text:
\begin{document}
\maketitle
\begin{center}
 Institut f\"ur Mathematik,

 Technische Universit\"at Berlin,\\
 Strasse~des 17.~Juni 136, 10623 Berlin, Germany\\
 \medskip e-mail:{\tt bobenko@math.tu-berlin.de}
\end{center}

\begin{abstract}
A new functional for simplicial surfaces is suggested. It is
invariant with respect to M\"obius transformations and is a
discrete analogue of the Willmore functional. Minima of this
functional are investigated. As an application a bending energy
for discrete thin-shells is derived.
\end{abstract}

\paragraph{Keywords:} Conformal energy, Willmore functional,
simplicial surfaces, discrete differential geometry \eject

\section{Introduction}
\label{s.intro}

In the variational description of surfaces the following
functionals are of primary importance:
\begin{itemize}
    \item The area ${\cal A}=\int dA$, where $dA$ is the area
    element, is preserved by isometries.
    \item The total Gaussian curvature ${\cal G}=\int K dA$,
    where $K$ is the Gaussian curvature, is topological invariant.
    \item The total mean curvature ${\cal M}=\int H dA$, where $H$ is the
    mean curvature, depends on external geometry of the surface.
    \item The Willmore energy ${\cal W}=\int H^2 dA$ is invariant with
    respect to M\"obius transformations.
\end{itemize}

Geometric discretizations of the first three functionals for
simplicial surfaces are well known. For the area functional it is
obvious. The local Gaussian curvature at a vertex $v$ is defined
as the angle defect
$$ G(v)=2\pi - \sum_i \alpha_i, $$
 where
$\alpha_i$ are the angles of all triangles (see Fig.
\ref{f.alpha-beta}) at vertex $v$. The total Gaussian curvature is
a sum over all vertices $G=\sum_v G(v)$. The local mean curvature
at an edge $e$ is defined as
$$ M(e)=l\theta, $$
where $l$ is the length of the edge and $\theta$ is the angle
between the normals to the adjacent faces at $e$ (see Fig.
\ref{f.bending}). The total mean curvature is the sum over all
edges $M=\sum_e M(e)$. These discrete functionals possess the
geometric symmetries of the smooth functionals mentioned above.

Until recently a geometric discretization of the Willmore
functional was missing. In this paper we introduce a M\"obius
invariant energy for simplicial surfaces and show that it should
be treated as a discrete Willmore energy.

\section{Conformal Energy}
\label{s.energy}

Let $S$ be a simplicial surface in 3-dimensional Euclidean space
with the set of vertices $V$, edges $E$ and (triangular) faces
$F$. We define a conformal energy for simplicial surfaces using
circumcircles of their faces. Each (internal) edge $e\in E$ is
incident to two triangles. Consistent orientation of the triangles
naturally induces an orientation of the corresponding
circumcircles. Let $\beta(e)$ be the external intersection angle
of the circumcircles of the triangles sharing $e$, which is the
angle between the tangent vectors of the oriented circumcircles.
Now we are ready to define a new energy functional for simplicial
surfaces.

\begin{definition}
Local conformal (discrete Willmore) energy at a vertex $v$ is
given by the sum over all incident edges
$$
W(v)=\sum_{e\ni v} \beta(e)-2\pi.
$$
The conformal (discrete Willmore) energy of the simplicial surface
is the sum over all vertices
$$
W(S)=\dfrac{1}{2}\sum_{v\in V}W(v)=\sum_{e\in E} \beta(e)-\pi\mid
V \mid,
$$
where $\mid V\mid$ is the number of vertices of $S$.
\end{definition}

\begin{figure}[h]
\begin{center}
\parbox[c]{0.4\textwidth}{\input{dwill2.pstex_t}}

\parbox[c]{0.3\textwidth}{\input{circles.pstex_t}}
\end{center}
\caption{Definition of conformal (discrete Willmore) energy
\label{f.dEnergy}}
\end{figure}

Fig.\ref{f.dEnergy} presents two neighboring circles with their
external intersection angle $\beta_i$ as well as a view ``from the
top'' at a vertex $v$ showing all $n$ circumcircles passing
through $v$ with the corresponding intersection angles
$\beta_1,\ldots,\beta_n$. For simplicity we will consider only
simplicial surfaces without boundary.

Note that the energy we defined is obviously invariant with
respect to M\"obius transformations. This invariance is an
important property of the classical Willmore energy defined for
smooth surfaces (see below).

Note also that the changing of the orientation of both circles
preserves the angle $\beta(e)$. This shows that the energy $W(S)$
is well defined for non-oriented simplicial surfaces as well.

The star $S(v)$ of the vertex $v$ is the subcomplex of $S$
comprised by the triangles incident with $v$. The vertices of
$S(v)$ are $v$ and all its neighbors. We call $S(v)$ convex if for
any its face $f\in F(S(v))$ the star $S(v)$ lies to one side of
the plane of $F$ and strictly convex if the intersection of $S(v)$
with the plane of $f$ is $f$ itself.

\begin{proposition} \label{p.non-negative}
The conformal energy is non-negative
$$W(v)\ge 0,$$
and vanishes if and only if the star $S(v)$ is convex and all its
vertices lie on a common sphere.
\end{proposition}

The proof of this proposition is based on the following elementary
lemma.
\begin{lemma} \label{l.alpha-beta}
Let $\cal P$ be a (not necessarily planar) $n$-gon with external
angles $\beta_i$. Choose a point $P$ and connect it to all
vertices of $\cal P$. Let $\alpha_i$ be the angles (see
Fig.\ref{f.alpha-beta}) of the triangles at the tip $P$ of the
obtained pyramid. Then
$$
\sum_{i=1}^n \beta_i\ge \sum_{i=1}^n \alpha_i,
$$
and the equality holds if and only if $\cal P$ is planar and
convex\footnote{The obtained pyramid is convex in this case. Note
that we distinguish convex and strictly convex polygons (and
pyramides). Some of the angles $\beta_i$ of a convex polygon may
vanish. The corresponding cite-triangles of the pyramid lie in one
plane.}
 and the vertex
$P$ lies inside $\cal P$.
\end{lemma}

\begin{figure}[h]
\begin{center}
\parbox[c]{0.4\textwidth}{\input{lemma3.pstex_t}}
\end{center}
\caption{Proof of Lemma \ref{l.alpha-beta} \label{f.alpha-beta}}
\end{figure}

{\it Proof}. Let us denote by $\gamma_i$ and $\delta_i$ the angles
of the cite-triangles at the vertices of $\cal P$ (see
Fig.\ref{f.alpha-beta}). The claim of Lemma \ref{l.alpha-beta}
follows from adding over all $i=1,\ldots,n$ two obvious
(in)equalities
\begin{eqnarray*}
\beta_{i+1}&\ge& \pi -(\gamma_{i+1}+\delta_i)\\
\pi -(\gamma_{i}+\delta_i)&=&\alpha_i.
\end{eqnarray*}
All inequalities become equalities only in the case when $\cal P$
is planar, convex and contains $P$. Note that some of the external
angles $\beta_i$ may vanish. As a corollary we obtain a polygonal
version of Fenchel's theorem \cite{Fenchel}.

\begin{corollary}  \label{c.non-negative}
$$
\sum_{i=1}^n \beta_i\ge 2\pi.
$$
\end{corollary}
{\it Proof.} For a given $\cal P$ choose the point $P$ varying on
a straight line encircled by $\cal P$. There always exist points
$P$ so that the star at $P$ is not strictly convex, and thus $\sum
\alpha_i \ge 2\pi$.

{\it Proof of Proposition \ref{p.non-negative}}. The claim of
Proposition \ref{p.non-negative} is invariant with respect to
M\"obius transformations. Applying a M\"obius transformation $M$
which maps the vertex $v$ to infinity, $M(v)=\infty$, one observes
that all the circles passing through $v$ become straight lines and
we arrive at the geometry shown in Fig.\ref{f.alpha-beta} with
$P=M(\infty)$. Now the claim follows immediately from Corollary
\ref{c.non-negative}.

\begin{theorem}
Let $S$ be a simplicial surface. Then
$$W(S)\ge 0,$$
and the equality holds if and only if $S$ is a (part of a) convex
polyhedron inscribed in a sphere.
\end{theorem}
{\it Proof.} Only the second statement needs to be proven. Due to
Proposition \ref{p.non-negative}  the equality $W(S)=0$ implies
that all vertices and edges of $S$ are convex (but not necessarily
strictly convex). Deleting the edges which separate triangles
lying in one plane one obtains a polyhedral surface $S_P$ with
circular faces and all strictly convex vertices and edges.
Proposition \ref{p.non-negative} implies that for every vertex $v$
there exists a sphere $S_v$ with all vertices of the star $S(v)$
lying on it. For any edge $(v_1,v_2)$ of $S_P$ two neighboring
spheres $S_{v_1}$ and $S_{v_2}$ share two different circles of
their common faces. This implies $S_{v_1}=S_{v_2}$ and finally the
coincidence of all the spheres $S_v$.

Discrete conformal energy $W$ defined above is a discrete analogue
of the Willmore energy for smooth surfaces \cite{Willmore}
$$
{\cal W(S)}=\frac{1}{4}\int_{\cal S} (k_1-k_2)^2 dA=\int_{\cal S}
H^2 dA- \int_{\cal S} K dA.
$$
Here $dA$ is the area element, $k_1, k_2$ the principal
curvatures, $H=\frac{1}{2}(k_1+k_2)$ the mean curvature, $K=k_1
k_2$ the Gaussian curvature of the surface. Here we prefer a
definition for $\cal W$ with a M\"obius invariant integrand. It
differs from the one in the introduction by a topological
invariant.

Let us mention two important properties of the Willmore energy:
\begin{itemize}
    \item  ${\cal W(S)}\ge 0$ and ${\cal W(S)}=0$ if and only if
    $\cal S$ is the round sphere.
    \item  $\cal W(S)$ (and the integrand $(k_1-k_2)^2 dA$) is M\"obius
    invariant \cite{Blaschke,Willmore}
\end{itemize}
Whereas the first claim almost immediately follows from the
definition, the second one is a non-trivial property. We have
shown that the same properties hold for the discrete energy $W$;
in the discrete case the M\"obius invariance is built into the
definition and the non-negativity of the energy is non-trivial.

In the same way one can define conformal
    (Willmore) energy for simplicial surfaces in
    Euclidean spaces of higher dimension and space forms.

The discrete conformal energy is well defined for
    polyhedral surfaces with circular faces (not necessarily
    simplicial).

\section{Computation of the Energy}

Consider two triangles with a common edge. Let $a, b, c, d\in
{{\mathbb R}^3}$ be their other edges oriented as in
Fig.\ref{f.formula}. Identifying vectors in ${\mathbb R}^3$ with
imaginary quaternions ${\rm Im}\ \mathbb{H}$ one obtaines for the
quaternionic product
\begin{equation}                \label{product}
ab=-<a,b>+a\times b,
\end{equation}
where $<a,b>$ and $a\times b$ are the scalar and vector products
in ${\mathbb R}^3$.

\begin{figure}[h]
\begin{center}
\parbox[c]{0.4\textwidth}{\input{formula.pstex_t}}
\end{center}
\caption{Formula for the angle between circumcircles
\label{f.formula}}
\end{figure}

\begin{proposition}
The external angle $\beta\in [0,\pi ]$ between the circumcircles
of the triangles in Fig.\ref{f.formula} is given by one of the
equivalent formulas:
\begin{eqnarray*}
\cos(\beta)&=&-\dfrac{{\rm Re}\ q}{\mid q\mid}=
            -\dfrac{{\rm Re}\ abcd}{\mid abcd\mid}= \\
           &=&\dfrac{<a,c><b,d>-<a,b><c,d>-<b,c><d,a>}
           {\mid a\mid \mid b\mid \mid c\mid \mid d\mid}.
\end{eqnarray*}
Here $q=ab^{-1}cd^{-1}$ is the cross-ratio of the quadrilateral.
\end{proposition}
{\it Proof.} Since ${\rm Re}\ q$, $\mid q\mid$ and $\beta$ are
M\"obius invariant it is enough to prove the first formula for the
planar case $a,b,c,d\in {\mathbb C}$, mapping all four vertices to
a plane by a M\"obius transformation. In this case $q$ becomes the
classical complex cross-ratio. Considering the arguments
$a,b,c,d\in {\mathbb C}$ one easily arrives at $\beta=\pi-\arg q$.
The second representation follows from the identity
$b^{-1}=-b/\mid b\mid$ for imaginary quaternions. Finally applying
(\ref{product}) we obtain
\begin{eqnarray*}
{\rm Re}\ abcd=<a,b><c,d>-<a\times b,c\times d>=\\
<a,b><c,d>+<b,c><d,a>-<a,c><b,d>. \end{eqnarray*}

\section{Minimizing Discrete Conformal Energy}

Similarly to the smooth Willmore functional $\cal W$, minimizing
the discrete conformal energy $W$ makes the surface as round as
possible.

Let us denote by $\bf S$ the combinatorial data of $S$. The
simplicial surface $S$ is called a geometric realization of the
abstract simplicial surface $\bf S$.

\begin{definition}
Critical points of $W(S)$ are called simplicial Willmore surfaces.
The conformal (Willmore) energy of an abstract simplicial surface
is the infimum over all geometric realizations
$$
W({\bf S})=\inf_{S\in {\bf S}} W(S).
$$
\end{definition}

\begin{figure}[h]
 \begin{center}
\begin{tabular}{c c c c c}
\includegraphics[width=0.25\textwidth]{sphere1.eps}&\hfill&
\includegraphics[width=0.25\textwidth]{nonsphere1.eps}&\hfill&
\includegraphics[width=0.25\textwidth]{boy1.eps}\\
\includegraphics[width=0.25\textwidth]{sphere4.eps}&\hfill&
\includegraphics[width=0.25\textwidth]{nonsphere3.eps}&\hfill&
\includegraphics[width=0.25\textwidth]{boy2.eps}

\end{tabular}
\end{center}
\caption{Discrete Willmore spheres of ``inscribable'' ($W=0$) and
``non-inscribable'' ($W>0$) type and discrete Boy surface
\label{f.minimizing}}
\end{figure}

Kevin Bauer implemented the proposed conformal functional with the
Brakke's evolver \cite{Brakke} and did some minimization numerical
experiments. Examples of those are presented in
Fig.\ref{f.minimizing}. In the first and the second lines the
initial configurations and the corresponding Willmore surfaces
minimizing the conformal energy respectively are shown. Let us
call the gradient flow of the energy $W$ the discrete Willmore
flow. By this flow the energy of the first simplicial sphere
decreases to zero and the surface evolves into a convex polyhedron
with all the vertices lying on a sphere. The abstract simplicial
surface of the central example is different and we obtain a
simplicial Willmore sphere with positive conformal energy. The
third example is a simplicial projective plane. The initial
configuration is made from squares divided into triangles (cf.
\cite{Petit}). We see that the minimum is close to the smooth Boy
surface known to minimize \cite{KarcherP} the Willmore energy for
projective planes.

The minimization of the conformal energy for simplicial spheres is
related to a classical result of Steinitz \cite{Steinitz} who has
shown that there exist abstract simplicial 3-polytopes without
geometric realizations all vertices of which belong to a sphere.
We call these combinatorial types non-inscribable.

The non-inscribable examples of Steinitz are constructed as
follows \cite{Gruenbaum}. Let $\bf S$ be an abstract simplicial
sphere with vertices colored in black and white. Denote the sets
of white and black vertices by $V_w$ and $V_b$ respectively,
$V=V_w\cup V_b$. Assume that the number of black vertices does not
exceed the number of white vertices, $\mid V_w\mid \ge \mid
V_b\mid $, and there are no edges connecting two white vertices
and there are edges connecting black vertices. It is easy to see
that $\bf S$ with these properties cannot be inscribed in a
sphere. Indeed, assume that we have constructed such inscribed
convex polyhedron. Then the equality of the intersection angles at
both ends of an edge (see left Fig.\ref{f.dEnergy}) implies
$$
2\pi \mid V_b\mid \ge  \sum_{e\in E} \beta(e)\ge 2\pi \mid
V_w\mid,
$$
and the equalities hold only if all edges connect vertices of
different color. The obtained contradiction to the assumed
inequality implies the claim.

To construct abstract polyhedra with $\mid V_w\mid \ge \mid
V_b\mid $ and with some edges connecting black points, take a
polyhedron $\bf P$ whose number of vertices does not exceed the
number of faces $\mid \hat{F}\mid \ge \mid \hat{V}\mid $. Color
all the vertices in black, add white vertices at the faces and
connect them to all black vertices of a face. We obtain a
polyhedron with black (original) edges and $\mid V_w\mid=\mid
\hat{F}\mid\ge\mid V_b\mid=\mid \hat{V}\mid$. The example with
minimal possible number of vertices $\mid V\mid =11$ is shown in
Fig.\ref{f.11}. The starting polyhedron $\bf P$ here are two
tetrahedra identified along a common face: $\hat{F}=6, \hat{V}=5$.

\begin{figure}[h]
 \begin{center}
\includegraphics[width=0.35\textwidth]{11.eps}
\end{center}
\caption{A discrete Willmore sphere of ``non-inscribable'' type
with 11 vertices and $W=2\pi$ \label{f.11}}
\end{figure}
Hodgson, Rivin and Smith \cite{HodgsonRS} found  a
characterization of inscribable combinatorial types, based on a
transfer to the Klein model of hyperbolic 3-space. It is not clear
whether there exist non-inscribable examples of non-Steinitz type.

Numerical experiments lead us to the following
\begin{conjecture}
The conformal energy of simplicial Willmore spheres is quantized
$$
W=2\pi N,\quad N\in {\mathbb N}.
$$
\end{conjecture}
Note that this claim belongs to differential geometry of discrete
surfaces. It would be interesting to find a (combinatorial)
meaning of the integer $N$.  Compare also with the famous
classification of smooth Willmore spheres by Bryant \cite{Bryant},
who has shown that the energy of Willmore spheres is quantized
${\cal W}= 4\pi N, N\in{\mathbb N}$.

The discrete Willmore energy is defined for the ambient spaces
($\mathbb{R}^n$ or $S^n$) of any dimension. This leads to
combinatorial Willmore energies
$$
W_n(S)=\inf_{S\in {\bf S}} W(S), \qquad S\subset S^n,
$$
where the infimum is taken over all realizations in the
$n-$dimensional sphere. Obviously these numbers build a
non-increasing sequence $W_n(S)\ge W_{n+1}(S)$ which becomes
constant for sufficiently large $n$.

Complete understanding of non-inscribable simplicial spheres is an
interesting mathematical problem. However the phenomenon of
existence of such spheres might be seen as a problem in using of
the conformal functional for applications in computer graphics,
such as fairing of surfaces. Fortunately the problem disappears
just after one refinement step: all simplicial spheres become
inscribable. Let $\bf S$ be an abstract simplicial sphere. Define
its refinement $\bf S_R$ as follows: split every edge of $\bf S$
in two by putting additional vertices and connect these new
vertices sharing a face of $\bf S$ by additional edges.

\begin{proposition}\label{p.refine}
The refined simplicial sphere $\bf S_R$ is inscribable, and thus
$W({\bf S_R})=0$.
\end{proposition}

{\it Proof.} Koebe's theorem (see, for example, \cite{Ziegler},
\cite{BobenkoS}) claims that every abstract simplicial sphere $\bf
S$ can be realized as a convex polyhedron $S$ all edges of which
touch a common sphere $S^2$. Starting with this realization $S$ it
is easy to construct a geometric realization $S_R$ of the
refinement $\bf S_R$ inscribed in $S^2$. Indeed choose the
touching points of the edges of $S$ with $S^2$ as additional
vertices of $S_R$ and project the original vertices of $S$ (which
lie outside of the sphere $S^2$) to $S^2$. One obtaines a convex
simplicial polyhedron $S_R$ inscribed in $S^2$.

Another interesting variational problem with the conformal energy
is the optimization of triangulations of a given simplicial
surface. Here one fixes the vertices and chooses an equivalent
triangulation (abstract simplicial surface $\bf S$) minimizing the
conformal functional. The minimum
$$
W(V)=\min_{{\bf S}\ni S} W(S)
$$
yields ``an optimal'' triangulation for a given vertex data. In
the case of $S^2$ this optimal triangulation is classical.

\begin{proposition}
Let $S$ be a simplicial surface with all vertices $V$ on a two
dimensional sphere $S^2$. Then $W(S)=0$ if and only if it is the
Delaunay triangulation on the sphere, i.e. $S$ is the boundary of
the convex hull of $V$.
\end{proposition}

In differential geometric applications like numerical minimizing
the Willmore energy of smooth surfaces (cf. \cite{HsuKS}) it is
not natural to preserve the triangulation by minimizing the
energy, and one should also change the combinatorial type
decreasing the energy.

Discrete conformal energy $W$ is not only a discrete analogue of
the Willmore energy. One can show that it approximates the smooth
Willmore energy although the smooth limit is very sensitive to the
refinement method and should be chosen in a very special way. A
computation which will be published elsewhere shows that if one
chooses the vertices of a curvature line net of a smooth surface
$\cal S$ for the vertices of $S$ and triangularizes it then $W(S)$
converges to $\cal W(S)$ by natural refinement. On the other hand
the infinitesimal equilateral triangular lattice gives in the
limit 3/2 times larger energy. Possibly the minimization of the
discrete Willmore energy with the vertices on the smooth surface
could be used for computation of the curvature line net. We are
going to investigate this interesting and complicated phenomenon.

\section{Bending of Simplicial Surfaces}

An accurate model for bending of discrete surfaces is important
for modelling in virtual reality.

Let ${\cal S}_0$ be a thin shell and ${\cal S}$ its deformation.
The bending energy of smooth thin shells is given by the integral
\cite{GrinspunHDS}
$$
E=\int (H-H_0)^2 dA,
$$
where $H_0$ and $H$ are the mean curvatures of the original and
deformed surface respectively. For $H_0=0$ it reduces to the
Willmore energy.

To derive the bending energy for simplicial surfaces let us
consider the limit of fine triangulation, i.e. of small angles
between the normals of neighboring triangles. Consider an
isometric deformation of two adjacent triangles. Let $\theta$ be
the complement of the dihedral angle of the edge $e$, or,
equivalently, the angle between the normals of these triangles
(see Fig.\ref{f.bending}) and $\beta(\theta)$ the external
intersection angle between the circumcircles of the triangles (see
Fig.\ref{f.dEnergy}) as a function of $\theta$.

\begin{proposition}  \label{p.bending}
Assume that the circumcenters of the circumcircles of two adjacent
triangles do not coincide. Then in the limit of small angles
$\theta\to 0$ the angle $\beta$ between the circles behaves as
follows:
$$
\beta(\theta)=\beta(0)+\dfrac{l}{L}\theta^2+o(\theta^3).
$$
Here $l$ is the length of the edge and $L\ne 0$ is the distance
between the centers of the circles.
\end{proposition}

This proposition and our definition of conformal energy for
simplicial surfaces motivate to suggest
$$
E=\sum_{e\in E}\dfrac{l}{L}\theta^2
$$
for the bending energy of discrete thin-shells.

\begin{figure}[h]
\begin{center}
\parbox[c]{0.4\textwidth}{\input{bending2.pstex_t}}
\end{center}
\caption{To definition of the bending energy for simplicial
surfaces \label{f.bending}}
\end{figure}

In \cite{BridsonMF, GrinspunHDS} similar representations for the
bending energy of simplicial surfaces were found empirically. They
were demonstrated to give convincing simulations and good
comparison with real processes. In \cite{GrinspunHDS} the distance
between the barycenters is used for $L$ in the energy expression
but possible numerical advantages in using circumcenters are
indicated.

Using the Willmore energy and Willmore flow is a hot topic in
computer graphics. Applications include fairing of surfaces and
surface restoration. We hope that our conformal energy will be
useful for these applications and plan to work on them.

{\bf Acknowledgements}. I would like to thank Ulrich Pinkall for
the discussion in which the idea of the discrete Willmore
functional was born. I am also grateful to G\"unter Ziegler, Peter
Schr\"oder, Boris Springborn, Yuri Suris and Ekkerhard Tjaden for
useful discussions and to Kevin Bauer for making numerical
experiments with the conformal energy.

config: arxiv


TITLE: How do I multiply n linear polynomials together in python?
QUESTION [2 upvotes]: I want to write some python code that finds the coefficients of the polynomial that comes from $(b_0z - a_0)*(b_1z - a_1)*...*(b_nz - a_n)$ when given two lists of $a_0, ..., a_n$ and $b_0, ..., b_n$.
I know that I need to multiply by the n brackets separately using recursion but I'm not getting the right outputs.
This is my attempt so far:
def p(a,b):
    p = [1 for j in range(len(a))]
    p[0] = -a[0]
    p[1] = b[0]
    for j in range(len(a)-1):
        p[j] = p[j-1] * b[j] - p[j] * a[j]
    return p

REPLY [0 votes]: I think the array p should be initialized with 0s and not 1s.
But in the product formula, you have to make two loops.
We have
$\left(\sum_{j=0}^{n} p[j]x^j\right)\left(b[k]x-a[k]\right)$
$=\sum_{j=0}^{n} p[j]b[k]x^{j+1}-\sum_{j=0}^{n} p[j]a[k]x^{j}$
$=\sum_{j'=1}^{n+1} p[j'-1]b[k]x^{j'}-\sum_{j=0}^{n} p[j]a[k]x^{j}$
$=p[n]b[k] x^{n+1} + \left(\sum_{j=1}^{n} (p[j-1]b[k]-p[j]a[k])x^{j}\right)+p[0]a[k]$
So you must have one loop on k and another on j  (in the non-recursive version)


TITLE: How to understand the function $o(h) + o(h) = o(h)$?
QUESTION [0 upvotes]: On page 150 of the book Introduction to Mathematical Statistics, there's the following passage:

Let the symbol $o(h)$ represent any function such that $lim_{h→0}[o(h)/h] = 0;$ for example, $h^2 = o(h)$ and $o(h) + o(h) = o(h)$.

I can understand how $h^2 = o(h)$ fits the description. But I couldn't understand the function $o(h) + o(h) = o(h)$.

REPLY [3 votes]: Big- and little-O notation can be confusing because it violates everything we know about using variables in equations, since each instance of $o(h)$ refers to a (potentially) different function and the quantification (i.e. 'for all' vs. 'there exists') has to be inferred from context.
When we write an equation involving $o(h)$ terms, you should read it as saying that each $o(h)$ can be replaced by $f(h)$ for some function $f$ satisfying $\lim_{h \to 0} \frac{f(h)}{h} = 0$, and that the function $f$ might be different each time.
So in your case, writing $o(h)+o(h)=o(h)$ means:

For all functions $f,g$ such that $\lim_{h \to 0} \frac{f(h)}{h} = 0$ and $\lim_{h \to 0} \frac{g(h)}{h} = 0$, there exists a function $k$ such that $\lim_{h \to 0} \frac{k(h)}{h} = 0$ and $f(h)+g(h)=k(h)$.

(Note that $f$ and $g$ are universally quantified and $k$ is existentially quantified.)
Proving this statement is now easy using elementary facts about limits of functions: just define $k(h)=f(h)+g(h)$ and prove that $\lim_{h \to 0} \frac{k(h)}{h} = 0$.


TITLE: Proving that an alternating sequence does not converge
QUESTION [2 upvotes]: I have the sequence $$(a_n)_{n \in {\mathbb{N}}} = \lim_{n\to\infty} \frac{\frac{n!}{n^n}+1}{\frac{n!}{n^n}+(-1)^n}$$ I can see intuitively why this doesn't converge as it acts like $(-1)^n$ for large $n$, which doesn't converge, but I'm unsure how to show formally that it doesn't converge?
I think assuming it has a limit, say $L$, and maybe choosing a certain $\epsilon$ so that $|a_n - L| < \epsilon$, then trying a few different cases of $a_n$'s to try and find a contradiction regarding the limit is the way to go, but I have no idea what $\epsilon$ to choose or how to construct the proof.

REPLY [3 votes]: The easiest way is to consider the subsequences $(a_{2n})$ and $(a_{2n+1})$, and show that the first one converges to $1$ and the second one to $-1$.


TITLE: Proving the identity $(\tan^2(x)+1)(\cos^2(-x)-1)=-\tan^2(x)$
QUESTION [1 upvotes]: Proving the trigonometric identity $(\tan{^2x}+1)(\cos{^2(-x)}-1)=-\tan{^2x}$ has been quite the challenge. I have so far attempted using simply the basic trigonometric identities based on the Pythagorean Theorem. I am unsure if these basic identities are unsuitable for the situation or if I am not looking at the right angle to tackle this problem.

REPLY [0 votes]: We know

$(\tan x)' = 1 + \tan^2 x = \frac{1}{\cos^2 x}$ and
$\cos (-x) = \cos x$

Now, it follows immediately
\begin{eqnarray*} (\tan^2(x)+1)(\cos^2(-x)-1)
& = & (\tan x)'\left(\frac{1}{(\tan x)'}-1\right) \\
& = & 1 - (\tan x)' \\
& = & 1 - (1 + \tan ^2 x) \\
& = & - \tan ^2 x \\
\end{eqnarray*}


TITLE: Stoke's Theorem Theory Question
QUESTION [1 upvotes]: If $C$ is the boundary of a surface $S$ and $\phi$ and $\psi$ are arbitrary smooth scalar fields, Show that
$$\int_C \phi \nabla \psi \cdot dr = - \int_C \psi \nabla \phi \cdot d r = \iint_S (\nabla \phi \times \nabla \psi) \cdot n dS$$
I've tried working backwards, letting $\phi$ and $\psi$ be arbitrary and then generating something, but I just get a mess of symbols. Can't seem to figure it out.

REPLY [2 votes]: Use $$\text{curl} (fX) = \nabla \times fX = f\, \nabla \times X + \nabla f \times X$$
and
$$\text{curl} (\nabla f) = \nabla \times \nabla f = 0 $$
for every smooth scalar field (function) $f$ and vector field $X$.
Then Stokes Theorem implies
\begin{align*}
\int_C \phi \nabla \psi \cdot dr & = \iint_S (\nabla\times (\phi \nabla \psi)) \cdot n dS = \iint_S (\nabla \phi \times \nabla\psi + \phi \nabla \times \nabla\psi) \cdot n dS = \\
%
& = \iint_S (\nabla \phi \times \nabla\psi) \cdot n dS = -\iint_S (\nabla \psi \times \nabla\phi) \cdot n dS = - \int_C \psi \nabla \phi \cdot dr
\end{align*}


TITLE: Finding the minimal number of elements present in a set
QUESTION [0 upvotes]: Let $S$ be a set of $n$ distinct real numbers, and $A_s$ the set of arithmetic means of two distinct numbers from $S$. For given $n \geq 2$ find the minimal number of elements in $A_s$.
I feel that it will probably happen when the set of elements in $S$ can be paired up in such a way that the sum of each pair is a constant, but I am unable to give a rigorous proof for it. Any help please?

REPLY [1 votes]: Let the elements be $a_1 < a_2 < .... < a_n$.
Now the $n-1$ means of $a_1$ and $a_i$ where $i \ne 1$ are all distinct as the $a_i$s are all distinct so there are at least $n-1$ distinct means.
Now the mean between $a_2$ and $a_i$ might be equal to one of the $a_1$ and $a_j$ means for a $j > i$.  But the $a_2, a_n$ must be larger than any mean we've done so far.  So that's at least one more distinct mean.
We can continue that the mean between each $a_k$ and $a_n$ must be larger than any mean betweeen $a_i, a_j$ where $i < k$.  So that means the means the $n-2$ means, $\frac{a_2 + a_n}2$ to $\frac {a_{n-1}+ a_n}2$ are each distinct and larger than the $n-1$ distinct means $\frac {a_1 + a-2}2$ to $\frac {a_1 + a_n}2$.
So at the very least there are $(n-1) + (n-2)=2n -3$ possible distinct means.
But is that possible to have that few?
And yes.  If the $a_i$ are in arithemetic progression (example if $a_i = i$) then the mean between $a_i$ and $a_j$ ($i <j$) will be the same as the mean between $a_{i-k}$ and $a_{j+k}$ for all $k; 0\le k < i$ and this is precisely such a case.
So the minimum is $2n -3$
....
A visual:  $a_1 < .... < a_n$ so
the means are
$\begin{cases} {\frac {a_1+a_2}2} & \frac {a_1 + a_3}2 &\frac {a_1 + a_4}2&...&\frac {a_1 + a_{n-2}}2&\frac {a_1 + a_{n-1}}2&\frac {a_1 + a_{n}}2 \\
=m_{1,2} & =m_{1,3} &m_{1,4}&...&=m_{1,n-2}&=m_{1,n-1}&=m_{1,n} \\
* & \frac {a_2 + a_3}2 &\frac {a_2 + a_4}2&...&\frac {a_2 + a_{n-2}}2&\frac {a_2 + a_{n-1}}2&\frac {a_2 + a_{n}}2\\
* & =m_{2,3} &=m_{2,4}&...&=m_{2,n-2}&=m_{2,n-1}&=m_{2,n}\\
* & * &\frac {a_3 + a_4}2&...&\frac {a_3 + a_{n-2}}2&\frac {a_3 + a_{n-1}}2&\frac {a_3 + a_{n}}2\\
* & * &=m_{3,4}&...&=m_{3,n-2}&=m_{3,n-1}&=m_{3,n}\\
...\\
* & * &*&...&\frac {a_{n-3} + a_{n-2}}2&\frac {a_{n-3} + a_{n-1}}2&\frac {a_{n-3} + a_{n}}2\\
* & * &*&...&=m_{n-3,n-2}&=m_{{n-3},n-1}&=m_{{n-3},n}\\
* & * &*&...&*&\frac {a_{n-2} + a_{n-1}}2&\frac {a_{n-2} + a_{n}}2\\
* & * &*&...&*&=m_{{n-2},n-1}&=m_{{n-2},n}\\
* & * &*&...&*&*&\frac {a_{n-1} + a_{n}}2\\
* & * &*&...&*&*&=m_{{n-1},n}\\
\end{cases}$
This makes at most $\frac {n(n-1)}2$ potentially distinct means.
Because $a_i > a_{i-1}$ we can see $m_{i,j} > m_{i-1, j}$ (every cell is larger than the cell to the left) and $m_{i,j} > m_{i,j-1}$ (every cell is larger than the cell above).
So $m_{1,2} < m_{1,3} < ... < m_{1,n} < m_{2,n} < .... < m_{n-1,n}$ make at least $2n -3$ distinct means.
By pigeon hole precision we can see that there can be exactly $2n -3$ distinct means if only if each $m_{j,k} = m{j+1, k-1}$.  I'll leave it to you to do the algebra to note that that occurs precisely if each $a_k = b + k*d$ for some $A$ and $d> 0$.  (i.e. if the $a_i$ are in an arithmetic progression.)
Hint.  $\frac {a_j + a_k}2 = \frac {a_{j+1} + a_{k-1}}2 \iff$
$a_{j+1} - a_j = a_k -a_{k-1}$.  Apply induction.

REPLY [1 votes]: First, the set $S = \{1,2,...,n\}$, has $A_S = \{\frac l2 : 3 \leq l \leq 2n-1\}$. I leave you to see this for yourself.

Now, let $n \geq 2$, and $S = \{a_1,...,a_n\}$ with $a_i < a_j$ for $i < j$. Clearly, $\frac{a_i + a_j}{2} < \frac{a_k + a_l}{2}$ for every $1 \leq i ,j,k,l \leq n$, with $\max\{i,j\} \leq \min\{k,l\}$ and at least two of $i,j,k,l$ distinct.
Denote $\frac{a_i + a_j}{2} = (i,j)$. Then $(1,2) < (2,3) < ... < (n-1,n)$ is an obvious chain, giving $n-1$ means. However, we can actually do better :
$$
(1,2) < (1,3) < (2,3) < (2,4) < (3,4) < (3,5) < (4,5) < \ldots < (n-2,n-1) < (n-2,n) < (n-1,n)
$$
are all members of $A_S$, where the inequality follows from what we said earlier. This shows that $|A_S| \geq 2n-3$ ($n-1$ elements from the earlier argument, and then one in between each gives $n-2$ more elements). Of course, we have found a set attaining equality.

For example, $\{1,2\}$ attains equality for the $n=2$ case. $\{1,2,3\}$ has three means and attains equality for the $n = 3$ case.


TITLE: How exactly is Newton's Binomial Theorem Proven Using Combinatorial Ideas?
QUESTION [0 upvotes]: Sorry for the childish question. I am in 7th grade and need some help. Thank You!

REPLY [4 votes]: Let's imagine what happens if we expand
$$
(x+y)^3 = \underbrace{(x+y)}_{\text{factor 1}} \underbrace{(x+y)}_{\text{factor 2}} \underbrace{(x+y)}_{\text{factor 3}}
$$
without collecting like terms. The result is
\begin{align}
(x + y)^3 &= x\cdot x \cdot x \\
& + x \cdot x \cdot y + x \cdot y \cdot x + y \cdot x \cdot x \\
&+ x \cdot y \cdot y + y \cdot x \cdot y + y \cdot y \cdot x \\
&+ y \cdot y \cdot y.
\end{align}
Each term in the sum contains a contribution from factor 1, a contribution from factor 2, and a contribution from factor 3. Each factor contributes either an $x$ or a $y$.
Now how many terms in this sum are equal to $x y^2$? Well, how many ways are there to choose two of the factors to contribute a $y$? The answer is $\binom{3}{2}$.
Likewise, the sum contains $\binom{3}{0}$ terms that are equal to $x^3$, $\binom{3}{1}$ terms that are equal to $x^2 y$, and $\binom{3}{3}$ terms that are equal to $y^3$.
That explains why
$$
(x + y)^3 = \binom{3}{0} x^3 + \binom{3}{1}x^2 y + \binom{3}{2} x y^2 + \binom{3}{3} y^3.
$$


TITLE: Find the radius increase to account for a segmented circle
QUESTION [0 upvotes]: I want to find the required radius increase in order to maintain circle area when the circle is segmented. Example below:

(please note the outer circle is off centre, it is meant to maintain the same central position as the inner circle)
I essentially want to find $R_2$. $R_2$ is chosen such that $A_1 + A_2 = \pi(R_1)^2$. The only unknowns are $R_2$ and $A_2$.
Does anyone know the standard way of doing this? The current way of going about it involves finding the segmented areas of the inner circle, to find what area is missing. But even then I'm not sure how to account for the fact that the new circle will also be segmented by these same lines.

REPLY [1 votes]: The key is to remember that we can parametrize a circle with radius $r$ centered at $(x_c, y_c)$ using
$$y(x) = y_c \pm \sqrt{r^2 - (x - x_c)^2} \quad \iff \quad x(y) = x_c \pm \sqrt{r^2 - (y - y_c)^2}$$
For simplicity, let's mirror the case so that $A_2$ is the area between the two circles in the positive quadrant, $x \ge 0$ and $y \ge 0$.
Let the first circle be centered at $(x_1, y_1)$, $x_1 \ge 0$ and $y_1 \ge 0$, with radius $r_1 \ge x_1$ and $r_1 \ge y_1$.  Then, area $A_1$ is divided into four quadrants: one is a quarter of the circle, one is the rectangle between origin and the center of the circle, and two are rectangles with one side a circular arc:
$$A_1 = \frac{\pi}{4}r_1^2 + x_1 y_1 + \int_0^{y_1} \sqrt{r_1^2 - y^2} d x + \int_0^{x_1} \sqrt{r_1^2 - x^2} d y \tag{1a}\label{EQ1a}$$
That looks nasty, but when $0 \le X \le R$,
$$\int_0^X \sqrt{R^2 - x^2} d x ~ = ~  \frac{X}{2}\sqrt{R^2 - X^2} + \frac{r^2}{2}\operatorname{arcsin}\left(\frac{X}{R}\right)\tag{2}\label{EQ2}$$
Therefore,
$$\begin{aligned}
A_1 & = \frac{\pi}{4}r_1^2 ~ + ~ x_1 y_1 \\
~ & + \frac{x_1}{2}\sqrt{r_1^2 - x_1^2} ~ + ~ \frac{y_1}{2}\sqrt{r_1^2 - y_1^2} \\
~ & + \frac{r_1^2}{2}\left(\operatorname{arcsin}\left(\frac{x_1}{r_1}\right) + \operatorname{arcsin}\left(\frac{y_1}{r_1}\right)\right) \\
\end{aligned} \tag{1b}\label{EQ1b}$$
Let's assume the outer circle is concentric with the first one, with radius $r_2 \gt r_1$. The area $A_2$ is divided into three sections: one is a full quarter annulus, plus two ring sections between the two circular arcs:
$$\begin{aligned}
A_2 & = \frac{\pi}{4}\left(r_2^2 - r_1^2\right) \\
~ & + \int_0^{x_1} \sqrt{r_2^2 - x^2} - \sqrt{r_1^2 - x^2} d x \\
~ & + \int_0^{y_1} \sqrt{r_2^2 - y^2} - \sqrt{r_1^2 - y^2} d y \\
\end{aligned} \tag{3a}\label{EQ3a}$$
In this case, we can split each integrals into two, and use $\eqref{EQ2}$. We get
$$\begin{aligned}
A_2 & = \frac{\pi}{4}\left(r_2^2 - r_1^2\right) \\
~ & + \frac{x_1}{2}\left(\sqrt{r_2^2 - x_1^2} - \sqrt{r_1^2 - x_1^2} \right) \\
~ & + \frac{y_1}{2}\left(\sqrt{r_2^2 - y_1^2} - \sqrt{r_1^2 - y_1^2} \right) \\
~ & + \frac{r_2^2}{2}\left(\operatorname{arcsin}\left(\frac{x_1}{r_2}\right) + \operatorname{arcsin}\left(\frac{y_1}{r_2}\right)\right) \\
~ & - \frac{r_1^2}{2}\left(\operatorname{arcsin}\left(\frac{x_1}{r_1}\right) + \operatorname{arcsin}\left(\frac{y_1}{r_1}\right)\right) \\
\end{aligned}\tag{3b}\label{EQ3b}$$
Note that if the circles are not concentric, you get three to five segments between the two arcs instead (five integrals); it solves the exact same way, except that you have eight possible cases (depending on how the two centers are located with respect to each other; this here case, concentric, being the ninth case).
Unfortunately, it seems that solving $A_1 = A_2$ for $r_2$ does not yield a nice algebraic expression, so you'll have to find the answer numerically.
Geometrically, we can see that when $r_2$ increases, $A_2$ increases as well; it is just that the ratio of the increases varies in a way that is very hard to express algebraically.  Mathematically, we can say that $A_2$ is a monotonically increasing function of $r_2$ at $r_2 \gt r_1 \ge 0$.  This means that we can use a very efficient binary search algorithm to find $r_2$:

Start with $r_2 = 2 r_1$.
Calculate $A_2$.
If $A_2 \lt A_1$, double $r_2$, and repeat step 2.
If $A_2 = A_1$, you have found $r_2$, and are done.
Set $r_{min} = r_2 / 2$ and $r_{max} = r_2$.
Set $r_2 = \frac{r_{min} + r_{max}}{2}$, and calculate $A_2$.
If $A_2 \lt A_1$, set $r_{min} = r_2$, and go back to step 6.
If $A_2 \gt A_1$, set $r_{max} = r_2$, and go back to step 6.
Found $r_2$, and are done.

If you know the precision at which you need $r_2$, add a test for that to step 6, to check if $r_{min}$ and $r_{max}$ are close enough, and if so skip to end.
If you know the precision at which you know the areas, add a test for that to step 6, checking if the two areas are close enough, and if so skip to end.
If you use floating-point numbers, at minimum you need to add a check to step 6, that verifies that $r_2 \ne r_{min}$ and $r_2 \ne r_{max}$ (i.e., that there is enough precision to split the range), and otherwise skips to end.
This converges nicely, adding one bit of precision at each iteration.


TITLE: Contour integral around the disk with radius $ r$
QUESTION [1 upvotes]: In an exercise, we are looking at $\int_0^{2\pi}\frac{1}{re^{i\phi}+w}d\phi$ with $ w \in \Bbb C$. Now I tried to write it as an integral around the disk with radius $1$ and midpoint $1$. However my answer differs by a $-$ sign from the solution. I tried to do the following. $\delta B_r(0): [0, 2\pi]\to \Bbb C,\; \phi \mapsto re^{i\phi}$ and thus I obtain $\int_{\delta B_r(0)}\frac{i}{z(z+w)}\,dz\:$ instead of the solution $\:-\int_{\delta B_r(0)}\frac{i}{z(z+w)}\,dz$. I do not seem to understand where the minus comes into play.

REPLY [2 votes]: You have the change of variable $\; z=r\mathrm e^{i\phi}$, so that
$$\mathrm d z=ir\mathrm e^{i\phi}\,\mathrm d\phi=iz\,\mathrm d\phi\iff \mathrm d\phi=\frac1{i}\frac{\mathrm d z}z,$$
and $\:\dfrac1i=-i$.

REPLY [1 votes]: If $z=re^{i\theta}$ then ${\rm d}z=ire^{i\theta}{\rm d}\theta=iz\,{\rm d}\theta$, and so
$${\rm d}\theta=\dfrac{1}{iz}{\rm d}z=\color{red}{-}\dfrac{i}{z}{\rm d}z$$


TITLE: Valid argument form for knaves problem (spoilers)
QUESTION [0 upvotes]: I would like to know if my argument is valid for the following problem.
Another two natives C and D approach you but only C
speaks.
C says: Both of us are knaves.
What are C and D?
My solution is as follows:
assume C is telling the truth (C means truth ~C means lie)
$$C$$
$$C \implies \neg C \wedge \neg D$$
$$\neg C \wedge \neg D$$
$$\therefore \neg C$$
Now I must assume $\neg C$ is true by elimination of cases for C.
$$\neg C \implies C \vee D$$
Which means C is lying, and so one of them is a knight.
$$\neg C$$
$$\therefore D$$
Therefore D is a knight and C is a knave.
The first part still looks strange to me, should it be written in this way? Would this argument hold up in a real mathematical proof? Is there something I should do better or not do at all?

REPLY [1 votes]: Let $C$ represent the statement “native $C$ is a knight”, and similarly $D$.  The conditions of the problem are (as usual) that knights always tell the truth, that knaves always lie, that everyone is either a knight or a knave.  We can render the utterance “we are both knaves” in this context as “$\lnot C \land \lnot D$”.
Since the problem tells us that $C$ actually said that, we know that $$C\iff \lnot C \land \lnot D.$$
From this we can derive the material that you used:
$$\begin{align}C&\implies \lnot C\land \lnot D\\
\lnot C& \implies C\lor D\end{align}$$
I have no fault with your reasoning, which is mathematically valid.  We know that either $C$ or $\lnot C$ is true, and since  both assumptions lead to the same conclusion $\lnot C$, that proves $\lnot C$.  This technique (of showing that both $X$ and $\lnot X$ imply the same conclusion) is frequently used in mathematics.


TITLE: Solution of differential equation $y' + c_1 y^2= c_2 e^{-c_2 x}$
QUESTION [0 upvotes]: I am curious about the solution of the 1st order ODE
$$y' + c_1 y^2= c_2 e^{-c_2 x}$$
where $c_1, c_2$ are constants. WolframAlpha gives a complicated solution
https://www.wolframalpha.com/input/?i=y%27%28x%29%2Bc_1+y%28x%29%5E2%3D+c_2+e%5E%7B-c_2+x%7D
involving modified Bessel function, but I cannot see how it is derived.
Any ideas or suggestions?

REPLY [2 votes]: $$y' = - c_1 y^2+ c_2 e^{-c_2 x}$$
This is a Riccati ODE. The usual way to solve it in the present case is the change of function :
$$y(x)=\frac{u'(x)}{c_1u(x)}$$
which transforms the non-linear ODE into a linear ODE :
$$u''-c_1c_2e^{-c_2x}u=0$$
Then the change of variable
$$X=e^{-c_2x/2}$$
transforms it into an ODE of Bessel kind (Generalized Bessel). To solve it see Eqs.(6-7) in https://mathworld.wolfram.com/BesselDifferentialEquation.html


TITLE: Explanation of several remarks of Gauss on representations of a given number as sum of four squares.
QUESTION [4 upvotes]: Remark 1: On p.384 of volume 3 of Gauss's Werke, which is a part of an unpublished treatise on the arithmetic geometric mean, Gauss makes the following remark:

On the theory of the division of numbers into four squares:
The theorem that the product of two sums of four squares is itself a sum of four squares, is most simply represented as follows: let $l,m,\lambda,\mu,\lambda',\mu'$ be six complex numbers such that $\lambda,\lambda'$ and $\mu,\mu'$ are conjugate. Let $N$ denote the norm, than $$(Nl+Nm)(N\lambda + N\mu)=N(l\lambda+m\mu)+N(l\mu'-m\lambda')$$ and also $$(N(n+in')+N(n''+in'''))(N(1-i)+N(1+i))=N((n+n'+n''-n''')+i(-n+n'+n''+n'''))+N((n+n'-n''+n''')-i(n-n'+n''+n'''))$$
From this it is easy to derive the following two propositions, in which different representation of a number by a sum of four squares refer to the different value systems of the four roots, taking into account both the signs and the sequence of the roots. 1. If the fourfold  of a number of the form $4k+1$ can be represented by four odd squares, then it can be represented half as often by one odd and three even squares, and vice versa, if a number can be represented in this way, that it can be represented twice as often in the first way. 2. If the fourfold of a number of the form $4k+3$ can be represented by four odd squares, then it can be represented half as often by one even and three odd squares, and vice versa, if a number can be represented in this way, than its quadruple can be represented twice as often in the first way.

Gauss than says that a certain identity of theta functions can be derived by these two theorems, namely the assertion (written here in modern notation):
$$\vartheta_{00}(0;\tau)^4 = \vartheta_{01}(0;\tau)^4 + \vartheta_{10}(0;\tau)^4$$
(Gauss denotes the three theta functions by $p(y),q(y),r(y)$).
Notes on remark 1:

The first identity in this passage was verifed algebrically by me, and I also suspect it might be intimately connected with quaternions.
The second identity follows directly from the first by substituting $\lambda = 1-i$ and $\mu = 1+i$. Since $N(1-i)+N(1+i)=4$, this identity enables one to generate new representations of an integer $4s$ as sum of four squares by simply changing the signs and order of the different numbers in the representation of $s$ as sum of four squares. For example, if $s = 13 = 2^2+2^2+2^2+1^2$ than this identity implies $52=4s = 5^2+3^2+3^2+3^2$.

Remark 2: On p. 1-2 of volume 8 of Gauss's Werke there is an additional note on the representation of numbers as sums of squares. According to Dickson's "history of the theory of numbers":

Gauss noted that every decomposition of a multiple of a prime $p$ into $a^2+b^2+c^2+d^2$ corresponds to a solution of $x^2+y^2+z^2\equiv 0 \pmod{p}$ proportional to $a^2+b^2,ac+bd,ad-bc$ or to the sets derived by interchanging $b$ and $c$ or $b$ and $d$. For $p\equiv 3 \pmod{4}$, the solutions of $1+x^2+y^2\equiv 0 \pmod{p}$ coincide with those of $1+(x+iy)^{p+1}\equiv 0$. From one value of $x+iy$ we get all by using: $$(x+iy)\frac{(u+i)}{(u - i)}$$ (where $u = 0,1,\cdots, p-1$). For $p\equiv 1 \pmod{4}, p = a^2+b^2$; then $b\frac{(u+i)}{a(u-i)}$ give all values of $x+iy$ if we exclude the values $a/b$ and $b/a$ of $u$.

Why is this interesting?
The reason why these passages interests me is that it might give a clue of answering a previous question I posted on math stack exchange Did Gauss know Jacobi's four squares theorem?.
Since these remarks show how to generate new resolutions into four squares in terms of known resolutions, Gauss seems to have developed an enumarting argument (of the amount of representations of a given number as a sum of four squares) in those remarks, an argument which might have guided him in the developement of the identity for the generation function of $\vartheta_{00}(0;\tau)^4$ (an identity which leads to Jacobi's four squares theorem); this generating function appears in p. 445 of volume 3 of his works, but with no specific reference to sums of four squares.
Questions

The two propositions which Gauss mentions in remark 1 are unclear to me, and I mean that the propositions themself are unclear, not their derivation. It is simply not formulated clearly. Therefore, I'd like to get an explanation of the two propositions which Gauss mentions.
I'd also like to understand how the theta function identity follows from the two propositions.
What is the meaning of the results in remark 2? they are very complicated and i don't have a clue of understanding its meaning.

REPLY [1 votes]: I use the following notation in what follows:
\begin{align}
\vartheta_2(q) &=\sum_{n\in\mathbb {Z}} q^{(n+(1/2))^2}\tag{1a}\\
\vartheta_3(q) &=\sum_{n\in\mathbb {Z}} q^{n^2}\tag{1b}\\
\vartheta_4(q) &=\vartheta_3(-q)\tag{1c}
\end{align}
Let $a(n) $ be the number of ways in which the positive integer $n$ can be expressed as sum of four squares. Then we have $$\vartheta_3 ^4(q)=1+\sum_{n=1}^{\infty} a(n) q^n\tag{2}$$
Let $b(n) $ represents the number of ways of expressing $n$ as a sum of four odd squares. Then $$\vartheta_2^4(q^4)=\sum_{n=1}^{\infty} b(n) q^n\tag{3}$$ Since $$\vartheta_2(q)=2q^{1/4}\sum_{n=1}^{\infty} q^{n(n+1)}$$ it follows that $$\vartheta_2^4(q)=16q\left(\sum_{n=1}^{\infty} q^{n(n+1)}\right)^4$$ and $\vartheta_2^4(q)$ is an odd function of $q$.
Hence we can rewrite $(3)$ as $$\vartheta_2^4(q^4)=\sum_{n=0}^{\infty} b(4(2n+1))q^{4(2n+1)}$$ or $$\vartheta_2^4(q)=\sum_{n=0}^{\infty} b(4(2n+1))q^{2n+1}\tag{4}$$ Next we can observe (using $(1c),(2)$) that $$\vartheta_4 ^4(q)=1+\sum_{n=1}^{\infty} (-1)^na(n)q^n\tag{5}$$ Therefore the coefficients of even powers of $q$ in $\vartheta_3^4(q),\vartheta_4 ^4(q)$ are equal and for odd powers of $q$ they differ in sign.
Now consider the target identity $$\vartheta_3 ^4(q)=\vartheta_2^4(q)+\vartheta_4^4(q)\tag{6}$$ which needs to be established. The first term on right side is an odd function and contains only odd powers of $q$. And we have noted that the coefficients of even powers of $q$ match on both sides. Thus our job is done if we can match the coefficients of odd powers of $q$ ie $$a(2n+1)=b(4(2n+1))-a(2n+1)$$ or $$b(4(2n+1))=2a(2n+1)\tag{7}$$ Next we observe that if an odd number $(2n+1)$ can be expressed as sum of four squares then either the expression consists of one odd square and three even squares or it consists of one even and three odd squares. Also the first case arises when $2n+1=4k+1$ and the second case arises when $2n+1=4k+3$.
For both these cases Gauss makes claims in his remark 1 and says that $$b(4(4k+1))=2a(4k+1),b(4(4k+3))=2a(4k+3)\tag{8}$$ which is same as identity $(7)$ and hence $(6)$ is established if we assume that both propositions in remark 1 are true.


TITLE: A die is rolled and a coin is flipped. What is wrong with the following reasoning?
QUESTION [9 upvotes]: A die is rolled and a coin is flipped. What is wrong with the following reasoning?:
Let $A=\{1,3,6\}$ (event from die roll), let $B=\{H\}$ (event of heads)
then, $A\cap B=\emptyset$ (since they have no common elements)
and $\Pr(A\cap B)=\Pr(\emptyset)=0$.
I realize that the probability of the intersection of two independent events is the product of their probabilities but I want to know what is 'wrong' with this reasoning.

REPLY [1 votes]: On the other hand note:
$$P(A\cap B)=P(A)\cdot P(B|A)=P(A)\cdot P(B)=\frac{3}{6}\cdot \frac{1}{2}=\frac{1}{4}.$$
The thing is this is a two step experiment with the first step of rolling a die and the second step of flipping a coin. The counting rule of the total number of experimental outcomes of the multiple step experiment is to multiply the numbers of outcomes in each step. In this case it is:
$$n(S)=n_1\cdot n_2=6\cdot 2=12.$$
Thus, all expermental outcomes make up the sample space, from which one should form the events (or subsets of the sample space). In your case, you should have:
$$X=\{1H,3H,6H\}, P(X)=\frac{n(X)}{n(S)}=\frac{3}{12}=\frac{1}{4}.$$
Note that all outcomes are equally likely.
Also one can form an event from within each step, however the intersection of the events of distinct steps in probability theory imply coexisting outcomes (unlike in set theory common elements). Moreover, the events or outcomes within a step can be mutually exclusive or not, while across steps dependent or independent. Thus, in the case of rolling a die and flipping a coin, the events $A$ and $B$ are across steps and independent.


TITLE: Prove a planar graph is unique up to isomorphisms
QUESTION [0 upvotes]: Find a planar graph with 12 vertices and such that all vertices have
degree 5. Show that such a graph is unique modulo isomorphism.
I know we can find the number of faces and edges using Euler's theorem, but how do I do the second part?

REPLY [2 votes]: Hint Prove that each face must be a triangle.
Next, if you have two graphs with this property, start from a triangle in each of them, call them $u_1,u_2,u_3$ and $v_1,v_2,v_3$. Define  $f(u_1)=v_1,..f(u_3)=v_3$.
Now each edge must belong to a second triangle, otherwise some vertex would have degree $2$, so you can extend your isomorphism, and repeat.


TITLE: Optimum of linear program from graph
QUESTION [4 upvotes]: Given a connected (undirected) graph $G$ with vertex set $V$ of size at least $2$, we are allowed to put a real number $x_v$ on each $v\in V$. The constraint is that, for any $W\subseteq V$ such that the induced subgraphs on both $W$ and $V\setminus W$ are connected, $\displaystyle\left|\sum_{v\in W}x_v\right|\le 1$. We want to maximize $\displaystyle\sum_{v\in V} |x_v|$.
Is it true that there is a maximizing solution where all $x_v$'s are integers summing to $0$?
Examples: If $G$ is a path of length $n$, a maximum occurs at $(1,-2,2,-2,\dots,-2,1)$ if $n$ is odd, and $(1,-2,2,\dots,2,-1)$ if $n$ is even.
If $G$ is a cycle of length $n$, a maximum occurs at $(1,-1,\dots,1,-1)$ if $n$ is even and $(1,-1,\dots,1,-1,0)$ if $n$ is odd.
The constraints and objective can be written as a linear program by taking out absolute values, and some linear programming facts may be useful.

REPLY [1 votes]: Let $G$ be the graph with vertex set $\{1,2,3,4,5\}$ and all edges except $12$. Here, we have a constraint on every subset $W$ except for $W = \{1,2\}$ and its complement $W = \{3,4,5\}$.
The LP optimum is $3$, achieved for example when $x_1 = 1$, $x_2 = \frac12$, and $x_3 = x_4 = x_5 = -\frac12$.
However, if $x_1,x_2,x_3,x_4,x_5$ are all required to be integers, we can't do better than $2$. First of all, $x_i \in \{-1,0,1\}$ for all $i$. Can we have $x_i = x_j = 1$ for some $i,j$? Only if $\{i,j\} = \{1,2\}$, otherwise the $|x_i + x_j| \le 1$ constraint is violated. But if $x_1 = x_2 = 1$, then $|x_1 + x_2 + x_k| \le 1$ for $k=3,4,5$ forces $x_3 = x_4 = x_5 = -1$, and then $|x_3 + x_4| = 2 > 1$. So we can't set two variables to $1$; similarly, we can't set two variables to $-1$. So the optimal integer solution is to set one variable equal to $1$, another to $-1$, and the rest to $0$.
If we do the same thing with $6$ vertices, it appears that the optimal value is $\frac83$, which is not even an integer (achieved by $x_1 = 1$, $x_2 = \frac13$, and $x_3 = x_4 = x_5 = x_6 = -\frac13$).


TITLE: Is the compact complement topology locally arc connected?
QUESTION [1 upvotes]: Let $\tau$ be the standard topology on $\mathbb{R}$ and let $\tau'$ be the compact complement topology on $\mathbb{R}$: $$\tau' = \{\mathbb{R} \setminus K \mid K \text{ is $\tau$-compact}\} \cup \{\varnothing, \mathbb{R}\}$$
π-Base claims that $(\mathbb{R},\tau')$ is locally arc-connected. I believe the proof that it is arc-connected: since $\tau' \subseteq \tau$, the identity map $id:(\mathbb{R}, \tau) \rightarrow (\mathbb{R},\tau')$ is a continuous injection and thus gives an arc between any two distinct points.
They give the same proof for local arc-connectedness, but it is insufficient since we must show that every nonempty open set contains an arc-connected open neighborhood of each of its points. In particular, I'm skeptical that any nonempty, proper, $\tau'$-open set can even be path-connected.

REPLY [2 votes]: You are correct that the given argument does not work.  However, $(\mathbb{R},\tau')$ is in fact locally arc-connected.  It suffices to show that if $a<b<c<d$ then the open set $U=(-\infty,a)\cup(b,c)\cup(d,\infty)$ is arc-connected, since such open sets form a basis.  I will show that if $x\in (d,\infty)$ and $y\in (b,c)$ then there is a continuous injection $f:[0,1]\to U$ with $f(0)=x$ and $f(1)=y$; the other cases are similar.
To define $f$, first let $g:[0,1)\to[x,\infty)$ be an order-isomorphism.  Then define $f(t)=g(t)$ for $t\in[0,1)$ and $f(1)=y$.  Clearly $f$ is injective, and it is continuous at any $t\in[0,1)$ since $g$ is continuous with respect $\tau$ and hence also with respect to $\tau'$.  So we just need to check continuity at $1$.  If $V\subseteq U$ is any $\tau'$-open set containing $y$, then $V$ must contain $(z,\infty)$ for some $z\in[x,\infty)$.  Thus $f^{-1}(V)$ contains the entire interval $(g^{-1}(z),1]$, and in particular contains a neighborhood of $1$.  Thus $f$ is continuous at $1$.


TITLE: Minimum boolean lattice containing all poset of fixed size
QUESTION [4 upvotes]: I need help with the following:
What is the minimum $n$ such that the boolean lattice $2^{[n]}$ contains all posets of size $m$?
I noticed that it should contain a chain of length $m$, and the longest chain in $2^{[n]}$ has length $n+1$, so $n \ge m - 1$.
Also it should contain an antichain of size $m$, so $n \geq n'$, where $n'$ is the least number such that ${n' \choose \lfloor n'/2\rfloor} \geq m$. By manually checking we can see that for $m = 1,2,3,4$; $n$ is respectively $0,2,3,4$, and from there the first condition starts to imply the second, and my conjecture is that if $m \geq 5$ then $n = m-1$.
I appreciate any help trying to prove (or disprove) this. Thanks.

REPLY [1 votes]: Hint: map each element of the poset to its down-set.


TITLE: Is Einstein notation universally applicable?
QUESTION [2 upvotes]: It gets surprisingly monotonous to write the sum symbol repeatedly, so I'm wondering if one can just swap in Einstein notation at will to avoid the monotony, assuming that the symbol that's being summed over appears exactly twice?
For example, the wiki definition of $\nabla$ uses the sum symbol:
$\nabla = \sum_{i=1}^n \vec e_i {\partial \over \partial x_i} $
Could one simply write
$\nabla = \vec e_i {\partial \over \partial x_i} $
instead, and if so, would it need to be prefaced with "Using Einstein summation notation... " or is even that necessary?
If this question is more appropriate for physics.se, I'll post it there. Math.se has more hits on the search "Einstein notation," so I thought I'd ask here first.
Thank you for any help.

REPLY [0 votes]: I'm studying mathematics and my favourite area of research is Differential and Riemann Geometry among Tensor and Exterior Algebra. I must say that in these fields, even though they can be heavily applied to physics, it's very common and actually nice to represent things using Einstein Notation and Tensorial Notation. Let's take a look at this example in Vector Calculus:
There's an identity that stands for (I'm not explaining here what it means, I'm only concerned about the notation that can be used in a pure maths scenario):
$$A \times (\nabla \times B) = \nabla_B\langle{A,B}\rangle - \langle{A, \nabla}\rangle B$$
You may ask, what the **** is this? Don't worry, so do I and all mathematicians on earth, what is $\nabla_B$ for example? No way it can be understood in a regular vectorial notation used in Vector Calculus. Again, I'm not explaining the meaning of it but take a look how I can rewrite the identity above using only einstein summation and tensor notation:
$$\epsilon^{rsi}A_s\epsilon_{ijk}\nabla^jB^k=(\delta^{r}_{j}\delta^{s}_{k} - \delta^{s}_{j}\delta^{r}_{k})A_s\nabla^jB^k=A_s\nabla^rB^s - A_s\nabla^sB^r$$
Ok, being quiet honest if you're not familiar with tensor and einstein notation the latter will make you tear apart your sheets, but trust me, once you learn these notations the latter identity will make you understand the first identity (which are actually the same) and you will see that it looks PRETTY MUCH harder than it actually is.


TITLE: different shaped open basis on $\mathbb{R}^2$
QUESTION [0 upvotes]: $J_1$ be the smallest topology on $\mathbb{R}^2$ containing the sets $(a,b)\times (c,d)\forall a,b,c,d\in\mathbb{R}$
$J_2$ be the smallest topology on $\mathbb{R}^2$ containing the sets $\{(x,y):(x-a)^2+(y-b)^2<c,\forall a,b\in\mathbb{R},c>0\}$
$J_3$ be the smallest topology on $\mathbb{R}^2$ containing the sets $\{(x,y):|(x-a)|+|(y-b)|<c,\forall a,b\in\mathbb{R},c>0\}$
My question is is true that $J_1=J_2=J_3$?or there is some containement relationship among them?
In exam hall I  thought that any open set of $J_1$,I can push it into some open set of $J_2,J_3$ and I can do the same thing for other two, so they are equal? Thank you.

REPLY [1 votes]: They will all be the same. You need to show that you can fit each basis element of $J_1$ inside a basis element of $J_2$ inside a (... think matryoshka dolls). If you draw a picture of the sets, you'll see why it's true intuitively, and then you just need to back it up with a little algebra.


TITLE: How can we interpret polarization and frequency when we are dealing with one single photon?
QUESTION [22 upvotes]: If polarization is interpreted as a pattern/direction of the electric-field in an electromagnetic wave and the frequency as the frequency of oscillation, how can we interpret polarization and frequency when we are dealing with one single photon?

REPLY [17 votes]: Maxwell's equations exactly define the propagation of a lone photon in free space. The state of a photon can be defined by a vector valued state in Hilbert space and this vector valued state is a precise mathematical analogy of the $\vec{E}$ and $\vec{H}$ fields of a macroscopic, classical field. That's not to say that, for a single photon, the $\vec{E}$ and $\vec{H}$ are to be interpreted as electric and magnetic field: the vector valued  $\vec{E}$ and $\vec{H}$ state is the unitarily evolving quantum state before any measurement is made. But:

There is a one-to-one, onto correspondence between every classical electromagnetic field for a given system and a one photon quantum state for a photon propagating in that system.

This is the first quantized description of the photon. To understand what measurements a photon state implies, one has to shift to a second quantized description where we have electric and magnetic field observables, whose measurements behave more and more like classical measurements as the number of photons gets bigger. A classical state is a coherent state of the second quantised field. But, given a photon can be described by a vector valued quantum state, it should be clear that polarization and all like "classical" attributes are meaningful for a lone photon.
In particular, a photon can be a quantum superposition of eigenstates, so:

One photon can be spread over a range of frequencies and wavelengths (i.e. it can be in a superposition of energy eigenstates), with possibly different polarisation for all components of the superposition.

One can even broaden this concept to propagation through dielectric mediums: the light becomes a quantum superposition of free photons and excited matter states, and the lone, first quantized quasiparticle that results from this superposition (strictly speaking a "polariton" rather than a true, fundamental, photon) has a quantum state which evolves following Maxwell's equations solved for the medium. Thus, for example, we talk about lone photons propagating in the bound modes of optical fibres.
Another take on the one photon state is given in the first chapter of Scully and Zubairy "Quantum Optics". The one photon state $\psi$ can be defined by the ensemble statistics derived from the second quantized electric and magnetic field observables:
$$\vec{E} = \left(\begin{array}{c}\left<0 | \hat{E}_x | \psi\right>\\\left<0 | \hat{E}_y | \psi\right>\\\left<0 | \hat{E}_z | \psi\right>\end{array}\right);\quad\quad
\vec{B} = \left(\begin{array}{c}\left<0 | \hat{B}_x | \psi\right>\\\left<0 | \hat{B}_y | \psi\right>\\\left<0 | \hat{B}_z | \psi\right>\end{array}\right)$$
where $\hat{E}_j$ is the $j^{th}$ component of the vector valued electric field observable and $\hat{B}_j$ that of the magnetic induction observables. ($[\hat{E}_j, \hat{B}_j]=0$ for $j\neq k$ and, in the right units, $[\hat{E}_j, \hat{B}_j]=i\,\hbar\,I$). For a one-photon state $\psi$, these statistics:

Propagate exactly following Maxwell's equations;
Unquely define the light field's quantum state for a one-photon state, even though they are not the state. This is in the same way that the mean of the classical Poisson probability distribution uniquely defines the distribution (even though it is a lone number, not a distribution).

Things are much more complicated for general, $N$ photon states so we need much more information than simple means to fully define the quantum state particularly with entangled states. Going back to our classical probability distribution analogy, the normal distribution needs two independent parameters, mean and variance, to wholly specify it, so it's a more complicated thing than the Poisson distribution, which is defined by only its mean (which equals the variance).So quantum fields are hugely more complicated things than classical ones. But a coherent state of any photon is again uniquely defined by the mean values of the field observables, which means again propagate following the same Maxwell equations as the one-photon means: hence the one-to-one, onto correspondence between classical and one-photon states I spoke of - I like to call this the one photon correspondence principle ("OpCoP"). Why our macroscopic EM fields seem to behave like coherent quantum states rather than hugely more general, entangled ones (unless one goes to considerable experimental effort to observe entanglement) is still an open question. It is interesting to note, though, that the class of coherent states is the unique class of quantum harmonic oscillator states that achieve the lower bound of the Heisenberg uncertainty inequality.
Also see my answers to:

If photons carry 1 spin unit, why does visible light seem to have no angular momentum? and
Electromagnetic radiation and quanta.

Incidentally, even though general, entangled light states are hugely more complicated than one-photon (and, equivalently through the OpCoP, classical) light states, in principle we can still decompose them into a quantum superposition of tensor products of coherent states and so represent a general state by a set of field observable means. This was one of the contributions of 2005 Nobel Laureate Roy Glauber, who showed the above in 1963 in:
R. Glauber, "Coherent and Incoherent States of the Radiation Field", Phys. Rev. 131, 2766–2788 (1963)
The coherent state tensor products are, however, overcomplete so the decomposition of a general quantum state into coherent states is highly not unique. Nonetheless, such a decomposition allows classical-like techniques to be brought to bear on entangled quantum states (in principle - in practice it is still complicated!).
If you google Iwo Bialynicki-Birula and his work on the photon wave function, he has heaps more to say about the one-photon wave function. He defines the photon wave function as the positive frequency part of left and right circularly polarized eigenfunctions $\vec{F}_\pm = \sqrt{\epsilon} \vec{E} \pm i \sqrt{\mu} \vec{H}$. Iwo Bialynicki-Birula's personal website is http://cft.edu.pl/~birula and all his publications are downloadable therefrom. $|\vec{F}_+|^2 + |\vec{F}_-|^2$ is the electromagnetic energy density. He defines the pair $(\vec{F}_+, \vec{F}_-)$, normalised so that $|\vec{F}_+|^2 + |\vec{F}_-|^2$ becomes a probability density to absorb the photon at a particular point, to be a first quantized photon wave function (without a position observable). There is special, nonlocal inner product to define the Hilbert space and in such a formalism the general Hamiltonian observable is $\hbar\, c\, \mathrm{diag}\left(\nabla\wedge, -\nabla\wedge\right)$. Please also see Arnold Neumaier's pithy summary (here) of a key result in section 7 of Bialynicki-Birula's "Photon wave function" in Progress in Optics 36 V (1996), pp. 245-294 also downloadable from arXiv:quant-ph/0508202. The Hilbert space of Riemann Silberstein vector pairs that Bialynicki-Birula defines is acted on by an irreducible unitary representation, defined by Bialynicki-Birula's observables $\hat{H}$, $\hat{\mathbf{P}}$, $\hat{\mathbf{K}}$ and $\hat{\mathbf{J}}$, of the full Poincaré group presented in the paper.

REPLY [0 votes]: Well the energy of the photon is simply $h\cdot\nu$ , so if you can determine the energy of a single photon, you can determine its frequency.
One way to determine the energy of a photon; assuming that you can generate one photon at a time, all of the same energy, would be to use the photo-electric effect, with adjustable band gap photo-cathode materials in a PMT, which can detect single photons.  Variable bandgap photo-cathodes, can be made over limited ranges from III-V Ternary or quaternary compounds, such as GaAsP or InGaAsP.   Smaller bandgap cathodes will emit a photo-electron; higher bandgap ones will not.
Now you didn't say you wanted to know a practical way of doing that, but if you have a need for single photons of known frequency and polarization, making the PMTs should be no problem for you.


text:
\begin{document}

\title{Galkin's Lower Bound Conjecture Holds for the Grassmannian}

\author{La'Tier Evans}
\address{ Department of Mathematics and Computer Science, Henson Science Hall, Salisbury University, Salisbury MD 21801 USA}\email{levans4@gulls.salisbury.edu}
\author{Lisa Schneider}
\address{ Department of Mathematics and Computer Science, Henson Science Hall, Salisbury University, Salisbury MD 21801 USA}\email{lmschneider@salisbury.edu}
\author{Ryan M. Shifler}
\address{ Department of Mathematics and Computer Science, Henson Science Hall, Salisbury University, Salisbury MD 21801 USA}\email{rmshifler@salisbury.edu}
\author{Laura Short}
\address{ Department of Mathematics and Computer Science, Henson Science Hall, Salisbury University, Salisbury MD 21801 USA}\email{lshort1@gulls.salisbury.edu}
\author{Stephanie Warman}
\address{ Department of Mathematics and Computer Science, Henson Science Hall, Salisbury University, Salisbury MD 21801 USA}\email{swarman2@gulls.salisbury.edu}

\subjclass[2010]{Primary 14N35; Secondary 15B48, 14N15, 14M15}

\begin{abstract}
Let $\Gr(k,n)$ be the Grassmannian. The quantum multiplication by the first Chern class $c_1(\Gr(k,n))$ induces an endomorphism $\hat c_1$ of the finite-dimensional vector space $\mathrm{QH}^*(\Gr(k,n))_{|q=1}$ specialized at $q=1$. Our main result is a case that a conjecture by Galkin holds. It states that the largest real eigenvalue of $\hat{c}_1$ is greater than or equal to $\dim \Gr(k,n)$+1 with equality if and only if $\Gr(k,n)=\mathbb{P}^{n-1}.$ \end{abstract}
\maketitle

\section{Introduction}\label{s:intro} Fix $1 \leq k \leq n-1$ and let $\Gr(k,n)$ be the Grassmannian. This is the parametrization of $k$ dimensional linear spaces $V \subset \mathbb{C}^n$. The purpose of this paper is to verify a conjecture by Galkin \cite{Galkin} for the Grassmannian. We recall the precise statements, following \cite{Galkin} and \cite[\S 3]{GGI}.

Let $F$ be a Fano manifold. Let $K:=K_{F}$ be the canonical bundle of $F$ and let $c_1(F):=c_1(-K) \in H^2(F)$ be the anticanonical class. The quantum cohomology ring $(\mathrm{QH}^*(F), \star)$ is a graded algebra over $\Z[q]$, where $q$ is the quantum parameter. Consider the specialization $H^\bullet(F):= \mathrm{QH}^*(F)_{|q=1}$ at $q=1$. The quantum multiplication by the first Chern class $c_1(F)$  induces an endomorphism $\hat c_1$ of the finite-dimensional vector space $H^\bullet(F)$:
 \[ y\in H^\bullet(F) \mapsto \hat c_1(y):= (c_1(F)\star y)|_{q=1} \/. \] Denote by $\delta_0:=\max\{|\delta|:\delta \mbox{ is an eigenvalue of } \hat c_1\}$. Galkin's lower bound conjecture \cite{Galkin} states the following:
 \[ \delta_0 \geq \dim_{\mathbb{C}}F+1 \mbox{ with equality if and only if } F=\mathbb{P}^N. \]
 The conjecture was verified for del Pezzo surfaces \cite{HKLY} and projective complete intersections \cite{Ke}.
\begin{remark}
The conjecture is related to Property $\mathcal{O}$ which states the following:
  \begin{enumerate}
    \item The real number $\delta_0$ is an eigenvalue of $\hat c_1$ of multiplicity one.
    \item If $\delta$ is any eigenvalue of $\hat c_1$ with $|\delta|=\delta_0$, then $\delta=\delta_0 \zeta$ for some $r$-th root of unity $\zeta\in \mathbb{C}$, where $r$ is the Fano index of $F$.
  \end{enumerate}

Property $\cO$ is conjectured to hold for any complex Fano manifold $F$. Property $\cO$ has been verified for the Grassmannian and several other cases of Fano manifolds; see \cite{Riet,GaGo,Cheo,ChLi,GaIr,SaSh,Ke,HKLY,LMS,BFSS,Withrow}. It is the main hypothesis needed for the statement of Gamma Conjectures I and II, which in turn are related to mirror symmetry on $F$ and generalize the Dubrovin conjectures; we refer the reader to \cite{GGI} for details.

\end{remark}

The (small) quantum cohomology of $\Gr(k,n)$, denoted by $\mbox{QH}^*(\Gr(k,n))$, has a graded $\mathbb{Z}[q]$-basis consisting of Schubert classes $\{\sigma_\lambda \}$ indexed by partitions in the set \[ \Lambda:=\{(\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_k): n-k \geq \lambda_1, \lambda_k \geq 0\} \] where $|\lambda|=\lambda_1+\lambda_2+\cdots+\lambda_k$ for $\lambda \in \Lambda$ and $\deg q=n$. The ring multiplication is given by \[ \sigma_\lambda \star \sigma_\mu = \sum_{\nu, d \geq 0} c_{\lambda, \mu}^{\nu,d}q^d \sigma_\nu \] where $c_{\lambda, \mu}^{\nu,d}$ is the Gromov-Witten invariant that enumerates degree $d$ rational curves intersecting general translates of $\sigma_{\lambda}, \sigma_{\mu},$ and the Poincar\'e dual of $\sigma_{\nu}$. We refer the reader to \cite{FP}, \cite{buch}, and \cite{kontsevich.manin:GW:qc:enumgeom} for additional details regarding the quantum cohomology of the Grassmanian. The quantum Chevalley formula is given by the multiplication of the divisor class $\sigma_{(1)}$ (\cite{bertram}). Let $\lambda^*=(\lambda_2-1 \geq \cdots \geq \lambda_k-1 \geq 0)$ if $\lambda_1=n-k$ and $\lambda_k>0$, otherwise $\lambda^*$ does not exist.

If $\lambda \in \Lambda$ then
\[\sigma_{(1)} \star \sigma_{\lambda}=q \sigma_{\lambda^*}+\sum \sigma_{\mu} \]
where the sum is over all partitions $\mu \in \Lambda$ such that $|\mu|=|\lambda|+1$ and $\lambda \subset \mu$ and the quantum term is omitted if $\lambda^*$ does not exist. The anticanonical class is $c_1(\Gr(k,n))=n \sigma_{(1)}$ and the Fano index is $n$. The linear operator $\hat c_1$ is given next.

If $\lambda \in \Lambda$ then
\[ \hat c_1( \sigma_{\lambda}) =n\sigma_{\lambda^*}+n\sum \sigma_{\mu} \]
where the sum is over all partitions $\mu \in \Lambda$ such that $|\mu|=|\lambda|+1$ and $\lambda \subset \mu$ and the term $\sigma_{\lambda^*}$ is omitted if $\lambda^*$ does not exist.

\begin{remark}
There is a graph theoretic context for $\delta_0$. We will now recall the notion of the (oriented) {\it quantum Bruhat graph} $\mathcal{G}$ of $\Gr(k,n)$; see \cite{brenti.fomin}. The vertices of this graph consist of partitions  $\lambda \in \Lambda$. There is an oriented edge $\lambda \rightarrow \mu$ if the class $\sigma_\mu$ appears with a positive coefficient (possibly involving $q$) in the quantum Chevalley multiplication $\sigma_{(1)} \star \sigma_{\lambda}$. See Figure \ref{graphs} for examples. By the Perron-Frobenius Theory of nonnegative matrices, $\delta_0$ is equal to the largest real eigenvalue of $n$ times the incidence matrix of $\mathcal{G}$; see \cite{Riet,ChLi}.
\end{remark}

\begin{figure}
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    \draw [edge] (1) to (2);
    \draw [edge] (1) to (3);
    \draw [edge] (2) to (4);
    \draw [edge] (3) to (4);
    \draw [edge] (4) to (5);
    \draw [edge] (4) to [bend right=100] (0);
    \draw [edge] (5) to [bend left=100](1);
\end{tikzpicture}
\caption{Quantum Bruhat graph for $\Gr(2,4)$}\label{fig1a}
\end{subfigure}\hfill
\begin{subfigure}[b]{.49\linewidth}
\centering
\begin{tikzpicture}[scale=.7]
\tikzset{edge/.style = {->,> = latex'}}
    \node (0) at (0,0) {$\tiny\emptyset$};
    \node (1) at (0,-1.5) {$\tiny\yng(1,0)$};
    \node (2) at (-1,-3) {$\tiny\yng(2,0)$};
    \node (3) at (1,-3) {$\tiny\yng(1,1)$};
    \node (4) at (0,-4.5) {$\tiny\yng(2,1)$};
    \node (5) at (-2,-4.5) {$\tiny\yng(3,0)$};
    \node (6) at (-1,-6){$\tiny\yng(3,1)$};
    \node (7) at (1,-6){$\tiny\yng(2,2)$};
    \node (8) at (0,-7.5){$\tiny\yng(3,2)$};
    \node (9) at (0,-9){$\tiny\yng(3,3)$};
    \draw [edge] (0) to (1);
    \draw [edge] (1) to (2);
    \draw [edge] (1) to (3);
    \draw [edge] (2) to (4);
    \draw [edge] (3) to (4);
    \draw [edge] (2) to (5);
    \draw [edge] (5) to (6);
    \draw [edge] (4) to (6);
    \draw [edge] (4) to (7);
    \draw [edge] (6) to (8);
    \draw [edge] (7) to (8);
    \draw [edge] (8) to (9);
    \draw [edge] (9) to [bend left=100] (2);
    \draw [edge] (8) to [bend right=100] (1);
    \draw [edge] (6) to [bend left=100] (0);
\end{tikzpicture}
\caption{Quantum Bruhat graph for $\Gr(2,5)$}\label{fig1b}
\end{subfigure}

\caption{Quantum Bruhat graphs}
\end{figure}
\label{graphs}

We are now ready to state our main theorem.
\begin{thm} \label{mainthm}
The largest real eigenvalue $\delta_0$ of $\hat c_1$ for $\Gr(k,n)$ is greater than or equal to $ \dim \Gr(k,n) +1 = k(n-k)+1$ with equality if and only if $\Gr(k,n)=\mathbb{P}^{n-1}$.
\end{thm}

We will give two proofs. One is due to the authors and the other is given to us by an anonymous referee. The value $\delta_0$ is explicitly computed by Rietsch in \cite{Riet} and will be the key ingredient to the first proof. The proof uses calculus in conjunction with the explicit computation of $\delta_0$. To the best of the author's knowledge, the study of $\delta_0$ and Conjecture $\mathcal{O}$ has occurred on a case-by-case basis by considering fixed Fano varieties. Our approach deviates from the case-by-case approach and uses the two parameters $k$ and $n$ to prove the main result. The second proof by the anonymous referee is shorter and makes use of elementary inequalities. The proof will be presented in the final section of the manuscript.

{\em Acknowledgements.} The third named author thanks Leonardo Mihalcea for useful discussions. We thank the anonymous referees for their comments and an alternative proof.

\section{Preliminaries}
 We follow \cite{Riet} to state the preliminary results. The eigenvalue $\delta_0$ of $\hat c_1$ is computed by evaluating a Schur function at a particular $k$-tuple. There are many definitions of Schur polynomials; we will use the version obtained from complete symmetric polynomials. Let $h_0,h_1, h_2, h_3, \cdots$ denote the complete symmetric polynomials in $k$ variables where $\deg h_i=i$ for $i \geq 0$ and let $h_i=0$ for $i <0$. Given $\lambda \in \Lambda$, the Schur polynomial $S_\lambda$ is given by
\[S_\lambda = \begin{vmatrix}
h_{\lambda_1} & h_{\lambda_1+1} & \cdots & h_{\lambda_1+k-1} \\
h_{\lambda_2-1} & h_{\lambda_2} & \cdots & h_{\lambda_2+k-2} \\
 &  & \ddots &  \\
 h_{\lambda_k-k+1} & \cdots & \cdots & h_{\lambda_k} \\
\end{vmatrix}. \]
We refer the reader to \cite{Macdonald} for further details on Schur functions.
We now give a useful definition for the computation of $\delta_0$.
\begin{defn}[The set $\mathcal{I}_{k,n}$]
We fix the primitive $n$-th root of unity $\zeta=e^{\frac{2\pi i}{n}}$. Let $\zeta^I:=(\zeta^{i_1}, \cdots, \zeta^{i_k})$ be an unordered $k$-tuple of distinct $n$-th roots of $(-1)^{k+1}$. Then $I=(i_1,\cdots,i_k)$ maybe chosen uniquely such that $-\frac{k-1}{2} \leq i_1 < i_2<\cdots<i_k \leq n-\frac{k+1}{2}$ and the $i_k$'s are all integers (respectively half-integers) if $k$ is odd (resp. even). Denote the set of all $k$-tuples $I$ by $\mathcal{I}_{k,n}$.

\end{defn}
It is known that $\mbox{QH}^\bullet(\Gr(k,n))$ is a finite dimensional vector space with basis given by (the restriction of) the Schubert basis $\{ \sigma_\lambda \}_{\lambda \in \Lambda}$. Rietsch presents another basis given by $\sigma_I:= \sum _{\nu \in \Lambda} \overline{S_{\nu}(\zeta^I)} \sigma_{\nu}$ for $I \in \mathcal{I}_{k,n}$. Consider the multiplication operator $[\sigma_\lambda]: \sigma \mapsto \sigma_\lambda \star \sigma$ on $\mbox{QH}^{\bullet}(\Gr(k,n))$. It follows that
\[\sigma_\lambda \star \sigma_{I}=S_{\lambda}(\zeta^I)\sigma_I \]
and $\sigma_I$ is an eigenvector of the multiplication operator $[\sigma_\lambda]$ with eigenvalue $S_\lambda(\zeta^I)$. In particular, $\hat c_1 = n[\sigma_{(1)}]$ and $\delta_0=n \max \{ |S_{(1)}(\zeta ^I)| I \in \mathcal{I}_{k,n} \}.$
This leads us to a Lemma by Rietsch which explicitly computes $\delta_0$.
\begin{lemma}[\cite{Riet} Proposition 11.1]
Let $I_0=(-\frac{k-1}{2}, \cdots, \frac{k-1}{2})$ and $I \in \mathcal{I}_{k,n}$ any other elements, and let $\zeta=e^{\frac{2 \pi i}{n}}$ as before. Then for any partitions $\lambda \in \Lambda$, \[|S_{\lambda}(\zeta^I)| \leq S_\lambda(\zeta^{I_0}). \]
In particular, $\delta_0=n S_\lambda(\zeta^{I_0}) \in \mathbb{R}.$
\end{lemma}
\begin{remark}
Another formulation is $\delta_0=n \frac{\sin(\pi k/n)}{\sin(\pi/n)}$. See \cite[Lemma 9.21]{GaIr}. This is the formulation that the anonymous referee uses in their proof.
\end{remark}
The number $\delta_0=n S_{(1)}(\zeta^{I_0})$ is computed directly for $\Gr(k,n)$. For notation, let $\delta_0^k(n)=\delta_0$ for $\Gr(k,n)$.  If $k$ is even then we have that
\begin{eqnarray*}
\delta^k_0(n)

&=&2n \sum_{j=1}^{\frac{k}{2}}  \cos\left( \frac{(k-(2j-1)) \pi }{n} \right).
\end{eqnarray*}
If $k$ is odd then we have that
\begin{eqnarray*}
\delta^k_0(n)=n+ 2n \sum_{j=1}^{\frac{k-1}{2}}  \cos\left( \frac{(k-(2j-1)) \pi }{n} \right).
\end{eqnarray*}
To prove Galkin's bound for $\Gr(k,n)$ it suffices to show that $\delta^k_0(n)-k(n-k)-1 \geq 0$ for all $k \in \mathbb{Z}^+$ with equality if and only if $k=1$ or $k=n-1$. Thus it is natural to consider the family of real valued functions $F^k(x): \mathbb{R} \rightarrow \mathbb{R}$, indexed by $k \in \mathbb{Z}^+$, that are defined by $F^k(x)=\delta_0^k(x)-k(x-k)-1$ where

\begin{eqnarray*}
\displaystyle \delta_0^{k}(x)= \begin{cases}
     \displaystyle 2x \sum_{j=1}^{\frac{k}{2}}  \cos\left( \frac{(k-(2j-1)) \pi }{x} \right), & k \mbox{ even} \\
    \displaystyle  x+ 2x \sum_{j=1}^{\frac{k-1}{2}}  \cos\left( \frac{(k-(2j-1)) \pi }{x} \right), & k \mbox{ odd.}
   \end{cases}
\end{eqnarray*}
The next Lemma is a technical Lemma that allows us to reduce the number of cases that need to be checked for Galkin's lower bound to hold.
\begin{lemma} \label{sufflemma}
Let $k \geq 4$. It suffices to show that Galkin's lower bound holds for $\Gr(k,n)$ where $n \geq 2(k-1)$.
\end{lemma}
\begin{proof}
Consider $\Gr(k,n)$ where $k+1 \leq n \leq 2(k-1).$ Then $\Gr(k,n)$ is isomorphic to $\Gr(\ell,n)$ where $\ell=n-k$. The Lemma follows from the observation that $2k+2>2k-2$ and $2k+2 \geq n \Leftrightarrow n \geq 2(l-1)$.

\end{proof}

\section{Proof of the Main Theorem}

\subsection{Calculus on the functions $F^k(x)$}
In this section, we use calculus to explore features of the function $F^k$ to further understand the number $\delta_0^k(n)$. We will prove Proposition \ref{propinc} by using Lemma \ref{techlemma} and Lemma \ref{lemmacalc}.

\begin{prop}\label{propinc}
Let $k \geq 2$. The function $F^k$ is increasing for $x \in (2(k-1),\infty).$
\end{prop}
The proof of the first Lemma is reserved until Section \ref{LEMMA}. The proof involves several applications of the Mean Value Theorem.

\begin{lemma} \label{techlemma}
Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a second differentiable function where $\displaystyle \lim_{x \rightarrow \infty}f(x)=L\in \mathbb{R}$. If $f''(x)<0$ for all $x \in (a,\infty)$ then $f'(x)>0$ for all $x \in (a,\infty)$.
\end{lemma}
The next Lemma follows by nice properties of the function $F^k.$ The ability to calculate the $\displaystyle \lim_{x \rightarrow \infty} F^k(x)$ is a straight forward application of l'H\^{o}pital's rule after slightly reorganizing the expression. The authors anticipate that similar asymptotic behavior will occur in at least the submaximal isotropic Grassmannians in types $B$, $C$, and $D$. The second derivative of $F^k(x)$ takes on a nice form since the $\sin$ functions that appear in the first derivative cancel out in the calculation of the second derivative.

\begin{lemma} \label{lemmacalc}

We have that
\begin{enumerate}
\item $\displaystyle \lim_{x \rightarrow \infty} F^k(x)=k^2-1$;
\item Let $k \geq 2$. The function $F^k(x)$ is concave down on the interval $(2(k-1), \infty)$;
\item The equality $F^k(2(k-1)) = F^{k-2}(2(k-1))$ holds for $k \geq 3$.
\end{enumerate}
\end{lemma}
\begin{proof}
For the first part observe that l'H\^{o}pital's rule implies that \[ \lim_{x \rightarrow \infty} 2x\left( \cos\left( \frac{(k-(2j-1)) \pi }{x} \right)-1\right)=0 \mbox{ for any integer }j, \mbox{ for }1 \leq j \leq \frac{k-1}{2}.\]
If $k$ is even then
\begin{eqnarray*}
\lim_{x\rightarrow \infty} F^k(x) &=& \lim_{x \rightarrow \infty} \left( \sum_{j=1}^{\frac{k}{2}}  2x\left( \cos\left( \frac{(k-(2j-1)) \pi }{x} \right)-1\right) \right)+k^2-1\\
&=&k^2-1.
\end{eqnarray*}
In a similar fashion, if $k$ is odd, it follows that $\displaystyle\lim_{x\to \infty}F^k(x)=k^2-1$ by a simple calculation.

For the second part a direct calculation shows that
\begin{eqnarray*}
\displaystyle \frac{d^2}{dx^2}F^k(x)&=&-\sum_{j=1}^{\frac{k}{2}}  \left(\frac{(k-(2j-1))^2 \pi^2}{x^3}\right)\cos\left( \frac{(k-(2j-1)) \pi }{x} \right).
\end{eqnarray*}
Since $0 < k+1-2j \le k-1$ for $j=1,2,\cdots,\frac{k}{2}$ then
\begin{eqnarray*}

0 < \frac{(k-(2j-1))\pi}{2(k-1)}\le \frac{\pi}{2}.
\end{eqnarray*}
It follows that $0 < \frac{(k-(2j-1))\pi}{x}< \frac{\pi}{2}$ for all $x \in (2(k-1),\infty)$. Thus $0< \cos\left( \frac{(k-(2j-1)) \pi }{x} \right) < 1$ for all $x\in (2(k-1),\infty)$. Therefore $\frac{d^2}{dx^2}F^k(x)<0$ for all $x \in (2(k-1),\infty)$.
The odd case can be shown in a similar fashion.

The third part of the proof follows from the isomorphism $\Gr(k,2(k-1)) \cong \Gr(k-2,2(k-1))$.
\end{proof}
\begin{remark}
Part 3) of the proof can easily be proved analytically since the equation $F^k(x)=F^{k-2}(x)$ simplifies to
\[x\left({\cos\left({\frac{(k-1)\pi}{x}}\right)}\right)+2(k-1)-x=0. \] The number $x=2(k-1)$ is a root.
\end{remark}

Proposition \ref{propinc} follows from Lemmas \ref{techlemma} and \ref{lemmacalc}.

\begin{figure}[h!]
    \centering

    \caption{The function $F^k(x)$ plotted for $k=2,4,6,8$.}
    \label{fig:my_label}
\begin{tikzpicture}[scale=.5]
\begin{axis}[
    axis lines = left,
    xlabel = $x$,
    ylabel = {$F^k$},
    xmin=3, xmax=100,
    ymin=0,ymax=70,
    width=.9\textwidth,
    height=0.6\textwidth,
    legend style={at={(axis cs:10,40)},anchor=south west}
]

\addplot [
    domain=3:100,
    samples=100,
    color=red,
]
{2*x*(cos(deg(((2-(2-1))*(3.14))/(x))))-2*(x-2)-1};

\addlegendentry{$F^{2}(x)$}

\addplot [
    domain=5:100,
    samples=100,
    color=blue,
]
{2*x*cos(deg(((4-(2-1))*(3.14))/(x)))+2*x*cos(deg(((4-(4-1))*(3.14))/(x)))- 4*(x-4)-1};

\addlegendentry{$F^{4}(x)$}

\addplot [
    domain=7:100,
    samples=100,
    color=green,
]
{2*x*cos(deg(((6-(2-1))*(3.14))/(x)))+2*x*cos(deg(((6-(4-1))*(3.14))/(x)))+2*x*cos(deg(((6-(6-1))*(3.14))/(x)))- 6*(x-6)-1};

\addlegendentry{$F^{6}(x)$}

\addplot [
    domain=9:100,
    samples=100,
    color=orange,
]
{2*x*cos(deg(((8-(2-1))*(3.14))/(x)))+2*x*cos(deg(((8-(4-1))*(3.14))/(x)))+2*x*cos(deg(((8-(6-1))*(3.14))/(x)))+2*x*cos(deg(((8-(8-1))*(3.14))/(x)))- 8*(x-8)-1};
\addlegendentry{$F^{8}(x)$}

\end{axis}
\end{tikzpicture}
\end{figure}

\subsection{}
The next proposition follows by using induction and the property $F^k$ is increasing for $x \in (2(k-1),\infty)$ when $k \geq 2$.
\begin{prop} \label{mainprop}
Galkin's lower bound holds for
\begin{enumerate}
\item $\Gr(1,n)$ (which is isomorphic to $\mathbb{P}^{n-1})$ where $n \geq 2$;
\item $\Gr(2,n)$ where $n \geq 3$;
\item $\Gr(3,n)$ where $n \geq 4$;
\item $\Gr(k,n)$ where $n \geq 2(k-1)$ and $k\geq 4$.
\end{enumerate}
\end{prop}
\begin{proof}
The first case is known and can easily be seen in this context by evaluating $F^1(x)=0$ for $x \geq 2$.

For the second case where $k=2$ observe that $\Gr(2,3)$ is isomorphic to $\mathbb{P}^2$ and a direct calculation shows that $F^2(3)=0$. Since $F^2(x)$ is increasing for $x>2$ it follows that $F^2(x)>0$ for $x>3$. Observe that $\Gr(2,n)$ is not a projective space for $n>3$. Case (2) follows.

 For the third case where $k=3$ observe that $\Gr(3,4)$ is isomorphic to $\mathbb{P}^3$ and a direct calculation shows that $F^3(4)=0$. Since $F^3(x)$ is increasing for $x>4$ it follows that $F^3(x)>0$ for $x>4$. Observe that $\Gr(3,n)$ is not a projective space for $n>4$. Case (3) follows.

For the fourth case, it suffices to show that $F^k(2(k-1))>0$ since $F^k(x)$ is increasing for $x >2(k-1)$. We will use induction to show this inequality. A direct calculation shows that $F^4(6)=6\sqrt{3}-9>0.$ Assume that $F^k(2(k-1))>0$ where $k>4$ is even. Since $F^k(x)$ is increasing for $x \geq 2(k-1)$ then we have that \[F^{k+2}(2((k+2)-1))=F^{k}(2((k+2)-1))>F^k(2(k-1))>0. \] Observe that $\Gr(k,n)$ where $k \geq 4$ and $n \geq 2(k-1)$ is not a projective space. The even case follows. The case where $k \geq 5$ is odd follows by implementing a similar strategy.
\end{proof}

Theorem \ref{mainthm} follows from Lemma \ref{sufflemma} and Proposition \ref{mainprop}.

\begin{figure}[h!]

    \centering

    \caption{Illustration of Proposition \ref{mainprop}}
    \label{fig:my_label}

\begin{tikzpicture}[scale=.5]
\begin{axis}[
    axis lines = left,
    xlabel = $x$,
    ylabel = {$F^k$},
    xmin=3, xmax=20,
    ymin=0,ymax=20,
    width=.9\textwidth,
    height=0.6\textwidth,
    legend style={at={(axis cs:6,13)},anchor=south west}
]

\addplot [
    domain=3:30,
    samples=100,
    color=red,
]
{2*x*(cos(deg(((2-(2-1))*(3.14))/(x))))-2*(x-2)-1};

\addlegendentry{$F^{2}(x)$}

\addplot [
    domain=5:30,
    samples=100,
    color=blue,
]
{2*x*cos(deg(((4-(2-1))*(3.14))/(x)))+2*x*cos(deg(((4-(4-1))*(3.14))/(x)))- 4*(x-4)-1};

\addlegendentry{$F^{4}(x)$}

\addplot [
    domain=7:30,
    samples=100,
    color=green,
]
{2*x*cos(deg(((6-(2-1))*(3.14))/(x)))+2*x*cos(deg(((6-(4-1))*(3.14))/(x)))+2*x*cos(deg(((6-(6-1))*(3.14))/(x)))- 6*(x-6)-1};

\addlegendentry{$F^{6}(x)$}

\addplot [
    domain=9:30,
    samples=100,
    color=orange,
]
{2*x*cos(deg(((8-(2-1))*(3.14))/(x)))+2*x*cos(deg(((8-(4-1))*(3.14))/(x)))+2*x*cos(deg(((8-(6-1))*(3.14))/(x)))+2*x*cos(deg(((8-(8-1))*(3.14))/(x)))- 8*(x-8)-1};

\addlegendentry{$F^{8}(x)$}
\addplot[smooth,mark=*,black] plot coordinates {
        (6,1.39389713287)
    };
    \addlegendentry{(6,$F^2(6))=(6,F^4(6)$)}

\addplot[smooth,mark=*,black] plot coordinates {
        (10,5.8)
    };
\addlegendentry{(10,$F^4(10))=(10,F^6(10)$)}
\addplot[smooth,mark=*,black] plot coordinates {
        (14,12.38)
    };
\addlegendentry{(14,$F^6(14))=(14,F^8(14)$)}

\end{axis}
\end{tikzpicture}

\end{figure}

\section{Proof of Lemma \ref{techlemma}} \label{LEMMA}
In this section we prove Lemma \ref{techlemma}. The authors were unable to find a source and we expect that this Lemma is well-known. We first recall a version of the Mean Value Theorem for the second derivative and prove Lemma $\ref{techlemma}$ by its contrapositive statement in Lemma \ref{LEMMA2}.

\begin{lemma} \label{MVTSD}
Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a second differentiable function. If $f'(a)\leq 0$ and $f'(b) \geq 0$ for some $a,b \in \mathbb{R}$ where $a<b$ then there exists a number $c \in (a,b)$ where $f''(c)\geq 0$.
\end{lemma}

\begin{lemma} \label{LEMMA2}
Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a second differentiable function where $\displaystyle \lim_{x \rightarrow \infty}f(x)=L\in \mathbb{R}$. If $f'(b) \leq 0$ for some $b \in (a, \infty)$ then there exists a number $c \in (a, \infty)$ where $f''(c) \geq 0$.
\end{lemma}

\begin{proof}
Case (I) Consider the case that $f(b) < L$. Since $f$ is continuous and $y=L$ is a horizontal asymptote of $f$, there exists a $c \in (b,\infty)$ where $f(b)<f(c)<L$. By the Mean Value Theorem there exists a $d \in (b,c)$ where $f'(d)\ge 0.$ By Lemma \ref{MVTSD}, there exists $e \in (b,d) \subset (a,\infty)$ where $f''(e) \geq 0$. The result follows.

Case (II) Consider the case that $f(b) \geq L$.
Subcase (1) If $f(b)=L$ and $f(x)=L$ is the constant function then we are done.

Subcase (2): Suppose that there exists a $c\in (b, \infty)$ where $f(c)<L.$ By the Mean Value theorem there exists $d \in (b,c)$ where $f'(d)=\frac{f(c)-f(b)}{c-b}<0.$ Since $f(c)<L$ and $y=L$ is a horizontal asymptote of $f$, there exists $e \in (c,\infty)$ where $f(c)<f(e)<L.$ By The Mean Value Theorem there exists $h \in (c,e)$ where $f'(h)=\frac{f(e)-f(c)}{e-c}>0$. The result follows from Lemma \ref{MVTSD} since $f'(d)<0$ and $f'(c)>0.$

Subcase (3): Suppose that $f(x) \geq L$ for all $x \in (b,\infty)$. We may assume that there is a $d \in (b,\infty)$ where $f(d)>L.$

Sub-subcase (i): Suppose that there exists $e \in (d,\infty)$ where $f(e)=L$. Choose $h \in (e, \infty)$. Then by the Mean Value Theorem there exits $k \in (e,h)$ where $f'(k)=\frac{f(h)-f(e)}{h-e} \geq 0$. The result follows from Lemma \ref{MVTSD}.

Sub-subcase (ii): Since $y=L$ is a horizontal asymptote of $f$ then there exists a $e \in (d,\infty)$ where $L<f(e)<f(d)$. By the Mean Value Theorem there exists an $h \in (d,e)$ where $f'(h)=\frac{f(e)-f(d)}{e-d}<0.$ Observe that \[ \lim_{x \rightarrow \infty} \frac{f(x)-f(e)}{x-e}=0. \] Thus one can choose $k \in (e,\infty)$ where $\frac{f(k)-f(e)}{k-e} \geq f'(h).$ In particular, there exists an $i \in (e,k)$ where $f'(i)=\frac{f(k)-f(e)}{k-e} \geq f'(h)$ by the Mean Value Theorem. By the Mean Value Theorem, again, there exists a $l \in (h,i) \subset (a, \infty)$ where $f''(l)=\frac{f'(i)-f'(h)}{i-h} \geq 0$. This completes the proof.
\end{proof}
Lemma \ref{techlemma} follows.

\section{A second proof}
The authors would like to thank the anonymous referee for sharing the following alternative proof. Recall that $\displaystyle \delta_0(x)=n\frac{\sin(\pi x/n)}{ \sin(\pi/n)}$. We begin with a Lemma.

\begin{lemma} \label{lemma}
Let $n \geq 6$ be a positive integer. Then $\displaystyle n\frac{\sin(\pi x/n)}{ \sin(\pi/n)} \geq x(n-x)+1$ for $x \in [3, n/2]$.
\end{lemma}

\begin{proof}
We use the elementary inequalities $\displaystyle x-\frac{x^3}{6}=x\left(1-\frac{x^2}{6} \right) \leq \sin(x) \leq x$ for $x \geq 0$ to see that \[ \displaystyle n\frac{\sin(\pi x/n)}{ \sin(\pi/n)} \geq \frac{n(\frac{\pi x}{n})\left(1-\frac{(\frac{\pi x}{n})^2}{6} \right)}{\frac{\pi}{n}} = nx-\frac{\pi^2x^3}{6n}. \]
Now for $3 \leq x \leq \frac{n}{2}$ we have $x^3 \leq (n/2)x^2$ so we have that \[ \displaystyle nx-\frac{\pi^2x^3}{6n} \geq nx-\frac{\pi^2x^2}{12}.\]
To prove the lemma, it suffices to show that $\displaystyle nx-\frac{\pi^2x^2}{12} \geq nx-x^2+1$ for $x \in [3,n/2]$. Equivalently, we must show that $x^2 \geq \left(1-\frac{\pi^2}{12} \right)^{-1}$. Using the fact that $\pi^2<10$, we see that $\left(1-\frac{\pi^2}{12}\right)^{-1}<6<x^2$ for $x \geq 3$. The inequality holds for $x \in [3,n/2]$. The result follows.
\end{proof}

\begin{prop}
Let $n \geq 4$ be a positive integer and let $k \in \{2,3, \cdots, \lfloor n/2 \rfloor \}$. Then $n\frac{\sin(\pi k/n)}{ \sin(\pi/n)} >k(n-k)+1$.
\end{prop}

\begin{proof}
For $k \geq 3$ and consequently $n \geq 6$, the result follows from Lemma \ref{lemma}. It suffices to consider the case when $k=2$. By the double angle formula, we have $\sin(2 \pi/n)=2 \cos(\pi/n)\sin(\pi/n)$, so the case $k=2$ reduces to the inequality \[\displaystyle 2n \cos(\pi/n) \geq 2n-3. \] Using the elementary inequality $\displaystyle \cos(x) \geq 1-\frac{x^2}{2}$ for $x \geq 0$, we have that \[\displaystyle 2n \cos \left(\pi/n \right) \geq 2n- \frac{2n \left (\frac{\pi}{n} \right)^2}{2}=2n-\frac{\pi^2}{n}. \] Since $n \geq 4$ and $\pi^2<10$ we see $2n-\frac{\pi^2}{n}>2n-3$. The result follows.
\end{proof}

Since $\Gr(k,n)$ is isomorphic to $\Gr(n-k,n)$ it is suffices to show that Galkin's conjectured lower bound holds for $k=1,2, \cdots, \lfloor n/2 \rfloor$ with equality only when $k=1$. This follows immediately from the above inequalities for $n \geq 4$. The bound obviously holds for the $n=2$ and $n=3$ cases. Theorem \ref{mainthm} follows.

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