{2}$. The well-
known, still unresolved, conjecture is that in this case the zero set of $F$
is finite. We give full details proving this conjecture in the Special Case
where the charges themselves lie on a line in $\mathbb{R}^{2}$.

While the Special Case may seem to be an easy case to analyze, it turns out
that the details are complicated, especially if one wants to know the
structure of not only the zero set of $F=(X,Y)$, but also the zero sets of $X$
and $Y$. We give different ways of seeing that the zero set of $F=(X,Y)$ is
finite in the Special Case when the point charges lie on a line. In addition,
we obtain structural information about the zero sets of $X$ and $Y$ at
infinity by analyzing power series expansions, using binomial series,
geometric series, and also critical moments that are given by the coefficients
of the field and the positions of the charges. Indeed, in order to complete
the analysis in the case where the point charges lie on a line, we need to
consider the structure of the zero set of $F=(X,Y)$ with the point charges in
general position in $\mathbb{R}^{2}$. Details will be discussed for what we
know and what yet remains to be shown.

The outline of this article is as follows. In Section 2, we set up the problem
and explain how it relates to a larger class of problems of distinguishing
and/or reconstructing electrical fields and also gravitational fields. In
Section 3, we show that the zero set of $F=(X,Y)$ cannot contain a non-trivial
curve when the point charges lie on a line, the Special Case. We show that it
follows that the zero set of $F=(X,Y)$ must be finite when the point charges
lie on a line. In Section 4, in our Special Case, we derive facts about the
asymptotic directions at infinity of the zero sets of $X$ and $Y$ and explain
why these directions do not overlap.

##  2 Finite Gravitational and Electrical Fields

The original problem we considered that gave rise to this work is how to
determine a gravitational field from a limited set of measurements. In
general, we would want to be able to have a calculus for this when the mass
generating this field is in motion. But for now we only consider the case
where we have a finite number of masses in fixed positions, and idealize these
to being point masses. So this means we have a finite set of points
$(x_{j},y_{j},z_{j}),j=1,2,\dots,M$, and corresponding masses
$m_{j},j=1,2,\dots,M$. We take a point $(x,y,z)$ and assume that there is a
unit mass at that point. Then the force at $(x,y,z)$ from the mass at
$(x_{j},y_{j},z_{j})$ is proportional to the mass $m_{j}$ and inversely
proportionally to the square of the Euclidean distance

$r_{j}=\|(x_{j},y_{j},z_{j})-(x,y,z)\|_{2}=((x_{j}-x)^{2}+(y_{j}-y)^{2}+(z_{j}-z)^{2})^{1/2}\,.$

There is a direction of this force too of course: from $(x,y,z)$ toward the
mass at $(x_{j},y_{j},z_{j})$. We take units so that the gravitational
constant $G$ is $1$. Hence the force field

$F(x,y,z)=(X(z,y,z),Y(x,y,z),Z(x,y,z))=\sum_{j=1}^{M}{\frac{m_{j}}{r_{j}^{2}}\,\frac{(x_{j},y_{j},z_{j})-(x,y,z)}{r_{j}}}\,.$

So we have these equations for the components of the force field:

$X(x,y,z)=\sum_{j=1}^{M}{\frac{m_{j}(x_{j}-x)}{r_{j}^{3}}}\,,\quad
Y(x,y,z)=\sum_{j=1}^{M}{\frac{m_{j}(y_{j}-y)}{r_{j}^{3}}\,,\quad
Z(x,y,z)=\sum_{j=1}^{M}\frac{m_{j}(z_{j}-z)}{r_{j}^{3}}}\,.$

There are actually two problems to deal with here. The first is to distinguish
two gravitational fields by a limited set of measurements, and the second is
to determine the full structure of an unknown gravitational field, by taking a
limited set of measurements. The first problem is the one we focus on in these
notes. The second one is in a class of problems of this type, many of which
seem to be very difficult to solve.

The problem of distinguishing gravitational fields can be phrased in terms of
finite (point) electrical fields (i.e. electrical fields given by a finite
number of point charges). Indeed, suppose we have two finite point mass
gravitational fields $F_{1}$ and $F_{2}$ and have a given set $\mathcal{V}$
such that $F_{1}(x,y,z)=F_{2}(x,y,z)$ for all $(x,y,z)\in\mathcal{V}$. That
is, $F_{1}(x,y,z)-F_{2}(x,y,z)=0$ for all $(x,y,z)\in\mathcal{V}$. But
$F_{1}-F_{2}$ has the form of the vector field given by a finite set of point
electrical charges in $\mathbb{R}^{3}$. Here the units have been chosen so
that the electrical constant $K$ is $1$. What we want to know is this:

Question: Taking an unknown finite electrical field, what method of taking a
limited set of measurements will guarantee that if the measurements are all
zero, then the field itself is zero everywhere?

To answer this question completely, we would want to known the geometry of the
zero set of a field given just as $F$ is above except that the coefficients
$m_{j}$ are allowed to be real numbers $a_{j}$.

We say that the electrical field is empty when there are actually no point
charges. Since a finite electrical field can always be seen as the difference
of two gravitational fields, this is the same as saying that the gravitational
fields are identical.

The structure of the zero set of such a field in $\mathbb{R}^{3}$ is certainly
different than in $\mathbb{R}^{2}$. We make some remarks about the three
dimensional case in the following, but this article actually focuses on the
two dimensional case, indeed for complete results on a Special Case of the two
dimensional case.

###  2.1 Finite Electrical Fields in Three Dimensions

We take our field $F=(X,Y,Z)$ to be a finite point charge electrical field. So
now, consistent with positive charges repulsing each other, rather than in the
case of gravitational fields where positive masses attract each other, we have

$X(x,y,z)=\sum_{j=1}^{M}{\frac{a_{j}(x-x_{j})}{r_{j}^{3}}}\,,\quad
Y(x,y,z)=\sum_{j=1}^{M}{\frac{a_{j}(y-y_{j})}{r_{j}^{3}}\,,\quad
Z(x,y,z)=\sum_{j=1}^{M}\frac{a_{j}(z-z_{j})}{r_{j}^{3}}}\,,$

with

$r_{j}=((x-x_{j})^{1/2}+(y-y_{j})^{2}+(z-z_{j})^{2})^{1/2}\,,\qquad
j=1,2,\dots,M\,.$

We know that in $\mathbb{R}^{3}$, $X,Y,$ and $Z$ can have zero sets consisting
of points, lines, curves, and even two dimensional surfaces. How these
intersect is not clear. The main question for the three dimensional case was
whether or not the common zeros must be only points or curves (perhaps even
just lines) of vectors? Also, how many points, and how many curves and lines,
will there be at most, say given only $M$? The basic structure is limited in a
more general setting by A. I. Yanushauskas [10]. Yanushauskas used the fact
that the force field in this case is the gradient of a harmonic function to
show that the zero set consists of a locally finite set of points and analytic
curves.

######  Proposition 2.1.

If a finite electrical field in $\mathbb{R}^{3}$ vanishes on a set containing
a two dimensional surface (or even as much as an open disc), then the field is
identically zero everywhere and the point charge set is empty.

But we still have structural issues to consider in order to understand the
geometry of such a zero set.

Conjecture: The zero set of $X$, $Y$, or $Z$ is asymptotically planar or
linear at infinity, with only a finite number of planes or lines being
possible. Moreover, these planes or lines are distinct among the three
components and the overlap of these three zero sets can only consist of a
finite set of points and asymptotically linear curves.

Of course, we could have the component zero sets being asymptotically a plane
or line in one direction and another plane or line in another direction. If
this conjecture is true then the common zero set would either be bounded or
consist of a finite number of curves each of which is asymptotically linear at
infinity.

######  Remark 2.2.

Examples show that the zero set in this case can contain curves (e.g. circles
and straight lines). But counting these and understanding what types of curves
one can get is a future project. The main question otherwise is whether or not
the unbounded portion is asymptotically linear. ∎

Without any additional work, we can address the Conjecture and the structural
issues by using the basic theory of real analytic varieties. See Whitney [6],
Whitney [7], and Bruhat and Whitney [8]. But this will only give us local
information about the zero set. Instead, we will use a product argument, and
the structure of real algebraic varieties, to get more information. See
Whitney [6] for an early advanced view, and Gibson [5] for details at a very
elementary level.

An algebraic curve in the Euclidean plane is the set of the points whose
coordinates are the solutions of a bivariate polynomial equation $P(x,y)=0$.
This equation is often called the implicit equation of the curve, in contrast
to the curves that are the graph of a function defining explicitly $y$ as a
function of $x$, or vice versa. With a curve given by such an implicit
equation, the first problem would be to determine the shape of the curve and
to “draw it”. These problems are not as easy to solve as in the case of the
graph of a function, for which $y$ may easily be computed for various values
of $x$. The fact that the defining equation is a polynomial implies that the
curve has some structural properties that may help in solving these problems.
Every algebraic curve may be uniquely decomposed into a finite number of
smooth monotone arcs (called branches) sometimes connected by some points
sometimes called remarkable points, and possibly a finite number of isolated
points called acnodes. A smooth monotone arc is the graph of a smooth function
which is defined and monotone on an open interval of the $x$-axis or the
$y$-axis. In each direction, an arc is either unbounded (usually called an
infinite arc) or has an endpoint which is either a singular point (this will
be defined below) or a point with a tangent parallel to one of the coordinate
axes. The singular points of a curve of degree $d$ defined by a polynomial
$P(x,y)$ of degree $d$ are the solutions of the system of equations

$\frac{\partial{P}}{\partial{x}}(x,y)=\frac{\partial{P}}{\partial{y}}(x,y)=P(x,y)=0\,.$

First, we need a basic algebraic principle. We take here $n=2,3$ for
concreteness. We are also thinking of applying this to the components of the
vector field. So when $n=2$, our functions below are
$A_{j}(x,y)=a_{j}(x-x_{j})$ or $A_{j}(x,y)=a_{j}(y-y_{j})$, and
$B_{j}(x,y)=((x-x_{j})^{2}+(y-y_{j})^{2})^{3}.$ When $n=3$, our functions
$A_{j}$ include also $A_{j}(x,y,z)=a_{j}(z-z_{j})$ and
$B_{j}(x,y,z)=((x-x_{j})^{2}+(y-y_{j})^{2}+(z-z_{j})^{2})^{3}.$ Let
$\mathbb{R}[x_{1},x_{2},\ldots,x_{n}]$ denote the ring of polynomials in the
variables $x_{1},x_{2},\ldots,x_{n}$ with real coefficients.

######  Proposition 2.3.

Let

$R=\sum_{j=1}^{M}{\frac{A_{j}}{B_{j}^{1/2}}}\,,\qquad
A_{j},B_{j}\in\mathbb{R}[x_{1},x_{2},\ldots,x_{n}]\,,$

with $n=2$ or $n=3$, where $B_{j}\geq 0$ on $\mathbb{R}^{n}$ for all
$j=1,2,\dots,M$. Then the zero set

$\\{(x_{1},x_{2},\ldots,x_{n}):R(x_{1},x_{2},\ldots,x_{n})=0\\}$

is a subset of a real algebraic variety in $\mathbb{R}^{n}$. As a consequence,
we have the following.

a) If $n=2$, the zero set $\\{(x,y):R(x,y)=0\\}$ is a subset of a finite
number of points and curves.

b) If $n=3$, the zero set $\\{(x,y,z):R(x,y,z)=0\\}$ is a subset of a finite
number of points, curves, and two dimensional surfaces.

###### Proof.

Take $R$ and put the terms over the common denominator $D^{1/2}$ where
$D=B_{1}B_{2}\cdots B_{M}$. This expresses
$R=D^{-1/2}\sum_{j=1}^{M}{A_{j}D_{j}^{1/2}}$ where each $D_{j}$ is $D$ with
$B_{j}$ divided out. We see then that $\\{R=0\\}$ is a subset of where
$S=\sum_{j=1}^{M}A_{j}D_{j}^{1/2}=0$. Let $\Sigma$ be the collection of the
$2^{M}$ functions $\sigma:\\{1,2,\ldots,M\\}\rightarrow\\{-1,1\\}$. Now we
take the product

$P=\prod_{\sigma\in\Sigma}{\Bigg{(}\sum_{j=1}^{M}{\sigma(j)A_{j}D_{j}^{1/2}}\Bigg{)}}\,.$

Here the product has $2^{M}$ factors. We show that $P$ is a polynomial (of
degree at most $3M\cdot 2^{M}$). Let

${\mathcal{P}}(\chi_{1},\chi_{2},\ldots,\chi_{M}):=\prod_{\sigma\in\Sigma}{\bigg{(}\sum_{j=1}^{M}{\sigma(j)\chi_{j}}\Bigg{)}}\,.$

Observe that for every fixed $k\in\\{1,2,\ldots,M\\}$ the product
${\mathcal{P}}$ is a polynomial in $\chi_{k}$ and its value remains the same
if we replace $\chi_{k}$ by $-\chi_{k}$, as the factors of the product remain
the same, they are only permuted. Hence ${\mathcal{P}}$ is a polynomial in
$\chi_{k}^{2}$ for every fixed $k\in\\{1,2,\ldots,M\\}$. Now apply this with
$\chi_{j}=A_{j}D_{j}^{1/2}$ to conclude that $P$ is a polynomial (of degree at
most $3M\cdot 2^{M}$), indeed. To finish the proof observe that
$\sum_{j=1}^{M}{A_{j}D_{j}^{1/2}}$ is a factor of this product defining $P$,
namely the sum when $\sigma(j)=1$ for each $j$. So $\\{R=0\\}$ is a subset of
the real algebraic variety given by $P=0$. ∎

######  Remark 2.4.

a) It is an interesting issue that deriving $P$ is not so straightforward.
Most elementary cases of equations with square roots are handled just by
isolating the terms on one side of the equality, squaring, rearranging the
terms, and continuing this process. But with multiple terms (five or more)
that are square roots, this does not work.

b) We believe that in fact in a) and b) above, not only is the zero set
contained in the associated structure, but is itself in fact of the same
structure. At this time, we do not have a complete proof of this conjecture. ∎

######  Remark 2.5.

The asymptotic directions of the zero sets of the component functions of the
electric field when $n=2$ is one structural aspect that is examined in detail
below. But the case of the surfaces being asymptotically linear in the case
that $n=3$ is not resolved here. Nonetheless, if this is true, then two
gravitational fields in $\mathbb{R}^{3}$ can be distinguished by taking a set
of three non-colinear vectors $v_{1},v_{2},$ and $v_{3}$, and then translating
them by a vector $w$ which is far from the origin. Now, the intersection of
the component zero sets can only consist of a finite set of curve sections of
an associated real algebraic variety containing where $F=(0,0,0)$. Also, these
curves are probably asymptotically linear. If so, this shows that if two
finite point gravitational fields $F_{1}$ and $F_{2}$ have
$F_{1}(v_{i}+w)=F_{2}(v_{i}+w)$ for $i=1,2,$ and $3$, and $w$ of large norm,
then the gravitational fields are the same everywhere. Note: there is the
additional ambiguity of how far from the origin one must choose $w$, but
taking these values in a sequence going to infinity would resolve that. ∎

###  2.2 Finite Electric Fields in the Plane

In $\mathbb{R}^{2}$, we consider a finite set of points $(x_{j},y_{j})$ and a
finite set of coefficients $a_{j}$ that are real numbers. We let the Euclidean
distance $r_{j}=\|(x,y)-(x_{j},y_{j})\|_{2}$, We take a vector field $F$ given
by

$F(x,y)=(X(z,y),Y(x,y))=\sum_{j=1}^{M}{\frac{a_{j}}{r_{j}^{2}}\,\frac{(x,y)-(x_{j},y_{j})}{r_{j}}}\,.$

So for the real-valued components of this vector field, we have:

$X(x,y)=\sum_{j=1}^{M}{\frac{a_{j}(x-x_{j})}{r_{j}^{3}}}\qquad\text{and}\qquad
Y(x,y)=\sum_{j=1}^{M}{\frac{a_{j}(y-y_{j})}{r_{j}^{3}}}\,.$

Examples suggest that now the zero sets of $X$ and $Y$ consists of points,
lines, and curves, both bounded and unbounded. However, the common zero set is
more limited.

Conjecture: The zero set of a two-dimensional finite electrical field consists
of a finite set of points.

Of course, we would want to know also what the corresponding Maxwell
Conjecture would be: how many points can there be given only $M$? See
Gabrielov, Novikov, and Shapiro [2]. See also Proposition 4.1 in [1] and
Killian [3].

This conjecture should not be so difficult to solve. However, despite a number
of attempts to find simple methods to resolve it, and to estimate the number
of points, the conjecture has not been proven to be correct. In any case, what
this result would say about distinguishing two finite gravitational fields in
an ecliptic is that a finite, possibly large, number of measurements suffice
to distinguish these fields. Certainly, measurement in an entire disc would
suffice because $F$ could not be zero on a disc without being zero throughout
the plane. Even showing that the same thing is true if a disc is replaced by a
line segment seems not so easy to prove.

Indeed, the conjecture for finite electrical fields in the plane says that if
you take a measurement far from the point charges, and it is zero in both
components, then the field is identically zero everywhere and there are
actually no point charges at all. In practice, since one does not know how far
out one must be, the right thing to do is to take a fixed sequence of points
tending to infinity and take the measurements there. The conjecture says that
these cannot all be zero without the field being identically zero everywhere.

######  Remark 2.6.

Another version of the measurement problem is not easy to answer. How can we
take a limited number of measurements of two gravitational fields and based on
these measurements conclude the fields must be a rotation and/or a translation
of one another? This question can be rephrased for finite electrical fields in
a suitable fashion. ∎

One can consider there being two separate issues with describing the zero set
of a finite electrical field in the plane . One issue is how to show that all
the zeros must be in a large disc i.e. there cannot be zeros arbitrarily far
out in the plane. This is not the case for $X=0$ or for $Y=0$ separately; they
typically do have portions arbitrarily far from the origin.

The second issue is how to show that there can only be a finite set of zeros
within a fixed distance of the origin. The second issue is not completely
resolved at this time unless we had a result like Proposition 4.1 in [1].

Actually, what we seek to prove, with some technical work that is much more
difficult to complete than it ought to be, is that at infinity the zero set of
$X$ or of $Y$ consists of a finite number of curves that are to some degree at
least asymptotically linear. In the process, or as a result, we would want to
find equations that give the slopes of these lines and show that they are
distinct from one another when we consider $X$ or when we consider $Y$. This
is how we could show that the common zeros of $X$ and $Y$ must lie in a large
disc.

The obvious approach to these questions would seem to be just a smart use of
implicit differentiation. For example, where $X=0$, if there is a solution
$y=y(x)$ or $x=x(y)$, we should be able to compute $dy/dx$ or $dx/dy$ and work
with this. In particular, if we believe that a given curve $y=y(x)$ in the
zero set of $X$ must be asymptotically linear at infinity, we would just need
to show that $y$ is asymptotically $\alpha x$, for some $\alpha$. At this
time, we cannot see how to carry this out in general.

So instead, what we do is consider the implicit equation $Y=0$ and $X=0$ and
expand this as an infinite series. This series has to be adapted to the region
in question; one series does not seem to suffice for the whole plane. Then we
consider these series asymptotically and get equations that determine the
directions of solutions to $X=0$ at large distances from the origin. These
equations only have a finite number of solutions. Then we carry out the same
thing for $Y=0$. We then show that the asymptotic directions are different for
the two zero sets. This approach gives us at least part of the results that we
want.

######  Remark 2.7.

One aspect of this analysis that is worth noting is that certain moments are
closely tied to the structure of the zero sets. In the Special Case we call
$\sum_{j=1}^{M}{a_{j}x_{j}^{k}}$ the $k$th moment. If $k=0$, then we get the
ground state moment $\sum_{j=1}^{M}{a_{j}}$. If $M$ is large, then many of
these could be zero without forcing all the charges $a_{j}$ to be zero. The
critical moment is the one given by $L$ such that
$\sum_{j=1}^{M}{a_{j}x_{j}^{k}}=0$ for all $0\leq k<L$ and
$\sum_{j=1}^{M}{a_{j}x_{j}^{L}}\neq 0$. As the number of point charges
increases, the structure of the zero sets of $X$ and $Y$ become more
complicated, and as $L$ increases, the structure of the zero sets of $X$ and
$Y$ at infinity become more complicated. We have found no way to see that
these moments play an important role in the structure of the zero sets without
doing asymptotic analysis using power series expansions. ∎

###  2.3 A Special Case in the Plane

The Special Case will mean that we take points in $\mathbb{R}^{2}$ that are
all on a line and consider the zeros of $F=(X,Y)$. The General Case will mean
that we take any points in $\mathbb{R}^{2}$ and consider the zeros of
$F=(X,Y)$. Oddly enough, the Special Case is not so easy to analyze, even
though it seems quite restrictive.

By rotating and translating, or by a appropriate choice of the coordinate
system, the Special Case becomes the following. Take distinct points
$P_{j}=(x_{j},0),j=1,2,\dots,M$, on the positive $x$-axis. Assume that
$0<x_{1}<x_{2}<\dots<x_{M}$. Put non-zero electric charges $a_{j}$ at these
points. Then the field in the plane is given by

$F(x,y)=\sum_{j=1}^{M}{\frac{a_{j}((x,y)-(x_{j},0))}{\left((x-x_{j})^{2}+y^{2}\right)^{3/2}}}\,.$

Where is $F(x,y)=(0,0)$? Here the function $F(x,y)=(X(x,y),Y(x,y))$ where

$X(x,y)=\sum_{j=1}^{M}{\frac{a_{j}(x-x_{j})}{\left((x-x_{j})^{2}+y^{2}\right)^{3/2}}}\quad\text{and}\quad
Y(x,y)=\sum_{j=1}^{M}{\frac{a_{j}y}{\left((x-x_{j})^{2}+y^{2}\right)^{3/2}}}\,.$

Can we prove that $X$ and $Y$ only vanish simultaneously at only a finite
number of values $(x,y)$?

###  2.4 Zeros on the Line of the Point Charges

######  Proposition 2.8.

The function $F(x,0)$ vanishes only at finitely many $x\in{\mathbb{R}}$.

###### Proof.

This is the subcase that $y=0$. This is actually the case of a finite
electrical field on a line. But now $Y(x,y)=Y(x,0)=0$ and

$X(x,y)=X(x,0)=\sum_{j=1}^{M}{\frac{a_{j}(x-x_{j})}{\left((x-x_{j})^{2}\right)^{3/2}}}=\sum_{j=1}^{M}{\frac{a_{j}(x-x_{j})}{|x-x_{j}|}\frac{1}{(x-x_{j})^{2}}}\,.$

Let $I_{0}:=(-\infty,x_{1})$, $I_{M}:=(x_{M},\infty)$, and
$I_{j}:=(x_{j},x_{j+1}),\enskip j=1,2,\ldots,M-1$. The function
$h(x):=\displaystyle{\frac{x-x_{j}}{|x-x_{j}|}}$ is identically $-1$ or
identically $1$ on any of these intervals $I_{j}$ for $j=0,1,\ldots,M$. Hence,
$X(x,y)=X(x,0)$ is a rational function on any of these $I_{j}$ and can only
have a finite number of zeros on $I_{j}$. This means that $F(x,0)$ can be zero
only finitely many times too. ∎

###  2.5 Zeros Off the Line of the Point Charges

Now we work to argue that the zeros must be in some disc. Suppose $y$ is not
zero but $F(x,y)=(0,0)$. Then $Y(x,y)=0$ implies that

$\sum_{j=1}^{M}{\frac{a_{j}}{\left((x-x_{j})^{2}+y^{2}\right)^{3/2}}}=0$

since $y\neq 0$. So

$0=X(x,y)=\sum_{j=1}^{M}{\frac{a_{j}(x-x_{j})}{\left((x-x_{j})^{2}+y^{2}\right)^{3/2}}}=\sum_{j=1}^{M}{\frac{a_{j}(-x_{j})}{\left((x-x_{j})^{2}+y^{2}\right)^{3/2}}}\,.$

Hence, if we have zeros at an arbitrarily large distance away from the origin
(necessarily not on the $x$-axis and so $y\neq 0$), then multiply

$0=\sum_{j=1}^{M}\frac{a_{j}}{\left((x-x_{j})^{2}+y^{2}\right)^{3/2}}\qquad\text{and}\qquad
0=\sum_{j=1}^{M}{\frac{a_{j}x_{j}}{\left((x-x_{j})^{2}+y^{2}\right)^{3/2}}}$

by $\left(x^{2}+y^{2}\right)^{3/2}$. Then take a limit along a sequence of
values $(x,y)$ whose distance from the origin is going to $\infty$, for which
the forces are zero, and conclude that $\sum_{j=1}^{M}{a_{j}}=0$ and
$\sum_{j=1}^{M}{a_{j}x_{j}}=0$.

We examine the cases $M=2$ and $M=3$ briefly in the rest of this section.
Suppose $M=2$ so that there are only two point charges. Then the existence of
zeros arbitrarily far from the origin, would give $a_{1}+a_{2}=0$ and
$a_{1}x_{1}+a_{2}x_{2}=0$. Since $x_{1}\neq x_{2}$, this mean that both
$a_{1}=a_{2}=0$. So if there are zeros, then they must lie in some large disc.

How do we argue that there are only finitely many of these even in this very
simple case? The zero set can be non-empty for sure. Actually, with two point
charges, it is not hard to see we cannot have zeros off the axis, and there is
only a finite number on the axis. The number is very limited too, maybe at
most two (or three?) zeros.

We can extend the argument above to handle three points, at least in some
Special Cases. Assume there are zeros arbitrarily far out. We then have
$\sum_{j=1}^{M}{a_{j}}=0$ and $\sum_{j=1}^{M}{a_{j}x_{j}}=0$. Using points
$(x,y)$ that are zeros, we have

$\sum_{j=1}^{M}{\frac{a_{j}x_{j}}{\left((x-x_{j})^{2}+y^{2}\right)^{3/2}}}=0\,.$

So using $-\sum_{j=2}^{M}{a_{j}x_{j}}=a_{1}x_{1}$, we have

$\sum_{j=2}^{M}{a_{j}x_{j}\left(\frac{1}{\left((x-x_{j})^{2}+y^{2}\right)^{3/2}}-\frac{1}{\left((x-x_{1})^{2}+y^{2}\right)^{3/2}}\right)}=0\,.$

Now we can use the Mean Value Theorem to rewrite

$\frac{1}{\left((x-x_{j})^{2}+y^{2}\right)^{3/2}}-\frac{1}{\left((x-p_{1})^{2}+y^{2}\right)^{3/2}}=(x_{j}-x_{1})\left(\frac{-3t_{j}}{(t_{j}^{2}+y^{2})^{5/2}}\right)$

for some $t_{j}$ between $x-x_{1}$ and $x-x_{j}$. Again, taking zeros of large
radius, that also have the $x$ value large, and multiplying by
$\displaystyle{\frac{(x^{2}+y^{2})^{5/2}}{-3x}}$ gives

$\sum_{j=2}^{M}{a_{j}x_{j}(x_{j}-x_{1})}=\sum_{j=2}^{M}{a_{j}x_{j}^{2}}-\sum_{j=2}^{M}{a_{j}x_{j}x_{1}}=0\,.$

Since $\sum_{j=2}^{M}{a_{j}x_{j}}=-a_{1}p_{1}$, we get
$\sum_{j=1}^{M}{a_{j}x_{j}^{2}}=0$. So in this Special Case we have
$\sum_{j=1}^{M}{a_{j}}=0$, $\sum_{j=1}^{M}{a_{j}x_{j}}=0$, and
$\sum_{j=1}^{M}{a_{j}x_{j}^{2}}=0$. The conclusion is that if
$a_{1}=a_{2}=a_{3}=0$. So there cannot be zeros arbitrarily far out, with $x$
values also large unless $a_{1}=a_{2}=a_{3}=0$. We need a separate argument to
treat the case where the $x$ values do not get large. ∎

##  3 Special Case: No Curves and Finiteness of the Zero Set

It turns out that there are a couple of ways, closely related to each other,
to show that in the Special Case there cannot be any non-trivial curves in the
zero set $F=(X,Y)=(0,0)$. We choose here one particular approach.

First, in Section 3.1, we prove that this zero set is, in fact, countable. The
advantage of this argument is that we do not even need to define what we mean
by “non-trivial curve”. Then in Section 3.2, we show that the zero set is in
fact finite.

###  3.1 The Zero Set $\\{X=Y=0\\}$ Is Countable

With $0<x_{1}<x_{2}<\cdots<x_{M}\,$, let

$X_{m}(x,y):=\sum_{j=1}^{M}{\frac{a_{j}(x-x_{j})}{((x-x_{j})^{2}+y^{2})^{(2m+1)/2}}}\,,\qquad
m=1,2,\ldots\,,$

and

$Y_{m}(x,y):=\sum_{j=1}^{M}{\frac{a_{j}y}{((x-x_{j})^{2}+y^{2})^{(2m+1)/2}}}\,,\qquad
m=1,2,\ldots\,.$

We have

$\begin{split}\frac{\partial{X_{m}}}{\partial
x}&=\,\sum_{j=1}^{M}{a_{j}\,\frac{((x-x_{j})^{2}+y^{2})-(2m+1)(x-x_{j})^{2}}{((x-x_{j})^{2}+y^{2})^{(2m+3)/2}}}\,,\cr\frac{\partial{X_{m}}}{\partial
y}&=\,\sum_{j=1}^{M}{a_{j}\,\frac{-(2m+1)y(x-x_{j})}{((x-x_{j})^{2}+y^{2})^{(2m+3)/2}}}\,,\cr\frac{\partial{Y_{m}}}{\partial
x}&=\,\sum_{j=1}^{M}{a_{j}\,\frac{-(2m+1)y(x-x_{j})}{((x-x_{j})^{2}+y^{2})^{(2m+3)/2}}}\,,\cr\frac{\partial{Y_{m}}}{\partial
y}&=\,\sum_{j=1}^{M}{a_{j}\,\frac{((x-x_{j})^{2}+y^{2})-(2m+1)y^{2}}{((x-x_{j})^{2}+y^{2})^{(2m+3)/2}}}\,.\cr\end{split}$

Let

$H:=\\{(x,y):\enskip X_{1}=Y_{1}=0,\enskip y\neq 0\\}$

and

$H_{m}:=\\{(x,y):X_{u}=Y_{u}=0,\enskip y\neq 0\enskip u=1,2,\ldots,m,\enskip
X_{m+1}^{2}+Y_{m+1}^{2}>0\\}\,,$

$m=1,2,\ldots$. Observe that for $(x,y)\in H_{m}$ we have

$\frac{\partial{X_{m}}}{\partial x}=-\frac{\partial{Y_{m}}}{\partial
y}=(2m+1)yY_{m+1}$

and

$\frac{\partial{X_{m}}}{\partial y}=\frac{\partial{Y_{m}}}{\partial
x}=-(2m+1)yX_{m+1}\,.$

In this section we use only undergraduate real analysis such as Vandermonde
determinants and the Implicit Function Theorem and prove a structural property
of the set $H$. In particular, we prove that $H$ is a countable set. The basic
idea is under the assumptions

$X_{1}(x_{0},y_{0})=Y_{1}(x_{0},y_{0})=0\,,\qquad\frac{\partial
X_{1}}{\partial x}(x_{0},y_{0})\neq 0$

the algebraic varieties

$\\{(x,y):X_{1}(x,y)=0\\}\qquad\text{and}\qquad\\{(x,y):Y_{1}(x,y)=0\\}\,,$

which are graphs of functions in a neighborhood of $(x_{0},y_{0})$ by the
Implicit Function Theorem, intersect each other in an orthogonal fashion.
However, proving our results in this section requires more technical details.

######  Proposition 3.1.

We have

$H=\bigcup_{m=1}^{M-1}{H_{m}}$

unless $a_{1}=a_{2}=\cdots=a_{M}=0$.

###### Proof.

Suppose $(x_{0},y_{0})\notin H$. For the sake of brevity let

$r_{j}:=((x_{0}-x_{j})^{2}+y_{0}^{2})^{1/2}\,,\qquad j=1,2,\ldots,M\,.$

We have

$\sum_{j=1}^{M}{\frac{a_{j}}{r_{j}}\frac{1}{(r_{j}^{2})^{k}}}=0\,,\qquad
k=1,2,\dots,M\,,$ (3.1)

and

$\sum_{j=1}^{M}{\frac{a_{j}x_{j}}{r_{j}}\frac{1}{(r_{j}^{2})^{k}}}=0\,,\qquad
k=1,2,\dots,M\,.$ (3.2)

Let $\\{R_{1}<R_{2}<\ldots<R_{m}\\}=\\{r_{1},r_{2},\ldots,r_{M}\\}$. Since our
point charges $(x_{j},0)$ are on a line, for every $u\in\\{1,2,\ldots,m\\}$
there are at most $2$ values of $j\in\\{1,2,\ldots,M\\}$ for which
$r_{j}=R_{u}$. Hence by (3.1) and (3.2) we have

$\sum_{u=1}^{m}{\frac{b_{u}}{R_{u}}\frac{1}{(R_{u}^{2})^{k}}}=0\,,\qquad
k=1,2,\dots,m\,,$ (3.3)

and

$\sum_{u=1}^{m}{\frac{c_{u}}{R_{u}}\frac{1}{(R_{u}^{2})^{k}}}=0\,,\qquad
k=1,2,\dots,m\,,$ (3.4)

where

$b_{u}=a_{j}\qquad\text{and}\qquad c_{u}=x_{j}a_{j}$ (3.5)

if there is only one value $j\in\\{1,2,\ldots,M\\}$ such that $r_{j}=R_{u}$,
and

$b_{u}=a_{j_{1}}+a_{j_{2}}\qquad\text{and}\qquad
c_{u}=x_{j_{1}}a_{j_{1}}+x_{j_{2}}a_{j_{2}}$ (3.6)

if there are two distinct values $j_{1},j_{2}\in\\{1,2,\ldots,M\\}$ such that
$r_{j_{1}}=r_{j_{2}}=R_{u}$. Using (3.3)–(3.6) and the well known non-
vanishing property of Vandermonde determinants we conclude that

$b_{u}=c_{u}=0\,,\qquad u=1,2,\ldots,m\,,$

from which

$a_{j}=0,\qquad j=1,2,\ldots,M\,,$

follows. ∎

######  Proposition 3.2.

Each point in $H_{m}$ is an isolated point in $H_{m}$ for every
$m=1,2,\ldots,M-1$, unless $a_{1}=a_{2}=\cdots=a_{M}=0$.

###### Proof.

Suppose $a_{1}^{2}+a_{2}^{2}+\cdots+a_{M}^{2}>0$. Let $m$ be a fixed positive
integer. Suppose $(x_{0},y_{0})\in H_{m}$, that is, either

$X_{u}(x_{0},y_{0})=Y_{u}(x_{0},y_{0})=0\,,\enskip y_{0}\neq 0\,,\enskip
u=1,2,\ldots,m\,,\enskip X_{m+1}(x_{0},y_{0})\neq 0\,,$ (3.7)

or

$X_{u}(x_{0},y_{0})=Y_{u}(x_{0},y_{0})=0\,,\enskip y_{0}\neq 0\,,\enskip
u=1,2,\ldots,m\,,\enskip Y_{m+1}(x_{0},y_{0})\neq 0\,.$ (3.8)

If (3.7) holds, then we have

$\frac{\partial{X_{m}}}{\partial
y}(x_{0},y_{0})=-(2m+1)y_{0}X_{m+1}(x_{0},y_{0})\neq 0$ (3.9)

and

$\frac{\partial{Y_{m}}}{\partial
x}(x_{0},y_{0})=-(2m+1)y_{0}X_{m+1}(x_{0},y_{0})\neq 0$ (3.10)

Also,

$\begin{split}\frac{\partial{Y_{m}}}{\partial
y}(x_{0},y_{0})&=\,\frac{Y_{m}(x_{0},y_{0})}{y_{0}}-(2m+1)y_{0}Y_{m+1}(x_{0},y_{0})\cr&=\,-(2m+1)y_{0}Y_{m+1}(x_{0},y_{0})\cr\end{split}$
(3.11)

and

$\begin{split}\frac{\partial{X_{m}}}{\partial
x}(x_{0},y_{0})&=\,(2m+1)y_{0}Y_{m+1}(x_{0},y_{0})-\frac{(2m)Y_{m}(x_{0},y_{0})}{y_{0}}\cr&=\,(2m+1)y_{0}Y_{m+1}(x_{0},y_{0})\,.\end{split}$
(3.12)

By the Implicit Function Theorem and (3.9), there are $\delta>0$,
$\varepsilon>0$, and a function $y=f(x)$ differentiable on
$(x_{0}-\delta,x_{0}+\delta)$ such that the set

$\begin{split}C_{1}:&=\,\\{(x,y):X_{m}(x,y)=0,\enskip
x\in(x_{0}-\delta,x_{0}+\delta),\enskip
y\in(y_{0}-\varepsilon,y_{0}+\varepsilon)\\}\cr&=\,\\{(x,f(x)):\enskip
x\in(x_{0}-\delta,x_{0}+\delta)\\}\cr\end{split}$

is the graph of a differentiable function $y=f(x)$ on
$(x_{0}-\delta,x_{0}+\delta)$, and the equation for the tangent line to the
curve $C_{1}$ at the point $(x_{0},y_{0})$ is

$y-y_{0}=\alpha(x-x_{0})\,,$

where

$\alpha=\left(\frac{\partial{X_{m}}}{\partial
x}\Big{/}\frac{\partial{X_{m}}}{{\partial
y}}\right)(x_{0},y_{0}=\frac{(2m+1)y_{0}Y_{m+1}(x_{0},y_{0})}{-(2m+1)y_{0}X_{m+1}(x_{0},y_{0})}=\frac{-Y_{m+1}(x_{0},y_{0})}{X_{m+1}(x_{0},y_{0})}\,.$

Similarly, by the Implicit Function Theorem and (3.10), there are $\delta>0$,
$\varepsilon>0$, and a function $x=g(y)$ differentiable on
$(y_{0}-\delta,y_{0}+\delta)$ such that the set

$\begin{split}C_{2}:&=\,\\{(x,y):Y_{m}(x,y)=0,\enskip
y\in(y_{0}-\delta,y_{0}+\delta),\enskip
x\in(x_{0}-\varepsilon,x_{0}+\varepsilon)\\}\cr&=\,\\{(g(y),y):\enskip
y\in(y_{0}-\delta,y_{0}+\delta)\\}\cr\end{split}$

is the graph of a differentiable function $x=g(y)$ on
$(y_{0}-\delta,y_{0}+\delta)$, and the equation for the tangent line to the
curve $C_{2}$ at the point $(x_{0},y_{0})$ is

$x-x_{0}=\beta(y-y_{0})\,,$

where by (3.11) and (3.12) we have

$\beta=\left(\frac{\partial{Y_{m}}}{\partial
y}\Big{/}\frac{\partial{Y_{m}}}{{\partial
x}}\right)(x_{0},y_{0})=\frac{-(2m+1)y_{0}Y_{m+1}(x_{0},y_{0})}{-(2m+1)y_{0}X_{m+1}(x_{0},y_{0})}=\frac{Y_{m+1}(x_{0},y_{0})}{X_{m+1}(x_{0},y_{0})}\,.$

Observe that $\alpha=-\beta$. We conclude that the curves $C_{1}$ and $C_{2}$
are orthogonal to each other at $(x_{0},y_{0})$. As $H_{m}\subset C_{1}\bigcap
C_{2}$, the point $(x_{0},y_{0})\in H_{m}$ is an isolated point of $H_{m}$.

If (3.8) holds, then we have

$\begin{split}\frac{\partial{Y_{m}}}{\partial
y}(x_{0},y_{0})&=\,\frac{Y_{m}(x_{0},y_{0})}{y_{0}}-(2m+1)y_{0}Y_{m+1}(x_{0},y_{0})\cr&=\,-(2m+1)y_{0}Y_{m+1}(x_{0},y_{0})\neq
0\cr\end{split}$ (3.13)

and

$\begin{split}\frac{\partial{X_{m}}}{\partial
x}(x_{0},y_{0})&=\,(2m+1)y_{0}Y_{m+1}(x_{0},y_{0})-\frac{(2m)Y_{m}(x_{0},y_{0})}{y_{0}}\cr&=\,(2m+1)y_{0}Y_{m+1}(x_{0},y_{0})\neq
0\,.\end{split}$ (3.14)

Also,

$\begin{split}\frac{\partial{X_{m}}}{\partial
y}(x_{0},y_{0})=\frac{\partial{Y_{m}}}{\partial
x}(x_{0},y_{0})=-(2m+1)y_{0}X_{m+1}(x_{0},y_{0})\end{split}$ (3.15)

By the Implicit Function Theorem and (3.13), there are $\delta>0$,
$\varepsilon>0$, and a function $y=f(x)$ differentiable on
$(x_{0}-\delta,x_{0}+\delta)$ such that the set

$\begin{split}C_{3}:&=\,\\{(x,y):Y_{m}(x,y)=0,\enskip
x\in(x_{0}-\delta,x_{0}+\delta),\enskip
y\in(y_{0}-\varepsilon,y_{0}+\varepsilon)\\}\cr&=\,\\{(x,f(x)):\enskip
x\in(x_{0}-\delta,x_{0}+\delta)\\}\cr\end{split}$

is the graph of a differentiable function $y=f(x)$ on
$(x_{0}-\delta,x_{0}+\delta)$, and the equation for the tangent line to the
curve $C_{3}$ at the point $(x_{0},y_{0})$ is

$y-y_{0}=\alpha(x-x_{0})\,,$

where

$\alpha=\left(\frac{\partial{Y_{m}}}{\partial
x}\Big{/}\frac{\partial{Y_{m}}}{{\partial
y}}\right)(x_{0},y_{0})=\frac{-(2m+1)y_{0}X_{m+1}(x_{0},y_{0})}{-(2m+1)y_{0}Y_{m+1}(x_{0},y_{0})}=\frac{X_{m+1}(x_{0},y_{0})}{Y_{m+1}(x_{0},y_{0})}\,.$

Similarly, by the Implicit Function Theorem and (3.14), there are $\delta>0$,
$\varepsilon>0$, and a function $x=g(y)$ differentiable on
$(y_{0}-\delta,y_{0}+\delta)$ such that the set

$\begin{split}C_{4}:&=\,\\{(x,y):X_{m}(x,y)=0,\enskip
y\in(y_{0}-\delta,y_{0}+\delta),\enskip
x\in(x_{0}-\varepsilon,x_{0}+\varepsilon)\\}\cr&=\,\\{(g(y),y):\enskip
y\in(y_{0}-\delta,y_{0}+\delta)\\}\cr\end{split}$

is the graph of a differentiable function $x=g(y)$ on
$(y_{0}-\delta,y_{0}+\delta)$, and the equation for the tangent line to the
curve $C_{4}$ at the point $(x_{0},y_{0})$ is

$x-x_{0}=\beta(y-y_{0})$

where by (3.15) we have

$\beta=\left(\frac{\partial{X_{m}}}{\partial
y}\Big{/}\frac{\partial{X_{m}}}{{\partial
x}}\right)(x_{0},y_{0})=\frac{-(2m+1)y_{0}X_{m+1}(x_{0},y_{0})}{(2m+1)y_{0}Y_{m+1}(x_{0},y_{0})}=\frac{-X_{m+1}(x_{0},y_{0})}{Y_{m+1}(x_{0},y_{0})}\,.$

Observe that $\alpha=-\beta$. We conclude that the curves $C_{3}$ and $C_{4}$
are orthogonal to each other at $(x_{0},y_{0})$. As $H_{m}\subset C_{3}\bigcap
C_{4}$, the point $(x_{0},y_{0})\in H_{m}$ is an isolated point of $H_{m}$. ∎

Now by combining Propositions 3.1, 3.2, and Section 2.4, our final conclusion
in this section is the following.

######  Proposition 3.3.

In the Special Case, the set $\\{(x,y):X(x,y)=Y(x,y)=0\\}$ is countable, and
hence the zero set of $F=(X,Y)$ does not contain a non-trivial curve.

###  3.2 Bézout and Finiteness of the Zero Set

Now we use the above result on the absence of curves in the joint zero set to
show that the zero set is actually finite in the Special Case. The argument is
along the lines of the proof of Proposition 2.3.

######  Proposition 3.4.

In the Special Case, the zero set of $F=(X,Y)$ is a finite set.

###### Proof.

Again, suppose $0<x_{1}<x_{2}<\cdots<x_{M}$ and the real numbers
$a_{1},a_{2},\ldots,a_{M}$ are not all zero. Let, as before,

$X(x,y)=\sum_{j=1}^{M}{\frac{a_{j}(x-x_{j})}{((x-x_{j})^{2}+y^{2})^{3/2}}}\,,\quad\text{and}\quad
Y(x,y)=\sum_{j=1}^{M}{\frac{a_{j}y}{((x-x_{j})^{2}+y^{2})^{3/2}}}\,,$

Let $\Sigma$ be the collection of the $2^{M}$ functions
$\sigma:\\{1,2,\ldots,M\\}\rightarrow\\{-1,1\\}$. Let

$X_{\sigma}(x,y):=\sum_{j=1}^{M}{\frac{\sigma(j)a_{j}(x-x_{j})}{((x-x_{j})^{2}+y^{2})^{3/2}}}\,,\quad\text{and}\quad
Y_{\sigma}(x,y):=\sum_{j=1}^{M}{\frac{\sigma(j)a_{j}y}{((x-x_{j})^{2}+y^{2})^{3/2}}}\,.$

Let

$D_{j}(x,y):=\prod_{k=1,k\neq j}^{M}{((x-x_{k})^{2}+y^{2}))^{3/2}}\,,\qquad
j=1,2,\ldots,M\,,$

and

$D(x,y):=\prod_{k=1}^{M}{((x-x_{k})^{2}+y^{2}))^{3/2}}\,.$

We have

$\begin{split}&\,X_{\sigma}(x,y)^{2}+Y_{\sigma}(x,y)^{2}\cr=&\,\Bigg{(}\sum_{j=1}^{M}{\frac{\sigma(j)a_{j}(x-x_{k})D_{j}(x,y)}{D(x,y)}}\Bigg{)}^{2}+\Bigg{(}\sum_{j=1}^{M}{\frac{\sigma(j)a_{j}yD_{j}(x,y)}{D(x,y)}}\Bigg{)}^{2}\cr=&\,\frac{1}{D(x,y)^{2}}\Bigg{(}\Bigg{(}\sum_{j=1}^{M}{\sigma(j)a_{j}(x-x_{j})D_{j}(x,y)}\Bigg{)}^{2}+\Bigg{(}\sum_{j=1}^{M}{\sigma(j)a_{j}yD_{j}(x,y)}\Bigg{)}^{2}\Bigg{)}\,.\cr\end{split}$

Observe that the function

$F(x,y):=\prod_{\sigma\in\Sigma}{\Bigg{(}\Bigg{(}\sum_{j=1}^{M}{\sigma(j)a_{j}(x-x_{j})D_{j}(x,y)}\Bigg{)}^{2}+\Bigg{(}\sum_{j=1}^{M}{\sigma(j)a_{k}yD_{j}(x,y)}\Bigg{)}^{2}\Bigg{)}}$

is an even polynomial in each of the variables

$\chi_{j}:=D_{j}(x,y)\,,\qquad j=1,2,\ldots,M\,,$

as it remains the same when $\chi_{j}$ is replaced by $-\chi_{j}$. Hence
$F(x,y)$ is a polynomial in each of the variables

$\chi_{j}^{2}=D_{j}(x,y)^{2},\qquad j=1,2,\ldots,M\,.$

We conclude that

$G(x,y):=\prod_{\sigma\in\Sigma}{\Bigg{(}X_{\sigma}(x,y)^{2}+Y_{\sigma}(x,y)^{2}\Bigg{)}}=\frac{P(x,y)}{D(x,y)^{2^{M+1}}}\,,$

where $P$ is a polynomial of degree at most $(3M)2^{M}$. We claim that the set

$E:=\\{(x,y):P(x,y)=0\\}$

cannot contain a non-trivial curve. Indeed, observe that $E$ is a finite union
of the countable sets $X_{\sigma}\cap Y_{\sigma}$, hence $E$ is also
countable. Thus $E=\\{(x,y):P(x,y)=0\\}$ is an algebraic variety not
containing a non-trivial curve. Such an algebraic variety contains only
finitely many points. As the zero set of $F=(X,Y)$ is a subset of $E$, it is
also finite. ∎

With the help of Bézout’s Theorem we can give an upper bound for the number of
points in the zero set of $F=(X,Y)$.

######  Proposition 3.5.

In the Special Case the zero set of $F=(X,Y)$ contains at most $9M^{2}4^{M}$
points.

To prove the above proposition we need the following classic version of
Bézout’s Theorem. Two polynomials $f\in{\mathbb{R}}[x,y]$ and
$g\in{\mathbb{R}}[x,y]$ are coprime if there is no non-constant polynomial
which is a factor of both $f$ and $g$. Associated with a polynomial
$f\in{\mathbb{R}}[x,y]$ we define its zero set $Z(f):=\\{(x,y):\enskip
f(x,y)=0\\}\,.$

######  Proposition 3.6.

(Bézout’s Theorem) Given any two coprime polynomials $f,g\in{\mathbb{R}}[x,y]$
of degrees $d_{1}$ and $d_{2}$, respectively, $Z(f)\cap Z(g)$ contains at most
$d_{1}d_{2}$ points.

A consequence of the above Bézout’s Theorem is the following.

######  Proposition 3.7.

Let $f\in{\mathbb{R}}[x,y]$ be a polynomial of degree $d$. If $Z(f)$ contains
only finitely many points, then $Z(f)$ contains at most $d(d-1)$ points.

###### Proof.

Without loss of generality we may assume that $f$ is square-free, that is, $f$
is a product of irreducible polynomial factors none of which is repeated, as
keeping only one of each repeated factors in a factorization of $f$ the set
$Z(f)$ remains the same. As $Z(f)$ contains only finitely many points, we may
also assume that $f$ does not have a factor that is a polynomial of $y$ of
degree at least $1$. Observe that if $Z(f)$ contains only finitely many
points, then each of these points are singular, that is,

$Z(f)=\left\\{(x,y):f(x,y)=\frac{\partial f}{\partial x}(x,y)=\frac{\partial
f}{\partial y}(x,y)=0\right\\}\,,$

otherwise the Implicit Function Theorem would imply that $Z(f)$ contains a
curve (graph of a function) with infinitely many points. For the sake of
brevity let

$g(x,y):=\frac{\partial f}{\partial x}(x,y)\,.$

As $f$ is square-free, $f$ and $g$ are coprime. Hence Bézout’s Theorem implies
that $Z(f)=Z(f)\cap Z(g)$ contains at most $d(d-1)$ points. ∎

Now we are ready to prove Proposition 3.5.

###### Proof.

Observe that the degree of $P$ in the proof of Proposition 3.4 is at most
$d:=(3M)2^{M}$. Hence, by Proposition 3.7 we can deduce that the zero set
$E=\\{(x,y):P(x,y)=0\\}$ contains at most $d^{2}=9M^{2}4^{M}$ points. As the
zero set of $F=(X,Y)$ is a subset of $E$, it contains at most
$d^{2}=9M^{2}4^{M}$ points as well. ∎

Observe that no general fact about analytic varieties has been used. All that
is needed is Bézout’s Theorem. See Gibson [5] for Bézout’s Theorem.

###  3.3 Impact of the Next Dimension

The zero sets of $X$ and $Y$ do not meet in an orthogonal fashion in general.
But under simple assumptions they do, and in fact aspects of the field in one
higher dimension can play a role.

Consider an analysis of this by taking charges in the $xy$-plane, with the
force field in $\mathbb{R}^{3}$. Now we have $F=(X,Y,Z)$ but our charges are
not in general position in $\mathbb{R}^{3}$. Instead they are restricted to
being in the $xy$-plane: that is, the charges are at points
$(x_{j},y_{j},0),j=1,2,\dots,M$.

Again, the force field in $\mathbb{R}^{3}$ has a potential

$\phi(x,y,z)=\frac{-1}{2}\sum_{j=1}^{M}{\frac{a_{j}}{r_{j}}}\,.$

But now, unlike in two dimensions, this potential is harmonic.

Suppose $X=Y=0$ at a point $(x_{0},y_{0},0)$. Suppose again as before that
both $y_{X}=y_{X}(x)$ and $y_{Y}=y_{Y}(x)$ can be implicitly determined by
$X=0$ and $Y=0$ in a neighborhood of $(x_{0},y,0)$. For the sake of this
argument, we assume that this happens because

$\frac{\partial X}{\partial y}\neq 0\qquad\text{and}\qquad\frac{\partial
Y}{\partial y}\neq 0$

at the point $(x_{0},y_{0},0)$. This means that

$\frac{\partial^{2}\phi}{\partial y\partial x}\neq
0\qquad\text{and}\qquad\frac{\partial^{2}\phi}{\partial y^{2}}\neq 0$

at the point $(x_{0},y_{0},0)$. We also have

$0=\frac{\partial X}{\partial x}+\frac{\partial X}{\partial
y}\frac{dy_{X}}{dx}=\frac{\partial^{2}\phi}{\partial
x^{2}}+\frac{\partial^{2}\phi}{\partial y\partial x}\frac{dy_{X}}{dx}$

and

$0=\frac{\partial Y}{\partial x}+\frac{\partial Y}{\partial
y}\frac{dy_{Y}}{dx}=\frac{\partial^{2}\phi}{\partial x\partial
y}+\frac{\partial^{2}\phi}{\partial y^{2}}\frac{dy_{Y}}{dx}$

at the point $(x_{0},y_{0},0)$. Hence

$\frac{dy_{X}}{dx}\frac{dy_{Y}}{dx}=\left(-\frac{\partial^{2}\phi}{\partial
x^{2}}\Big{/}\frac{\partial^{2}\phi}{\partial y\partial
x}\right)\left(-\frac{\partial^{2}\phi}{\partial x\partial
y}\Big{/}\frac{\partial^{2}\phi}{\partial
y^{2}}\right)=\left(\frac{\partial^{2}\phi}{\partial
x^{2}}\Big{/}\frac{\partial^{2}\phi}{\partial y^{2}}\right)$

holds at the point $(x_{0},y_{0},0)$. We have just proved the following
result.

######  Proposition 3.8.

If $X=Y=0$,

$\frac{\partial X}{\partial y}\neq 0\,,\qquad\frac{\partial Y}{\partial y}\neq
0\,,\qquad\text{and}\qquad\frac{\partial^{2}\phi}{\partial
x^{2}}+\frac{\partial^{2}\phi}{\partial y^{2}}=0$

at the point $(x_{0},y_{0},0)$, then

$\frac{dy_{X}}{dx}\frac{dy_{Y}}{dx}=-1$

holds at the point $(x_{0},y_{0},0)$, and the zero sets $\\{X=0\\}$ and
$\\{Y=0\\}$ in the $xy$-plane meet at the point $(x_{0},y_{0},0)$ in an
orthogonal fashion.

However, in general $\displaystyle{\frac{\partial^{2}\phi}{\partial
x^{2}}+\frac{\partial^{2}\phi}{\partial y^{2}}}$ is not zero. Rather

$\frac{\partial^{2}\phi}{\partial x^{2}}+\frac{\partial^{2}\phi}{\partial
y^{2}}+\frac{\partial^{2}\phi}{\partial z^{2}}=0\,.$

But

$\frac{\partial^{2}\phi}{\partial
z^{2}}=\sum_{j=1}^{M}{\frac{a_{j}}{r_{j}^{3}}}-3\sum_{j=1}^{M}{\frac{a_{j}z^{2}}{r_{j}^{5}}}\,.$

So when $z=0$, we have $\displaystyle{\frac{\partial^{2}\phi}{\partial
z^{2}}=\sum_{j=1}^{M}{\frac{a_{j}}{r_{j}^{3}}}}$. But
$\displaystyle{Z=\sum_{j=1}^{M}{\frac{a_{j}z}{r_{j}^{3}}}}$. So at a point
$(x_{0},y_{0},0)$ which is a limit of points $(x,y,z)$ with $z\neq 0$ and
$Z(x,y,z)=0$ we have $\displaystyle{\frac{\partial^{2}\phi}{\partial
z^{2}}=0}$, and so

$\frac{\partial^{2}\phi}{\partial x^{2}}+\frac{\partial^{2}\phi}{\partial
y^{2}}=0\,.$

Hence, as above, this means that

$\frac{dy_{X}}{dx}\frac{dy_{Y}}{dx}=-1$

at the point $(x_{0},y_{0},0)$, and so again we have the same result as in
Proposition 3.8, that is, the zero sets of $\\{X=0\\}$ and $\\{Y=0\\}$ meet at
the point $x_{0},y_{0},0)$ in an orthogonal fashion.

However, examples should show that there are are points $(x,y,0)$ in the
$xy$-plane where both $X$ and $Y$ are zero, but

$\sum_{j=1}^{M}{\frac{a_{j}}{r_{j}^{3}}}=\sum_{j=1}^{M}{\frac{a_{j}}{((x-x_{j})^{2}+(y-y_{j})^{2})^{3/2}}}$

is not zero. So at such points the zero sets $\\{X=0\\}$ and $\\{Y=0\\}$ would
not necessarily meet in an orthogonal fashion.

##  4 Special Case: Asymptotic Directions of Component Zero Sets

Let $0\neq\alpha\in(-\infty,\infty)$. The line $y=\alpha x$ is called an
asymptotic direction for a set $A\subset{\mathbb{R}}^{2}$ if there are
$(p_{m},q_{m})\in A$ such that

$\lim_{|p_{m}|\rightarrow\infty}{\frac{q_{m}}{p_{m}}}=\lim_{|q_{m}|\rightarrow\infty}{{\frac{q_{m}}{p_{m}}}}=\alpha\,.$

The $x$-axis is called an asymptotic direction for a set
$A\subset{\mathbb{R}}^{2}$ if there are $(p_{m},q_{m})\in A$ such that

$\lim_{|p_{m}|\rightarrow\infty}{\frac{q_{m}}{p_{m}}}=0\,.$

The $y$-axis is called an asymptotic direction for a set
$A\subset{\mathbb{R}}^{2}$ if there are $(p_{m},q_{m})\in A$ such that

$\lim_{|q_{m}|\rightarrow\infty}{\frac{p_{m}}{q_{m}}}=0\,.$

For the sake of brevity we will use the notation $\beta:=1/\alpha$. Note that
the case $\beta=0$ corresponds to the asymptotic direction given by the
$y$-axis. The goal in this section is is simple: to show that the set of
possible asymptotic directions for $\\{X=0\\}$ and the set of the possible
asymptotic directions for $\\{Y=0\\}$ are distinct. We succeed in showing this
with the exception of the asymptotic directions $y=\pm x$. We call the domains
$\\{(x,y):|x|<|y|\\}$ and $\\{(x,y):|y|<|x|\\}$ Type I domain and Type II
domain, respectively. To find all possible asymptotic directions for the zero
sets $\\{Y=0\\}$ and $\\{X=0\\}$ we proceed differently in Type I and Type II
domains, but the cases of the zero sets $\\{Y=0\\}$ and $\\{X=0\\}$ are quite
similar to each other in both domains. Our results in this section have their
own intrinsic interest, even though the boundedness of the zero set
$\\{X=Y=0\\}$ contained in Propositions 3.4 and 3.5 is not completely
recaptured in this section as we cannot show that the lines $|y|=|x|$ cannot
be a common asymptotic direction to both of the zero sets $\\{Y=0\\}$ and
$\\{X=0\\}$.

###  4.1 Notation

Let, as before, $0<x_{1}<x_{2}<\cdots<x_{M}$ and

$X(x,y):=\sum_{j=1}^{M}{\frac{a_{j}(x-x_{j})}{((x-x_{j})^{2}+y^{2})^{3/2}}}\qquad\text{and}\qquad
Y(x,y):=\sum_{j=1}^{M}{\frac{a_{j}y}{((x-x_{j})^{2}+y^{2})^{3/2}}}\,.$

By the Binomial Theorem the Taylor series expansion

$(1+z)^{-3/2}=\sum_{n=0}^{\infty}{\binom{-3/2}{n}{z^{n}}}\,,\qquad
z\in(-1,1)\,,$

converges uniform on the interval $[-1+\delta,1-\delta]$ for every $\delta>0$.
Let $\mu_{u}:=(-1)^{u}\sum_{j=1}^{M}{a_{j}x_{j}^{u}}$. By the non-vanishing
property of the Vandermonde determinants there is an integer $0\leq L\leq M$
such that $\mu_{u}=0$ for $u=0,1,\ldots,L-1$, and $\mu_{L}\neq 0$. This is the
choice of $L$ throughout Sections 4.1-4.6. Typically, as $L$ gets larger, the
complexity to detect possible asymptotic directions for the zero sets
$\\{X=0\\}$ and $\\{Y=0\\}$ increases.

###  4.2 Type I Domain, Possible Asymptotes for $Y=0$

######  Proposition 4.1.

Let $\delta\in(0,1/2)$ be fixed. If

$Y(p_{m},q_{m})=0,\quad\left|\frac{p_{m}}{q_{m}}\right|<1-\delta,\quad
q_{m}\neq
0,\quad\lim_{|q_{m}|\rightarrow\infty}{\frac{p_{m}}{q_{m}}}=\beta\,,$

then

$\frac{d^{L}}{d\beta^{L}}\left(\frac{1}{(1+\beta^{2})^{3/2}}\right)=0\,.$

###### Proof.

We have

$\begin{split}0&=\,q_{m}^{L+1}|q_{m}|Y(p_{m},q_{m})=q_{m}^{L}\sum_{j=1}^{M}{a_{j}\left(1+\left(\frac{p_{m}-x_{j}}{q_{m}}\right)^{2}\right)^{-3/2}}\cr&=\,q_{m}^{L}\sum_{j=1}^{M}{a_{j}\sum_{n=0}^{\infty}{\binom{-3/2}{n}\left(\frac{p_{m}-x_{j}}{q_{m}}\right)^{2n}}}\cr\end{split}$

for all sufficiently large $m$. Note that we have adjusted with the factor
$q_{m}^{L+1}|q_{m}|$. We also have

$\begin{split}\sum_{j=1}^{M}{a_{j}(p_{m}-x_{j})^{2n}}&=\,\sum_{j=1}^{M}{a_{j}\sum_{u=0}^{2n}{\binom{2n}{u}(-1)^{u}x_{j}^{u}p_{m}^{2n-u}}}=\sum_{u=0}^{2n}{\sum_{j=1}^{M}{\binom{2n}{u}(-1)^{u}a_{j}x_{j}^{u}p_{m}^{2n-u}}}\cr&=\,\sum_{u=L}^{2n}{\sum_{j=1}^{M}{\binom{2n}{u}(-1)^{u}x_{j}^{u}p_{m}^{2n-u}}}=\sum_{u=L}^{2n}{\binom{2n}{u}\mu_{u}p_{m}^{2n-u}}\,,\cr\end{split}$
(4.1)

and hence

$\begin{split}0&=\,q_{m}^{L}\sum_{j=1}^{M}{a_{j}\sum_{n=0}^{\infty}{\binom{-3/2}{n}\left(\frac{p_{m}-x_{j}}{q_{m}}\right)^{2n}}}=q_{m}^{L}\sum_{n=0}^{\infty}{\binom{-3/2}{n}\sum_{j=1}^{M}{a_{j}\left(\frac{p_{m}-x_{j}}{q_{m}}\right)^{2n}}}\cr&=\,\sum_{n=0}^{\infty}{\binom{-3/2}{n}\sum_{u=L}^{2n}{\binom{2n}{u}\mu_{u}\left(\frac{p_{m}}{q_{m}}\right)^{2n-u}q_{m}^{L-u}}}\cr\end{split}$

for all sufficiently large $m$. Separating the term for which $u=L$, we obtain

$\begin{split}0&=\,\sum_{n=0}^{\infty}{\binom{-3/2}{n}{\binom{2n}{L}\mu_{L}\left(\frac{p_{m}}{q_{m}}\right)^{2n-L}}}+\sum_{n=0}^{\infty}{\binom{-3/2}{n}\sum_{u=L+1}^{2n}{\binom{2n}{u}\mu_{u}\left(\frac{p_{m}}{q_{m}}\right)^{2n-u}q_{m}^{L-u}}}\cr&=\,\frac{\mu_{L}}{L!}\frac{d^{L}}{dz^{L}}\left(\left(1+z^{2}\right)^{-3/2}\right)\Big{|}_{z=p_{m}/q_{m}}+\sum_{u=L+1}^{\infty}{\frac{\mu_{u}}{u!}\frac{d^{u}}{dz^{u}}\left(\left(1+z^{2}\right)^{-3/2}\right)\Big{|}_{z=p_{m}/q_{m}}}q_{m}^{L-u}\cr\end{split}$
(4.2)

for all sufficiently large $m$. Here we interchanged the order of summations
which, is legal as one can easily check that under our conditions the double
sum converges absolutely. See Section 4.6. Observe that

$|\mu_{u}|=\left|\sum_{j=1}^{M}{a_{j}x_{j}^{u}}\right|\leq\left(\sum_{j=1}^{M}{|a_{j}|}\right)x_{M}^{u}=Ax_{M}^{u}\,.$
(4.3)

The Cauchy Integral Formula and $|p_{m}/q_{m}|<1-\delta$ imply that

$\begin{split}\Bigg{|}\frac{1}{u!}\frac{d^{u}}{dz^{u}}\left(\left(1+z^{2}\right)^{-3/2}\right)\Big{|}_{z=p_{m}/q_{m}}\Bigg{|}\leq\frac{\delta}{2}\,\max_{|z|=1-\delta/2}{\Bigg{|}\left(1+z^{2}\right)^{-3/2}\Bigg{|}}\left(\frac{\delta}{2}\right)^{-(u+1)}\leq\left(\frac{2}{\delta}\right)^{u+2}\,.\end{split}$
(4.4)

Observe also that

$|q_{m}|^{L-u}=|q_{m}|^{L-u+1/2}|q_{m}|^{-1/2}\leq|q_{m}|^{-u/(2L+2)}|q_{m}|^{-1/2}\,,\qquad
u\geq L+1\,,\enskip|q_{m}|>1\,.$ (4.5)

Combining 4.3, 4.4, and 4.5, we obtain

$\begin{split}&\,\Bigg{|}\sum_{u=L+1}^{\infty}{\frac{\mu_{u}}{u!}\frac{d^{u}}{dz^{u}}\left(\left(1+z^{2}\right)^{-3/2}\right)\Big{|}_{z=p_{m}/q_{m}}}q_{m}^{L-u}\Bigg{|}\cr\leq&\,\sum_{u=L+1}^{\infty}{|\mu_{u}|\Bigg{|}\frac{1}{u!}\frac{d^{u}}{dz^{u}}\left(\left(1+z^{2}\right)^{-3/2}\right)\Big{|}_{z=p_{m}/q_{m}}\Bigg{|}|q_{m}|^{L-u}}\cr\leq&\,\sum_{u=L+1}^{\infty}{Ax_{M}^{u}\left(\frac{2}{\delta}\right)^{u+2}|q_{m}|^{L-u}}\leq|q_{m}|^{-1/2}\sum_{u=L+1}^{\infty}{Ax_{M}^{u}\left(\frac{2}{\delta}\right)^{u+2}|q_{m}|^{-u/(2L+2)}}\cr\leq&\,|q_{m}|^{-1/2}A\left(\frac{2}{\delta}\right)^{2}\sum_{u=L+1}^{\infty}{\left(\frac{2x_{M}}{\delta|q_{m}|^{1/(2L+2)}}\right)^{u}}\,.\cr\end{split}$
(4.6)

As $|q_{m}|\rightarrow\infty$, 4.6 implies that

$\lim_{|q_{m}|\rightarrow\infty}{\left(\sum_{u=L+1}^{\infty}{\frac{\mu_{u}}{u!}\frac{d^{u}}{dz^{u}}\left(\left(1+z^{2}\right)^{-3/2}\right)\Big{|}_{z=p_{m}/q_{m}}}q_{m}^{L-u}\right)}=0\,.$

Now we are ready to take the limit in 4.2 when $|q_{m}|\rightarrow\infty$. We
obtain

$\frac{\mu_{L}}{L!}\frac{d^{L}}{d\beta^{L}}\left(\frac{1}{(1+\beta^{2})^{3/2}}\right)+0=0\,.$

∎

###  4.3 Type I domain, Possible Asymptotes for $X=0$

######  Proposition 4.2.

Let $\delta\in(0,1/2)$ be fixed. If

$X(p_{m},q_{m})=0,\quad\left|\frac{p_{m}}{q_{m}}\right|<1-\delta,\quad
q_{m}\neq
0,\quad\lim_{|q_{m}|\rightarrow\infty}{\frac{p_{m}}{q_{m}}}=\beta\,,$

then

$\frac{d^{L}}{d\beta^{L}}\left(\frac{\beta}{(1+\beta^{2})^{3/2}}\right)=0\,.$

###### Proof.

We have

$\begin{split}0&=\,q_{m}^{L+1}|q_{m}|X(p_{m},q_{m})=q_{m}^{L-1}\sum_{j=1}^{M}{a_{j}(p_{m}-x_{j})\left(1+\left(\frac{p_{m}-x_{j}}{q_{m}}\right)^{2}\right)^{-3/2}}\cr&=\,q_{m}^{L-1}\sum_{j=1}^{M}{a_{j}\sum_{n=0}^{\infty}{\binom{-3/2}{n}\frac{(p_{m}-x_{j})^{2n+1}}{q_{m}^{2n}}}}\cr\end{split}$

for all sufficiently large $m$. Note that we have adjusted with the factor
$q_{m}^{L+1}|q_{m}|$. Replacing $2n$ by $2n+1$ in 4.1, we also have

$\sum_{j=1}^{M}{a_{j}(p_{m}-x_{j})^{2n+1}}=\sum_{u=L}^{2n+1}{\binom{2n+1}{u}\mu_{u}p_{m}^{2n+1-u}}\,,$

and hence

$\begin{split}0&=\,q_{m}^{L-1}\sum_{j=1}^{M}{a_{j}\sum_{n=0}^{\infty}{\binom{-3/2}{n}\frac{(p_{m}-x_{j})^{2n+1}}{q_{m}^{2n}}}}=q_{m}^{L-1}\sum_{n=0}^{\infty}{\binom{-3/2}{n}\sum_{j=1}^{M}{\frac{(p_{m}-x_{j})^{2n+1}}{q_{m}^{2n}}}}\cr&=\,\sum_{n=0}^{\infty}{\binom{-3/2}{n}\sum_{u=L}^{2n+1}{\binom{2n+1}{u}\mu_{u}\left(\frac{p_{m}}{q_{m}}\right)^{2n+1-u}q_{m}^{L-u}}}\,.\cr\end{split}$

for all sufficiently large $m$. Separating the term for which $u=L$, we obtain

$\begin{split}&\sum_{n=0}^{\infty}{\binom{-3/2}{n}{\binom{2n+1}{L}\mu_{L}\left(\frac{p_{m}}{q_{m}}\right)^{2n+1-L}}}+\sum_{n=0}^{\infty}{\binom{-3/2}{n}\sum_{u=L+1}^{2n+1}{\binom{2n+1}{u}\mu_{u}\left(\frac{p_{m}}{q_{m}}\right)^{2n+1-u}q_{m}^{L-u}}}\cr=&\,\frac{\mu_{L}}{L!}\frac{d^{L}}{dz^{L}}\left(z\left(1+z^{2}\right)^{-3/2}\right)\Big{|}_{z=p_{m}/q_{m}}+\sum_{u=L+1}^{\infty}{\frac{\mu_{u}}{u!}\frac{d^{u}}{dz^{u}}\left(z\left(1+z^{2}\right)^{-3/2}\right)\Big{|}_{z=p_{m}/q_{m}}}q_{m}^{L-u}\cr\end{split}$
(4.7)

vanishes for all sufficiently large $m$. Here we interchanged the order of
summations, which is legitimate as one can easily check that under our
conditions the double sum converges absolutely. See Section 4.6. The Cauchy
Integral Formula and $|p_{m}/q_{m}|\leq 1-\delta$ imply that

$\begin{split}\Bigg{|}\frac{1}{u!}\frac{d^{u}}{dz^{u}}\left(z\left(1+z^{2}\right)^{-3/2}\right)\Big{|}_{z=p_{m}/q_{m}}\Bigg{|}\leq\frac{\delta}{2}\,\max_{|z|=1-\delta/2}{\Bigg{|}\left(1+z^{2}\right)^{-3/2}\Bigg{|}}\left(\frac{\delta}{2}\right)^{-(u+1)}\leq\left(\frac{2}{\delta}\right)^{u+2}\,.\end{split}$
(4.8)

Combining 4.3, 4.8, and 4.5, we obtain

$\begin{split}&\,\Bigg{|}\sum_{u=L+1}^{\infty}{\frac{(-1)^{u}\mu_{u}}{u!}\frac{d^{u}}{dz^{u}}\left(z\left(1+z^{2}\right)^{-3/2}\right)\Big{|}_{z=p_{m}/q_{m}}}q_{m}^{L-u}\Bigg{|}\cr\leq&\,\sum_{u=L+1}^{\infty}{|\mu_{u}|\Bigg{|}\frac{1}{u!}\frac{d^{u}}{dz^{u}}\left(z\left(1+z^{2}\right)^{-3/2}\right)\Big{|}_{z=p_{m}/q_{m}}\Bigg{|}|q_{m}|^{L-u}}\cr\leq&\,\sum_{u=L+1}^{\infty}{Ax_{M}^{u}\left(\frac{2}{\delta}\right)^{u+2}|q_{m}|^{L-u}}\leq|q_{m}|^{-1/2}\sum_{u=L+1}^{\infty}{Ax_{M}^{u}\left(\frac{2}{\delta}\right)^{u+2}|q_{m}|^{-u/(2L+2)}}\cr\leq&\,|q_{m}|^{-1/2}A\left(\frac{2}{\delta}\right)^{2}\sum_{u=L+1}^{\infty}{\left(\frac{2x_{M}}{\delta|q_{m}|^{1/(2L+2)}}\right)^{u}}\,.\cr\end{split}$
(4.9)

As $|q_{m}|\rightarrow\infty$, 4.9 implies that

$\lim_{|q_{m}|\rightarrow\infty}{\left(\sum_{u=L+1}^{\infty}{\frac{\mu_{u}}{u!}\frac{d^{u}}{dz^{u}}\left(z\left(1+z^{2}\right)^{-3/2}\right)\Big{|}_{z=p_{m}/q_{m}}}q_{m}^{L-u}\right)}=0\,.$

Now we are ready to take the limit in 4.7 when $|q_{m}|\rightarrow\infty$. We
obtain

$\frac{\mu_{L}}{L!}\frac{d^{L}}{d\beta^{L}}\left(\frac{\beta}{(1+\beta^{2})^{3/2}}\right)+0=0\,.$

∎

###  4.4 Type II domain, Possible Asymptotes for $Y=0$

######  Proposition 4.3.

Let $\delta\in(0,1/4)$ be fixed. If

$Y(p_{m},q_{m})=0,\quad|p_{m}|>x_{M},\quad q_{m}\neq
0,\quad\left|\frac{q_{m}}{p_{m}}\right|<1-2\delta,\quad\lim_{|p_{m}|\rightarrow\infty}{\frac{q_{m}}{p_{m}}}=\alpha\,,$

then $\alpha\neq 0$ and

$\frac{d^{L}}{d\alpha^{L}}\left(\frac{\alpha^{L+2}}{(1+\alpha^{2})^{3/2}}\right)=0\,.$

###### Proof.

The Binomial Theorem gives

$\frac{1}{(x-x_{j})^{2n+3}}=\,\frac{1}{x^{2n+3}}\left(1-\frac{x_{j}}{x}\right)^{-(2n+3)}=\frac{1}{x^{2n+3}}\sum_{u=0}^{\infty}{\binom{-(2n+3)}{u}\left(\frac{-x_{j}}{x}\right)^{u}}\,,\qquad|x|>x_{M}\,,$

for each $j=1,2,\ldots,M$, and hence

$\begin{split}\sum_{j=1}^{M}{\frac{a_{j}}{(x-x_{j})^{2n+3}}}&=\,\sum_{j=1}^{M}\frac{a_{j}}{x^{2n+3}}\sum_{u=0}^{\infty}{\binom{-(2n+3)}{u}\left(\frac{-x_{j}}{x}\right)^{u}}\cr&=\,\sum_{j=1}^{M}\frac{a_{j}}{x^{2n+3}}\left(\sum_{u=0}^{L}{\binom{-(2n+3)}{u}\left(\frac{-x_{j}}{x}\right)^{u}}+R_{L,n}\left(\frac{-x_{j}}{x}\right)\right)\cr&=\,\mu_{L}\binom{-(2n+3)}{L}x^{-(2n+3)-L}+A_{L,n}(x)\,,\qquad|x|>|x_{M}|\,,\cr\end{split}$
(4.10)

with

$A_{L,n}(x):=x^{-(2n+3)}\sum_{j=1}^{M}{a_{j}R_{L,n}\left(\frac{-x_{j}}{x}\right)}\,,$

where $R_{L,n}(x)$ is the $L$th remainder term in the Taylor series expansion
of the function $f(t):=(1+t)^{-(2n+3)}$ centered at $0$, that is,

$(1+t)^{-(2n+3)}=T_{L,n}(t)+R_{L,n}(t)\,,\qquad|t|<1\,,$

where $T_{L,n}(t)$ it the $L$th Taylor polynomial centered at $0$ associated
with the function $f(t):=(1+t)^{-(2n+3)}$. Estimating by using the Cauchy form
of the remainder term $R_{L,n}(t)$ in the Taylor series expansion of the
function $f(t):=(1+t)^{-(2n+3)}$, we obtain

$|R_{L,n}(t)|\leq\,(L+1)\left|\binom{-(2n+3)}{L+1}\right|(1-|t|)^{-(2n+4)}|t|^{L+1}\,,\qquad|t|<1\,,$

and hence with $A:=\sum_{j=1}^{M}{|a_{j}|}$ we have

$\begin{split}|A_{L,n}(x)|&=\,\left|x^{-(2n+3)}\sum_{j=1}^{M}{a_{j}R_{L,n}\left(\frac{-x_{j}}{x}\right)}\right|\cr&\leq\,AM|x|^{-(2n+3)}(L+1)\left|\binom{-(2n+3)}{L+1}\right|\left(1-\left|\frac{x_{M}}{x}\right|\right)^{-(2n+4)}\left|\frac{x_{M}}{x}\right|^{L+1}\,,\cr&\leq\,AM(L+1)x_{M}^{L+1}\left|\binom{-(2n+3)}{L+1}\right|\left(|x|-x_{M}\right)^{-(2n+4)}|x|^{-L}\,,\qquad|x|>x_{M}\,.\cr\end{split}$
(4.11)

A simple algebra shows that

$\frac{1}{y}Y(x,y)=\sum_{j=1}^{M}{\frac{a_{j}}{((x-x_{j})^{2}+y^{2})^{3/2}}}=\sum_{j=1}^{M}{\frac{a_{j}}{|x-x_{j}|^{3}}\left(1+\left(\frac{y}{x-x_{j}}\right)^{2}\right)^{-3/2}}\,,\quad|x|>x_{M},\enskip
y\neq 0\,.$

Observe that $\displaystyle{\left|\frac{q_{m}}{p_{m}-x_{j}}\right|\leq
1-\delta}$ for all sufficiently large $m$, and hence the Binomial Theorem
gives

$\begin{split}0&=\,\pm\frac{p_{m}^{L+3}}{q_{m}}Y(p_{m},q_{m})=p_{m}^{L+3}\sum_{j=1}^{M}{\frac{a_{j}}{(p_{m}-x_{j})^{3}}\left(1+\left(\frac{q_{m}}{p_{m}-x_{j}}\right)^{2}\right)^{-3/2}}\cr&=\,p_{m}^{L+3}\sum_{j=1}^{M}{\frac{a_{j}}{(p_{m}-x_{j})^{3}}\sum_{n=0}^{\infty}{\binom{-3/2}{n}\left(\frac{q_{m}}{p_{m}-x_{j}}\right)^{2n}}}\cr\end{split}$

for all sufficiently large $m$. Note that we have adjusted with the factor
$p_{m}^{L+3}$. Combining this with 4.10, we obtain

$\begin{split}0=&\,p_{m}^{L+3}\sum_{j=1}^{M}{\frac{a_{j}}{(p_{m}-x_{j})^{3}}\sum_{n=0}^{\infty}{\binom{-3/2}{n}\left(\frac{q_{m}}{p_{m}-x_{j}}\right)^{2n}}}\cr=&\,p_{m}^{L+3}\sum_{n=0}^{\infty}{\sum_{j=1}^{M}{\binom{-3/2}{n}\frac{a_{j}q_{m}^{2n}}{(p_{m}-x_{j})^{2n+3}}}}\cr=&\,\sum_{n=0}^{\infty}{\binom{-3/2}{n}q_{m}^{2n}\left(\mu_{L}\binom{-(2n+3)}{L}p_{m}^{-2n}+p_{m}^{L+3}A_{L,n}(p_{m})\right)}\cr=&\,\sum_{n=0}^{\infty}{\binom{-3/2}{n}\left(\mu_{L}\binom{-(2n+3)}{L}\left(\frac{q_{m}}{p_{m}}\right)^{2n}+q_{m}^{2n}p_{m}^{L+3}A_{L,n}(p_{m})\right)}\end{split}$
(4.12)

for all sufficiently large $m$. It follows from 4.11 that

$\begin{split}&\,\left|\sum_{n=0}^{\infty}{\binom{-3/2}{n}q_{m}^{2n}p_{m}^{L+3}A_{L,n}(p_{m})}\right|\cr\leq&\,AM(L+1)x_{M}^{L+1}\sum_{n=0}^{\infty}{\left|\binom{-3/2}{n}\binom{-(2n+3)}{L+1}\right|\left(\frac{|q_{m}|}{|p_{m}|-x_{M}}\right)^{2n}\left(\frac{|p_{m}|}{|p_{m}|-x_{M}}\right)^{3}\frac{1}{|p_{m}|-x_{M}}}\cr\leq&\,AM(L+1)x_{M}^{L+1}\sum_{n=0}^{\infty}{\left|\binom{-3/2}{n}\binom{-(2n+3)}{L+1}\right|\left(|\alpha|+\delta\right)^{2n}\left(\frac{|p_{m}|}{|p_{m}|-x_{M}}\right)^{3}\frac{1}{|p_{m}|-x_{M}}}\end{split}$

for all sufficiently large $m$, where the right-hand side converges to

$AM(L+1)x_{M}^{L+1}\left(\sum_{n=0}^{\infty}{\left|\binom{-3/2}{n}\binom{-(2n+3)}{L+1}\right|\left(|\alpha|+\delta\right)^{2n}}\right)\left(\lim_{|p_{m}|\rightarrow\infty}{\frac{|p_{m}|^{3}}{(|p_{m}|-x_{M})^{4}}}\right)=0\,.$

Therefore taking the limit in 4.12 as $|p_{m}|\rightarrow\infty$ gives

$\sum_{n=0}^{\infty}{\binom{-3/2}{n}\mu_{L}\binom{-(2n+3)}{L}\alpha^{2n}}=0\,.$

That is,

$\sum_{n=0}^{\infty}{\binom{-3/2}{n}\mu_{L}\binom{-(2n+L+2)}{L}\alpha^{2n}}=0\,.$

As $|\alpha|\leq 1-\delta$ we conclude that $\alpha\neq 0$ and

$\frac{\mu_{L}}{L!}\frac{d^{L}}{d\alpha^{L}}\left(\frac{\alpha^{L+2}}{(1+\alpha^{2})^{3/2}}\right)=0\,.$

∎

###  4.5 Type II domain, Possible Asymptotes for $X=0$

######  Proposition 4.4.

Let $\delta\in(0,1/4)$ be fixed. If

$Y(p_{m},q_{m})=0,\quad|p_{m}|>x_{M},\quad\left|\frac{q_{m}}{p_{m}}\right|<1-2\delta,\quad\lim_{|p_{m}|\rightarrow\infty}{\frac{q_{m}}{p_{m}}}=\alpha\,,$

then $\alpha\neq 0$ and

$\frac{d^{L}}{d\alpha^{L}}\left(\frac{\alpha^{L+1}}{(1+\alpha^{2})^{3/2}}\right)=0\,.$

###### Proof.

Replacing $2n+3$ by $2n+2$ in 4.10, we obtain

$\sum_{j=1}^{M}{\frac{a_{j}}{(x-x_{j})^{2n+2}}}=\mu_{L}\binom{-(2n+2)}{L}x^{-(2n+3)-L}+A_{L,n}(x)\,,\qquad|x|>|x_{M}|\,,$
(4.13)

with

$A_{L,n}(x):=x^{-(2n+2)}\sum_{j=1}^{M}{a_{j}R_{L,n}\left(\frac{-x_{j}}{x}\right)}\,,$

where $R_{L,n}(x)$ is the $L$th remainder term in the Taylor series expansion
of the function $f(t):=(1+t)^{-(2n+2)}$ centered at $0$, that is,

$(1+t)^{-(2n+2)}=T_{L,n}(t)+R_{L,n}(t)\,,\qquad|t|<1\,,$

where $T_{L,n}(t)$ it the $L$th Taylor polynomial centered at $0$ associated
with the function $f(t):=(1+t)^{-(2n+2)}$. Estimating by using the Cauchy form
of the remainder term $R_{L,n}(t)$ in the Taylor series expansion of the
function $f(t):=(1+t)^{-(2n+3)}$, we obtain

$|R_{L,n}(t)|\leq\,(L+1)\left|\binom{-(2n+2)}{L+1}\right|(1-|t|)^{-(2n+3)}|t|^{L+1}\,,\qquad|t|<1\,,$

and hence, replacing $2n+3$ with $2n+2$ in 4.11, we have

$|A_{L,n}(x)|\leq
AM(L+1)x_{M}^{L+1}\left|\binom{-(2n+2)}{L+1}\right|\left(|x|-x_{M}\right)^{-(2n+3)}|x|^{-L}\,,\qquad|x|>x_{M}\,.$
(4.14)

A simple algebra shows that

$X(x,y)=\sum_{j=1}^{M}{\frac{a_{j}(x-x_{j})}{((x-x_{j})^{2}+y^{2})^{3/2}}}=\sum_{j=1}^{M}{\frac{a_{j}(x-x_{j})}{|x-x_{j}|^{3}}\left(1+\left(\frac{y}{x-x_{j}}\right)^{2}\right)^{-3/2}}\,,\quad|x|>x_{M},\enskip
y\neq 0\,.$

Observe that

$\left|\frac{q_{m}}{p_{m}-x_{j}}\right|\leq 1-\delta$

for all sufficiently large $m$, and hence the Binomial Theorem gives

$\begin{split}0&=\,\pm
p_{m}^{L+2}X(p_{m},q_{m})=p_{m}^{L+2}\sum_{j=1}^{M}{\frac{a_{j}(p_{m}-x_{j})}{(p_{m}-x_{j})^{3}}\left(1+\left(\frac{q_{m}}{p_{m}-x_{j}}\right)^{2}\right)^{-3/2}}\cr&=\,p_{m}^{L+2}\sum_{j=1}^{M}{\frac{a_{j}}{(p_{m}-x_{j})^{2}}\sum_{n=0}^{\infty}{\binom{-3/2}{n}\left(\frac{q_{m}}{p_{m}-x_{j}}\right)^{2n}}}\cr\end{split}$

for all sufficiently large $m$. Note that we have adjusted with the factor
$p_{m}^{L+2}$. Hence, similarly to 4.12, we get

$\begin{split}0&=\,p_{m}^{L+2}\sum_{j=1}^{M}{\frac{a_{j}}{(p_{m}-x_{j})^{2}}\sum_{n=0}^{\infty}{\binom{-3/2}{n}\left(\frac{q_{m}}{p_{m}-x_{j}}\right)^{2n}}}\cr&=\,\sum_{n=0}^{\infty}{\binom{-3/2}{n}\left(\mu_{L}\binom{-(2n+2)}{L}\left(\frac{q_{m}}{p_{m}}\right)^{2n}+q_{m}^{2n}p_{m}^{L+2}A_{L,n}(p_{m})\right)}\end{split}$
(4.15)

for all sufficiently large $m$. It follows from 4.14 that

$\begin{split}&\left|\sum_{n=0}^{\infty}{\binom{-3/2}{n}q_{m}^{2n}p_{m}^{L+2}A_{L,n}(p_{m})}\right|\cr\leq&AM(L+1)x_{M}^{L+1}\sum_{n=0}^{\infty}{\left|\binom{-3/2}{n}\binom{-(2n+2)}{L+1}\right|\left(\frac{|q_{m}|}{|p_{m}|-x_{M}}\right)^{2n}\left(\frac{|p_{m}|}{|p_{m}|-x_{M}}\right)^{2}\frac{1}{|p_{m}|-x_{M}}}\cr\leq&AM(L+1)x_{M}^{L+1}\sum_{n=0}^{\infty}{\left|\binom{-3/2}{n}\binom{-(2n+2)}{L+1}\right|\left(|\alpha|+\delta\right)^{2n}}\left(\frac{|p_{m}|}{|p_{m}|-x_{M}}\right)^{2}\frac{1}{|p_{m}|-x_{M}}\cr\end{split}$

for all sufficiently large $m$, where the right-hand side converges to

$AM(L+1)x_{M}^{L+1}\left(\sum_{n=0}^{\infty}{\left|\binom{-3/2}{n}\binom{-(2n+2)}{L+1}\right|\left(|\alpha|+\delta\right)^{2n}}\right)\left(\lim_{|p_{m}|\rightarrow\infty}{\frac{|p_{m}|^{2}}{(|p_{m}|-x_{M})^{3}}}\right)=0\,.$

Therefore taking the limit in 4.15 as $|p_{m}|\rightarrow\infty$ gives

$\sum_{n=0}^{\infty}{\binom{-3/2}{n}\mu_{L}\binom{-(2n+2)}{L}\alpha^{2n}}=0\,.$

That is,

$\sum_{n=0}^{\infty}{\binom{-3/2}{n}\mu_{L}\binom{2n+L+1}{L}\alpha^{2n}}=0\,.$

As $|\alpha|\leq 1-2\delta$ we conclude that $\alpha\neq 0$ and

$\frac{\mu_{L}}{L!}\frac{d^{L}}{d\alpha^{L}}\left(\frac{\alpha^{L+1}}{(1+\alpha^{2})^{3/2}}\right)=0\,.$

∎

‘

###  4.6 Interchanging the Order of Summation

In the analysis in Section 4.2 to find the possible asymptotes for $\\{Y=0\\}$
in the Type I domain we interchanged the order of summations in

$\sum_{n=0}^{\infty}{\binom{-3/2}{n}{\sum_{u=L}^{2n}{\binom{2n}{u}\mu_{u}\left(\frac{p_{m}}{q_{m}}\right)^{2n-u}q_{m}^{L-u}}}}=\sum_{u=L}^{\infty}{\sum_{n=0}^{\infty}{\binom{2n}{u}\mu_{u}\left(\frac{p_{m}}{q_{m}}\right)^{2n-u}q_{m}^{L-u}}}$

for all sufficiently large $m$, or equivalently

$\sum_{n=0}^{\infty}{\binom{-3/2}{n}{\sum_{u=L}^{2n}{\binom{2n}{u}\mu_{u}\left(\frac{p_{m}}{q_{m}}\right)^{2n-u}q_{m}^{-u}}}}=\sum_{u=L}^{\infty}{\sum_{n=0}^{\infty}{\binom{2n}{u}\mu_{u}\left(\frac{p_{m}}{q_{m}}\right)^{2n-u}q_{m}^{-u}}}$

for all sufficiently large $m$. To see that this is legitimate, recall that
$\mu_{u}:=(-1)^{u}\sum_{j=1}^{u}{a_{j}x_{j}^{u}}$ and $\mu_{u}=0$ for each
$u=0,1,\ldots,L-1$, and hence we have

$\begin{split}&\,\sum_{n=0}^{\infty}{\sum_{u=L}^{2n}{\left|\binom{-3/2}{n}\binom{2n}{u}\mu_{u}\left(\frac{p_{m}}{q_{m}}\right)^{2n-u}q_{m}^{-u}\right|}}=\sum_{n=0}^{\infty}{\sum_{u=0}^{2n}{\left|\binom{-3/2}{n}\binom{2n}{u}\mu_{u}\left(\frac{p_{m}}{q_{m}}\right)^{2n-u}q_{m}^{-u}\right|}}\cr=\,&\sum_{n=0}^{\infty}{\sum_{u=0}^{2n}{\left|\binom{-3/2}{n}\binom{2n}{u}\sum_{j=1}^{M}{a_{j}(-x_{j})^{u}}\left(\frac{p_{m}}{q_{m}}\right)^{2n-u}q_{m}^{-u}\right|}}\leq\sum_{j=1}^{M}{\sum_{n=0}^{\infty}{|a_{j}|\left|\binom{-3/2}{n}\right|\left(\frac{|p_{m}|+x_{j}}{|q_{m}|}\right)^{2n}}}\cr\leq\,&\sum_{j=1}^{M}{\sum_{n=0}^{\infty}{|a_{j}|\left|\binom{-3/2}{n}\right|\left(|\beta|+\delta/2\right)^{2n}}}<\infty\cr\end{split}$

for all sufficiently large $m$, as $|\beta|\leq 1-\delta$ and the power series
$\sum_{n=0}^{\infty}{\binom{-3/2}{n}z^{2n}}$ converges absolutely for all
$z\in(-1,1)$.

A similar argument shows that interchanging the order of summation in our work
in Section 4.2 to find the possible asymptotes for $\\{X=0\\}$ in the Type I
domain is legitimate.

###  4.7 Interlacing properties of zeros of polynomials I

We now address how we can conclude that the asymptotic directions of the zero
sets of $Y$ and $X$ are distinct. Proving this would show that there are no
zeros for $F=(X,Y)$ outside some large disc, in the Special Case. From the
analysis in Section 3, we already know this because the zero set is finite.
But our argument may have independent value because it gives interlacing
properties for a class of functions that has not previously been considered.
In addition, this interlacing is needing for applications of the General Case
that we will address later. Moreover, this interlacing sets up a context in
which some of the challenges in the General Case can be understood better. We
thank Vilmos Totik who early on gave us methods for getting results like the
ones in this section.

We define

$A_{0}(x):=\frac{x}{(1+x^{2})^{3/2}}\,,\qquad
B_{0}(x):=\frac{1}{(1+x^{2})^{3/2}}\,,$
$A_{L}(x):=\frac{d^{L}A_{0}}{dx^{L}}\,,\qquad
B_{L}(x):=\frac{d^{L}B_{0}}{dx^{L}}\,,\qquad L=1,2,\ldots\,.$

It is easy to see that there are polynomials $P_{L}$ of degree $L+1$ and
$Q_{L}$ of degree $L$ such that

$A_{L}(x)=\frac{P_{L}(x)}{(1+x^{2})^{(2L+3)/2}}=\frac{(1+x^{2})P_{L-1}^{\prime}(x)-(2L+1)xP_{L-1}(x)}{(1+x^{2})^{(2L+3)/2}}$

and

$B_{L}(x)=\frac{Q_{L}(x)}{(1+x^{2})^{(2L+3)/2}}=\frac{(1+x^{2})Q_{L-1}^{\prime}(x)-(2L+1)xQ_{L-1}(x)}{(1+x^{2})^{(2L+3)/2}}\,.$

As the degree $L+1$ of the polynomial $P_{L}$ and the degree $L$ of the
polynomial $Q_{L}$ are less than $(2L+3)/2$ we have

$\lim_{x\rightarrow\pm\infty}{\frac{P_{L}(x)}{(1+x^{2})^{(L+3)/2}}}=\lim_{x\rightarrow\pm\infty}{\frac{Q_{L}(x)}{(1+x^{2})^{(L+3)/2}}}=0$
(4.16)

######  Proposition 4.5.

Let $L\geq 1$ be an integer.

(a) The polynomial $P_{L}$ of degree $L+1$ has $L+1$ distinct real zeros, and
the zeros of $P_{L}$ and $P_{L-1}$ strictly interlace.

(b) The polynomial $Q_{L}$ of degree $L$ has $L$ distinct real zeros, and the
zeros of $Q_{L}$ and $Q_{L-1}$ strictly interlace.

###### Proof.

We prove only (a), the proof of (b) is identical. The proof is a simple
induction on $L$. The statements is obviously true for $L=1$. Assume that
statement (a) is true for some integer $L-1\geq 1$. Note that the zeros of
$P_{L-1}$ and $A_{L-1}$ are the same. Let us denote the distinct real zeros of
$P_{L-1}$ and $A_{L}$ by $x_{1}<x_{2}<\cdots<x_{L}$. The limit relations 4.16
and Rolle’s theorem imply that $A_{L+1}=A_{L}^{\prime}$ has zeros
$y_{j}\in(x_{j},x_{j+1})$, $j=0,1,\ldots,L$, where $x_{0}:=-\infty$ and
$x_{L+1}:=\infty$. However, the zeros of $A_{L}$ are the same as the zeros of
$P_{L}$, which means that statement (a) is true for $L$. ∎

######  Proposition 4.6.

Let $L\geq 1$ be an integer.

(a) The zeros of zeros of $P_{L}$ and $P_{L}^{\prime}$ strictly interlace.

(b) The zeros of $Q_{L}$ and $Q_{L}^{\prime}$ strictly interlace.

###### Proof.

Suppose a polynomial $R$ of degree $L$ has $L$ distinct zeros
$x_{1}<x_{2}<\cdots<x_{L}$. Then Rolle’s Theorem implies that $R^{\prime}$ has
zeros $y_{j}\in(x_{j},x_{j+1})$, $j=1,2,\ldots,L-1$. However, the degree of
$R^{\prime}$ is $L-1$, so $R^{\prime}$ has $L-1$ distinct zeros. ∎

######  Proposition 4.7.

Let $L\geq 1$ be an integer. The zeros of $A_{L}$ and $B_{L}$ strictly
interlace. Equivalently, the zeros of $P_{L}$ and $Q_{L}$ strictly interlace.

###### Proof.

Observe that

$A_{L}(\beta)=\frac{d^{L}A_{0}}{d\beta^{L}}\qquad\text{and}\qquad
A_{0}(\beta)=\beta B_{0}(\beta)\,,$

so using the Leibniz formula we obtain

$A_{L}(\beta)=LB_{L-1}(\beta)+\beta B_{L}(\beta)\,,$

or equivalently

$P_{L}(\beta)=LQ_{L-1}(\beta)(1+\beta^{2})+\beta Q_{L}(\beta)$ (4.17)

The fact that $P_{L}$ and $Q_{L}$ have no common zeros already follows from it
simply, as any such common zero of $P_{L}$ and $Q_{L}$ is a common zero of
$Q_{L}$ and $Q_{L-1}$ which is impossible by Proposition 1, as the zeros of
$Q_{L}$ and $Q_{L}^{\prime}$ strictly interlace. It is somewhat more subtle to
see that the zeros $P_{L}$ and the zeros of $Q_{L}$ interlace Denoting the
zeros of $Q_{L}$ by $\beta_{1}<\beta_{2}<\cdots<\beta_{L}$, we can deduce from
4.17 that

$P_{L}(\beta_{j-1})P_{L}(\beta_{j})<0\,,\qquad j=2,3,\ldots,L\,,$

since by Proposition 1.1 we already know that $Q_{L-1}$ has exactly one zero
in $(\beta_{j-1},\beta_{j})$. Hence $P_{L}$ has at least one zero in each of
the intervals $(\beta_{j-1},\beta_{j})\,,\enskip j=2,3,\ldots,L\,.$ Also, it
is easy to see by induction that the sign of the leading coefficient of
$Q_{L}$ is $(-1)^{L}$, hence

$\lim_{\beta\rightarrow\pm\infty}{\frac{Q_{L}(\beta)}{\beta^{L}}}=(-1)^{L}\,.$
(4.18)

It follows from 4.17 and 4.18 that $P_{L}$ has a sign change, and hence at
least one zero in $(\beta_{L},\infty)$, and $P_{L}$ has a sign change, and
hence at least one zero in $(-\infty,\beta_{1})$. Finally recall that the
degree of $P_{L}$ is $L+1$ and the degree of $Q_{L}$ is $L$. In conclusion,
there is exactly one zero of $Q_{L}$ strictly between any two consecutive real
zeros of $P_{L}$. ∎

###  4.8 The Inversion Formula and Interlacing Properties of Polynomials

There is an interesting connection between the values of $\beta$ in the Type I
domains, and the values of $\alpha$ in the Type II domains. This connection is
not at first evident because of constraints that need to be made in order that
we have convergence of the power series that we consider. But we can observe
this now.

Let $L\geq 0$ be an integer. We define

$\begin{split}A_{0}(x)&:=\,\frac{x}{(x^{2}+1)^{3/2}}\,,\qquad
B_{0}(x):=\,\frac{1}{(x^{2}+1)^{3/2}}\,,\cr
A_{L}(x)&:=\,\frac{d^{L}}{dx^{L}}\frac{x}{(x^{2}+1)^{3/2}}\,,\qquad
B_{L}(x):=\,\frac{d^{L}}{dx^{L}}\frac{1}{(x^{2}+1)^{3/2}}\,,\cr
C_{L}(x)&:=\,\frac{d^{L}}{dx^{L}}\frac{x^{L+2}}{(x^{2}+1)^{3/2}}\,,\qquad
D_{L}(x):=\,\frac{d^{L}}{dx^{L}}\frac{x^{L+1}}{(x^{2}+1)^{3/2}}\,.\cr\end{split}$

######  Proposition 4.8.

(Inversion Formula) We have

$\text{\rm sgn}(x)x^{L+1}B_{L}(x)=(-1)^{L}C_{L}(1/x)\,,\qquad
x\in\mathbb{R}\setminus\\{0\\}\,,\qquad L=0,1,\ldots\,.$

###### Proof.

As both sides are even functions of $x\in\mathbb{R}\setminus\\{0\\}$, without
loss of generality we may assume that $x>0$. For an integer $L\geq 0$ we
define

$\widetilde{B}_{L}(x):=\,x^{L+1}B_{L}(x)\,,\qquad
C_{L}^{*}(x):=\,C_{L}(1/x)\,,\qquad\widetilde{C}_{L}(x):=\,(-1)^{L}C_{L}^{*}(x)\,.$

Observe that for $L\geq 1$ we have

$\begin{split}\widetilde{B}_{L}(x)&=\,x^{L+1}B_{L}(x)=xx^{L}B_{L-1}^{\prime}(x)\cr&=\,x(x^{L}B_{L-1}^{\prime}(x))+x(Lx^{L-1}B_{L-1}(x))-Lx^{L}B_{L-1}(x)\cr&=\,x\widetilde{B}_{L-1}^{\prime}(x)-L\widetilde{B}_{L-1}(x)\,,\cr\end{split}$

that is, the functions $\widetilde{B}_{L}$ satisfy the recursion

$\widetilde{B}_{0}(x)=\frac{x}{(1+x^{2})^{3/2}}\,,$ (4.19)
$\widetilde{B}_{L}(x)=x\widetilde{B}_{L-1}^{\prime}(x)-L\widetilde{B}_{L-1}(x)\,,\qquad
L=1,2,\ldots\,.$ (4.20)

Observe that

$\widetilde{C}_{0}(x)=\frac{x}{(x^{2}+1)^{3/2}}\,,$ (4.21)

and the Leibniz formula yields

$C_{L}(x)=\frac{d^{L}}{dx^{L}}\frac{x^{L+1}x}{(x^{2}+1)^{3/2}}=xC_{L-1}^{\prime}(x)+LC_{L-1}(x)\,,\qquad
L=1,2,\ldots\,.$ (4.22)

Replacing $x$ by $1/x$ we get

$C_{L}(1/x)=(1/x)C_{L-1}^{\prime}(1/x)+LC_{L-1}(1/x)\,,\qquad L=1,2,\ldots\,,$

that is,

$C_{L}^{*}(x)=-xC_{L-1}^{*\prime}(x)+LC_{L-1}^{*}(x)\,,\qquad L=1,2,\ldots\,,$

and hence

$\widetilde{C}_{L}(x)=x\widetilde{C}_{L-1}^{\prime}(x)-L\widetilde{C}_{L-1}(x)\,,\qquad
L=1,2,\ldots\,.$ (4.23)

Now observe that by 4.19, 4.20, 4.21, and 4.23 the functions
$\widetilde{B}_{L}$ and $\widetilde{C}_{L}$ satisfy the same recursion. In
conclusion

$\widetilde{B}_{L}(x)=\widetilde{C}_{L}(x)\,,\qquad L=0,1,\ldots\,,$

and the lemma follows. ∎

######  Proposition 4.9.

Let $L\geq 0$ be an integer. The functions $C_{L}$ and $D_{L}$ do not have a
common zero different from $0$.

###### Proof.

The statement is obvious for $L=0$. Let $L\geq 1$ be an integer. Observe that

$D_{L}(x)=C_{L-1}^{\prime}(x)\,,$

and hence 4.22 can be written as

$C_{L}(x)=xD_{L}(x)+LC_{L-1}(x)\,,\qquad L=1,2,\ldots\,.$

Therefore if $\alpha\neq 0$ is a common zero of $C_{L}$ and $D_{L}$, then
$\alpha\neq 0$ is a common zero of $C_{L}$ and $C_{L-1}$, and hence
$\alpha\neq 0$ is a common zero of $C_{L-1}$ and $C_{L-1}^{\prime}$. By the
Inversion Formula this means that $a$ is a common zero of

$\text{\rm sgn}(x)\widetilde{B}_{L-1}(1/x)=x^{-L}B_{L-1}(1/x)$

and

$\frac{d}{dx}\widetilde{B}_{L-1}(1/x)=-x^{-L-2}B_{L-1}^{\prime}(1/x)+(-L)x^{-L-1}B_{L-1}(1/x)\,.$

This means that $1/\alpha$ is a common zero of $B_{L-1}$ and $B_{L}$. However,
this is impossible. Indeed, as as we have seen before, the Leibniz formula
implies that

$A_{L}(x)=LB_{L-1}(x)+xB_{L}(x)\,,$

and hence any common zero of $B_{L-1}$ and $B_{L}$ is a common zero of $A_{L}$
and $B_{L}$. However, the zeros of $A_{L}$ and $B_{L}$ strictly interlace by
Proposition 1.3 of the previous section. ∎

###  4.9 The Boundary Case $y=\pm x$

In a forthcoming paper we may be able to prove that $y=\pm x$ cannot be a
common asymptotic direction to both of the zero sets $\\{X=0\\}$ and
$\\{Y=0\\}$. Originally we needed this to argue that the zero set
$\\{X=Y=0\\}$ is bounded. However, Proposition 3.4 already implies that in the
Special Case the zero set $\\{X=Y=0\\}$ is bounded, as it is a finite set of
points.

###  4.10 Final Conclusion

Combining the results in Section 4.2 through Section 4.9, we conclude that in
the Special Case the set of possible asymptotic directions of the zero set
$\\{(x,y):X(x,y)=0\\}$ and the set of possible asymptotic directions of the
zero set $\\{(x,y):Y(x,y)=0\\}$ are distinct, except possibly the lines $y=\pm
x$. Whether or not the lines $y=\pm x$ can be a common asymptotic direction to
both of the zero sets $\\{X=0\\}$ and $\\{Y=0\\}$ remains open in this paper.

Acknowledgments: We thank Bruce Reznick for showing us the product method for
eliminating square roots in our field component equations in Section 2 and
Section 3. We thank Vilmos Totik for helpful remarks about the interlacing of
zeros of in our formulas for asymptotic directions in Section 4.

## References

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Authors:

T. Erdélyi, Department of Mathematics, Texas A & M University, College
Station, TX,

<EMAIL_ADDRESS>
J. Rosenblatt, Department of Mathematics, University of Illinois at Urbana-
Champaign, Urbana, IL,

<EMAIL_ADDRESS>
R. Rosenblatt, Education and Human Resources at the National Science
Foundation, Alexandria, VA,

<EMAIL_ADDRESS>
Date: March, 2021