TITLE: How can we derive the polarity of the induced voltage in an inductor?
QUESTION [1 upvotes]: As we know, the voltage-current relation of a two-terminal constant-inductance inductor is:
$v_L(t) = L \, \dfrac{\mathrm di_L(t)}{\mathrm dt} \tag 1$
I've managed to prove that equation without considering the signs nor the polarity and direction of the voltage and current. However, in many textbooks on circuit analysis and physics that equation is valid for any sign of the voltage, current, and rate of change of current, as long as the reference polarity of the voltage and the reference direction of the conventional current are chosen as shown in the following figure.

Figure 1. Reference direction for the current and reference polarity for the voltage in an inductor. Image source: Engineering Circuit Analysis (8th ed.) by Hayt, Kemmerly and Durbin.
So I'm trying to prove if equation (1) indeed considers the signs and the reference polarity and reference direction of the voltage, current, and rate of change of current. Specifically, as you'll see below, I have troubles while determining the polarity of the induced EMF or voltage (such that its numerical value is positive), given the current and its rate of change.
Basics
Before showing my attempt, I'll briefly mention two basic facts I'll use: the right-hand rule and the direction of the induced magnetic field.
Right-hand rule
As we know there are many right-hand rules used in electromagnetism (for Lorentz's force law, for Ampère's law, Flemming's right-hand rule, etc.). The one I'm going to use is the one used to determine the direction of the magnetic field induced by a positive conventional current through a cylindrical inductor/solenoid/coil, as shown in the following figure.

Figure 2. The right-hand rule #1: for the magnetic field induced by a positive conventional current in a cylindrical inductor. Image source: link.
The previous rule can be used to determine the direction of the magnetic field if given the direction of the current, or the vice versa.
Change in the magnetic field and the direction of the induced magnetic field
As we know, Faraday's law is based on the change in the magnetic flux/field, and Lenz's law says the induced EMF will be such that it opposes the change in the magnetic flux/field, or in other words the induced magnetic field is the negative of the change in the original magnetic field. The change in some quantity is defined as the final value minus the initial value. And to find the difference $\mathbf u − \mathbf v$ of two vectors geometrically, first we find the negative $− \mathbf v$ of the second vector $\mathbf v$, then move it such that its tail is at the head of the first vector $\mathbf u$, and then the difference of the two original vectors is the sum $\mathbf u + (− \mathbf v)$, which is the vector that points form the tail of the first vector $\mathbf u$ to the head of the vector $− \mathbf v$. With these facts we can determine the direction of the induced magnetic field (the one that opposes the original magnetic field), as shown below.

Figure 3. Lenz’s law: direction of induced magnetic field due to an increasing magnetic field. Image source: own.

Figure 4. Lenz’s law: direction of induced magnetic field due to a decreasing magnetic field. Image source: own.
My attempt
We’re going to use the following notation: $i(t)$ will be the conventional current through the inductor imposed by the external active circuit connected to the terminals of the inductor, whose direction and numerical sign will be assumed as known/given; $\mathbf B(t)$ will be the magnetic field produced by the current $i$ as described by Ampère’s law, whose direction is described by the RHR (figure 2); $v_\text{ind}(t)$, $i_\text{ind}(t)$ and $\mathbf B_\text{ind}(t)$ will be the induced voltage/emf, induced conventional current and induced magnetic fields, respectively, by the change in the magnetic field $\mathbf B$ as described by Faraday’s law, and whose polarities/directions are described by Lenz’s law (figures 3 and 4) and the RHR (figure 2); $v_L(t)$ and $i_L(t)$ will be the voltage across and current through the inductor respectively such that their reference polarity and reference direction are as shown in figure 1.
I think many of us will agree that an intuitive way to think about the order in which the "events" occur is as follows: the current $i$ creates a changing magnetic field $\mathbf B$ that induces a voltage $v_\text{ind}$, which produces a current $i_\text{ind}$, which produces a magnetic field $\mathbf B_\text{ind}$. So, in my attempt I'll proceed as follows: given $i$ and its rate of change, I find the $\mathbf B$, then I find the change in $\mathbf B$, then I find $\mathbf B_\text{ind}$, then I find $i_\text{ind}$, and then I find $v_\text{ind}$. As you'll see below, I managed to find all of those quantities and understand why they had the direction they had, except $v_\text{ind}$.
Now, I'll explain my attempt.
Consider a two-terminal constant-inductance cylindrical inductor. Let’s consider four different scenarios: when the positive conventional current is flowing in one direction and then in the other direction, and when the positive conventional current is increasing and then decreasing. The inductor is connected to an external active circuit that establishes a time-varying current $i(t)$ in each of the four cases. We assume such current is conventional, is positive, we know its direction and we know whether it is increasing or decreasing. This is shown in the following figure.

Figure 5. Image source: own.
Since there is current flowing through the inductor, from Ampère’s law a magnetic field $\mathbf B(t)$ is produced due to that current $i$, whose direction we can determine applying the RHR (figure 2) to each of the four cases; to do this, we assume we grab the inductor with our right hand, such that our fingers point in the direction the current $i$ is flowing throughout the inductor, then our thumb points in the direction of the magnetic field $\mathbf B$ inside the inductor. Applying this rule to each of the four cases of the previous figure:

In cases a) and b), when we look at the inductor as shown in the previous figure (and not, say, from the top or from the bottom or from the rear side), we see the current $i$ is flowing from left to right, so we grab the inductor with our right hand such that our fingers point to the right. Then our thumb will point upwards, so that’s the direction the magnetic field $\mathbf B$ points inside the inductor.

In cases c) and d), when we look at the inductor as shown in the previous figure, we see the current $i$ is flowing from right to left, so we grab the inductor with our right hand such that our fingers point to the left. Then our thumb will point downwards, so that’s the direction the magnetic field $\mathbf B$ points inside the inductor.

Thus we get the direction of the magnetic field shown in the following figure.

Figure 6. Image source: own.
Because we assume the current $i$ is time-varying, the magnetic field $\mathbf B$ produced by it is also time-varying, so the magnetic flux $\Phi(t)$ is also time-varying, so by Faraday’s law an EMF $v_\text{ind}(t)$ is induced across the inductor. To find the polarity of such EMF or voltage, we first find the direction of the current $i_\text{ind}(t)$ such EMF tries to establish, which we determine using Lenz’s law. So while our goal is to find the polarity of the induced EMF, we first find the direction of the induced magnetic field, secondly we find the direction of the induced current that produced such induced magnetic field, and thirdly we find the polarity of the induced EMF produced by the original changing magnetic field. First, we find the change $\mathrm d \mathbf B$ in the magnetic field.

In case a), the magnetic field points upwards inside the inductor, and the current is increasing so the magnetic field is also increasing, to the change points upwards (figure 3).

In case b), the magnetic field points upwards inside the inductor, and the current is decreasing so the magnetic field is also decreasing, to the change points downwards (figure 4).

In case c), the magnetic field points downwards inside the inductor, and the current is increasing so the magnetic field is also increasing, to the change points downwards (figure 3).

In case d), the magnetic field points downwards inside the inductor, and the current is decreasing so the magnetic field is also decreasing, to the change points upwards (figure 4).

Thus we get the direction of the change in the magnetic field shown in the following figure.

Figure 7. Image source: own.
The voltage $v_\text{ind}$ induced across the inductor according to Faraday’s law due to the time-varying magnetic field $\mathbf B$ will try to establish a current $i_\text{ind}$, that from Ampère’s law would produce a magnetic field $\mathbf B_\text{ind}(t)$ whose direction according to Lenz’s law will try to oppose the change $\mathrm d \mathbf B$ in the original magnetic field $\mathbf B$. In order for $\mathbf B_\text{ind}$ to oppose $\mathbf B$, the former must point in the opposite direction of the latter.

In case a), the change in the original magnetic field points upwards, so the induced magnetic field points downwards.

In case b), the change in the original magnetic field points downwards, so the induced magnetic field points upwards.

In case c), the change in the original magnetic field points downwards, so the induced magnetic field points upwards.

In case d), the change in the original magnetic field points upwards, so the induced magnetic field points downwards.

Thus we get the direction of the induced magnetic field shown in the following figure.

Figure 8. Image source: own.
Secondly we find the direction of the induced current, by applying the RHR (figure 2), except that now, unlike figure 5, we know the direction of the magnetic field and we want to find the direction of the current. To apply this rule, we assume we grab the inductor with our right hand, such that our thumb points in the direction of the induced magnetic field $\mathbf B_\text{ind}$ inside the inductor, then our fingers point in the direction the induced current $i_\text{ind}$ is flowing throughout the inductor. Applying this rule to each of the four cases of the previous figure:

In cases a) and d), when we look at the inductor as shown in the previous figure, we see the induced magnetic field $\mathbf B_\text{ind}$ is flowing downwards inside the inductor, so we grab the inductor with our right hand such that our thumb points downwards. Then our fingers will point to the left, so that’s the direction the induced current $i_\text{ind}$ is flowing throughout the inductor.

In cases b) and c), when we look at the inductor as shown in the previous figure, we see the induced magnetic field $\mathbf B_\text{ind}$ is flowing upwards inside the inductor, so we grab the inductor with our right hand such that our thumb points upwards. Then our fingers will point to the right, so that’s the direction the induced current $i_\text{ind}$ is flowing throughout the inductor.

Thus we get the direction of the induced current shown in the following figure.

Figure 9. Image source: own.
So far so good; I don't have any doubts. The only step missing is to find the polarity of the induced voltage $v_\text{ind}$. I don't know how to do this, this is where I'm stuck.
However, I realized/observed something. It looks like we must assign the reference polarity of $v_\text{ind}$ such that the induced current $i_\text{ind}$ flows from higher potential (the terminal marked with the “$+$” sign of $v_\text{ind}$) to lower potential (the terminal marked with the “$-$” sign of $v_\text{ind}$) through the external active circuit. (Equivalently, the induced current $i_\text{ind}$ flows from lower potential [the terminal marked with the “$-$” sign of $v_\text{ind}$] to higher potential [the terminal marked with the “$+$” sign of $v_\text{ind}$] through the inductor.) But I don't know why it is correct, and I'd like to know it.
You may ask how I'm sure my observation above is correct. The reason is because if we assume it is correct, then, as I'll show below, equation (1) indeed considers the sign of $v$ and the sign of the rate of change of $i$.
So let's suppose the above observation is correct. Then, in the previous figure:

In cases a) and d), the induced current $i_\text{ind}$ flows into the active circuit through the upper terminal, so the induced voltage $i_\text{ind}$ has the positive reference polarity at the upper terminal.

In cases b) and c), the induced current $i_\text{ind}$ flows out of the active circuit through the upper terminal, so it flows into the active circuit through the lower terminal, so the induced voltage $i_\text{ind}$ has the positive reference polarity at the lower terminal.

In this way we get the polarity of the induced voltage shown in the following figure.

Figure 10. Image source: own.
Now that we’ve obtained the polarity of the induced voltage $v_\text{ind}$ (such that its numerical value is positive) and we know the direction of the current $i$ (such that its numerical value is positive) as well as the sign of the rate of change of $i$, let’s get rid of the irrelevant details in the previous figure (magnetic fields, induced current). Also, let’s change the reference direction of the current (which also changes its sign) and the reference polarity of the voltage (which also changes its sign) such that the reference direction of the current always points towards the inductor in the upper terminal and such that the reference polarity of the voltage is positive at the upper terminal. Thus we get the following figure.

Figure 11. Image source: own.
Notice that each case of the previous figure has the reference direction for the current and the reference polarity for the voltage such that the arrowhead points into the inductor towards the positive reference terminal. In other words, the four cases satisfy the reference direction and reference polarity shown in figure 1.

In case a), $i_L(t) = i(t)$ and $v_L = v_\text{ind}(t)$.

In case b), $i_L(t) = i(t)$ and $v_L = -v_\text{ind}(t)$.

In case c), $i_L(t) = -i(t)$ and $v_L = -v_\text{ind}(t)$.

In case d), $i_L(t) = -i(t)$ and $v_L = v_\text{ind}(t)$.

Thus we get the following figure.

Figure 12. Image source: own.
At last, from equation (1), since $L$ is always a positive parameter, the sign of the right-hand side depends only on (and is the same as) the sign of $\mathrm di_L/ \mathrm dt$. Thus this equation requires that $v_L$ must have the same sign as $\mathrm di_L/ \mathrm dt$. Let’s see if this is satisfied in each case of the previous figure:

In case a), $\mathrm di_L(t)/\mathrm dt > 0$ and $v_L > 0$.

In case b), $\mathrm di_L(t)/\mathrm dt < 0$ and $v_L < 0$.

In case c), $\mathrm di_L(t)/\mathrm dt > 0$ and $v_L < 0$.

In case d), $\mathrm di_L(t)/\mathrm dt < 0$ and $v_L > 0$.

Notice indeed $v_L$ and $\mathrm di_L(t)/\mathrm dt$ have the same sign for each case. Therefore, the equation $v_L(t) = L \, \mathrm di_L(t)/ \mathrm dt$ indeed takes into account the signs of $v_L$, $i_L$ and $\mathrm di_L/ \mathrm dt$, provided the reference direction for $i_L$ and the reference polarity for $v_L$ are chosen as shown in figure 1.
By the way, notice in figure 10 that another way to formulate my question is: why does the current $i$ flow from higher potential to lower potential through the inductor when it is increasing (cases a) and c)), and why does it flow from lower potential to higher potential through the inductor when it is decreasing (cases b) and d))?

I found this question, which is essentially the same as mine. The user Farcher explained in his answer how to obtain the direction of $i_\text{ind}$ (such that its numerical value would be positive). They assumed an increasing (and positive) increasing current $i$, and got the same result as me (figure 10, cases a) and c)), namely that the current $i$ flows from positive to negative inside the inductor. When obtaining the polarity of $v_\text{ind}$ (such that its numerical value is positive), they said in the comments the induced current $i_\text{ind}$ flows from negative to positive inside the inductor, whether the current $i$ is increasing or decreasing, which is the same as my result.
Their explanation for the polarity of the induced voltage was based on an analogy with a battery: that inside a battery, current flows from negative to positive. However, that's true only when the battery is supplying energy to the circuit (it's been discharged/de-energized), it's not true when the battery is consuming energy from the circuit (it's been recharged/re-energized) in which case current flows from positive to negative inside the battery. While in the case of the inductor, in cases a) and c) the inductor consumes/stores energy (and $i_\text{ind}$ flows from negative to positive inside the inductor, as expected from the comparison with the battery), but in cases b) and d) the inductor supplies/releases energy (and yet $i_\text{ind}$ stills flows from negative to positive inside the inductor, unlike in the battery). So I'm not satisfied/convinced with Farcher's reasoning.

REPLY [2 votes]: Your thought train is fine up to Fig.6. But then you get into a needlessly complicated descriptions of signs. You state

"by Faraday’s law an EMF $v_\text{ind}(t)$ is induced across the inductor. To find the polarity of such EMF or voltage, we first find the direction of the current $i_\text{ind}(t)$ such EMF tries to establish, which we determine using Lenz’s law. So while our goal is to find the polarity of the induced EMF"

It seems you are using the word "voltage" and symbol $v_{ind}$ for two different concepts: 1) induced EMF in the ideal inductor - that is due to induced electric field present in the coils; 2) potential drop across the ideal inductor terminals (when moving from one terminal to the other in the designated positive direction) - that is due to the fact that electric field has electrostatic component, and potential drop is integral of this electrostatic field.
This (using "voltage" for two different concepts and getting confused as a result) is common in many people's understanding of electricity, even those with high credentials. I think mostly because many textbooks and teachers do not understand this either, because they don't understand the general concept of electromotive force and its variants. Old papers and textbooks on electricity (pre-WWII) did not suffer from this confusion.
Understanding the difference between EMF and potential difference in physics is important. It also resolves the question of sign of potential drop across the ideal inductor which you are interested in.
Both concepts - EMF and potential drop - are valid and useful, and both depend on the sign convention. The sign convention is that current intensity $i$ is positive when it flows in the designated positive direction in the loop, and emf and potential drops are positive when their effect is to act on current in the circuit element to increase it in that same direction.
Induced EMF
Induced emf for the path defined by coils of the inductor is defined as integral of induced electric field over that path.
When we have single inductor of self-inductance $L$ whose interaction with other currents/inductors can be neglected, induced EMF always obeys the equation (whether the inductor is ideal or not):
$$
emf = -L\frac{di}{dt},
$$
the minus sign making sure that emf acts against changes of current. This follows from Faraday's law and the mentioned sign convention for current and emf. It is not a good idea to call this quantity "voltage", but it is often being done, with various variations (electromotive voltage, induced voltage, etc). Although induced EMF has the unit Volt just as voltage in electrostatics has, EMF is a path-dependent quantity and in general one cannot associate two points of space (such as terminals of an inductor) with unique EMF. We need to state the path as well, but because of this, it is always better to say EMF instead. It is also almost never the quantity of interest when measuring in practice; instead, we want to measure potential differences, which do not depend on paths.
Potential drop
Potential drop across the inductor has (unfortunately) become a more complicated thing to explain. It is not always the same as EMF, not even in magnitude. Potential drop is integral of the electrostatic part of electric field when going from one terminal to the other in the positive direction, the exact path not being important. This quantity is not, in general, determined merely by induced part of electric field; the value of current $i$, resistance of the inductor $R_i$ and its internal capacitance $C_i$ is also important. So in general, potential drop across inductor cannot be determined from induced EMF alone.
However, in the special case where the inductor is ideal (zero internal resistance $R$, capacitance $C$), there is a simple relation: potential drop is exactly minus induced EMF. This is because electric field in ideal inductor coils has to be zero, and this implies that electrostatic field completely cancels the induced field inside the conductive coils, and that implies integral of electrostatic field from one terminal to the other has to be minus integral of induced field over path from one terminal to the other that goes inside the coils. So, for a perfect inductor, we have
$$
potential~drop = - emf = L\frac{di}{dt}.
$$
This quantity is also often called voltage drop across the inductor, or simply voltage across the inductor. This quantity is useful when writing down the so-called Kirchhoff's voltage law for any closed loop in a lumped model of AC circuit. Then it is simply denoted $V$ or $v$. This is the preferred meaning of the word "voltage"; it is used in electrostatics, and is used also in AC circuits. It is also what we often want to measure in a complicated real circuit via  oscilloscope. To make sure we really measure this and not some path dependent quantity like EMF, the probes and wires have to be made with good field insulation (coax cables) and during measurements, we prevent the wires from arranging themselves in loops.
From the above expression all signs are evident; when current increases in positive direction, emf is negative so the potential drop has to be positive, to counteract the induced EMF. When current decreases in that same direction, the induced EMF is positive, so the potential drop is negative to counteract the induced EMF.
What would be different for a real inductor with internal resistance? Here, net electric field must exist in the coil to push against resistance, so we cannot assume effect of potential drop cancels effect of the induced EMF. Let the real inductor be connected directly to a source of varying potential drop (more often called "source of voltage") $V(t)$:
     ---------------
    |      ->       |
    |               )
   (V)              )  real inductor
    |               )
    |      <-       |
     ---------------

Then by assumption, potential drop on the inductor is $V(t)$. We cannot find induced EMF or current $i$ from these assumptions alone.
However, if we assume this real inductor behaves as ideal inductor with self-inductance $L$ with resistor $R$ in series (ignoring capacitance effects), we can use Kirchhoff's second circuital law (for its formulation, see my answer here: Using Faraday's law twice ). The replacement lumped model circuit looks like this:
     ---------------
    |      ->       |
    |               |
   (V)              L
    |               |
    |               |
    |               R
    |               |
    |      <-       |
     ---------------

Now we can write down Kirchhoff's second circuital law for this circuit:
$$
Ri =  V - L\frac{di}{dt}.
$$
Here our voltage source contributes to the circuit effective electromotive force of magnitude V, and the other r.h.s. term is induced emf.
From this equation, we can find that potential drop on the real inductor is
$$
V = Ri + L\frac{dI}{dt} = Ri - emf.
$$
So potential drop on the inductor is not the same as induced EMF, not merely due to opposite sign but also due to magnitude. It depends on emf and current $i$ and internal resistance of the inductor $R$. In general it has different phase from EMF, and in special cases it can have the same sign as induced EMF has, something that can't happen with ideal inductor.
This was still a simplification, and a more realistic model would include contribution due to capacitive interactions between the inductor's coils (ideal capacitor $C$ in parallel with the ideal inductor $L$).


text:
\begin{document}

\title [On the global Tjurina numbers for line arrangements]
{On the minimal value of global Tjurina numbers for line arrangements}

\author[Alexandru Dimca]{Alexandru Dimca$^{1}$}
\address{Universit\'e C\^ ote d'Azur, CNRS, LJAD, France and Simion Stoilow Institute of Mathematics,
P.O. Box 1-764, RO-014700 Bucharest, Romania}
\email{dimca@unice.fr}

\thanks{$^1$ This work has been partially supported by the French government, through the $\rm UCA^{\rm JEDI}$ Investments in the Future project managed by the National Research Agency (ANR) with the reference number ANR-15-IDEX-01 and by the Romanian Ministry of Research and Innovation, CNCS - UEFISCDI, grant PN-III-P4-ID-PCE-2016-0030, within PNCDI III}

\subjclass[2010]{Primary 14H50; Secondary  14B05, 32S22}

\keywords{Tjurina number, Milnor number, line arrangement}

\begin{abstract}
We show that a general lower bound for the global Tjurina number of a reduced complex projective plane curve, given by A. A. du Plessis and C.T.C. Wall, can be improved when the curve is a line arrangement.

\end{abstract}

\maketitle

\section{Introduction}

Let $S=\C[x,y,z]$ be the graded polynomial ring in three variables $x,y,z$ and let $C:f=0$ be a reduced curve of degree $d$ in the complex projective plane $\PP^2$. The minimal degree of a Jacobian relation for $f$ is the integer $mdr(f)$
defined to be the smallest integer $m\geq 0$ such that there is a nontrivial relation
\begin{equation}
\label{rel_m}
 af_x+bf_y+cf_z=0
\end{equation}
among the partial derivatives $f_x, f_y$ and $f_z$ of $f$ with coefficients $a,b,c$ in $S_m$, the vector space of  homogeneous polynomials of degree $m$. We say that the plane curve has type $(d,r)$ if $d= \deg(f)$ and $r=mdr(f)$.

When $mdr(f)=0$, then $C$ is a pencil of lines, i.e. a union of lines passing through one point, a situation easy to analyze. We assume from now on that
$$ mdr(f)\geq 1.$$
Denote by $\tau(C)$ the global Tjurina number of the curve $C$, which is the sum of the Tjurina numbers of the singular points of $C$. When $C$ is a line arrangement,  its global Tjurina number, coincides with its global Milnor number $\mu(C)$, and is given by
$$\tau(C)=\sum_p(n(p)-1)^2,$$
where the sum is over all multiple points $p$ of $C$ and $n(p)$ denotes the multiplicity of $C$ at $p$.
The following result is due to  A.A. du Plessis and C.T.C. Wall, see \cite[Theorem 3.2]{duPCTC}.
\begin{thm}
\label{thmCTC}

For a positive integer $r$, define two integers by
$$\tau(d,r)_{min}=(d-1)(d-r-1) \text { and } \tau(d,r)_{max}=  (d-1)(d-r-1)+r^2.$$
For any reduced plane curve $C:f=0$ of type $(d,r)$, one has the inequalities
$$\tau(d,r)_{min} \leq \tau(C) \leq \tau(d,r)_{max}.$$
Moreover, suppose that  $r=mdr(f)>(d-1)/2$. Then one has the stronger inequality
$$\tau(C) \leq \tau(d,r)_{max}-{2r+2-d \choose 2}.$$
In particular, if $d$ is even and $r=d/2$, then
$ \tau(C) \leq \tau(d,r)_{max}-1.$
\end{thm}
It is easy to see that the lower bound is strict, i.e. there are curves $C$ of type $(d,r)$ such that
$\tau(C)=\tau(d,r)_{min}$ for any degree $d$ and $1 \leq r \leq d-1$, see \cite[Example 4.5]{DSt3syz} in general and \cite[Lemma 21]{E} for $(d,r)=(d,d-2)$.
The curves $C$ for which the maximal values for $\tau(C)$ are attained have special properties. Indeed, we have the following result, see \cite{Dmax}.

\begin{thm}
\label{thmDmax}
Let $C:f=0$ be a reduced plane curve of type $(d,r)$. Then the following hold.

\begin{itemize}
		\item[i)]  One has
$ \tau(C) =\tau(d,r)_{max}$
if and only if $C:f=0$ is a free curve with exponents $(r,d-1-r)$, and then $r<d/2$.
				\item[ii)]  One has
$ \tau(C) =\tau(d,r)_{max}-1$
if and only if $C:f=0$ is a nearly free curve with exponents $(r,d-r)$, and then $r \leq d/2$.

	\end{itemize}

\end{thm}
The free and nearly free curves, as well as related objects, are  actively investigated since some time, see \cite{ Abe18, AD, B+,Dcurves, Dmax,Drcc, DStExpo,DStFD,DStRIMS,DStMos,E,MaVa,ST,Yo}. Note that there are free  (resp. nearly free line arrangements)  in $\PP^2$ for any fixed type $(d,r)$ with $r<d/2$
(resp. with $r \leq d/2$), i.e. the maximal values for $\tau(C)$ can be attained with $C$ a line arrangement, at least when $r \leq d/2$.

In this note we consider the minimal values of $\tau(C)$ when $C:f=0$ is a {\it line arrangement}, once we fix its type $(d,r)$. Let $m(C)$ be the maximal multiplicity of a point in $C$,
and $n(C)$ the maximal multiplicity of a point in $C \setminus \{p\}$, where $p$ is any point in $C$ of multiplicity $m(C)$. Note that
$$ 1 \leq n(C) \leq m(C) \leq d.$$
Moreover $m(C)=d$ if and only if $mdr(f)=0$, and $m(C)=d-1$ if and only if $mdr(f)=1$,
see \cite[Proposition 4.7]{DIM}. In addition, the case
$2=n(C)\leq m(C) \leq d-2$,  corresponds to the intersection lattice $L(C)$ being the lattice $L(d,m(C))$ discussed in \cite[Proposition 4.7]{DIM}, i.e. the intersection lattice of an arrangement obtained from $m(C)$ concurrent lines by adding $d-m(C)$ lines in general position.
The main result of this note is the following.

\begin{thm}
\label{thmA}
Let $C:f=0$ be an arrangement of $d\geq 4$ lines in $\PP^2$ which is not free. If we set $r=mdr(f)\geq 2 $, then  the following hold.

\begin{itemize}
		\item[i)]  With the above notation, one has
		$$\tau(C) \geq \tau'(d,r)_{min}:= \tau(d,r)_{min}+{r \choose 2}+{n(C) \choose 2}+1.$$
\item[ii)] If $r \ne d-m(C)$, then the possibly stronger inequality
$$\tau(C) \geq \tau''(d,r)_{min}:=\tau(d,r)_{min}+{r \choose 2} +{m(C) \choose 2}+1$$ holds.

\end{itemize}

\end{thm}
The line arrangements such that $r=mdr(f) \in \{0,1,2\}$ are classified, see for instance \cite[Theorem 4.11]{DIM} for the case $r=2$, and for the remaining line arrangements we have the following.

\begin{cor}
\label{corA} Let $C:f=0$ be an arrangement of $d$ lines in $\PP^2$ which is not free and such that $r=mdr(f)\geq 3 $ and $n(C) \geq 3$. Then
$$\tau(C) \geq \tau^N(d,r)_{min}:=\tau(d,r)_{min}+{r \choose 2} +4 \geq \tau(d,r)_{min}+7.$$
\end{cor}

\begin{ex}
\label{exA}
Consider the line arrangements $C:f=0$ of $d=7$ lines such that $r=mdr(f)=3$.
Then $\tau(7,3)_{min}=6\cdot 3=18$ and
$$\tau'(7,3)_{min}=18+3+{n(C) \choose 2}+1=22+{n(C) \choose 2}.$$
The value $\tau=25$ is obtained for the following three line arrangements.
Two of them, say
$$C_1:f_1=xyz(x+y)(x+3y)(x+2y+z)(4x+8y+z)=0$$ and
$$C_2:f_2=xyz(x+2y+z)(y+z)(x-2y)(x-y)=0,$$
satisfy $m(C_i)=4$ and $n(C_i)=3$, since each of them has a quadruple point, a triple point and 12 double points. The two intersection lattices $L(C_1)$ and $L(C_2)$ are distinct, since only in $C_2$ there is a line in the arrangement containing both  points of multiplicity $>2$. Our lower  bound for both $C_1$ and $C_2$ is $\tau^N(7,3)_{min}=22+3=25$, i.e. a sharp lower bound.
The third line arrangement, say
$$C_3:f_3=xyz(2x-3y+z)(x-y)(x+z)(y+z)=0,$$
 satisfy $m(C_3)=n(C_3)=3$, since it has 4 triple points and 9 double points. Hence we can use the bounds
$\tau'(d,r)_{min}=\tau''(d,r)_{min}=25$, since $3=mdr(C_3) \ne d-m(C_3)=4$. Therefore our various bounds are sharp in all these three situations.
\medskip

\end{ex}

\begin{ex}
\label{exA2}
Consider the line arrangements $C:f=0$ of $d$ lines, with intersection lattice $L(C)$ of type $\tilde L(m_1,m_2)$ as in \cite[Proposition 4.9]{DIM}, where $2 \leq m_1 \leq m_2$.
This line arrangement is the union of two pencils of lines, one consisting of $m_1$ lines, the other of $m_2$ lines, in general position to each other.
Then we have $d=m_1+m_2$, $r=mdr(f)=m_1$, $m(C)=m_2$, $n(C)=m_1$,
$$\tau(C)=(d-1)^2-m_1m_2+1.$$
One has $\tau(d,r)_{min}=(d-1)^2-m_1m_2-(m_1^2-m_1)$,
while the lower bound given by Theorem \ref{thmA} i) is in this case
$$\tau'(d,r)_{min}=(d-1)^2-m_1m_2+1,$$
again a sharp bound.
\end{ex}

 Let $I_f$ denote the saturation of the ideal $J_f$ with respect to the maximal ideal ${\bf m}=(x,y,z)$ in $S$ and consider the local cohomology group
 $$N(f)=I_f/J_f=H^0_{\bf m}(M(f)).$$
 The graded $S$-module  $N(f)$ satisfies a Lefschetz type property with respect to multiplication by generic linear forms, see  \cite{DPop}. This implies in particular the inequalities
$$0 \leq n(f)_0 \leq n(f)_1 \leq ...\leq n(f)_{[T/2]} \geq n(f)_{[T/2]+1} \geq ...\geq n(f)_T \geq 0,$$
where $T=3d-6$ and $n(f)_k=\dim N(f)_k$ for any integer $k$. We set
$$\nu(C)=\max _j \{n(f)_j\},$$
and call $\nu(C)$ the {\it freeness defect}  of the curve $C$. It is known that a curve $C$ is free (resp. nearly free) if and only if $\nu(C)=0$ (resp. $\nu(C)=1$), see \cite{Dmax}.
The relation between the invariants $\nu(C)$ and $mdr(f)$ is given by the following result, see \cite[Theorem 1.2]{Drcc}.
\begin{thm}
\label{thmN}
Let $C:f=0$ be a reduced plane curve of degree $d$ and let $r=mdr(f)$.
Then the following hold.
\begin{enumerate}
\item If $r < d/2$, then
$$\nu(C)=(d-1)^2-r(d-1-r)-\tau(C)=\tau(d,r)_{max}-\tau(C).$$

\item If $r \geq (d-2)/2$, then
$$\nu(C)= \lceil   \  \frac{3}{4}(d-1)^2 \    \rceil  -\tau (C).$$

\end{enumerate}
\end{thm}
Here, for any real number $u$,  $\lceil     u    \rceil $ denotes the round up of $u$, namely the smallest integer $U$ such that $U \geq u$. Theorem \ref{thmA} yields the following.

\begin{cor}
\label{corB}
Let $C:f=0$ be an arrangement of $d$ lines in $\PP^2$. If  we have $2 \leq r=mdr(f) <d/2$, then
$$\nu(C) \leq \frac{r(r+1)}{2}-{n(C) \choose 2}-1 \leq \frac{r(r+1)}{2}-2.$$
\end{cor}

It is known that a line arrangement with $r=2$ is either free or nearly free, see \cite[Theorem 4.11]{DIM}. Hence in this case $\nu(C) \in \{0,1\}$, and the upper bound given by Corollary \ref{corB} is 1, hence this bound is sharp in this case. Another upper bound for the freeness defect $\nu(C)$ of a line arrangement is discussed in \cite{DStbound}.

\section{Proof of Theorem \ref{thmA} and of Corollary \ref{corB}}
Consider the graded $S-$submodule $AR(f) \subset S^{3}$ of {\it all relations} involving the derivatives of $f$, namely
$$\rho=(a,b,c) \in AR(f)_q$$
if and only if  $af_x+bf_y+cf_z=0$ and $a,b,c$ are in $S_q$, the space of homogeneous polynomials of degree $q$.
We have the following.
\begin{lem}
\label{L0} Let $C:f=0$ be a reduced plane curve of degree $d$, such that any irreducible component of $C$ is rational. Then
$$\tau(C)=\dim AR(f)_{d-2}+\frac{(d-1)(d-2)}{2}.$$
\end{lem}
\proof
Use \cite[Formula (3.4)]{DSe}
and the vanishing of $N(f)_{2d-3}$, to be defined below in the last section,  given in \cite[Theorem 2.7]{DStbound}.

\endproof

Let $\rho_1 \in AR(f)_r$ be a nontrivial relation of minimal degree. The inclusion
$S\cdot \rho_1 \subset AR(f)$ implies that
$$\dim AR(f)_{d-2}=  \dim S_{d-r-2}+ \dim \overline {AR(f)}_{d-2},$$
where $\overline {AR(f)}=AR(f)/(S\rho_1)$. This implies
$$\tau(C)=\dim AR(f)_{d-2}+\frac{(d-1)(d-2)}{2}=(d-1)^2-r(d-\frac{r+1}{2})+\dim \overline {AR(f)}_{d-2}.$$
The exact sequence (3.3) in \cite{Dmax} shows that the graded $S$-module  $\overline {AR(f)}$ is torsion free. Let $p \in C$ (resp. $q \in C$) be two distinct points such that $p$ (resp. $q$) has multiplicity $m=m(C)$ (resp. $n=n(C)$).  Proceeding as in \cite[Section (2.2)]{Dcurves}, we get two syzygies, namely $\rho_p \in AR(f)_{d-m}$ and $\rho_q \in AR(f)_{d-n}$. We have the following.

\begin{lem}
\label{L1} Assume that $C$ is a line arrangement of $d$ lines.
Then  $\rho_q \notin S_{m-n} \cdot \rho_p$.
\end{lem}

\proof
Following \cite[Section (2.2)]{Dcurves}, we recall the construction of the two syzygies $\rho_p $ and $\rho_q $. First choose the coordinates on $\PP^2$ such that $p=(1:0:0)$ and $q=(0:1:0)$. To construct $\rho_p$, we write
$f=g(y,z)h(x,y,z)$, where $g=0$ is the equation of the $m=m(C)$ lines passing through $p$, and $h=0$ is the equation of the remaining $d-m$ lines in $C$. With this notation, one has
\begin{equation}
\label{eqA}
\rho_p=(xh_x-d\cdot h,yh_x,zh_x),
\end{equation}
where $h_x$ denotes the partial derivative of $h$ with respect to $x$. Moreover, one has
$f_x=gh_x$. Similarly, one can write $f=\tilde g(x,z)\tilde h(x,y,z)$, where $\tilde g=0$ is the equation of the $m=n(C)$ lines passing through $q$, and $\tilde h=0$ is the equation of the remaining $d-m$ lines in $C$. With this notation, one has
\begin{equation}
\label{eqB}
\rho_q=(x\tilde h_y,y\tilde h_y-d\cdot \tilde h,z\tilde h_y),
\end{equation}
where $\tilde h_y$ denotes the partial derivative of $\tilde h$ with respect to $y$. Moreover, one has
$f_y=\tilde g \tilde h_y$.
Assume there is a polynomial $A \in S_{m-n}$ such that $\rho_q =A \rho_p$. Looking at the third coordinate, we get $\tilde h_y=Ah_x$.
Then the equality on the first coordinate implies $Ah=0$, which is a contradiction.
\endproof

Now we complete the proof of Theorem \ref{thmA}, and in view of Example \ref{exA2} we may assume that $C$ is not of type $\tilde L(m_1,m_2)$.
We apply \cite[Theorem 1.2]{Dcurves} and discuss the following cases, in order to prove the claim (i).

\medskip

\noindent {\bf Case 1}: $r=mdr(f)=d-m$. Then we can take $\rho_1=\rho_p$ and the class of $\rho_q$ in $\overline {AR(f)}$ is nonzero, because of Lemma \ref{L1} and of the fact that the syzygies constructed in \cite[Section (2.2)]{Dcurves} are primitive, i.e. they are not multiple of strictly lower degree syzygies. It follows that the vector space
$S_{n-2}\rho_q$ is naturally embedded in $\overline {AR(f)}_{d-2}$.

We recall now the construction of the Bourbaki ideal $B(C,\rho_1)$ associated to a degree $d$ reduced curve $C:f=0$ and to a minimal degree non-zero syzygy $\rho_1 \in AR(f)$, see \cite{DStJump}.
For any choice of the syzygy $\rho_1=(a_1,b_1,c_1)\in AR(f) $ with minimal degree $r=d_1$, we have a morphism of graded $S$-modules
\begin{equation} \label{B1}
S(-r)  \xrightarrow{u} AR(f), \  u(h)= h \cdot \rho_1.
\end{equation}
For any homogeneous syzygy $\rho=(a,b,c) \in AR(f)$, consider the determinant $\Delta(\rho)=\det M(\rho)$ of the $3 \times 3$ matrix $M(\rho)$ which has as first row $x,y,z$, as second row $a_1,b_1,c_1$ and as third row $a,b,c$. Then it turns out that $\Delta(\rho)$ is divisible by $f$, see \cite{Dmax}, and we define thus a new morphism of graded $S$-modules
\begin{equation} \label{B2}
 AR(f)  \xrightarrow{v}  S(r-d+1)   , \  v(\rho)= \Delta(\rho)/f,
\end{equation}
and a homogeneous ideal $B(C,\rho_1) \subset S$ such that $\im v=B(C,\rho_1)(r-d+1)$.
It is known that the ideal $B(C,\rho_1)$, when $C$ is not a free curve, defines a $0$-dimensional subscheme $Z(C,\rho_1)$ in $\PP^2$, which is locally a complete intersection, see \cite[Theorem 5.1]{DStJump}.
This construction yields the following exact sequence.
$$0 \to S(m-d)  \xrightarrow{u} AR(f) \xrightarrow{v}  B(C,\rho_1)(-m+1) \to  0.$$
Hence, with this notation,  $h_q=v(\rho_q) \in B_{d-n-m+1}$ and
$$S_{n-2}h_q  \subset \overline {AR(f)}_{d-2}  = B_{d-m-1}.$$
On the other hand we know, see \cite[Corollary 3.5]{SchCMH} or \cite[Corollary 3.6]{DIM}
, that  the Castelnuovo-Mumford regularity $reg(AR(f))$ of the graded $S$-module $AR(f)$ satisfies $reg(AR(f))\leq d-2$ as soon as $d \geq 4$. In particular, all the generators in a minimal set of generators for $AR(f)$ have degrees $\leq d-2$.
It follows that our ideal $B$ is generated in degrees $\leq d-m-1$. Moreover we know that the subscheme in $\PP^2$ defined by the ideal $B$ is 0-dimensional, \cite[Theorem 4.1]{DStJump}, and hence $B$ cannot be generated by a single polynomial $h_q$. It follows that there is at least one polynomial $\hat h \in B_{d-m-1} \setminus S_{n-2}h_q$ and hence
$$\dim AR(f)_{d-2} \geq \dim S_{n-2} +1.$$
This completes the proof of the claim $(i)$ in this case.
\medskip

\noindent {\bf Case 2}: $r=mdr(f)=m-1$ and $C$ is free, case which is discarded by our hypotheses.

\medskip

\noindent {\bf Case 3}: $m\leq r=mdr(f) \leq d-m-1$. Then let $\rho_1 \in AR(f)_r$ by a minimal degree non zero syzygy. Exactly as in the Case 1 above,
it follows that the vector space
$S_{m-2}\rho_p$ is naturally embedded in $\overline {AR(f)}_{d-2}=B_{r-m-1}$, and this proves the claims (i) and (ii) in this case as well, exactly as above.

\subsection{Proof of Corollary \ref{corB}}

 One has
$$\nu(C) = \tau(d,r)_{max}-\tau(C) \leq \tau(d,r)_{max}-\tau'(d,r)_{min}=$$
$$=\frac{r(r+1)}{2}-{n(C) \choose 2}-1\leq \frac{r(r+1)}{2}-2.$$

\section{An application to Terao's conjecture}

H. Terao has conjectured that if $\A$ and $\A'$ are hyperplane arrangements in $\PP^n$  with isomorphic intersection lattices $L(\A) =L(\A')$, and if $\A$ is free, then $\A'$ is also free, see  for details
\cite{DHA, OT, Yo}.
Using Theorem \ref{thmA}, we get the following partial positive answer in the case of line arrangements, which is an improvement of \cite[Corollary 2.3]{Dmax}.

\begin{cor}
\label{cor2} Let $C:f=0$ and $C':f'=0$ be two arrangements of $d$ lines  in $\PP^2$  with isomorphic intersection lattices $L(C) =L(C')$.  If $C$ is free, then there is  a unique integer $r\geq 0$ such that $r<d/2$ and
$$\tau(\A) =(d-1)(d-r-1)+r^2.$$
If this integer $r$ satisfies
$$r<\frac{-3+\sqrt {8d+41}}{2},$$
 then the line arrangement $C'$ is also free.

\end{cor}

\proof We can assume that $n(C) \geq 3$, since in the other cases Terao's conjecture is obvious. Indeed, for $n(C) \leq 2$, the intersection lattice $L(C)$ has the type $L(d,m(C))$, and the freeness of $C$ depends only on $m(C)$, see \cite[Proposition 4.7]{DIM}.
When $n(C) \geq 3$, a direct  computation shows that for $s<\frac{-3+\sqrt {8d+41}}{2}$, the intervals
$[\tau^N(d,s)_{min},\tau(d,s)_{max}]$ and $[\tau^N(d,s-1)_{min},\tau(d,s-1)_{max}]$ are disjoint, since
$$\tau(d,s)_{max} <\tau^N(d,s-1)_{min}.$$
It follows that each value $\tau(d,s)_{max}$ uniquely determines the corresponding $s$ when $s<\frac{-3+\sqrt {8d+41}}{2}$.
Moreover, $\tau(C)=\tau(C')$, since this number depends only on the intersection lattice  $L(C) =L(C')$. It follows that $mdr(f')=mdr(f)=r$ and hence $C'$ is free by Theorem  \ref{thmDmax} $i)$.

\endproof

\begin{ex}
\label{exE} When $d=100$, then we have
$$\frac{-3+\sqrt {8d+41}}{2}=\frac{-3+29}{2}=13.$$
Hence Terao's conjecture holds for any free arrangement $C:f=0$ of 100 lines with $r\leq 13$. The old result  \cite[Corollary 2.3]{Dmax} when $d=100$ gives the bound $r \leq \sqrt {d-2}$, hence $ r \leq 9$.

\end{ex}

config: arxiv


TITLE: Application of Grönwall's inequality
QUESTION [3 upvotes]: Let $F: \mathbb{R} \rightarrow \mathbb{R}$ be a Lipschitz function and $\varphi, \psi: \mathbb{R} \rightarrow \mathbb{R} $ be continuous differentiable functions.
Suppose that $\forall t \in \mathbb{R}$:
$$\varphi'(t)\geq F(\varphi(t))$$
$$\psi'(t) = F(\psi(t)) $$
Show that if $\psi(t_0)=\varphi(t_0)$, then $\psi(t) \leq \varphi(t), \forall t \geq t_0$.
I should use Grönwall's inequality:

Let $u: [a,b] \rightarrow \mathbb{R_+}$  be a continuous function.
  If there is a function $v: [a,b] \rightarrow \mathbb{R_+}$ and a constant $C>0$ such that:
  $$u(t) \leq C + \int_{a}^{t}{v(s)u(s)ds}  \quad \forall t \in [a,b]$$
  Then $u(t) \leq C \exp\bigg({\int_{a}^{t}{v(s) ds}} \bigg)$

My attempt:
Let $K>0$  be the Lipschitz constant of $F$. We have:
$$\varphi'(t)\geq F(\varphi(t)) \Rightarrow -\varphi'(t)\leq -F(\varphi(t))$$
$$\Rightarrow \psi'(t)-\varphi'(t)\leq F(\psi(t))-F(\varphi(t)) \leq K|\psi(t)-\varphi(t)|$$
$$\Rightarrow \int_{t_0}^{t}{(\psi'(s)-\varphi'(s))ds} \leq \int_{t_0}^{t}{K|\psi(s)-\varphi(s)|ds}$$
$$\Rightarrow \psi(t)-\varphi(t) -(\psi(t_0)-\varphi(t_0) \leq \int_{t_0}^{t}{K|\psi(s)-\varphi(s)|ds}$$
Since $\psi(t_0)=\varphi(t_0)$:
$$\Rightarrow \psi(t)-\varphi(t) \leq \int_{t_0}^{t}{K|\psi(s)-\varphi(s)|ds}$$
Let $u(t) = \psi(t)-\varphi(t)$, $v(s) = K$ and $C = 0$. Thus
$$u(t) \leq \int_{a}^{t}{v(s)|u(s)|ds}$$
It looks like the inequality in  Grönwall's inequality hypothesis. However, there's a modulus on the right hand side.
How could I fix it?

REPLY [1 votes]: There doesn't seem to be any need for the full force of Grönwall's inequality here. The much simpler differential form will suffice.
But we'll need one trick: Let $u(t)=\psi(t)-\varphi(t)$. We need to show that $u\le0$. So assume instead that $u(t_1)>0$, for some $t_1>t_0$. Now define
$$a=\sup\{\,t\in[t_0,t_1]: u(t)\le0\,\}.$$
Then $u(a)=0$, and $u(t)>0$ for $a<t\le t_1$.
Let $L$ be the Lipschitz constant of $F$. Then
$$u'(t)=\psi'(t)-\varphi'(t)\le F(\psi(t))-F(\varphi(t))\le L|\psi(t)-\varphi(t)|=Lu$$
for $a<t<t_1$. (Note that the absolute value is not needed because $u>0$.)
Now we can invoke the differential form of Grönwall – or much simpler, just differentiate $e^{-Lt}u(t)$ and find that this function is non-increasing on $[a,t_t]$. And since $u(a)=0$, that implies $u(t)\le0$ for $a\le t\le t_1$, which is a contradiction.


text:
\begin{document}

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\begin{titlepage}
\begin{flushright}
\date{\today}
\end{flushright}
\begin{center}
{\LARGE {\bf AREAL OPTIMIZATION OF POLYGONS}}\footnote{submitted for publication}\\
\vskip 0.5 true cm \vspace{2cm}
\renewcommand{\thefootnote}
{\fnsymbol{footnote}}
 \large{\textsl{\textsf{Erica Walker}}$^{a}$, \textsl{\textsf{Raza M.
 Syed}}$^{a,b}$, and \textsl{\textsf{Achille Corsetti}}$^{a,b}$}
\vskip 0.5 true cm
\end{center}

\noindent{$a$. Health Careers Academy, 110 The Fenway, Cahners Hall, Boston, MA 02115\\
{ $b$. Department of Physics, Northeastern University,
360 Huntington Ave., Boston\\$~~~~~$MA 02115-5000}} \\

\vskip 1.0 true cm
\centerline{\bf Abstract}
\medskip
\noindent
 We will first solve the following problem analytically:
given a piece of wire of specified length, we will find where the
wire should be cut and bent to form two regular polygons not
necessarily having the same number of sides, so that the combined
area of the polygons thus formed is maximized, minimized, greater
than, and less than a specified area. We will extend the results
to the cases where the wire is divided into three and finally into
an arbitrary number of segments. The second problem we will solve
is as follows: two wires of specified length are to be bent into
two regular polygons whose total number of sides is fixed.  We
will determine how the total number of polygonal sides are to be
allocated between the wires so that the total area of the polygons
is maximized. We will extend the results found here to the case
where we are given any number of wires of specified length.

\end{titlepage}

\section{Introduction}
A segment of length, $x$, is to be cut from a wire of length, $L$.
This segment, along with the remaining segment of length $L-x$,
are then bent to form two regular polygons not necessarily having
the same number of sides. We will begin by finding the total area,
$A_{_{total}}$, of these two polygons in terms of $x$ and $L$.
From this general form, we will solve the extrema problem; that
is, we will compute $x$-- the place where the wire must be cut and
divided into the perimeter of the polygons so that the total area
of the polygons are maximized and minimized.  Next, we will solve
the inequality problems by again computing $x$, the place where
the wire should be severed so that the total area of these two
polygons is greater than or less than a specified area, ${\mathsf
A}$. In addition to the above 1-cut (2-partitioned) problem, we
will also analyze the above four cases for the 2-cut problem and
finally generalizing it to a ${\mathsf k}$-cut (${\mathsf
k+1}$-partitioned problem).  We will illustrate the above four
cases by providing examples of specific polygons.

    In our second problem, we are given two wires of specified
lengths which we will bend to form two regular polygons with the
total number of polygonal sides fixed.  We will then calculate
ways in which the total number of sides can be allocated between
these two wires so that the combined area of the polygons is
maximized. This will generate a non-linear coupled equation in the
number of polygonal sides. This equation will be solved
numerically using a computer program.  Finally, we will generalize
this problem to ${\mathrm k}$-wires of specified lengths.

     The outline of the paper is as follows: in section 2 we will
examine the 2-partitioned case.  Here we will devote sub-sections
to deriving a general expression for $A_{_{total}}$ in terms of
$x$ and $ L$, computing the exact form for $x$ for which
$A_{_{total}}$ is  maximized and minimized, and considering lower
and upper bound problems as mentioned above. Here, we also mention
constraints on ${\mathsf A}$ and $L$. Finally, we illustrate each
of the above scenarios with specific examples. In sections 3 and 4
we will consider an extension of section 2 examining the
3-partition and the ${\mathsf (k+1)}$-partition cases,
respectively. In section 5 we consider the problem where two wires
of specified length are bent into two regular polygons where the
total number of the polygonal sides is fixed.  Here, we find how
the total number of sides must be divided between the polygons so
that the combined area of the two polygons is maximized.  Next, we
generalize these results to the case of ${\mathrm k}$ wires of
specified lengths. Section 6 is devoted to the conclusion. In
appendix A we show that the maximum area of a polygon for a fixed
perimeter occurs in the limiting case of a circle.  Appendix B
contains the computer program which looks for the right
combination of polygonal sides that would maximize the total area.
.

\section{2-partitioned wire}
{\bf{\textsf{I.~~The total area of two polygons}}}\\
In this sub-section we will derive $A_{_{total}}$, the total
combined area of two regular $m$- and $n$-sided polygons. We will
start by finding the area of each polygon using the formula:
$Area_{_{polygon}} =\frac{1}{2}ap$, where $a$ and $p$ are the
apothem and perimeter of the polygon, respectively. This is
equivalent to finding the area of one triangle (see accompanying
figure) and multiplying it by the number of sides since the number
of congruent triangles is equal to the number of sides.

Cutting length $x$ from a total length, $L$, we form a regular
$n$-gon with side length $\frac{x}{n}$. Dropping a perpendicular
of length $a$ from the center of this $n$-gon to a side will
bisect the side. Further, the radius of the polygon will bisect
the interior angle which is given by $\theta_{_{n}} =
(\frac{1}{2}- \frac{1}{n})\pi$. From the right triangle AOB we
find $a = \frac{x}{2n}\tan{\theta_{_{n}}}$. Hence, the area of the
$n$-sided polygon is given by
$A_{_{n}}(x)=\frac{x^2}{4n}\tan{\theta_{_{n}}}$. Since the
remaining length is $y\equiv L-x$, the area of the $m$-sided
polygon will be given by $A_{_{m}}(y)= \frac{y^2}
{4m}\tan{\theta_{_{m}}}$ where $\theta_{_{m}}
=(\frac{1}{2}-\frac{1}{m})\pi$.  Finally, the total area of the
two polygons is given by the expression:
\begin{equation}
A_{_{total}}(x,y) =
A_{_{n}}(x)+A_{_{m}}(y)=\frac{x^2}{4n}\tan{\theta_{_{n}}} +
\frac{(L-x)^2}{4m}\tan{\theta_{_{m}}}
\end{equation}

In the case of identical polygons ($m=n$ and therefore
$\theta_{_{m}}=\theta_{_{n}}$), Eq.1 reduces to
\begin{equation}
A_{_{total}}^{^{(m=n)}} =
\frac{2x^2-2xL+L^2}{4n}\tan{\theta_{_{n}}}
\end{equation}
{\bf\textsf{II.~~Minimum total area of two polygons}}\\
In this sub-section we consider the minima problem. We begin by
finding the critical values of $A_{_{total}}$. From Eq.(1) we
obtain,
\begin{equation}
\frac{dA_{_{total}}}{dx} = \frac{x\tan{\theta_{_{n}}}}{2n}-
\frac{(L-x)\tan{\theta_{_{m}}}}{2m}=0
\end{equation}
which results in a minimum point:
\begin{equation}
 x_{_{min}} = \frac{nL\tan{\theta_{_{m}}}}{m\tan{\theta_{_{n}}}+ n\tan{\theta_{_{m}}}}
\end{equation}
because,
\begin{equation}
\frac{d^2A_{_{total}}}{dx^2}=
\frac{\tan{\theta_{_{n}}}}{2n}+\frac{\tan{\theta_{_{m}}}}{2m} >0
\end{equation}
This is true since,
$\pi/6\leq\theta_{_{n}},\theta_{_{m}}\leq\pi/2$. Further,
\begin{equation}
 y_{_{min}} = \frac{mL\tan{\theta_{_{n}}}}{m\tan{\theta_{_{n}}}+ n\tan{\theta_{_{m}}}}
\end{equation}
With this result of $x_{_{min}}$ and $ y_{_{min}}$, we now find
expressions for $A_{_{n}}(x_{_{min}})$, $A_{_{m}}(y_{_{min}})$,
and  the total minimum area of the two polygons,
$A_{_{min-total}}$:
\begin{eqnarray}
A_{_{n}}(x_{_{min}})=\frac{L^2n\tan{\theta_{_{n}}}\tan^2{\theta_{_{m}}}}{4(m\tan{\theta_{_{n}}}+
n\tan{\theta_{_{m}}})^2}\nonumber\\
A_{_{m}}(y_{_{min}})=\frac{L^2m\tan^2{\theta_{_{n}}}\tan{\theta_{_{m}}}}{4(m\tan{\theta_{_{n}}}+
n\tan{\theta_{_{m}}})^2}\nonumber\\
A_{_{min-total}} \equiv A_{_{total}}(x_{_{min}},y_{_{min}})=
\frac{L^2\tan{\theta_{_{n}}}\tan{\theta_{_{m}}}}{4(m\tan{\theta_{_{n}}}+
n\tan{\theta_{_{m}}})}
\end{eqnarray}
 Further, in the case when $m=n$, we get:
\begin{eqnarray}
x^{^{(m=n)}}_{_{min}}=y^{^{(m=n)}}_{_{min}}=\frac{L}{2}\nonumber\\
A_{_{n}}(x_{_{min}})=A_{_{m}}(y_{_{min}})=\frac{L^2}{16n}\tan{\theta_{_{n}}}\nonumber\\
 A_{_{min-total}}^{^{(m=n)}}= \frac{L^2}{8n}\tan{\theta_{_{n}}}
\end{eqnarray}
{\bf\textsf{III.~~Maximum total area of two polygons}}\\
This sub-section is devoted to the corresponding maxima problem.
The feasible domain of $x$ is [$0$,$L$] and the only critical
value is $x_{_{min}}$. Therefore, using
\begin{eqnarray}
A_{_{total}}(0,L)=\frac{L^2}{4m}\tan{\theta_{_{m}}}\nonumber\\
A_{_{total}}(x_{_{min}},y_{_{min}})=\frac{L^2}{4m}\tan{\theta_{_{m}}}\left[
\frac{1}{1+\frac{n}{m}\frac{\tan{\theta_{_{m}}}}{\tan{\theta_{_{n}}}}}\right]\nonumber\\
A_{_{total}}(L,0)=\frac{L^2}{4n}\tan{\theta_{_{n}}}
\end{eqnarray}
we can conclude that maximum area occurs when $x=0$ or $x=L$
(endpoint maxima), depending on the integers $m$ and $n$. This is
equivalent to saying all the wire is used for the $m$-sided
polygon (if $m>n$) or $n$-sided polygon (if $n>m$). Thus,
\begin{eqnarray}
A_{_{max-total}}^{^{(m>n)}}=\frac{L^2}{4m}\tan{\theta_{_{m}}}\nonumber\\
A_{_{max-total}}^{^{(n>m)}}=\frac{L^2}{4n}\tan{\theta_{_{n}}}
\end{eqnarray}
{\bf\textsf{IV.~~Lower bound on the area of two polygons}}\\
In this sub-section we will use Eq.(1) to find the possible range
of values of $x$ for which the combined area of the two polygons
exceeds a given area, ${\mathsf A}$. This translates to the
following problem:
\begin{equation}
 \frac{x^2}{4n}\tan{\theta_{_{n}}} +
\frac{(L-x)^2}{4m}\tan{\theta_{_{m}}}>{\mathsf A}
\end{equation}
 which simplifies to
 \begin{equation}
\left(x-\left[x_{_{min}}-\hat{x}\right]\right)\left(x-\left[x_{_{min}}+\hat{x}\right]\right)>0
\end{equation}
 where we have defined for brevity
 \begin{equation}
\hat{x}=\sqrt{\frac{mx_{_{min}}}{nL\tan{\theta_{_{m}}}}\left(4n{\mathsf
A}-Lx_{_{min}}\tan{\theta_{_{n}}}\right)}
\end{equation}
and $x_{_{min}}$ is defined as before. Therefore,
\begin{equation}
x\in
\left(0,x_{_{min}}-\hat{x}\right)\cup\left(x_{_{min}}+\hat{x},L\right)
\end{equation}
\noindent
{\emph{{\textbf{Constraints:}}}\\
Firstly, $\hat{x}$ must be real. This implies
\begin{equation}
L^2-4{\mathsf
A}\left(\frac{n}{\tan{\theta_{_{n}}}}+\frac{m}{\tan{\theta_{_{m}}}}
\right)\leq 0
\end{equation}
Thus,
\begin{equation}
0<L\leq L_1
\end{equation}
where $L_1$ is defined by
\begin{equation}
L_1=2\sqrt{{\mathsf
A}\left(\frac{n}{\tan{\theta_{_{n}}}}+\frac{m}{\tan{\theta_{_{m}}}}
\right)}
\end{equation}
Secondly, $x$ must be non-negative, i.e.,
\begin{equation}
L^2-\frac{4m{\mathsf A}}{\tan{\theta_{_{m}}}}\geq 0
\end{equation}
The solution to this inequality is
\begin{equation}
L\geq L_2
\end{equation}
with $L_2$ given by
\begin{equation}
L_2=2\sqrt{\frac{m{\mathsf A}}{\tan{\theta_{_{m}}}}}
\end{equation}
Finally, noting that $L_1>L_2$ and taking the intersection of the
inequalities \\ (16) and (19) we find that $L$ must satisfy,
\begin{equation}
L_2\leq L \leq L_1
\end{equation}
This is equivalent to
\begin{equation}
{\mathsf A}_2\leq {\mathsf A}\leq {\mathsf A}_1
\end{equation}
where
\begin{eqnarray}
{\mathsf A}_1= \frac{L^2\tan{\theta_{_{m}}}}{4m}\equiv
A_{_{max-total}}^{^{(m>n)}}\nonumber\\
{\mathsf
A}_2=\frac{L^2\tan{\theta_{_{m}}}\tan{\theta_{_{n}}}}{4(m\tan{\theta_{_{n}}}+n\tan{\theta_{_{m}}})}
\equiv A_{_{min-total}}\end{eqnarray}

For completeness, we state the main results of this section (Eqs.
(14), (21), (22)) for the case identical polygons:
\begin{equation}
x\in\left(0,\frac{1}{2}\left(L-\sqrt{\frac{8n{\mathsf
A}-L^2\tan{\theta_{_{n}}}}{\tan{\theta_{_{n}}}}}\right)\right)\bigcup\left(\frac{1}{2}\left(L+\sqrt{\frac{8n{\mathsf
A}-L^2\tan{\theta_{_{n}}}}{\tan{\theta_{_{n}}}}}\right),L\right)
\end{equation}
subject to the constraint
\begin{equation}
2\sqrt{\frac{n{\mathsf A}}{\tan{\theta_{_{n}}}}}\leq L \leq
2\sqrt{\frac{2n{\mathsf A}}{\tan{\theta_{_{n}}}}}
\end{equation}
or equivalently
\begin{eqnarray}
\frac{L^2\tan{\theta_{_{n}}}}{8n}\leq {\mathsf A} \leq
\frac{L^2\tan{\theta_{_{n}}}}{4n}~~~~~~~~~~~~~~~~~~~~~~\nonumber\\
\textnormal{i.e.,}~~~~~~~~~~~~~~~~~~~~~~~~~~~A_{_{min-total}}^{^{(m=n)}}\leq
{\mathsf A}\leq A_{_{max-total}}^{^{(m=n)}}~~~~~~~~~~~~~~~~~~~~~~
\end{eqnarray}
{\bf\textsf{V.~~Upper bound on the area of two polygons}}\\
Now, the problem at hand is as follows:
\begin{equation}
 \frac{x^2}{4n}\tan{\theta_{_{n}}} +
\frac{(L-x)^2}{4m}\tan{\theta_{_{m}}}<{\mathsf A}
\end{equation}
Clearly, the solution to this inequality is
\begin{equation}
x\in(x_{_{min}}-\hat{x},x_{_{min}}+\hat{x})
\end{equation}
with the same constraint as before (see Eqs. 21 and 22)\\
\\
\\
{\bf\textsf{VI.~~Examples}}\\
In this sub-section we will work out examples that will illustrate
the results established in the previous sub-sections.
\\

{\bf(a.)} \texttt{Minimum area problem}

 ~~~~~~($\alpha$.)~~~\textit{hexagon} ($n=6,
~\theta_{_{n}}=\frac{\pi}{3})$ and \textit{square}
($m=4,~\theta_{_{m}}=\frac{\pi}{4})$
\begin{eqnarray}
x_{_{min}} = (2\sqrt {3}-3)L\nonumber\\
A_{_{min-total}}=\frac{(2-\sqrt{3})L^2}{8}
\end{eqnarray}
~~~~~~~~~~($\beta$.)~~\textit{circle }($n\rightarrow\infty,
~\theta_{_{n}}=\frac{\pi}{2}$ (see appendix A)) and
\textit{square}($m=4,~\theta_{_{m}}=\frac{\pi}{4})$
\begin{eqnarray}
x_{_{min}}=\frac{\pi L}{\pi+4}\nonumber\\
A_{_{min-total}}=\frac{L^2}{4(\pi+4)}
\end{eqnarray}
\\

{\bf(b.)} \texttt{Maximum area problem}

 ~~~~~~($\alpha$.)~~~\textit{hexagon}($n=6,
~\theta_{_{n}}=\frac{\pi}{3})$ and \textit{square}
($m=4,~\theta_{_{m}}=\frac{\pi}{4})$
\begin{eqnarray}
x_{_{max}}=L\nonumber\\
A_{_{max-total}}=\frac{L^2\sqrt 3}{24}
\end{eqnarray}
~~~~~~~~~~~($\beta$.)~~~\textit{pentagon} and \textit{pentagon}
($m=n=5, ~\theta_{_{m}}=\theta_{_{n}}=\frac{3\pi}{10})$
\begin{eqnarray}
x_{_{max}}=0,L\nonumber\\
A_{_{max-total}}=\frac{L^2\sqrt{5(5+2\sqrt 5)}}{100}
\end{eqnarray}
~~~~~~~~~~~~~~~~~~Note: $\tan {\frac{3\pi}{10}}=
\sqrt{\frac{5+2\sqrt 5}{5}}$
\\

 {\bf(c.)} \texttt{Lower bound problem}

  ~~~~~~($\alpha$.)~~~\textit{hexagon} ($n=6,
~\theta_{_{n}}=\frac{\pi}{3})$ and \textit{square}
($m=4,~\theta_{_{m}}=\frac{\pi}{4})$
\begin{eqnarray}
~~~~~~~~~~~x\in\left(0, \frac{3L - \sqrt{6[8(3 +2\sqrt{3}){\mathsf
A}-L^2\sqrt{3}]}}{3+ 2\sqrt{3}}\right)\bigcup \left(\frac{3L
+\sqrt{6[8(3 +2\sqrt{3}){\mathsf A}-L^2\sqrt{3}]}}{3+
2\sqrt{3}},L\right)
 \nonumber\\
4\sqrt{\mathsf A}\leq L \leq 2\sqrt{2{\mathsf
A}(2+\sqrt{3})}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \nonumber\\
\frac{(2-\sqrt 3)L^2}{8}\leq {\mathsf A}\leq
\frac{L^2}{16}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{eqnarray}

~~~~~~($\beta$.)~~~\textit{dodecagon}($n=12,
\theta_{_{n}}=\frac{5\pi}{12})$ and \textit{triangle}
($m=3,\theta_{_{m}}=\frac{\pi}{6})$
\begin{eqnarray}
x\in\left(0, \frac{4L -2 \sqrt{12(6 +7\sqrt{3}){\mathsf
A}-L^2(3+2\sqrt{3})}}{7+ 2\sqrt{3}}\right)\bigcup \left(\frac{4L
+2 \sqrt{12(6 +7\sqrt{3}){\mathsf A}-L^2(3+2\sqrt{3})}}{7+
2\sqrt{3}},L\right)
 \nonumber\\
2\sqrt{3{\mathsf A}\sqrt{3}}\leq L \leq
2\sqrt{3{\mathsf A}(8-3\sqrt{3})}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \nonumber\\
\frac{L^2(8+3\sqrt{3})}{444}\leq {\mathsf A}\leq
\frac{L^2\sqrt{3}}{36}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{eqnarray}
~~~~~~~~~~~~~~~~~~Note: $\tan {\frac{5\pi}{12}}= 2+\sqrt 3$
\\

 {\bf(d.)} \texttt{Upper bound problem}

  ~~~~~($\alpha$.)~~~\textit{hexagon }($n=6,
~\theta_{_{n}}=\frac{\pi}{3})$ and \textit{square}
($m=4,~\theta_{_{m}}=\frac{\pi}{4})$
\begin{eqnarray}
~~~~~~~~~~~~~~~~~~x\in\left(\frac{3L -\sqrt{6[8(3
+2\sqrt{3}){\mathsf A}-L^2\sqrt{3}]}}{3+ 2\sqrt{3}},\frac{3L
+\sqrt{6[8(3 +2\sqrt{3}){\mathsf A}-L^2\sqrt{3}]}}{3+
2\sqrt{3}}\right)
 \nonumber\\
4\sqrt{\mathsf A}\leq L \leq 2\sqrt{2{\mathsf
A}(2+\sqrt{3})}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \nonumber\\
\frac{(2-\sqrt 3)L^2}{8}\leq {\mathsf A}\leq
\frac{L^2}{16}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{eqnarray}

~~~~~($\beta$.)~~~\textit{circle}($n\rightarrow\infty,
\theta_{_{n}}=\frac{\pi}{2})$ and \textit{octagon}
($m=8,\theta_{_{m}}=\frac{3\pi}{8})$
\begin{eqnarray}
~~~~~~~~~~~~~~~~x\in\left(\frac{8L
-\sqrt{8\pi[4(8+\pi(1+\sqrt{2})){\mathsf A}-L^2]}}{8+
\pi(1+\sqrt{2})}, \frac{8L
-\sqrt{8\pi[4(8+\pi(1+\sqrt{2})){\mathsf A}-L^2]}}{8+
\pi(1+\sqrt{2})}\right)
 \nonumber\\
2\sqrt{\pi\mathsf A}\leq L \leq 2\sqrt{{\mathsf
A}[8(\sqrt{2}-1)+\pi]}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \nonumber\\
\frac{L^2}{4[8(\sqrt {2}-1)+\pi]}\leq {\mathsf A}\leq
\frac{L^2}{4\pi}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{eqnarray}
~~~~~~~~~~~~~~~~~~Note: $\tan {\frac{3\pi}{8}}= 1+\sqrt 2$
\\

{\bf(e.)} \texttt{An explicit numerical example}

~~~~~~Let the length of the wire be 12 units ($=L$). We would
  like to analyze the

  ~~~~~~above four problems in the case of a square ($n=4$) and a
  triangle ($m=3$).

~~~~~Firstly, to minimize the area of the polygons, the wire
should be severed

~~~~~at $x_{_{min}}=\frac{48}{4+3\sqrt 3}\approx 5.220$ units so
that the minimum total area of the

~~~~~polygons is $A_{_{min-total}}=\frac{36}{4+3\sqrt 3}\approx
3.915$ square units. This area is allocated

~~~~~between the square and triangle as
$A_{_{4}}|_{x=5.220}=\frac{144}{43+24\sqrt 3}\approx 1.703$ square

~~~~~units and $A_{_{3}}|_{x=5.220}=\frac{108\sqrt 3}{43+24\sqrt
3}\approx 2.212$ square units.

~~~~~~~Secondly, the total maximum value of the area of the
polygons is $A_{_{max-total}}=$

~~~~~$9$ square units, which means that all of the wire is used in
the perimeter of

~~~~~the square.

~~~~~For the inequality problems we first find the constraint:
$\frac{36}{4+3\sqrt 3}\leq{\mathsf A}\leq 4\sqrt 3$.

~~~~~We choose ${\mathsf A}= 5$ square units. Then,
$x_{_{min}}\pm\hat{x}=\frac{48\pm 4\sqrt{3(45-16\sqrt
3)}}{4+3\sqrt 3}\approx 8.352$,

~~~~~$2.087$ units. Thus, in the case where we want the total area
of the two

~~~~~polygons to exceed $5$ square units the wire can be cut
 either in the open

 ~~~~~interval $(0,2.087)$ or $(8.352,12)$ units. It is
 interesting to note that the

 ~~~~~corresponding areas of the square
 and triangle vary as $A_{_{4}}\in (0,0.272)\cup$

 ~~~~~$(4.360,5)$ square units and $A_{_{3}}\in (5,4.728)\cup(0.640,0)$
square units. For the

~~~~~upper bound problem, the cutting range lies in the bounded
open

~~~~~interval $(2.087,8.352)$ units.

\section{3-partitioned wire}
{\bf\textsf{I.~~The total area of three polygons}}\\
In this sub-section we consider the extrema problem when the wire
is cut at two places; the first of length, $x$, and the second
piece of length, $y$. The third segment will obviously be of
length $L-x-y\equiv z$. The total area of the $n$-, $m$- and
$p$-sided polygons is
\begin{equation}
A_{_{total}}(x,y,z) =
A_{_{n}}(x)+A_{_{m}}(y)+A_{_{p}}(z)=\frac{x^2}{4n}\tan{\theta_{_{n}}}
+\frac{y^2}{4m}\tan{\theta_{_{m}}}
+\frac{z^2}{4p}\tan{\theta_{_{p}}}
\end{equation}
In the case of identical polygons,
\begin{equation}
A_{_{total}}^{^{(m=n=p)}}=\frac{2x^2+2y^2-2L(x+y)+2xy+L^2}{4n}\tan{\theta_{_{n}}}
\end{equation}
\\
{\bf\textsf{II.~~Minimum total area of three polygons}}\\
For critical values,
\begin{eqnarray}
\frac{\partial A_{_{total}}}{\partial
x}=x\left(\frac{\tan{\theta_{_{n}}}}{2n}+\frac{\tan{\theta_{_{p}}}}{2p}\right)
+y\frac{\tan{\theta_{_{p}}}}{2p} -L\frac{\tan{\theta_{_{p}}}}{2p}
=0\nonumber\\
\frac{\partial A_{_{total}}}{\partial
y}=x\frac{\tan{\theta_{_{p}}}}{2p}+y\left(\frac{\tan{\theta_{_{m}}}}{2m}+\frac{\tan{\theta_{_{p}}}}{2p}\right)
 -L\frac{\tan{\theta_{_{p}}}}{2p}
=0
\end{eqnarray}
Solving these two simultaneous linear equations for $x$ and $y$
results in the minimum value for $A_{_{total}}$:
\begin{equation}
x_{_{min}}=\frac{nL\tan{\theta_{_{m}}}\tan{\theta_{_{p}}}}
{n\tan{\theta_{_{m}}}\tan{\theta_{_{p}}}+m\tan{\theta_{_{n}}}\tan{\theta_{_{p}}}
+p\tan{\theta_{_{m}}}\tan{\theta_{_{n}}}}
\end{equation}
\begin{equation}
y_{_{min}}=\frac{mL\tan{\theta_{_{n}}}\tan{\theta_{_{p}}}}
{n\tan{\theta_{_{m}}}\tan{\theta_{_{p}}}+m\tan{\theta_{_{n}}}\tan{\theta_{_{p}}}
+p\tan{\theta_{_{m}}}\tan{\theta_{_{n}}}}
\end{equation}
because
\begin{eqnarray}
\frac{\partial^2A_{_{total}}}{\partial x^2}\frac{\partial^2
A_{_{total}}}{\partial y^2}-\left( \frac{\partial^2
A_{_{total}}}{\partial x\partial
y}\right)^2=\frac{n\tan{\theta_{_{m}}}\tan{\theta_{_{p}}}+m\tan{\theta_{_{n}}}\tan{\theta_{_{p}}}
+p\tan{\theta_{_{m}}}\tan{\theta_{_{n}}}}{4nmp}>0 \nonumber\\
\textnormal{and}~~~~~
 \frac{\partial^2A_{_{total}}}{\partial
x^2}+\frac{\partial^2 A_{_{total}}}{\partial
y^2}=\frac{\tan{\theta_{_{n}}}}{2n}+\frac{\tan{\theta_{_{m}}}}{2m}+\frac{\tan{\theta_{_{p}}}}{p}>0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{eqnarray}
Further,
\begin{equation}
z_{_{min}}=\frac{pL\tan{\theta_{_{n}}}\tan{\theta_{_{m}}}}
{n\tan{\theta_{_{m}}}\tan{\theta_{_{p}}}+m\tan{\theta_{_{n}}}\tan{\theta_{_{p}}}
+p\tan{\theta_{_{m}}}\tan{\theta_{_{n}}}}
\end{equation}
 Therefore,
\begin{eqnarray}
A_{_{n}}(x_{_{min}})=\frac{L^2n\tan^2{\theta_{_{m}}}\tan^2{\theta_{_{p}}}\tan{\theta_{_{n}}}}
{4\left(n\tan{\theta_{_{m}}}\tan{\theta_{_{p}}}+m\tan{\theta_{_{n}}}\tan{\theta_{_{p}}}
+p\tan{\theta_{_{m}}}\tan{\theta_{_{n}}}\right)^2}\nonumber\\
A_{_{m}}(y_{_{min}})=\frac{L^2m\tan^2{\theta_{_{n}}}\tan^2{\theta_{_{p}}}\tan{\theta_{_{m}}}}
{4\left(n\tan{\theta_{_{m}}}\tan{\theta_{_{p}}}+m\tan{\theta_{_{n}}}\tan{\theta_{_{p}}}
+p\tan{\theta_{_{m}}}\tan{\theta_{_{n}}}\right)^2}\nonumber\\
A_{_{p}}(z_{_{min}})=\frac{L^2p\tan^2{\theta_{_{n}}}\tan^2{\theta_{_{m}}}\tan{\theta_{_{p}}}}
{4\left(n\tan{\theta_{_{m}}}\tan{\theta_{_{p}}}+m\tan{\theta_{_{n}}}\tan{\theta_{_{p}}}
+p\tan{\theta_{_{m}}}\tan{\theta_{_{n}}}\right)^2}\nonumber\\
A_{_{min-total}} \equiv
A_{_{total}}(x_{_{min}},y_{_{min}},z_{_{min}})
=\frac{L^2\tan{\theta_{_{n}}}\tan{\theta_{_{m}}}\tan{\theta_{_{p}}}}
{4\left(n\tan{\theta_{_{m}}}\tan{\theta_{_{p}}}+m\tan{\theta_{_{n}}}\tan{\theta_{_{p}}}
+p\tan{\theta_{_{m}}}\tan{\theta_{_{n}}}\right)}
\end{eqnarray}

In the case of identical polygons
\begin{eqnarray}
x^{^{(m=n=p)}}_{_{min}}=y^{^{(m=n=p)}}_{_{min}}=z^{^{(m=n=p)}}_{_{min}}=\frac{L}{3}\nonumber\\
A_{_{n}}(x_{_{min}})=A_{_{m}}(y_{_{min}})=A_{_{p}}(z_{_{min}})=
\frac{L^2}{36n}\tan{\theta_{_{n}}}\nonumber\\
A_{_{min-total}}^{^{(m=n=p)}}= \frac{L^2}{12n}\tan{\theta_{_{n}}}
\end{eqnarray}
\\
{\bf\textsf{III.~~Maximum total area of three polygons}}\\
To compute the maximum total area, we must consider the boundary
values at $x=0$ or $y=0$ or $z=0$. We begin by finding the
critical values $y_{_{cr}}$ and $z_{_{cr}}$ of
$A_{_{total}}(0,y,z)$. This total area, when evaluated at these
critical values, becomes a candidate for the maximum total area.
The results are
\begin{eqnarray}
A_{_{total}}(0,y_{_{cr}},z_{_{cr}})=
\frac{L^2\tan{\theta_{_{p}}}\tan{\theta_{_{m}}}}{4(p\tan{\theta_{_{m}}}+
m\tan{\theta_{_{p}}})}\nonumber\\
y_{_{cr}} = \frac{mL\tan{\theta_{_{p}}}}{m\tan{\theta_{_{p}}}+
p\tan{\theta_{_{m}}}}\nonumber\\
z_{_{cr}} = \frac{pL\tan{\theta_{_{m}}}}{m\tan{\theta_{_{p}}}+
p\tan{\theta_{_{m}}}}
\end{eqnarray}
Similarly,
\begin{eqnarray}
A_{_{total}}(x_{_{cr}},0,z_{_{cr}}')=
\frac{L^2\tan{\theta_{_{n}}}\tan{\theta_{_{p}}}}{4(n\tan{\theta_{_{p}}}+
p\tan{\theta_{_{n}}})}\nonumber\\
x_{_{cr}} = \frac{nL\tan{\theta_{_{p}}}}{n\tan{\theta_{_{p}}}+
p\tan{\theta_{_{n}}}}\nonumber\\
z_{_{cr}}' = \frac{pL\tan{\theta_{_{n}}}}{n\tan{\theta_{_{p}}}+
p\tan{\theta_{_{n}}}}
\end{eqnarray}
and
\begin{eqnarray}
A_{_{total}}(x_{_{cr}}',y_{_{cr}}',0)=
\frac{L^2\tan{\theta_{_{n}}}\tan{\theta_{_{m}}}}{4(n\tan{\theta_{_{m}}}+
m\tan{\theta_{_{n}}})}\nonumber\\
x_{_{cr}}' = \frac{nL\tan{\theta_{_{m}}}}{n\tan{\theta_{_{m}}}+
m\tan{\theta_{_{n}}}}\nonumber\\
y_{_{cr}}' = \frac{mL\tan{\theta_{_{n}}}}{n\tan{\theta_{_{m}}}+
m\tan{\theta_{_{n}}}}
\end{eqnarray}
Finally, the maximum total area is given by
\begin{equation}
A_{_{max-total}}=\max\left\{A_{_{total}}(0,y_{_{cr}},z_{_{cr}}),~
A_{_{total}}(x_{_{cr}},0,z_{_{cr}}'),~
A_{_{total}}(x_{_{cr}}',y_{_{cr}}',0)\right\}
\end{equation}
\\
{\bf\textsf{IV.~~Upper and lower bound on the area of three
polygons}}
\\
We consider a simple inequality problem in which the first two
polygons have the same perimeter $x$. As before, we look for
values of $x$ for which the combined area of these polygons is
greater than or less than a specified area, ${\mathsf A}$.  Thus,
for the lower bound problem, we can write
\begin{equation}
 \frac{x^2}{4n}\tan{\theta_{_{n}}} +\frac{x^2}{4m}\tan{\theta_{_{m}}}
+\frac{(L-2x)^2}{4p}\tan{\theta_{_{p}}}>{\mathsf A}
\end{equation}
The solution to this inequality is given by
\begin{equation}
x\in(0,x_-) \cup (x_+,L)
\end{equation}
where
\begin{eqnarray}
x_{\pm}=\frac{1}{{mp\tan{\theta_{_{n}}}+np\tan{\theta_{_{m}}}+4mn\tan{\theta_{_{p}}}}}\times
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\nonumber\\
\{2Lmn\tan{\theta_{_{p}}}
 \pm (mnp[4{\mathsf A}(mp\tan{\theta_{_{n}}}+np\tan{\theta_{_{m}}}+4mn\tan{\theta_{_{p}}})\nonumber\\
-L^2\tan{\theta_{_{p}}}(m\tan{\theta_{_{n}}}+n\tan{\theta_{_{m}}})])^{\frac{1}{2}}\}
\end{eqnarray}
Obviously, the solution to the upper bound problem is
\begin{equation}
x\in(x_-, x_+)
\end{equation}
The above solutions are valid provided that
\begin{equation}
\frac{L^2\tan{\theta_{_{p}}}(m\tan{\theta_{_{n}}}+n\tan{\theta_{_{m}}})}{4(mp\tan{\theta_{_{n}}}
+np\tan{\theta_{_{m}}}+4mn\tan{\theta_{_{p}}})}\leq {\mathsf
A}\leq \frac{L^2\tan{\theta_{_{p}}}}{4p}
\end{equation}
\\
{\bf\textsf{V.~~Example}}\\
A wire 10 units ($=L$) long is cut into three pieces. The first
piece is bent to form a square ($n=4$), the second to form a
triangle ($m=3$), and the third to form a circle
($p\rightarrow\infty$). We are interested in knowing where the
cuts should be made so that the sum of the three areas is a
minimum and also the manner of cutting that
will produce a maximum total area.\\
For the minima problem, the length of the first piece is
$x_{_{min}}=\frac{40}{4+3\sqrt 3+\pi}\approx 3.242$ units, the
second piece is of length $y_{_{min}}=\frac{30\sqrt 3}{4+3\sqrt
3+\pi}\approx 4.212$ units, and the third piece automatically
comes out to be $z_{_{min}}=\frac{10\pi}{4+3\sqrt 3+\pi}\approx
2.546$ units. This gives a total minimum area of
$A_{_{min-total}}=\frac{25}{4+3\sqrt 3+\pi}\approx 2.026$ square
units with the assignment
$A_{_{4}}|_{x=3.242}=\frac{100}{(4+3\sqrt 3+\pi)^2}\approx 0.657$
square units, $A_{_{3}}|_{y=4.212}=\frac{75\sqrt 3}{(4+3\sqrt
3+\pi)^2}\approx 0.853$ square units and
$A_{_{\infty}}|_{z=2.546}=\frac{25\pi}{(4+3\sqrt 3+\pi)^2}\approx
0.516$ square units.\\
The maximum total area of the polygons is
$A_{_{max-total}}=\frac{25\pi}{4+\pi}\approx 10.998$ square units
and occurs at the boundary $y=0$ with critical values
$x_{_{c}}=\frac{40}{4+\pi}\approx 5.601$ units and
$z_{_{c}}=\frac{10\pi}{4+\pi}\approx 4.399$ units.\\
In the case of the inequality problem, the area ${\mathsf A}$ must
be chosen so that \\ $\frac{25(4+3\sqrt 3)}{3\pi\sqrt
3+4\pi+48\sqrt 3}\leq {\mathsf A}\leq \frac{25}{\pi}$ square
units. We choose ${\mathsf A}$=5 square units. This gives
$x_{\pm}=\frac{240\sqrt 3\pm \sqrt{240\pi(9\pi+4\pi\sqrt
3+99-20\sqrt 3)}}{3\pi\sqrt 3+4\pi+48\sqrt 3}\approx 6.332, 1.089$
units. Thus, in the case of the upper bound problem $x\in
(1.089,6.332)$ units while for the lower bound problem $x\in
(0,1.089)\cup (6.332,10)$.

\section{(${\mathsf k}$\bf {+1})-partitioned wire}
{\bf\textsf{I.~~The total area of the (k+1)-polygons}}\\
In this sub-section we consider the extrema problem where the wire
is cut at $\mathsf {k}$ different places with lengths, $x_{_1}$,
$x_{_2}$,...., $x_{_{\mathsf k}}$. The last segment being of
length, $x_{_{\mathsf k+1}}=L-x_{_1}-x_{_2}-.....-x_{_{\mathsf
k}}$. Then, the total area of the $n_{_1}$-, $n_{_2}$-,.....,
$n_{_{\mathsf k+1}}$-sided polygons is

\begin{eqnarray}
A_{_{total}}(x_{_1},x_{_2},...,x_{_{\mathsf k+1}})
=\sum_{i=1}^{{\mathsf k}+1}A_{_{n_{_i}}}(x_{_i})=
\frac{x_{_1}^2}{4n_{_1}}\tan{\theta_{_{n_{_1}}}}
+\frac{x_{_2}}{4n_{_2}}\tan{\theta_{_{n_{_2}}}}
+.....+\frac{x_{_{\mathsf k+1}}^2}{4n_{_{\mathsf
k+1}}}\tan{\theta_{_{n_{_{\mathsf
k+1}}}}}\nonumber\\
=\sum_{i=1}^{{\mathsf
k}+1}\frac{x_{_i}^2}{4n_{_i}}\tan{\theta_{_{n_{_i}}}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\end{eqnarray}
In the case of identical polygons,
\begin{equation}
A_{_{total}}^{^{(n_{_1}=n_{_2}=...=n_{_{\mathsf k+1}}\equiv n)}}
=\frac{\tan{\theta_{_{n}}}}{4n}\left[L^2+2\sum_{i=1,i\neq
j}^{\mathsf k}\left(x_{_i}^2+x_{_i}x_{_j}-Lx_{_i}\right) \right]
\end{equation}
\\
{\bf\textsf{II.~~Minimum total area of (k+1)-polygons}}\\
For critical values
\begin{equation}
\frac{\partial A_{_{total}}}{\partial x_{_1}}=\frac{\partial
A_{_{total}}}{\partial x_{_2}}=.....=\frac{\partial
A_{_{total}}}{\partial x_{_{\mathsf k}}}=0
\end{equation}
with
\begin{equation}
\frac{\partial A_{_{total}}}{\partial x_{_i}}= x_{_i}\left(
\frac{\tan{\theta_{_{n_{_i}}}}}{2n_{_i}}+\frac{\tan{\theta_{_{n_{_{\mathsf
k+1}}}}}}{{2n_{_{\mathsf k+1}}}}
\right)+\frac{\tan{\theta_{_{n_{_{\mathsf k+1}}}}}}{{2n_{_{\mathsf
k+1}}}} \left(-L
 +\sum_{j=1,j\ne i}^{\mathsf k}x_{_j}\right);~~~(i=1~
{\textnormal{to}}~{\mathsf k})
\end{equation}

 Solving these ${\mathsf k}$ simultaneous linear equations for $x_{_1}$,
$x_{_2}$,...., $x_{_{\mathsf k}}$ results in the minimum value for
$A_{_{total}}$:
\begin{equation}
x_{_{i_{_{min}}}}=L\left(\frac{n_{_i}}{\tan{\theta_{_{n_{_i}}}}}\right)\left(\sum_{j=1}^{\mathsf
k+1}\frac{n_{_j}}{\tan{\theta_{_{n_{_j}}}}}\right)^{-1}=L\left(\frac{n_{_i}\prod_{j=1,j\ne
i}^{\mathsf k+1}\tan{\theta_{_{n_{_j}}}}} {\sum_{l=1}^{\mathsf
k+1}n_{_l}\prod_{j=1,j\ne l}^{\mathsf
k+1}\tan{\theta_{_{n_{_j}}}}}\right);~~~(i=1~
{\textnormal{to}}~{\mathsf k})
\end{equation}

because
\begin{eqnarray}
\det{\left(\matrix{\frac{\partial^2A_{_{total}}}{\partial x_{_1}
^2} & \frac{\partial^2 A_{_{total}}}{\partial x_{_1}\partial
x_{_2}} &.&.&.&.&\frac{\partial^2 A_{_{total}}}{\partial
x_{_1}\partial x_{_{\mathsf k}}}\cr
 \frac{\partial^2 A_{_{total}}}{\partial
x_{_2}\partial x_{_1}} & \frac{\partial^2A_{_{total}}}{\partial
x_{_2} ^2}&.&.&.&.& \frac{\partial^2 A_{_{total}}}{\partial
x_{_2}\partial x_{_{\mathsf k}}} \cr .&&&&&&. \cr .&&&&&&. \cr
.&&&&&&.\cr .&&&&&&. \cr \frac{\partial^2 A_{_{total}}}{\partial
x_{_{\mathsf k}}\partial x_{_1}}&.&.&.&.&.&\frac{\partial^2
A_{_{total}}}{\partial x_{_{\mathsf k}}^2}} \right)}>0\nonumber\\
\textnormal{and}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\sum_{i=1}^{\mathsf
k}\frac{\partial^2A_{_{total}}}{\partial x_{_i}
^2}>0~~~~~~~~~~~~~~
\end{eqnarray}
Therefore,
\begin{eqnarray}
A_{_{n_{_i}}}(x_{_{i_{_{min}}}})=\frac{L^2}{4}\left(\frac{n_{_i}\tan{\theta_{_{n_{_i}}}}\prod_{j=1,j\ne
i}^{\mathsf k}\tan^2{\theta_{_{n_{_j}}}}}
{\left[\sum_{l=1}^{\mathsf k+1}n_{_l}\prod_{j=1,j\ne l}^{\mathsf
k+1}\tan{\theta_{_{n_{_j}}}}\right]^2}\right);~~~(i=1~
{\textnormal{to}}~{\mathsf k})\nonumber\\
A_{_{min-total}}=\frac{L^2}{4}\left(\sum_{j=1}^{\mathsf
k+1}\frac{n_{_j}}{\tan{\theta_{_{n_{_j}}}}}\right)^{-1}=\frac{L^2}{4}\left(\frac{\prod_{j=1}^{\mathsf
k+1}\tan{\theta_{_{n_{_j}}}}} {\sum_{l=1}^{\mathsf
k+1}n_{_l}\prod_{j=1,j\ne l}^{\mathsf
k+1}\tan{\theta_{_{n_{_j}}}}}\right)
\end{eqnarray}

In the case of identical polygons
\begin{eqnarray}
x^{^{(n_{_1}=n_{_2}=...=n_{_{\mathsf
k+1}})}}_{_{i_{_{min}}}}=\frac{L}{{\mathsf k}+1};~~~(i=1~
{\textnormal{to}}~{\mathsf k})\nonumber\\
A_{_{n_{_i}}}^{^{(n_{_1}=n_{_2}=...=n_{_{\mathsf k+1}}\equiv
n)}}|_{{_{x=x_{_{i_{_{min}}}}}}}= \frac{L^2}{4({\mathsf
k}+1)^2n}\tan{\theta_{_{n}}};~~~(i=1~
{\textnormal{to}}~{\mathsf k})\nonumber\\
A_{_{min-total}}^{^{(n_{_1}=n_{_2}=...=n_{_{\mathsf k+1}}\equiv
n)}}= \frac{L^2}{4({\mathsf k}+1)n}\tan{\theta_{_{n}}}~~~~~~~~~~
\end{eqnarray}
\\
{\bf\textsf{III.~~Maximum total area of (k+1)-polygons}}\\
Here we need to consider the total area at the boundary values,
$x_{_b}=0;~b=1$ to ${\mathsf k+1}$:
\begin{equation}
A_{_{total}}(x_{_1},x_{_2},..,x_{_b}=0,..,x_{_{\mathsf k+1}})
=\sum_{i=1,i\neq b}^{{\mathsf
k}+1}\frac{x_{_i}^2}{4n_{_i}}\tan{\theta_{_{n_{_i}}}};~~~(b=1~
{\textnormal{to}}~{\mathsf k}+1)
\end{equation}

The critical value, $x_{_{i_{_{cr}}}}$, of this $A_{_{total}}$ is
given by
\begin{equation}
x_{_{i_{_{cr}}}}=L\left(\frac{n_{_i}\prod_{j=1,j\ne i,b}^{\mathsf
k+1}\tan{\theta_{_{n_{_j}}}}} {\sum_{l=1,l\ne b}^{\mathsf
k+1}n_{_l}\prod_{j=1,j\ne l,b}^{\mathsf
k+1}\tan{\theta_{_{n_{_j}}}}}\right);~~~(i=1~
{\textnormal{to}}~{\mathsf k}~{\textnormal{except}}~b)
\end{equation}

The above $A_{_{total}}$, when evaluated at these critical values,
becomes a candidate for the maximum total area:
\begin{equation}
A_{_{total}}(x_{_{1_{_{cr}}}},x_{_{2_{_{cr}}}},..,x_{_b}=0,..,x_{_{{\mathsf
({k}+1)}_{_{cr}}}})=\frac{L^2}{4}\left(\frac{\prod_{j=1,j\neq
b}^{\mathsf k+1}\tan{\theta_{_{n_{_j}}}}} {\sum_{l=1,l\neq
b}^{\mathsf k+1}n_{_l}\prod_{j=1,j\ne l,b}^{\mathsf
k+1}\tan{\theta_{_{n_{_j}}}}}\right);~~~(b=1~
{\textnormal{to}}~{\mathsf k}+1)
\end{equation}
and
\begin{eqnarray}
 A_{_{max-total}}=\max_{b=1}^{{\mathsf
k}+1}\left\{A_{_{total}}(x_{_{1_{_{cr}}}},x_{_{2_{_{cr}}}},..,x_{_b}=0,..,x_{_{{\mathsf
({k}+1)}_{_{cr}}}})\right\}
\end{eqnarray}
\\
{\bf\textsf{IV.~~Upper and lower bound on the area of (k+1)-polygons}}\\
As in the 2-cut case, we consider the simple case in which the
first ${\mathsf k}$ polygons have the same perimeter, $x$.
Therefore, the last one will have a perimeter of $L-{\mathsf k}x$.
Hence, for the lower bound problem
\begin{equation}
\frac{x^2}{4n_{_1}}\tan{\theta_{_{n_{_1}}}}
+\frac{x^2}{4n_{_2}}\tan{\theta_{_{n_{_2}}}}
+.....+\frac{x^2}{4n_{_{\mathsf k}}}\tan{\theta_{_{n_{_{\mathsf
k}}}}}+\frac{(L-{\mathsf k}x)^2}{{4n_{_{\mathsf
k+1}}}}\tan{\theta_{_{n_{_{\mathsf k+1}}}}}>{\mathsf A}
\end{equation}
This can be simplified to
\begin{eqnarray}
x^2\left[\sum_{l=1}^{\mathsf
k}\tan{\theta_{_{n_{_l}}}}\prod_{j=1,j\ne l}^{\mathsf
k+1}n_{_j}+{\mathsf k}^2 \tan{\theta_{_{n_{_{\mathsf
k+1}}}}}\prod_{j=1}^{\mathsf k}n_{_j}\right]-2L{\mathsf
k}x\tan{\theta_{_{n_{_{\mathsf k+1}}}}}\prod_{j=1}^{\mathsf
k}n_{_j}\nonumber\\
+L^2\tan{\theta_{_{n_{_{\mathsf k+1}}}}}\prod_{j=1}^{\mathsf
k}n_{_j}-4{\mathsf A}\prod_{j=1}^{\mathsf k+1}n_{_j}>0 \nonumber\\
\end{eqnarray}
The solution is
\begin{equation}
x\in (0,x_-)\cup (x_+,L)
\end{equation}
where
\begin{eqnarray}
x_{\pm}=\frac{L{\mathsf k}\tan{\theta_{_{n_{_{\mathsf
k+1}}}}}\prod_{j=1}^{\mathsf k}n_{_j}\pm
\sqrt{\prod_{j=1}^{\mathsf k+1}n_{_j}\sum_{l=1}^{{\mathsf
k+1}}(4{\mathsf A}\alpha_l-L^2\tan{\theta_{_{n_{_{\mathsf
k+1}}}}}\beta_l) \tan{\theta_{_{n_{_l}}}}\prod_{i=1,i\ne
l}^{\mathsf k+1}n_{_i}}} {\sum_{l=1}^{\mathsf
k+1}\alpha_l\tan{\theta_{_{n_{_l}}}}\prod_{i=1,i\ne l}^{\mathsf
k+1}n_{_i}}\nonumber\\
\end{eqnarray}
and
\begin{eqnarray}
\alpha_{\mathsf k+1}={\mathsf k}^2,~~~~~\beta_{\mathsf
k+1}=0\nonumber\\
 \alpha_i=\beta_i=1,~~~~\forall i\ne {\mathsf k}+1
\end{eqnarray}
Finally, the above solutions are valid if
\begin{equation}
\frac{L^2\tan{\theta_{_{n_{_{\mathsf k+1}}}}}\sum_{l=1}^{\mathsf
k}\tan{\theta_{_{n_{_l}}}}\prod_{j=1,j\ne l}^{\mathsf k}n_{_j}}
{4\sum_{l=1}^{\mathsf
k+1}\alpha_l\tan{\theta_{_{n_{_l}}}}\prod_{j=1,j\ne l}^{\mathsf
k+1}n_{_j}}\leq {\mathsf A}\leq \frac{L^2
\tan{\theta_{_{n_{_{\mathsf k+1}}}}}}{4n_{_{\mathsf k+1}}}
\end{equation}
The upper bound solution is
\begin{equation}
x\in(x_-, x_+)
\end{equation}
\\
{\bf\textsf{V.~~Example}}\\
We wrap-up this section with one final concrete example. To that
end, consider a wire of length 20 units. Further, assume that the
wire is cut into six segments, with the first one bent into a
triangle ($n_{_1}=3$), the second into a square ($n_{_2}=4$), the
third into a hexagon ($n_{_3}=6$), the fourth into an octagon
($n_{_4}=8$), the fifth into a dodecagon ($n_{_5}=12$), and the
sixth into a circle ($n_{_6}\rightarrow\infty$). As before, we
consider the mannerism in which the wire should be cut so that the
total area enclosed by the above six polygons is a minimum. Here
we also deal with the
corresponding maxima problem.\\
For the minima problem, the wire should be severed at
$x_{_{1_{_{min}}}}=4.654$ units, $x_{_{2_{_{min}}}}=3.582$ units,
$x_{_{3_{_{min}}}}=3.103$ units, $x_{_{4_{_{min}}}}=2.968$ units,
$x_{_{5_{_{min}}}}=2.880$ units, and $x_{_{6_{_{min}}}}=2.814$
units giving a $A_{_{min-total}}=4.478$ square units with the
allocation $A_{_{3}}|_{x_{_1}=4.654}=1.042$ square units,
$A_{_{4}}|_{x_{_2}=3.852}=0.802$ square units,
$A_{_{6}}|_{x_{_3}=3.103}=0.695$ square units,
$A_{_{8}}|_{x_{_4}=2.968}=0.665$ square units,
$A_{_{12}}|_{x_{_5}=2.880}=0.645$ square units,
$A_{_{\infty}}|_{x_{_6}=2.814}=0.630$ square units.\\
The maximum total area of the polygons is calculated to be
$A_{_{max-total}}=5.836$ square units and occurs at the boundary
point $x_{_1}=0$ with critical values, $x_{_{2_{_{cr}}}}=4.669$
units, $x_{_{3_{_{cr}}}}=4.043$ units, $x_{_{4_{_{cr}}}}=3.868$
units, $x_{_{5_{_{cr}}}}=3.753$ units, $x_{_{6_{_{cr}}}}=3.667$
units.\\
For the inequality problem, we must choose ${\mathsf A}$ between
4.599 and 31.831 square units. Take ${\mathsf A}= 23$ square
units. Then, $x_{\pm}\approx 6.235,~0.609$ units. Thus, for the
upper bound problem $x\in (0.609, 6.235)$ units, while for lower
bound problem $x\in(0,0.609)\cup(6.235,20)$ units.

\section{Extrema problem of the distribution of a fixed number of polygonal sides among an arbitrary number of
wires} In this section we first consider two wires of specified
lengths which are bent to form two regular polygons. The total
number of sides of the polygons, $I$, is fixed. The question we
answer is how the total number of polygonal sides, $I$, should be
divided between the two wires so that the total area of the two
polygons is maximized.

To solve the problem at hand, we consider two wires of lengths
$L_n$ and $L_m$. These are bent into $n$- and $m$-sided regular
polygons with $n+m(=I)$ fixed. The area of the two polygons are
given by
$\frac{L_n^2}{4n}\tan\left(\frac{\pi}{2}-\frac{\pi}{n}\right)$ and
$\frac{L_m^2}{4(I-n)}\tan\left(\frac{\pi}{2}-\frac{\pi}{I-n}\right)$.
Thus,
\begin{equation}
A_{_{total}}=\frac{L_n^2}{4n}\tan\left(\frac{\pi}{2}-\frac{\pi}{n}\right)+
\frac{L_m^2}{4(I-n)}\tan\left(\frac{\pi}{2}-\frac{\pi}{I-n}\right)
\end{equation}
Differentiating with respect to the number of polygonal sides,
$n$, we get,
\begin{eqnarray}
\frac{dA_{_{total}}}{dn}=\frac{L_n^2}{4n^2}\left[\frac{\pi}{n}+\frac{\pi}{n}
{\textnormal{cot}}^2\left(\frac{\pi}{n}\right)-{\textnormal{cot}}\left(\frac{\pi}{n}\right)\right]
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\nonumber\\
+\frac{L_m^2}{4(I-n)^2}\left[-\frac{\pi}{I-n}-\frac{\pi}{I-n}
{\textnormal{cot}}^2\left(\frac{\pi}{I-n}\right)+{\textnormal{cot}}\left(\frac{\pi}{I-n}\right)\right]=0
\end{eqnarray}
This gives a nonlinear equation in $n$:
\begin{equation}
(\alpha L_n)^2\left[\alpha
{\textnormal{cot}}^2\alpha-{\textnormal{cot}}\alpha +\alpha\right]
=(\beta L_m)^2\left[\beta
{\textnormal{cot}}^2\beta-{\textnormal{cot}}\beta +\beta\right]
\end{equation}
where
\begin{eqnarray}
\alpha=\frac{\pi}{n}\nonumber\\
\beta=\frac{\alpha \pi}{\alpha I-n}=\frac{\pi}{I-n}
\end{eqnarray}

Finally, we generalize the above analysis for the case when there
are $\mathrm{k}$ wires with lengths $L_1$, $L_2$,....,
$L_{\mathrm{k}}$. As before, we bent each of these wires into
regular $n_{_1}$-, $n_{_2}$-,...., $n_{_{\mathrm{k}}}$-gons with
$n_{_1}+n_{_2}+....+n_{_{\mathrm{k}}}=I$. We look for values of
$n_{_1}$, $n_{_2}$,...., $n_{_{\mathrm{k}}}$ which would maximize
the total area of the polygons. In this case, the total area of
the polygons is given by
\begin{equation}
A_{_{total}}=\frac{1}{4}\sum_{i=1}^{\mathrm
{k}}\frac{L_i^2}{n_{_i}}\tan{\left(\frac{\pi}{2}-\frac{\pi}{n_{_i}}\right)}
\end{equation}
To solve the extrema problem we set
\begin{equation}
\frac{\partial A_{_{total}}}{\partial n_{_1}}=\frac{\partial
A_{_{total}}}{\partial n_{_2}}=.....=\frac{\partial
A_{_{total}}}{\partial n_{_{\mathrm{k}-1}}}=0
\end{equation}
obtaining
\begin{eqnarray}
(\alpha _1L_1)^2\left[\alpha_1
{\textnormal{cot}}^2\alpha_1-{\textnormal{cot}}\alpha_1
+\alpha_1\right]~~~~~~~~~~~~~~~~~~\nonumber\\
=(\alpha_2 L_2)^2\left[\alpha_2
{\textnormal{cot}}^2\alpha_2-{\textnormal{cot}}\alpha_2
+\alpha_2\right]~~~~~~~~~~~~~~~~~\nonumber\\
=........~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\nonumber\\
=(\alpha_{\mathrm{k}-1}
L_{\mathrm{k}-1})^2\left[\alpha_{\mathrm{k}-1}
{\textnormal{cot}}^2\alpha_{\mathrm{k}-1}-{\textnormal{cot}}\alpha_{\mathrm{k}-1}
+\alpha_{\mathrm{k}-1}\right]\nonumber\\
=(\gamma L_{\mathrm{k}})^2\left[\gamma
{\textnormal{cot}}^2\gamma-{\textnormal{cot}}\gamma
+\gamma\right]~~~~~~~~~~~~~~~~~~~~~~~~
\end{eqnarray}
where
\begin{equation}
\alpha_i=\frac{\pi}{n_{_i}}~~{\textnormal{and}}~~\gamma=\frac{\pi}{I-\sum_{j=1}^{\mathrm{k}-1}n_{_j}}
\end{equation}

Appendix B contains a C program that singles out the $\{n_{_1},
n_{_2},...,n_{_k}\}$ set for which the total area (Eq.(78))
attains its maximum. Rather than solving Eqs.(80) the program
scans the entire subspace of permissible partitions ($n_{_1},
n_{_2},...,n_{_k}$;$~n_{_1}+n_{_2}+...+n_{_k}=I$) while it keeps
track of the largest value of $A_{total}$. Statements printing out
each set and the corresponding $A_{total}$ are included in the
code, but have been commented out for the sake of execution speed.

\section{Conclusions}
We began this paper with a single cut of a piece of wire and
computed the lengths of the segments formed so that the total
combined area of the arbitrary sided polygons bent from these two
segments is maximized and minimized.  After solving this extrema
problem, we went on to find the cutting range so that the combined
area of these polygons is greater than or less than a
user-specified area.  Here we also mention the constraint on the
selection of this user-specified area. For the reader we have
worked out examples in exact form, leaving the answer in terms of
the length of the wire.  These examples consist of all kinds of
combinations of polygons, however, for each of the above four
cases, a hexagon and a square is examined to get a feel of how
this combination carries itself out for each of the four cases.

We have extended the results of the above four cases to three
partitioned wires and finally to any number of partitioned wires.
However, for the inequality problem with more than one cut, we
considered the simplified case where all of the polygons are
assigned the same variable perimeter except the last one which is
automatically determined from the total length of the wire. To
illustrate to the reader the use of these results, we considered a
5-cut case (6-partitioned) of a wire of length 20 units . The
segments are bent into a triangle, square, hexagon, octagon,
dodecagon, and a circle. We found the following assignment for the
perimeter of the polygons: 4.654, 3.582, 3.103, 2.968, 2.880, and
2.814 units respectively, so that the combined area of these
polygons is minimized with this area given by 4.478 square units.
For the total area of the polygons to be maximized, we found the
following allocation for the perimeter of the polygons: 4.669 for
the square, 4.043 for the hexagon, 3.868 for the octagon, 3.753
for the dodecagon, and 3.667 for the circle. Note that for this
case none of the wire is used for the triangle. For the inequality
problem, we chose the user-specified area to be 23 square units
from a possible range of 4.599 to 31.831 square units. So, for the
upper bound problem, this result gives a cutting range between
0.609 and 6.235 units for the first five polygons and for the
lower bound problem a cutting range between 0 and 0.609 or 6.235
and 20 units.

In section 5 we explored the case where the number of sides of the
polygons formed is fixed and we calculated how this fixed number
of sides can be allocated firstly, among two wires, and then an
arbitrary number of wires so that the total area of the polygons
is maximized.

\section{Acknowledgements}
We wish to thank Health Careers Academy for allowing and
encouraging us to carry on many fruitful discussions surrounding
this paper. Two of us (RMS \& AC) wishes to also thank the
Department of Physics at Northeastern University for providing an
environment conducive to research.
\\

\section{Appendix A}
In this appendix we will show that the maximum area, ${\mathcal
A}_{_{n}}$, of an $n$-sided polygon (with perimeter, $P$) is
attained in the limiting case of a circle: $n\rightarrow\infty,
\theta_{_{n}}=\pi/2$.
\begin{eqnarray}
{\mathcal
A}_{_{n}}=\frac{P^2}{4n}\tan{\theta_{_{n}}};~~~\theta_{_{n}} =
(\frac{1}{2}-
\frac{1}{n})\pi \nonumber\\
\frac{d{\mathcal
A}_{_{n}}}{d\theta_{_{n}}}=\left[\frac{P(\pi-2\theta_{_{n}})}{4\pi}
    \right]^2\left[(\frac{\pi}{2}-\theta_{_{n}})\frac{1}{\cos^2\theta_{_{n}}}-\frac{\sin\theta_{_{n}}}{\cos\theta_{_{n}}}
    \right]=0
\end{eqnarray}
 so that
 \begin{eqnarray}
\textnormal{either}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\pi-2\theta_{_{n}}=0\nonumber\\
\textnormal{or}~~~~~~~~~~~~~~~~~~\sin^4\theta_{_{n}}-\sin^2\theta_{_{n}}+(\pi/2-\theta_{_{n}})^2=0
\end{eqnarray}
Clearly, the only solution to these equations is
\begin{equation}
\theta_{_{n}}=\frac{\pi}{2}
\end{equation}
Therefore,
 \begin{eqnarray}
\frac{\pi}{2}=(\frac{1}{2}- \frac{1}{n})\pi\nonumber\\
\Rightarrow n\rightarrow\infty
\end{eqnarray}
The acceptable values for $n$ are $3\leq n<\infty$, this implies
$\pi/6\leq \theta_{_{n}}\leq \pi/2$. $d{\mathcal
A}_{_{n}}/d\theta_{_{n}}<0$ for $\theta_{_{n}}$ in [$\pi/3$,
$\pi/2$) and $d{\mathcal A}_{_{n}}/d\theta_{_{n}}> 0$ for
$\theta_{_{n}}$ in ($\pi/2$, $\infty$). This implies $A_{_{n}}$
attains its maximum at $\theta_{_{n}}= \pi/2$.

Next, we calculate the corresponding maximum area,
\begin{eqnarray}
{\mathcal
A}_{_{n-max}}=\frac{P^2}{8\pi}(\pi-2\theta_{_{n}})\tan{\theta_{_{n}}}|_{\theta_{_{n}}=\frac{\pi}{2}}\nonumber
\end{eqnarray}
This is in the indeterminate form: $0\times\infty$. We put the
above expression in the form $\frac{\infty}{\infty}$ and apply
L'Hopital's rule successively  to carry out the limiting
procedure:
\begin{eqnarray}
 {\mathcal
A}_{_{n-max}}=\frac{P^2}{8\pi}\lim _{{\theta_{_{n}}\rightarrow
\frac{\pi}{2}}}
\frac{\tan{\theta_{_{n}}}}{\frac{1}{\pi-2\theta_{_{n}}}}~~~~~~~\nonumber\\
=\frac{P^2}{16\pi}\lim _{{\theta_{_{n}}\rightarrow \frac{\pi}{2}}}
\frac{(\pi-2\theta_{_{n}})^2}{\cos^2\theta_{_{n}}}\nonumber\\
=\frac{P^2}{4\pi}\lim _{{\theta_{_{n}}\rightarrow \frac{\pi}{2}}}
\frac{(\pi-2\theta_{_{n}})}{\sin 2\theta_{_{n}}}~~~\nonumber\\
=-\frac{P^2}{4\pi}\lim _{{\theta_{_{n}}\rightarrow
\frac{\pi}{2}}}\frac{1}{\cos 2\theta_{_{n}}}~~~~\nonumber\\
=\frac{P^2}{4\pi}~~~~~~~~~~~~~~~~~~~~~~~\nonumber\\
=\pi r^2~~~(P=C=2\pi r)
\end{eqnarray}
Obviously the minimum area occurs for the case of a triangle:
\begin{eqnarray}
{\mathcal
A}_{_{n-min}}=\frac{P^2}{4n}\tan{\left[\left(\frac{1}{2}-
\frac{1}{n}\right)\pi\right]}_{n=3}\nonumber\\
=\frac{P^2}{12\sqrt 3}~~~~~~~~~~~~~~~~~~~~~~~
\end{eqnarray}

\section{Appendix B}
The following is the computer code that looks for the $\{n_{_1},
n_{_2},...,n_{_k}\}$ set of polygonal sides which maximizes the
value of total area in Eq.(78). This appendix also contains four
outputs of the program.
\newpage

\section{References}
1.~~~Varberg, D. and Purcell, E. J., \emph{Calculus with Analytic
Geometry}. 6th ed. New $~~~~~~$Jersey:Prentice Hall, 1992. \\
2.~~~Embry, M. R., Schell, J. F. and Thomas, J. P., \emph{Calculus
and Linear Algebra}. $~~~~~~$Philadelphia:W.B. Saunders Company, 1972.\\
3.~~~Apostol, T. M., \emph{Mathematical Analysis}.
Reading:Addison-Wesley, 1958.

\end{document}


TITLE: Power Series (Laurent Series)
QUESTION [0 upvotes]: I need some help with this exercise:
I need to obtain the power series development of this function:
$$f(z)=\frac{\cos(z+1)}{(z^2-1)z}$$
Centered in $z_0=-1$ and valid in $z_1=\frac{1}{2}-i$
I know how to develop the power series of $\cos$ or $\frac{1}{1-\text{anything}}$ but this exercise is too difficult, I suppose.

REPLY [0 votes]: you can use the tree poles to calculate your series -1,0,1 XE
around the zero
$$\frac{x^2}{2 (x+1)}+\frac{x^2 \cos (2)}{2 (x-1)}-\frac{1}{2} x \cos (1)-\frac{\cos (1)}{x}+\sin (1)$$
or around the x=1
$$\frac{x^2}{8 (x+1)}+\frac{3 x^2 \sin (2)}{4 (x-1)}+\frac{5 x^2 \cos (2)}{8 (x-1)}-\frac{x}{4 (x+1)}+\frac{1}{8 (x+1)}-\frac{2 x \sin (2)}{x-1}+\frac{5 \sin (2)}{4 (x-1)}-x \cos (1)-\frac{2 x \cos (2)}{x-1}+\frac{15 \cos (2)}{8 (x-1)}-\frac{\cos (1)}{x}+2 \cos (1)$$
compare the solution of the taylor series around the zero
$$\frac{5}{6} x^2 \sin (1)-\frac{1}{2} x \cos (1)-\frac{\cos (1)}{x}+\sin (1)$$

the gree one is the taylor series therst color is the function with the other teo series.


TITLE: Entanglement across a bi-partition of a many particle system
QUESTION [0 upvotes]: Alice and Bob each have two qubits labelled $A', A$ and $B, B'$ respectively. The density operator of the total system is $\rho_{A'ABB' } \in \mathcal{L}(\mathcal{H}^2 \otimes \mathcal{H}^2 \otimes \mathcal{H}^2 \otimes \mathcal{H}^2) $.
I want to find the amount of entanglement in the $A'A : BB'$ bi-partition. Is it correct to think of Alice's and Bob's total system as a system of two qudits so that the density operator of the total system is now given by $\rho_{\mathbf{AB}} \in \mathcal{L}(\mathcal{H}^4 \otimes \mathcal{H}^4)$. Disregarding any structure of Alice's and Bob's subsystems,  as I am not interested in the amount of entanglement present in the subspaces of Alice and Bob individually? Where$\mathcal{L}(\mathcal{H})$ represents the set of linear operators on a Hilbert space \mathcal{H}.
What would be a suitable entanglement measure for the above scenario apart from Rungta's $I \  concurrence$ defined in https://journals.aps.org/pra/abstract/10.1103/PhysRevA.64.042315?
cross-posted at https://quantumcomputing.stackexchange.com/questions/26260/entanglement-across-a-bi-partition-of-a-many-particle-system

REPLY [2 votes]: Short answer: yes it is correct to think about qudits. In general, if you are looking at the amount of entanglement across a bipartition, you are ignoring the structure of whatever is within one side of the bipartition. Then you can apply any bipartite entanglement measure to your bipartite system (respectively combining each of $AA^\prime$ and $BB^\prime$ into one system) and you'll have succeeded.
The key is that YOU are the one choosing the bipartition. There may be entanglement properties across different bipartitions, there may be entanglement across all bipartitions, etc., and you get to choose what kind of entanglement you're quantifying when you inspect this system.


TITLE: Define $a \sim b$ if $a - b$ is an integer in $\Bbb R$. Show that ${}\sim{} $ is an equivalence relation. Show the classes of equivalence as well.
QUESTION [0 upvotes]: Define $a  \sim  b$ if $a - b$ is an integer in $\Bbb R$. Show that  ${}\sim{} $  is an equivalence relation. Show the classes of equivalence as well.
Here's my work. Am I correct? I also do not understand how to find equivalence classes.
Reflexive: $ a\sim  a$ -> a - a; so  ${}\sim{} $ is reflexive.
Symmetric: $a  \sim   b -> a - b$ then $b  \sim  a -> b - a$; so  ${}\sim{} $  is symmetric.
Transitive: $a  \sim   b -> a - b$ and $b \sim  c -> b - c$ then $a - c$ so $a  \sim  c$; so   ${}\sim{} $    is transitive.

REPLY [0 votes]: For the equivalence classes, each has a unique representative in $[0,1)$, so the the set of equivalence classes is
$$\{x+\mathbf Z\mid 0\le x<1 \}. $$
THis set is usually denoted $\mathbf R/\mathbf Z$ and is called the one-dimensional torus.