TITLE: How do I prove this set is completely bounded?
QUESTION [0 upvotes]: Let $K \in C([0,1]\times[0,1])$. For $f \in C([0,1])$, define:
$$Tf(x) = \int_{0}^{1} K(x,y)\ f(y) \ dy, \ \ \ \ x \in [0,1]$$.
With $||f||_{\infty} := sup_{x \in [0,1]} |f(x)|$ norm, prove that the set
$$\{Tf : f \in C ([0,1]), \ ||f||_{\infty} \le 1 \}$$
is completely bounded in $C([0,1])$.
I tried using Arzela-Ascoli but couldn't prove it.

REPLY [1 votes]: You mean "totally bounded" as in "can be covered by finitely many subsets of any fixed 'size'". Which is a given if the set is pre-compact. Whence the use of Arzela-Ascoli makes sense.
As
\begin{align}
|(Tf)(y)-(Tf)(x)|&\le\int_0^1|K(y,s)-K(x,s)|·|f(s)|\,ds\\
&\le\int_0^1|K(y,s)-K(x,s)|\,ds
\end{align}
the uniform continuity of $K$ on the compact set $[0,1]^2$ translates into the equi-continuity of the image of the unit ball.


TITLE: Why don't we take the universal gravitational constant $G$ to be equal to 1 in $F= \frac{Gm_{1}m_{2}}{r^2}$?
QUESTION [1 upvotes]: In the derivation of Newton's Second Law, we get to an equation $F=kma$. Since this equation is essentially defining force, Newton could have taken the value of $k$ to be anything. For the sake of convenience, he took it to be unity and hence we get $F=ma$. Going by the same argument, in Newton's law of gravitation we get the equation  $F= \frac{Gm_{1}m_{2}}{r^2}$. Why cannot we take the value of $G$ to be unity for sake of convenience. Why does it have such a complicated value of $6.626× 10^{-11}$. How is it different from Newton's second law?

REPLY [5 votes]: In fact, in the Planck system of units G is taken to be 1. The trouble is that Newton was working with units of mass, distance, and time, that had already been chosen (feet, pounds, seconds), and they were chosen without respect to making the gravitational constant simple. The Planck system of units is very handy for cosmology, but the implied units for mass, force, distance, and time, are very awkward for day to day use. For example, the Planck unit of length is $1.616255(18)\times10^{-35}$ m.

REPLY [2 votes]: Ok let's try it. Say mass $m_1 = 1000$ kg is at distance $1$ metre from mass $m_2 = 0.5$ kg. How will mass $m_2$ move?
With $G=1$ the force on it is $F = m_1 m_2/r^2 = 500$ newtons. So the acceleration will be $a = F/m_2 = 1000$ m/s$^2$.
Now we do the experiment, and we observe that the acceleration is $6.67 \times 10^{-8}$ m/s$^2$. Clearly the above theoretical calculation is wrong!
The point is that the experiment here tells us the value of $G$. It is not an independent parameter. Upon observing $6.67 \times 10^{-8}$ m/s$^2$ in this experiment, we deduce that the value of $G$ must be $6.67 \times 10^{-11}$  m$^3$ / kg s$^2$.
An early experiment of this type was the one done by Cavendish. He did not measure the acceleration directly (it is too small) but rather he inferred the value of the force $F$ from the movement of a torsion balance.


TITLE: Prove that a subset of an integral domain is relatively prime
QUESTION [1 upvotes]: This is a statement that appears in the text of D. J. H. Garling's book A Course in Galois Theory (page 30) without proof. I am trying to understand why it is true.
Assume that $R$ is an integral domain, and that $B$ is a non-empty subset of $R$. An element $a\in R$ is called a greatest common divisor of $B$ if $a$ is a common divisor of every element in $B$, and is divisible by every common divisor of all the elements in $B$. Assume that $a$ is the greatest common divisor of $B$, and define $$C=\{c\in{R}:ca\in B\}.$$ Prove that the greatest common divisor of $C$ is equal to $1$.
Here is what I have so far. Since $a$ is a greatest common divisor of $B$, it follows that $B\subset (a)$ where $(a)$ is the ideal generated by $a$. If $d$ is the greatest common divisor of $C$, then since every element of $B$ is a multiple of $a$, $d$ must be a common divisor of $B$, and must therefore divide $a$. If I can prove that there exists $c\in R$ that is prime to $a$ such that $ca\in R$, then I could conclude that $d$ must be a common divisor of two coprime elements and must therefore be equal to $1$. I haven't used the fact that $R$ is an integral domain so far. Wondering whether this helps me find such an element $c$, such that $c$ and $a$ are coprime and $ca\in B$.

REPLY [1 votes]: Hint: $\,\ \forall i\!:\ d\mid b_i/a\iff \forall i\!:\ ad\mid b_i\!\!\!\overset{\ \,\rm\color{#90f} U}\iff ad\mid\overbrace{\gcd(b_1,b_2,\ldots)}^{\textstyle \gcd(B)=\color{#0a0}a}\iff ad\mid\color{#0a0} a\iff d\mid\color{#c00} 1$
therefore $\, \gcd\{b_i/a\} =\color{#c00}1,\ $ by $\,\rm \color{#90f} U = $ GCD Universal Property.


TITLE: Ratio test for convergence of a given series
QUESTION [1 upvotes]: If $a_n=\frac{n!}{n^n}$ and $a_{n+1}=\frac{(n+1)!}{(n+1)^{n+1}}$, then applying the ratio test we get:
$$\frac{(n+1)!}{(n+1)^{n+1}}:\frac{n!}{(n+1)^n}=\frac{(n+1)!\cdot n^n}{n^{n+1}\cdot n!}=\frac{(n+1)n^n}{(n+1)(n+1)^n}=\left( \frac{n}{n+1} \right) ^n\to\frac{1}{e}.$$
So $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}<1 \Rightarrow $ $\sum_{n=1}^{\infty}a_n$ converges.
I understand the reasoning behind this example, but I did it in a different way and I would like somebody to show me where I did an error:
$$\frac{(n+1)!\cdot n^n}{n^{n+1}\cdot n!}=\frac{n!(n+1)\cdot n^n}{n^n \cdot n \cdot n!}=\frac{(n+1)}{n}=1,$$ so $\sum_{n=1}^{\infty}a_n$ diverges.

REPLY [1 votes]: I suppose it is only a typo while writing it down, but $a_{n+1}$ is given by $\frac{(n+1)!}{(n+1)^{n+1}}$ (as you noted earlier by yourself) but you used $\frac{(n+1)!}{\color{red}{n}^{n+1}}$ instead in your calculations. Then the ratio test is inconclusive as you obtain $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim\limits_{n\to\infty}1+\frac1n=1$ (this limit does not approach $e$ by the way). This mistake was of course already pointed out in the comments.


TITLE: Show that $29^{41}+41^{29}$ is divisible by 35
QUESTION [0 upvotes]: How con I Show that $29^{41}+41^{29}$ is divisible by 35.
attempt: I think it shouldn’t be difficult but not sure how to prove it.
29 is not a multiple of 5 nor of 7 same for 41.

REPLY [0 votes]: Hint
$$29\equiv -6(\mod 35)$$
$$41\equiv 6(\mod 35)$$
and
$$6^2\equiv 1(\mod 35)$$

REPLY [0 votes]: The diract way:
$29^{41} + 41^{29}$ mod $35$ = $(-6)^{41} + 6^{29}$ mod $35$ = $-6 \cdot (-6)^{40} + 6\cdot6^{28}$ mod $35$ = $-6 \cdot 36 ^{20} + 6 \cdot 36^{14}$ mod $35$ = $- 6 + 6$ mod $35$ = $0$ mod $35$
So we can see that $35$ divides $29^{41} + 41^{29}$.


TITLE: Charakterisation of Homomorphismen from $\mathbb{R} $ to $\mathbb{T}$
QUESTION [0 upvotes]: I wonder if there is a characterisation of all continous Homomorphismen from $(\mathbb{R}_+, \cdot)$ to $(\mathbb{T},\cdot)$. With $\mathbb{R}_+=(0, \infty)$ and $ \mathbb{T} $ the unit Circle in $ \mathbb{C}$. And $ \cdot $ the usual multiplication.

REPLY [0 votes]: Sure.$$\begin{array}{ccc}\mathbb{R}_+&\longrightarrow&\mathbb T\\x&\mapsto&e^{in\log(x)}\text,\end{array}$$with $n\in\mathbb Z$.


TITLE: Why can I remove the first term from euler equation for the shortest path between two points?
QUESTION [1 upvotes]: I'm studying the basic example of Euler's equation that is to obtain the shortest path between two points.
$$
f = \sqrt  {1 + y' ^ 2}
$$
$$
\frac{\partial f}{\partial y} - \frac{d}{dx}(\frac{\partial f}{\partial y'})
   = 0
$$
My textbook said that I could replace the first term
$ \frac{\partial f}{\partial y} $
to $ 0 $, because there is no $ y$ in the definition of $ f $.
I got no intuition to do so, so I tried to calculate $ \frac{\partial f}{\partial y} $.
$$
\frac{\partial f}{\partial x} \frac{\partial x}{\partial y}
 = \frac{\partial f}{\partial x} \frac{1}{y'} = {(y'^2 + 1)}^{-1/2}
 = \frac{1}{2}{(y'^2 + 1) ^ {-1/2}}{2y'y''}\frac{1}{y'} = {(y'^2+1)}^{-1/2}y''
$$
And I don't think $ {(y'^2+1)}^{-1/2}y'' = 0 $.
This derivation doesn't seem to give the same result of the textbook.
What did I do wrong?

REPLY [1 votes]: What you did wrong is computing the partial derivatives, for instance in your case  $\frac{\partial f}{\partial x}=0$ (see this).

For example if $f(x,u)=x^{2}+u(x)$, then $\frac{\partial f}{\partial x}=2x,\frac{\partial f}{\partial u}=1$ and $\frac{df}{dx}=2x+\frac{d}{dx}u.$

Continuing from your example, since $\frac{\partial f}{\partial y}=0$, then we have $- \frac{d}{dx}(\frac{\partial f}{\partial y'})
   = 0$ or $\frac{d}{dx}(\frac{\partial f}{\partial y'})
   =0.$
So $$\frac{d}{d x}(\frac{\partial f}{\partial y'})=\frac{d}{dx}(\frac{\partial }{\partial y'}\big((1+y'^{2})^{\frac{1}{2}}\big)$$
$$=\frac{d }{dx}(\frac{y'}{\sqrt{1+y'^{2}}})=0$$
(You may think of $\frac{\partial}{\partial y'}\big((1+y'^{2})^{\frac{1}{2}}\big)$ the same as computing $\frac{d}{dx}\sqrt{1+x^2}$).
Then you have $$\frac{y'}{\sqrt{1+y'^{2}}}=c_{1}$$ and since the denominator is never $0$ we have $y'^{2}=c_{1}^2(1+y'^{2})$ so re-arranging gives
$$y'^{2}=\frac{c_1^{2}}{1-c_{1}^2}$$
(Can $c_{1}=\pm 1$?) so you have $$y'=\sqrt{\frac{c_1^{2}}{1-c_{1}^2}}=\text{constant}=a$$
Thus $y(x)=ax+b.$


TITLE: General solution to differential problem with $dy/dx$
QUESTION [0 upvotes]: I am given $\frac{dy}{dx} + 6y=e^x$ and I am trying to find the general solution but I am struggling. Can someone solve it out? Thanks

REPLY [0 votes]: If
$$
y'+6y=e^x
$$
then we can multiply both sides by the integrating factor $e^{6x}$
$$y'e^{6x}+6ye^{6x}=e^{7x}
$$
Now observe that
$$
(ye^{6x})'=y'e^{6x}+6ye^{6x}
$$
Can you take it from here?


text:
\begin{document}

\title{Generalized Small Cancellation Presentations for Automatic Groups}

\author{Robert H. Gilman}
\address{Department of Mathematical Sciences\\Stevens Institute of Technology}
\email{rgilman@stevens.edu}
\thanks{Partially supported by NSF grant 1318716}

\subjclass[2010]{20F05, 20F65 }

\keywords{small cancellation, automatic group, pregroup}

\date{June 28, 2014}

\begin{abstract}
By a result of Gersten and Short finite presentations satisfying the
usual non-metric small cancellation conditions present biautomatic groups.
We show that in the case in which all pieces have length one, a generalization
of the C(3)-T(6) condition yields a larger collection of biautomatic groups.
\end{abstract}

\maketitle

\newcommand{\inv}{^{-1}}
\newcommand{\ra}{\rightarrow} \newcommand{\tto}{\buildrel * \over
\rightarrow }
\newcommand{\edge}[1]{\buildrel #1 \over \rightarrow }
\newcommand{\abs}[1]{\vert #1\vert} \newcommand{\Abs}[1]{\Vert #1 \Vert}
\newcommand{\set}[1]{\{ #1 \}}
\newcommand{\subgroup}[1]{\langle #1 \rangle}
\newcommand{\ovr}[1]{\overline #1}
\newcommand{\G}{\Gamma}
\newcommand{\K}{\mathcal K}
\newcommand{\PP}{P^{(2)}}

\section{Introduction}
Gersten and Short show in~\cite{GS1} that groups satisfying any of the usual non-metric small cancellation conditions C(6), or C(4)-T(4), or C(3)-T(6) are biautomatic. Most of their argument with respect to C(3)-T(6) applies verbatim to a more general situation and yields small cancellation type presentations for a larger class of automatic groups. These presentations are essentially finite prees (in the sense of Frank Rimlinger~\cite{Ri}) which satisfy two axioms introduced by Reinhold Baer~\cite{Ba}.

A pree, defined below, is a set with a partial multiplication satisfying certain axioms. An example is an amalgam of two groups; the multiplication is defined for pairs of elements within each group but is not defined for pairs which are not within one group or the other. Amalgams obey an additional axiom which make them pregroups as defined by John Stallings~\cite{St1}. See the recent survey~\cite{GL2012} for an introduction to prees and their axioms.

The axioms we employ, A(4) and A(5), are given in Section~\ref{prees}. Their relation to the usual small cancellation conditions is discussed at the beginning of Section~\ref{cancellation}, where we develop a variation of small cancellation theory. There are many such variations, e.g.~\cite{J1, J2, J3, J4, MW1, MW2, O}. The one which which seem closest to ours is by Uri Weiss~\cite{W}, who shows that V(6) groups with all pieces of length 1 are biautomatic. V(6) denotes a generalization of the C(6) and C(4)-T(4) conditions.

Section 4 contains the proof of our main theorem, Theorem~\ref{A45}, which says that the universal groups of finite prees satisfying Axioms A(4) and A(5) are biautomatic. As every finite group satisfies these two axioms and is its own universal group, the class of groups covered by Theorem~\ref{A45} includes all finite groups. Finite C(3)-T(6) groups, on the other hand, are all cyclic~\cite{EH}. It would be interesting to have other examples of groups from Theorem~\ref{A45} which are not C(3)-T(6).

\section{Prees}\label{prees}

\begin{definition}\label{pree}
A pree is a set equipped with a partial multiplication which affords an identity, denoted $1$, two-sided inverses, and the following form of associative law.
\end{definition}

It is not hard to show that inverses are unique in a pree. Further if $ab=c$, then $b = a\inv c$ etc. It follows that products $ab=c$ may be thought of as triangles. More precisely the six products obtained by reading the edge labels around the boundary of the triangle on the left-hand side of Figure~\ref{triangles} in either direction and starting at any vertex are defined if any one of them is. As usual an edge read against its orientation contributes the inverse of its label.
\begin{figure}
\includegraphics[width=.5\textwidth]{triangles}
\caption{The product $ab=c$ (left) and the associative law (right).\label{triangles}}
\end{figure}
\begin{definition}[The associative law]\label{associative}
If $ab$ and $bc$ are defined, then $(ab)c$ is defined if and only if $a(bc)$ is; and when they are defined, $(ab)c=a(bc)$.
\end{definition}

The associative law is equivalent to a geometric condition, namely that if three triangles fit around a common vertex as in the right-hand side of Figure~\ref{triangles}, then the perimeter is also a valid triangle in
$P$.

\begin{definition} The universal group of a pree $P$ is the group $U(P)$
with generators $P$ and relators $abc\inv$ for every product $ab=c$
defined in $P$. We write $\overline a$ for the image of $a$ in $U(P)$.
Elements of $U(P)$ are represented by words over the alphabet $P$. Note that $1$ is a letter in this alphabet.
\end{definition}
\begin{definition}\label{reducible}
A word over $P$ is reducible if the product of two successive letters is defined in $P$. Otherwise the word is called irreducible or reduced.
\end{definition}

It is straightforward to use Tietze transformations to change any finite
presentation into one given by a finite partial multiplication table, and with more
Tietze transformations one can insure that this multiplication table
satisfies the conditions of a pree. Thus every finitely presented group is the universal group of a finite pree, and the pree affords a finite presentation for the group.

\begin{theorem}\label{embedding} It is undecidable whether or not the canonical morphism $\pi: P \to U(P)$ from a pree $P$ to its universal group is an embedding.
\end{theorem}

This theorem is a special case of a result of Trevor Evans~\cite{Ev}
which says that if the embedding problem is solvable for a class of
finite partial algebras, then the word problem is solvable for the
corresponding class of algebras.

Now we present the additional axioms we need. There are two of them, and they have the geometric incarnations given in Figure~\ref{spree}

\begin{figure}
\includegraphics[width=.9\textwidth]{axioms}
\caption{Axioms A(4) and A(5) with $b_2b_3$ defined. \label{spree}}
\end{figure}

\begin{description}
\item [Axiom A(4)] If the products $a_1\inv a_2=b_1$, $a_2\inv a_3=b_2$, $a_3\inv a_4=b_3$, and $a_4\inv a_1=b_4$ are defined, then at least one of the products $b_ib_{i+1}$ (indices read modulo 4) is defined.
\item [Axiom A(5)] If the products $a_1\inv a_2=b_1$, $a_2\inv a_3=b_2$, $a_3\inv a_4=b_3$, $a_4\inv a_5=b_4$ and $a_5\inv a_1=b_5$ are defined, then at least one of the products $b_ib_{i+1}$ (indices read modulo 5) is defined.
\end{description}

\section{Small Cancellation Theory}\label{cancellation}

We require a simple extension of small cancellation theory for diagrams constructed from triangles

 All relations in the presentation $P$ for $U(P)$ have length 3, and there are no pieces of length 2. Indeed $ab$ can occur in at most one relator, namely the relator $abc\inv$ corresponding to a product $ab=c$ defined in $P$. Thus the small cancellation condition C(3) holds.

Similarly the condition T(6) says that the perimeter of any diagram formed by fitting 2, 3, 4 or 5 triangles around a common vertex (as in Figure~\ref{spree} and the right-hand side of Figure~\ref{triangles}) contains a subword $aa\inv$ for some $a\in P$. Axioms A(4) and A(5) on the other hand together with the Associative law enforce the weaker condition that the perimeter is reducible.

\begin{definition}
A diagram is a planar directed labeled graph whose boundary is a simple
closed curve and whose faces are triangles each with 3 distinct
vertices. The label of the boundary of each face is a relator in $P$.
The number of triangular faces of a diagram is its area. The boundary word, determined up to cyclic permutation and inverse, is the label of the boundary path. Boundaries of triangles are sometimes called perimeters. For brevity boundary words are sometimes themselves called boundaries.
\end{definition}

The label of a path in a diagram is the product of the labels of edges
except that an edge traversed against its orientation contributes the
inverse of its label. We may change the orientation of an edge as long
as we invert its label. To remove a triangle with one or more edges on
the boundary we remove the edges on the boundary except we keep any
vertices which began with degree more than 2.

\begin{lemma} If $D$ is a diagram with area greater than one, there is a
triangle in $D$ with one or more edges on the boundary such that
removing that triangle results in a diagram.
\end{lemma}

\begin{proof}
If there is a triangle with two edges on the boundary, then because the boundary is a simple closed curve, the common vertex has degree 2, and the triangle may be removed. Likewise a triangle with only one edge on the
boundary and the opposite vertex in the interior of the diagram may be removed.

Suppose
then that all triangles meeting the boundary have exactly one edge and
its opposing vertex on the boundary. For each such triangle the
complement of the diagram is divided into two parts. Since the triangle
has two edges not on the boundary, both parts have positive area. Pick a
triangle such that one of the complementary parts has minimum possible
area. Since that area is positive, it must contain another triangle
meeting the boundary. But then that triangle would give a complementary
part of smaller area. Thus this last case cannot arise.
\end{proof}

The preceding lemma has the following consequence.

\begin{lemma}\label{recursive}
Every diagram is generated by starting with a single triangle and attaching subsequent triangles in the following way. A triangle is attached to a diagram by identifying one or two edges of the triangle with an edge or two successive edges respectively of the boundary of the diagram so that identified edges have the
same label and are oriented in the same direction or have inverse labels
and are oriented in opposite directions. In addition two vertices of a
triangle may not be identified when it is attached.
\end{lemma}

\begin{lemma}\label{boundary} If a diagram has two triangles with the same vertices, then it has an internal vertex of degree two.
\end{lemma}

\begin{proof}
Let $D$ be a diagram. If $D$ has area 1, there is nothing to prove. Otherwise $D$ is formed by attaching a triangle $T$ to a diagram $D'$ of smaller area. By induction we may assume $D'$ does not have two triangles with the same vertices. Consequently the vertices of $T$ are also the vertices of some triangle $T'$ of $D'$. But then these vertices must be the vertices of a path of length 2 on the boundary of $D'$. It follows that the middle vertex is an internal vertex of degree 2 in $D$.
\end{proof}

\begin{definition}
 A minimal diagram is one which has minimum area among all diagrams with the same boundary word (up to cyclic permutation and inverse.).
\end{definition}

\begin{lemma}\label{minimal}
Let $D$ be a minimal diagram with boundary $w$. If $D$ has an internal vertex of degree 2, then $D$
consists of two triangles joined along two edges and $w=aa\inv$.
\end{lemma}

\begin{proof}
It is easy to see that no vertex of any diagram has degree 1. If $p$ is an internal vertex of degree 2, then $p$ is a vertex of two triangles both of which contain the two edges incident to $p$. Hence both edges opposite $p$ have label $c=a\inv b$. Unless both of these edges are on the boundary of $D$, they can be combined to make a diagram of smaller area. If they are on the boundary, they must be the whole boundary because the boundary of $D$ is a simple closed curve. Thus $D$ consists of two triangles joined along two edges, and $w=cc\inv$.
\end{proof}

\begin{lemma}\label{le:curvature} Consider any diagram, and let $d(p)$ be the degree of each
vertex $p$, then
\begin{equation} \label{eq:curvature}
\sum^{\circ} 4-d(p) =6 + \sum^{\bullet} d(p)-6
\end{equation}
where the first sum is over all boundary vertices and the second is over
all internal ones.
\end{lemma}
\begin{proof}
Argue by induction on area.
\end{proof}

Let $L(P)$ be the language of all labels of boundaries of diagrams. Each
diagram determines multiple labels as we may start at any vertex
on the perimeter and proceed clockwise or counterclockwise. Notice that
because we do not identify vertices when building a diagram by attaching
triangles, the boundary of every diagram has length at least two.

\begin{lemma} $L(P)$ is the set of all words $w$ over the alphabet $P$ of length at least 2 such that $\overline w$ is the identity in $U(P)$.
\end{lemma}

\begin{proof}
$U(P)$ is the quotient of the free semigroup $S$ over $P$ by the congruence $\sim$ with generators $ab\sim c$ for all products $ab=c$ defined in $P$. Say that $v, w\in S$ are {\em neighbors} if one is obtained from the other by replacing a subword $ab$ with $c$. Observe that if  $v$ and $w$ are neighbors of length at least 2, and $D$ is a diagram for $w$, then a diagram for $v$ can be obtained by attaching a triangle to $D$.

Suppose $|w|\ge 2$ and $w$ defines the identity in $U(P)$. Then there is a sequence of neighbors
\[ w=u_1, u_2, \ldots, abc\inv \]
where $ab=c$ is any product defined in $P$. As all elements of the sequence
\[ w, 1w, 1u_1, 1u_2, \ldots, 1abc\inv, abc\inv\]
have size at least 2, it follows that there is a diagram for $w$.

Conversely if $w$ is a perimeter label for a diagram $D$, then by removing triangles we can reduce $D$ to a single triangle $T$. The corresponding sequence of neighbors begins with $w$ and ends with a label of the perimeter of $T$. It follows that $\overline w = 1$.
\end{proof}

\section{Biautomaticity}\label{biautomaticity}

We consider a fixed pree $P$ satisfying Axioms A(4) and A(5) and show that it is biautomatic. Much of our argument follows that in~\cite{GS1}.

\begin{lemma}\label{d6} If $D$ is a minimal diagram without internal vertices of degree 2, then all internal vertices have degree at least 6.
\end{lemma}

\begin{proof}
If an internal vertex $p$ has degree 3, then by the associative law from
Definition~\ref{associative} we can remove $p$ and all incident edges and still have a diagram over $P$. Likewise if $p$ has degree 4 or 5, then Axioms A(4) and A(5) guarantee that one can discard $p$ and construct a diagram of smaller area by adding one or two internal edges.
We must check that each new triangle created by the addition of edges has all vertices distinct. But if it did not, $D$ would have two triangles sharing the same three vertices, contrary to Lemma~\ref{boundary}.
See Figure~\ref{spree}.
\end{proof}

\begin{lemma} \label{galleries} Let $w$ be a word of length at least 3
defining the identity in $U(P)$, and let $D$ be a diagram of minimal area
with boundary $w$. Let $\delta_2$ and $\delta_3$ be the number of
vertices of degree 2 and 3 respectively on the boundary of $D$, and let
$\delta_5$ be the number of degree greater than 4, then $2\delta_2 +
\delta_3 \ge 6 + \delta_5$. Further, if equality holds, then all
internal vertices have degree 6.
\end{lemma}

\begin{proof}
Each vertex of degree 2 contributes 2 to the left-hand side of the
Equation~\ref{eq:curvature} in Lemma~\ref{le:curvature}. Likewise each vertex of degree 3 contributes
1, and vertices of degree greater than 4 contribute -1 or less. As all
internal vertices have degree at least 6 by Lemmas~\ref{minimal} and~\ref{d6},
Equation~\ref{eq:curvature} yields the desired result.
\end{proof}

\begin{lemma} \label{45} Words of length 4 or 5 which define the
identity in $U(P)$ are reducible.
\end{lemma}

\begin{proof}
Let $w$ be such a word and apply Lemma~\ref{galleries} to a minimal
diagram for $w$. If $\delta_2\ge 2$, then some boundary vertex in the
interior of $w$ has degree 2, and we are done. The only other possibility
is $\abs w = 5$, $\delta_2=1$, and $\delta_3=4$. But it is easy to see
that there is no diagram with these parameters.
\end{proof}

\begin{theorem}\label{A45} If $P$ is a pree satisfying Axioms A(4) and
A(5), then $P$ embeds in $U(P)$, and the multiplication in $P$ is
induced by the multiplication in $U(P)$.
\end{theorem}
\begin{proof}
The theorem is proved by the same small cancellation argument used
in~\cite{GS1}. Let $w$ be a word of length at least two over $P$ which
defines the identity in $U(P)$ and consider a diagram $D$ of minimum
area for $w$. We must show that if $w=ab$, then $b=a\inv$; and if
$w=abc$, then $ab=c\inv$ in $P$. By Lemmas~\ref{minimal} and~\ref{d6} we may assume
that $w=abc$ and all internal vertices of $D$ have degree at least 6. By
Lemma~\ref{galleries} there are either three boundary vertices of degree
2 or at least 4 boundary vertices. It follows that $D$ is a triangle.
\end{proof}

The proof of the next theorem is given in a sequence of lemmas. It is
useful to keep in mind the following example
$$
P=\set{(0,0),(0,1), (0,-1), (1, 0), (-1,0), (1,1), (-1,-1) }
$$
with $U(P)=Z\times Z$. The partial multiplication in $P$ is inherited from the usual addition in $Z\times Z$.

\begin{theorem}If $P$ is a finite pree satisfying Axioms A(4) and A(5),
then $U(P)$ is biautomatic.
\end{theorem}

First we observe that Lemma~\ref{galleries} implies that in any diagram
$D$ of minimum area there must be a number of intervals along the
boundary consisting of two vertices of degree 2 or 3 either adjacent to
each other or separated by vertices of degree 4. As in~\cite{GS1} we
refer to these intervals as galleries. See Figure~\ref{gallery}.
\begin{figure}
\includegraphics[width=.7\textwidth]{gallery}
\caption{Some galleries. \label{gallery}}
\end{figure}
Note that some of the vertices in this figure might be identified in
$D$. In other words there is a morphism of diagrams from the gallery as
illustrated to $D$. Under this morphism the labeled vertices have
distinct images and their degrees are preserved.

The minimum number of galleries varies as in Table~\ref{g} depending on
the value of $\delta_2$.

\begin{table}
\begin{tabular}{ccc}
$\delta_2$ & $\delta_3$ & Galleries\\ \hline
 $\ge 3$ &$\ge 0$& 3 \\
2 & $\ge 2$ & 4 \\
1 & $\ge 4$ & 5 \\
0 & $\ge 6$ & 6 \\
\end{tabular}
\caption{Effect of $\delta_2$ on $\delta_3$ and the minimum number of galleries. \label{g}}
\end{table}

Now we augment Definition~\ref{reducible}.

\begin{definition} A word $w=a_1\cdots a_n$ over $P$ is irreducible if
for all $i$ the product $a_ia_{i+1}$ is not defined in $P$, and it is
strongly irreducible if in addition it cannot be shortened by attaching
to it a gallery of type $34^k3$ as in Figure~\ref{gallery} and replacing
$abc$ by $gh$ or $abcdef$ by $ghijk$ etc. Reductions of this latter type are called strong reductions.
\end{definition}

Both reductions and strong reductions shorten a word's length by 1.
There are also reductions corresponding to other types of galleries.
Reductions corresponding to galleries of type $24^k3$ shorten the length
of a word by 2.

Since galleries are diagrams over $P$, it is clear that the above
reductions do not change the element of $U(P)$ represented by a word,
over $P$. Thus every word can be reduced to a strongly irreducible word
representing the same group element.

\begin{lemma}\label{geodesic} A word $w$ over $P$ is the label of a
geodesic in $U(P)$ if and only if it is strongly reduced and not the
word $1$. Further the set of strongly irreducible words is a regular
language.
\end{lemma}
\begin{proof}
Clearly the set of irreducible words not equal to $1$ is regular.
Further it is straightforward to construct a finite automaton over the
alphabet $P\times P \cup P\times \set{\$}$ which accepts a pair of words
$(w,v\$)$ if and only if $w$ and $v$ are words over $P$ and $v$ is
obtained from $w$ from a single reduction corresponding to attaching a
gallery to $w$. Here $\$$ is a padding symbol not in $P$. The triangles
of $P$ can serve as the vertices of a suitable automaton. It follows by
standard techniques that the set of all words over $P$ which are strongly
reducible forms a regular language. Hence its complement is regular and
the intersection of the complement with the irreducible words not equal
to $1$ is regular too. Thus the second assertion of the lemma holds.

Clearly geodesics are strongly irreducible lest there be a shorter word
denoting the same group element. Suppose that $w$ is strongly
irreducible but not geodesic. In particular $w \ne aa\inv$. Also $\abs
w\ne 1$ by Theorem~\ref{A45}, so $\abs w \ge 3$. By our hypothesis there
must be a shorter word $v$ representing the same group element as $w$.
Thus there will be a diagram $D$ of minimum area with boundary label
$wv\inv$. Pick $v$ so that $D$ has smallest possible area.

Consider the possibilities in Table~\ref{g}. There cannot be 3 vertices
of degree 2, because then at least one of them would be in the interior
of $w$ or $v$ contradicting the fact that both are strongly irreducible.
Likewise there can be no gallery all of whose vertices of degree 3 or 4
lie in the interior of $w$ or in the interior of $v$. Thus there are 4
galleries and 2 vertices of degree 2. The only way they all fit into $D$
is if $\abs v \ge 1$, and the two boundary vertices separating the
boundary segments with labels $w$ and $v$ are of degree 2 and are each
part of two galleries. Each gallery has a vertex of degree 2 or 3 at its
other end. Since these vertices are in the interior of $w$ or $v$, they
must be of degree 3. In particular $\abs v \ge 2$.

Let $p$ and $q$ be the vertices separating $w$ and $v$ so that the
boundary of $D$ consists of a path with label $w$ from $p$ to $q$ and a
path with label $v$ also from $p$ to $q$. Let $w=a_1a_2\cdots w_m$, and
$v=b_1b_2\cdots b_n $, and denote by $r$ the boundary vertex between
$a_1$ and $a_2$.  By the previous paragraph there is a gallery of type
$24^k3$ attached to the boundary segment with label $a_1\inv b_1\cdots
b_{k+2}$. This gallery affords a reduction of $a_1\inv b_1 \cdots
b_{k+2}$ to $u=c_1\cdots c_{k+1}$. It follows that the path from $p$ to
$q$ with label $a_1c_1\cdots c_{k+1}b_{k+3}\cdots b_m$ is a geodesic. If
this path is $w$, then we are done. Otherwise there are paths from $r$
to $q$ with labels $a_2\cdots a_m$ and $c_1\cdots c_{k+1}b_{k+3}\cdots
b_n$. The first label is clearly strongly irreducible, and the second is
a geodesic. By induction on length, $m-1=n-1$.
\end{proof}

In order to show that $U(P)$ is biautomatic, we must find a set $L$ of
words over $P$ which maps onto $U(P)$ and such that for some constant
$K$ two paths which begin and end a distance at most one apart and which
have labels in $L$ $K$-synchronously fellow travel. $L$ will consist of some
geodesics, but not all of them. At this point in~\cite{GS1} use is made
of the C3-T6 small cancellation hypothesis, which need not hold in our situation. We must proceed differently.

\begin{figure}
\includegraphics[width=.7\textwidth]{maximal}
\caption{ Defining $L$ \label{maximal}}
\end{figure}

\begin{definition} \label{combing} Let $L$ be the set of strongly
reduced words $w=a_1\cdots a_n$ over $P$ with the property that whenever
there is a valid diagram of the type illustrated in Figure~\ref{maximal}
with $k\ge 3$, then the product $bc$ is defined.
\end{definition}

\begin{lemma}
$L$ is a regular set which maps onto $U(P)$.
\end{lemma}
\begin{proof}
The set of words which admit a valid diagram as in Figure~\ref{maximal}
such that $bc$ is not defined is clearly regular. It follows from this
observation and from Lemma~\ref{geodesic} that $L$ is regular. To show
that $L$ maps onto $U(P)$ let $g$ be any element of $U(P)$ and pick a
geodesic $w=a_1\cdots a_n$ from $1$ to $g$ with the property that the
number of words $ef$ representing the same group element as $a_1a_2$ is
maximal, and subject to that the number representing the same group
element as $a_3a_4$ is maximal, and so forth.

Suppose $w$ admits a diagram as in Figure~\ref{maximal} for which $bc$
is not defined. Let $ef$, representing the same group element as
$a_{2i+1}a_{2i+2}$. As $efc\inv b\inv a\inv$ represents the identity in
$U(P)$, it is reducible by Lemma~\ref{45}. On the other hand $b\inv
a\inv$ and $ef$ are subwords of geodesics and so are irreducible.
Further $bc$ is irreducible by assumption. Thus $fc\inv=f'$ for some
$f'\in P$. We conclude that $ef'$ represents the same group element as
$ab$.

But no $ef'$ obtained as above can have $e=a$ and $f'=b$. For if so,
$1=efc\inv b\inv a\inv = afc\inv b\inv a\inv$ would imply that $bc=f$ in
$P$. Thus the geodesic obtained by substituting $ab\cdots d$ for
$w_{2i+1}\cdots w_{2i+k}$ in $w$ contradicts the choice of $w$.
\end{proof}

It remains to show that for some constant $K$ two paths which begin and
end a distance at most one apart and which have labels in $L$,
$K$-synchronously fellow travel. Suppose $w = a_1\cdots a_n$ and $v =
b_1\cdots b_m$ are labels of two such paths. Let $g_i$ be the group
element reached by $a_1\cdots a_i$, and define $h_i$ likewise with
respect to $v$. We claim that for each $i$ there exists a $j$ such that
$\rho(g_i,h_j)$, the distance in $U(P)$ between $g_i$ and $h_j$, is at
most 2; and likewise with $i$ and $j$ reversed. Since $w$ and $v$ are
both geodesics beginning a distance at most one apart, it follows by a
straightforward argument that $\abs {i-j}\le 3$; and hence that $K=5$
suffices.

We use induction on $m+n$. It $m$ and $n$ are both at most 3, then the
desired conclusion is immediate. Thus we assume $m\ge 4$ or $n\ge 4$.
Since $w$ and $v$ are geodesics, it follows that $m\ge 2$ an $n\ge 2$.
Because of the way $L$ is defined, it suffices to show that
$\rho(g_2,h_2)\le 1$.

By hypothesis there are letters $a,b\in P$ such that $awb\inv v\inv$
represents the identity in $U(P)$. Note that $a=1$ and $b=1$ are
possible. Consider a corresponding diagram $D$ of minimum area with
vertices labeled as in Figure~\ref{rectangle}.
\begin{figure}
\includegraphics[width=.5\textwidth]{rectangle}
\caption{ Two geodesics \label{rectangle}}
\end{figure}

\begin{figure}
\includegraphics[width=.5\textwidth]{d2333}
\caption{ The case $\delta_2=1$. \label{D2333}}
\end{figure}
If $g_n$ is joined to $h_{m-1}$ or $h_m$ to $g_{n-1}$ in $D$, then we
are done by induction on $m+n$. Thus we may assume that only $g_0$ or
$h_0$ can have degree 2. Hence either $\delta_2=1$, and there are at
least 5 galleries or $\delta_2=0$ and there are at least 6 galleries. As
the vertices supporting a gallery cannot all lie on any one side, 6
galleries is the maximum there can be. Consequently the inequality
in Lemma~\ref{galleries} is an equality, and all internal vertices must
have degree 6.

Suppose $\delta_2=1$. By symmetry we may assume $g_0$ has degree 2.
Because there are at least 5 galleries, none attached to a single side,
it must be that the other corner vertices all have degree 3. The
situation is illustrated in Figure~\ref{D2333}.

Now consider the subdiagram $D'$ with corners $g_1$, $g_n$, $h_1$,
$h_m$. Of course $D'$ is a diagram of minimum area for its boundary.
If either $g_1$ or $h_1$ has degree 2 with respect to this subdiagram,
then there is an edge from $g_2$ to $h_2$, and we are done. The
alternative is that the corners of $D'$ have degree 3, there are 6
galleries, and all internal vertices of $D'$ have degree 6. See
Figure~\ref{D2333P} where the degree of $g_2$ is 3 or 4 depending on
whether or not $p=g_3$ and likewise for $h_2$ and $q$.
\begin{figure}
\includegraphics[width=.5\textwidth]{d2333p}
\caption{ The case $\delta_2=1$ continued. \label{D2333P}}
\end{figure}
We know that the vertices $h_1,h_2, \ldots$ are part of a gallery in
$D'$. That is, in $D'$,  $h_1$ and $h_k$ have degree 3 for some $k \ge
2$ and all intervening boundary vertices have degree 4. But now
Definition~\ref{combing} implies that we can remove the edge from $h_1$
to $r$ and replace it with an edge from $g_1$ to $h_2$. With this change
we obtain a diagram of minimum area with an internal vertex of degree 5
in contradiction to Lemma~\ref{minimal}.

It remains to consider the case in which $D$ has $\delta_2=0$. Here we
begin with the situation of Figure~\ref{D3333} and use the preceding
argument.
\begin{figure}
\includegraphics[width=.5\textwidth]{d3333}
\caption{ The case $\delta_2=0$. \label{D3333}}
\end{figure}
If $q \ne h_2$, then we replace the edge from $h_1$ to $q$ by one from
$r$ to $h_2$ thereby decreasing the degree of $q$ to 5 and obtaining a
contradiction. We obtain a similar contradiction if $p \ne g_2$. Finally
if $q=h_2$ and $p=g_2$, then we are done.

config: arxiv


TITLE: Partial differential equations $u u_{xy} + u_{x} = y$
QUESTION [1 upvotes]: I am self studying partial differential equations and I can do every problem in the first chapter except for this one.  Can anyone help me?

Solve this partial differential equation.
$u u_{xy} + u_{x} = y$

REPLY [2 votes]: Upon inspection: the book does not say "Solve this partial differential equation". It says "Look, this is a partial differential equation." You are supposed to look at it and say "Yes, I see, this is a partial differential equation." Nobody expects you to solve it, and you should not expect that there is a formula for its solutions.

Having a formula for solutions of a PDE (or even of an ODE) is an exception, not a rule. Textbooks present as exercises only carefully designed equations for which the methods they present work. If you compose a PDE at random by multiplying and adding a bunch of derivatives, none of those methods will yield a solution.


TITLE: If $T$ and $T^2$ have equal rank then $V=\ker T\oplus {\rm im}\, T$ for $V$ finite dimensional.
QUESTION [6 upvotes]: I am trying to prove the following:
Let $V$ be a finite-dimensional vector space. Consider an operator $T$ on $V$ such that $\text{dim range}(T)=\text{dim range}(T^2)$. Show that $V=\text{null}(T)\oplus \text{range}(T)$.
I have been given the following hint:
$\textit{Hint:}$ Show that there does not exist a $y\in V$ such that $y\notin \text{null}(T)$ and $y\notin \text{range}(T)$.
I decided to try and prove the hint by contradiction...
Suppose that there exists a $y\in V$ such that $y\notin \text{null}(T)$ and $y\notin \text{range}(T)$. Then
$$Ty\neq 0$$
and for every $v\in V$ we have $$Tv\neq y.$$
I have NO idea what to do with this information! I would greatly appreciate some hints/help! Thank you.

REPLY [1 votes]: By rank-nullity $ \dim V = \dim \ker T + \dim \textrm{Im}\, T $, so it suffices to check that $ \ker T \cap \textrm{Im}\, T $ is trivial. Assume that $ x $ is in the intersection, then $ x = Tv $ for some $ v $, and $ Tx = T^2 v = 0 $. Since the given condition implies $ \ker T = \ker T^2 $ ($ \ker T $ is a subspace of $ \ker T^2 $ of equal dimension), $ v \in \ker T $ and $ x = 0 $. Thus, the desired conclusion $ V = \ker T \oplus \textrm{Im}\, T $ follows.


TITLE: How do we find the surface charge density of a charged conductor?
QUESTION [1 upvotes]: It is mentioned in Feynman volume 2 that it is quite algorithmic , where the surface charge density is first guessed then check whether it is equipotential at the metal surface..
My question is for a better way.

REPLY [0 votes]: That’s how nature does it. The charges sense the field configuration of the surface: if its not an equipotential, then there are components of the electric force in the surface. Since it’s a conductor, the charges can and do move in response to those. That reduces the potential differences. Repeat for a few tiny fractions of a second until done.


text:
\section{Primitive of Reciprocal of x squared by Root of x squared minus a squared cubed}
Tags: Primitives involving Root of x squared minus a squared

\begin{theorem}
:$\displaystyle \int \frac {\d x} {x^2 \paren {\sqrt {x^2 - a^2} }^3} = \frac {-\sqrt {x^2 - a^2} } {a^4 x} - \frac x {a^4 \sqrt {x^2 - a^2} } + C$
\end{theorem}

\begin{proof}
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^2 \paren {\sqrt {x^2 - a^2} }^3}
      | r = \int \frac {a^2 \rd x} {a^2 x^2 \paren {\sqrt {x^2 - a^2} }^3}
      | c =
}}
{{eqn | r = \int \frac {\paren {x^2 + a^2 - x^2} \rd x} {a^2 x^2 \paren {\sqrt {x^2 - a^2} }^3}
      | c =
}}
{{eqn | r = \frac 1 {a^2} \int \frac {-\paren {x^2 - a^2} \rd x} {a^2 x^2 \paren {\sqrt {x^2 - a^2} }^3} + \frac 1 {a^2} \int \frac {x^2 \rd x} {x^2 \paren {\sqrt {x^2 - a^2} }^3}
      | c = [[Linear Combination of Integrals]]
}}
{{eqn | r = \frac {-1} {a^2} \int \frac {\d x} {x^2 \sqrt {x^2 - a^2} } + \frac 1 {a^2} \int \frac {\d x} {\paren {\sqrt {x^2 - a^2} }^3}
      | c = simplifying
}}
{{eqn | r = \frac {-1} {a^2} \frac {\sqrt {x^2 - a^2} } {a^2 x} + \frac 1 {a^2} \int \frac {\d x} {\paren {\sqrt {x^2 - a^2} }^3} + C
      | c = [[Primitive of Reciprocal of x squared by Root of x squared minus a squared|Primitive of $\dfrac 1 {x^2 \sqrt {x^2 - a^2} }$]]
}}
{{eqn | r = \frac {-\sqrt {x^2 - a^2} } {a^4 x} + \frac 1 {a^2} \frac {-x} {a^2 \sqrt {x^2 - a^2} } + C
      | c = [[Primitive of Reciprocal of Root of x squared minus a squared cubed|Primitive of $\dfrac 1 {\paren {\sqrt {x^2 - a^2} }^3}$]]
}}
{{eqn | r = \frac {-\sqrt {x^2 - a^2} } {a^4 x} - \frac x {a^4 \sqrt {x^2 - a^2} } + C
      | c = simplification
}}
{{end-eqn}}
{{qed}}
\end{proof}

config: wiki


text:
\section{Introduction}
\label{introduction}

We consider the well-studied compressed sensing problem of recovering an unknown signal $x^*\in \R^n$ by observing a set of noisy measurements $y \in \R^m$ of the form
\begin{equation}\label{eqn:y=Ax}
   y=Ax^*+ \eta.
\end{equation}

Here  $A\in \R^{m\times n}$ is a known measurement matrix, typically generated with random independent Gaussian entries. Since the number of measurements $m$ is smaller than the dimension $n$ of the unknown vector $x^*$, this is an under-determined system of noisy linear equations and hence ill-posed. There are many solutions, and some structure must be assumed on $x^*$ to have any hope of recovery. Pioneering research~\cite{donoho2006compressed,candes2006robust,candes2005decoding}
established that if $x^*$ is assumed to be sparse in a known basis, a small number of measurements will be provably sufficient to recover the unknown vector in polynomial time using methods such as Lasso~\cite{tibshirani1996regression}.

Sparsity approaches have proven successful, but more complex models with additional structure have been recently proposed such as model-based compressive sensing~\cite{baraniuk2010model} and manifold models~\cite{hegde2008random,hegde2012signal,eftekhari2015new}.
Bora et al.~\cite{bora2017compressed} showed that deep generative models can be used as excellent priors for images. They also showed that backpropagation can be used to solve the signal recovery problem by performing gradient descent in the generative latent space. This method enabled image generation with significantly fewer measurements compared to Lasso for a given reconstruction error.
Compressed sensing using deep generative models was further improved in very recent work~\cite{tripathi2018correction,grover2018amortized,kabkab2018task,shah2018solving,fletcher2017inference, DBLP:journals/corr/abs-1802-04073}. Additionally a theoretical analysis of the nonconvex gradient descent algorithm~\cite{bora2017compressed} was proposed by Hand et al.~\cite{hand2017global} under some assumptions on the generative model.

Inspired by these impressive benefits of deep generative models, we chose to investigate the application of such methods for medical imaging, a canonical application of compressive sensing.
A significant problem, however, is that all these previous methods require the existence of \emph{pre-trained} models. While this has been achieved for various types of images, e.g. human faces of CelebA~\cite{liu2015faceattributes} via DCGAN~\cite{radford2015unsupervised}, it remains significantly more challenging for medical images~\cite{wolterink2017generative,schlegl2017unsupervised,nie2017medical, schlemper2017deep}. Instead of addressing this problem in generative models, we found an easier way to circumvent it.

Surprising recent work by Ulyanov et al.~\cite{ulyanov2017deep} proposed Deep Image Prior (DIP), which uses \emph{untrained} convolutional neural networks. In DIP-based schemes, a convolutional neural network generator (e.g. DCGAN) is initialized with random weights; these weights are subsequently optimized to make the network produce an output as close to the target image as possible. This procedure is unlearned, using no prior information from other images. The prior is enforced only by the fixed convolutional structure of the generator network.

Generators used for DIP are typically over-parameterized, i.e. the number of network weights is much larger compared to the output dimension. For this reason DIP
has empirically been found to overfit to noise if run for too many iterations~\cite{ulyanov2017deep}.
In this paper we theoretically prove that this phenomenon occurs with gradient descent and justify the use of early stopping and other regularization methods.

\textbf{Our Contributions:}
\begin{itemize}
    \item In Section~\ref{methods} we propose DIP for compressed sensing (CS-DIP). Our basic method is as follows. Initialize a DCGAN generator with random weights; use gradient descent to optimize these weights such that the network produces an output which agrees with the observed measurements as much as possible. This unlearned method can be improved with a novel \textit{learned regularization} technique, which regularizes the DCGAN weights throughout the optimization process.
     \item
     In Section~\ref{theory} we theoretically prove that DIP will fit any signal to zero error with gradient descent. Our result is established for a network with a single hidden layer and sufficient constant fraction over-parametrization. While it is expected that over-parametrized neural networks can fit any signal, the fact that gradient descent can provably solve this non-convex problem is interesting and provides theoretical justification for early stopping.
    \item In Section~\ref{experiments_results} we empirically show that CS-DIP outperforms previous unlearned methods in many cases. While pre-trained or ``learned'' methods will likely perform better~\cite{bora2017compressed}, we have the advantage of not requiring a generative model trained over large datasets.
    As such, we can apply our method to various medical imaging datasets for which data acquisition is expensive and generative models are difficult to train.
\end{itemize}


TITLE: probability of exactly, atleast, and expected number
QUESTION [1 upvotes]: The people discovered that 54% of refugees who ask for asylum in the New York immigration court win asylum, but only 12% are granted asylum in the Florida immigration court.
Assume that you randomly select 20 refugees who are asking for asylum in the Florida immigration court.
A. find the probability that exactly five asylum seekers are granted asylum.
$${20}\choose{5}$$ $$= 15504$$
$$(15504)(.12)^5$$ $$= 0.385789133$$
$$(0.385789133)(.88)^{(20-5)}$$ $$= 0.056700916$$
B. find the probability that at least three asylum seekers are granted asylum.
(I used the same formula above to find the probability for each x value)
x | P(x)

0 | 0.077562794
1 | 0.211534892
2 | 0.274033847
3 | 0.224209503
$$ P(x=0) + P(x=1) + P(x=2) + P(x=3) $$ $$= 0.787341026$$
$$1 - 0.787341026 = 0.212658974$$
C. what is the expected number on asylum seekers who win their cases?
$$(n)(p)(q)$$ $$(20)(.12)(.88) = 2.112$$
are A,B,C correct?
for b, since probabilities must add up to 1, it is ok to find the probability for 0 - 3 and subtract that number from 1 to get the probability of atleast 3? (3,4,5,6,7,...) ?

REPLY [1 votes]: If the random variable $X$ is the number accepted, then the probability that $X$ is at least $3$ is $\Pr(X\ge 3)$. This is $1-[\Pr(X=0)+\Pr(X=1)+\Pr(X=2)]$.
You were right to subtract from $1$, but should not have used $\Pr(X=3)$.
The expected number $E(X)$ is $np$, not $npq$.
The calculation of $\Pr(X=5)$ is correct.


TITLE: Prove a function is one to one/injective and a correspondence/bijective
QUESTION [0 upvotes]: I'm having trouble proving the following:
1) Prove that a function f is one-one if and only if ∀xy∈A,x≠y→f(x)≠f(y).
I was told to do this by the contrapositive, so I show that f(x)=f(y) → x=y. However, how do I do this? Can I make an example?

2) Assume f:A→B is a correspondence (or bijection), i.e., f is one-one and onto. Show that f has a unique inverse (which we'll denote by f^(−1)), and that ∀x∈B,f(f^(−1)(x))=x, i.e., that f is f^(−1)'s inverse.
I'm unsure how to approach this one.

REPLY [1 votes]: For the first part, prove the theorem directly from the definition of an injective function: a function is injective if (and only if, though the latter phrase is often left out in definitions) $f(x)=f(y)$ implies $x=y$.
But $f(x)=f(y)$ implies $x=y$ if and only if $x \not = y$ implies $f(x) \not = f(y).$ This is the application of contrapositive to the definition of one-to-one (injective) functions.
For the second part, you must first show that $f(x)$ has an inverse, then show that this inverse is unique. Since $f(x)$ is injective onto its range, $f(x)=f(y)$ implies $x=y.$
Designate a function $f^{-1}$ on the range element $f(x)$ in such a way that $f^{-1}(f(x))=x.$ We must show the function is well-defined, that is, there is no more than just $x$ as the answer. But $f(x)=f(y)$ if and only if $x=y,$ so $f^{-1}(f(x))=f^{-1}(f(y))=x.$ Hence $f^{-1}$ is well-defined. Similarly, $f^{-1}(f(x))=f^{-1}f(y))$ implies $x=y,$ from the definition of inverse, but from the one-to-one property of $f,$ we have $f(x)=f(y)$ so that $f^{-1}$ is itself also a one-to-one function. Finally, $f(x)$ maps onto all the range, so that $f^{-1}(x)$ maps onto all the domain, showing $f^{-1}$ is a bijection.
Now suppose $f(f^{-1})(x)=y.$ Then applying $f^{-1}$ to both sides gets $f^{-1}(y)=f^{-1}(x)$ which means $y=x$ since $f^{-1}$ is one-to-one. Hence $f(f^{-1})(x)=x.$
For uniqueness, suppose $f^{-1}(f(x))=x=g(f(x))$ for all $x.$ Then $f^{-1}(f(x))-g(f(x))=0$ for all $x,$ meaning $f^{-1}$ and $g$ are the same function. So the inverse must be unique.


TITLE: Invert Integral
QUESTION [1 upvotes]: I need to invert I need to reverse the integration order $$\int_{0}^{1}[\int_{0}^{x}f(x,y) dy]dx$$
I know how to reverse the order through the graph, which will look like this:
$$\int_{0}^{1}[\int_{y}^{1}f(x,y) dx]dy$$
but how can I formalize it step by step through writing?

REPLY [1 votes]: This type of "inversion" can be carried out without resorting to a graph by a direct application of Fubini's theorem after an indicator function is introduced. Let $S = \{(x,y)\subset [0,1]^2:x \geqslant y\}$ and define the indicator function
$$\mathbb{1}_S(x,y) = \begin{cases}1,&x\geqslant y\\ 0, &x < y \end{cases}$$
Note that even if we are working with Riemann integrals here, the product function  $(x,y) \mapsto f(x,y)\mathbb{1}_S(x,y)$ is integrable since a discontinuity is introduced only on a set of measure zero where $x = y$.
The inner integral becomes
$$\int_0^x f(x,y) dy = \int_0^1f(x,y)\mathbb{1}_S(x,y)  \, dy $$
Now we can apply Fubini's theorem to the iterated inetgral over the fixed region $[0,1]\times [0,1]$ to obtain
$$\int_0^1\left(\int_0^x f(x,y)\, dy\right) \, dx \\=  \int_0^1\left(\int_0^1 f(x,y)\mathbb{1}_S(x,y) \,dy\right) \, dx \underbrace{=}_{\text{Fubini}}\int_0^1\left(\int_0^1 f(x,y)\mathbb{1}_S(x,y) \,dx\right) \, dy\\ =  \int_0^1\left(\int_y^1 f(x,y)\,dx\right) \, dy$$


TITLE: UMVUE related problem for $Bernoulli(p)$
QUESTION [0 upvotes]: $X_1,X_2,...,X_n$ follows iid $Bernoulli(p)$, $0<p<1$
Let $$ \tau(p)={(p+ qe^3)}^2 $$ is a function of parameter $p$ where $q=1-p$.
Q.1 $\:$ Write an unbiased estimator $T$ for $\tau(p)$
Q.2 $\:$ If $U=\sum_{i=1}^{n}X_i$ is the complete sufficient statistic, then calculate $E_p[T|U=u]$ using Lehmann-Schefee theorems
Q.3 $\:$ Find the UMVUE for $\tau(p)$.
($Hint:$ Use the concept of mgf of the Xs)
$\underline{\Large Solution}$:

I have chosen $T=e^{3(1-X_1)} \: e^{3(1-X_2)}$ which is unbiased for $\tau(p)$.

I have calculated $E_p[T|U=u]$ which is $$E_p[T|U=u]= e^6\: \frac{(n-u)(n-u-1)}{n(n-1)} + 2 e^3 \: \frac{u(n-u)}{n(n-1)} + \frac{u(u-1)}{n(n-1)}  $$

Now my questions are :
i.) Are my founding results correct?
ii.) Is it possible to simplify the resulting quantity of $E_p[T|U=u]$?
iii.) What is the answer to Q.3?
iv.) Please show me the Expectation of at least one of the three parts of $E_p[T|U=u]$
v.)  Should I choose $T=[X_1 + (1-X_1)e^3] \: [X_2 + (1-X_2)e^3]$ ?

REPLY [1 votes]: Your procedure is correct and you calculations look fine to me.
I did check only the first addend of your conditional expectation:
$$\mathbb{P}[T=e^6|U=u]=\frac{\mathbb{P}[T=e^6]\mathbb{P}[\sum_{i=1}^{n}X_i=u|X_1=0,X_2=0]}{\mathbb{P}[\sum_{i=1}^{n}X_i=u]}=$$
$$=\frac{q^2\binom{n-2}{u}p^u q^{n-2-u}}{\binom{n}{u}p^uq^{n-u}}=\dots=\frac{(n-u)(n-u-1)}{n(n-1)}=(1-\overline{X}_n)(1-\frac{n}{n-1}\overline{X}_n)$$
Which is exactly your result and thus I have no doubts that the rest is correct too.
Finally calculate
$$\mathbb{E}[T|U]=1\cdot\mathbb{P}[T=1|U=u]+e^3\cdot\mathbb{P}[T=e^3|U=u]+e^6\cdot\mathbb{P}[T=e^6|U=u]$$
Q3
Using Lehmann Scheffé and Rao Blackwell results together, $\mathbb{E}[T|U]$ is  UMVUE for $\tau(p)$


TITLE: Filling a cone part way, find formula for height expressed in V
QUESTION [2 upvotes]: I have an ice cream cone, the pointy side downwards. It's 6 cm tall and the diameter of the opening is 3 centimeter. I fill it to some height $h$ . Now how do I find a formula for its height as a function of volume.
I understand the volume of a cone is calculated by $V = \frac{\pi}{3}r^2 h$ and I know how to isolate $h$. But I'm totally in the dark as to how to approach this. If I only fill it partly,  let's just say I fill it to 5 centimeters, the diameter of the surface of the ice cream is smaller then the diameter of the opening of the cone itself. How do I calculate the volume then? Since I don't know the diameter of the cone at that point...
How does the diameter of the surface of the ice cream depend on the amount of ice cream in the cone? How do I approach this?

REPLY [0 votes]: Draw a right angle triangle representing the half of the vertical cross section of the cone. Draw a smaller triangle inside the previous triangle, this represents the partially filled ice cream. By using ratios of $\text{opening radius}:\text{cone height}=\text{radius if ice cream surface}:\text{height of the ice cream}$, you'll get the radius of the ice cream surface, hence u can get the volume.


TITLE: Find the Cartesian equation corresponding to $r = \frac{5}{3-2\cos(\theta)}$
QUESTION [2 upvotes]: Find the Cartesian equation corresponding to $r =
 \frac{5}{3-2\cos(\theta)}$

I got it into the form:
$(x^2 + y^2)(3-2x)^2 = 25$
and can see that maybe the equation of a circle will appear, but I can't seem to get much further without complicating things. I have a note saying that completing the square was used but I can't see how or where.

REPLY [4 votes]: Rearranging we get that $3r - 2r\cos(\theta) = 5$. Note that $r = \sqrt{x^2 + y^2}$ and $r \cos(\theta) = x$.
This gives us $3 \sqrt{x^2+y^2} -2x = 5$.
Hence,
\begin{align}
9(x^2+y^2) & = (2x+5)^2\\
9x^2+9y^2 & = 4x^2 + 20x + 25\\
5x^2 -20x + 9y^2 & = 25\\
5(x-2)^2 + 9y^2 & = 45\\
\frac{(x-2)^2}{3^2} + \frac{y^2}{\left(\sqrt{5} \right)^2} & = 1
\end{align}
The above is an ellipse with center $(2,0)$ with semi-major axis along $x$ of length $3$ and semi-minor axis along $y$ of length $\sqrt{5}$.
EDIT
In general, $$r(\theta) = \dfrac{a(1-e^2)}{1 \pm e \cos(\theta)}$$ represents an ellipse whose semi-major axis is $a$ and eccentricity $e$.


TITLE: What do the different topologies of the set {a,b} mean?
QUESTION [1 upvotes]: Let us consider the set $\{a,b\}$ containing only two elements. I can think of four different topological spaces associated to this set:

$\{ \{\emptyset\}, \{a,b\} \}$.
$\{ \{\emptyset\}, \{a\}, \{a,b\} \}$.
$\{ \{\emptyset\}, \{b\}, \{a,b\} \}$.
$\{ \{\emptyset\}, \{a\}, \{b\}, \{a,b\} \}$.

My question is: Geometrically, what do these different topological spaces mean?
For case 1, I am supposing that a and b are so close to each other that there is no open sets containing only $a$ or $b$. The points are lumped together. But if this reasoning is correct then I do not know what case 2 means. If there is an open set containing only $a$ it means that the points $a$ and $b$ are not not lumped together. Then why is not there an open set containing only $b$? Similarly, for case 3, why is not there an open set containing only $a$?
I would appreciate a simple or intuitive answers since I do not have a formal background in topology.

REPLY [5 votes]: You have the right idea for how to think about (1).  For (2) and (3), here is an intuition which I find very helpful.  Think of (2) as just the ordinary real line $\mathbb{R}$, except that all the nonzero numbers have been collapsed down to a single point $a$ and $b$ is $0$.  So $\{a\}$ is open, since it corresponds to the entire set $\mathbb{R}\setminus\{0\}$ in $\mathbb{R}$, which is open.  But $\{b\}$ is not open: any open interval around $0$ must contain nonzero points, so any open set containing $b$ needs to also contain $a$.
To put it another way, the set $\{a\}$ is open because the point $a$ is a really "fat" point--it secretly represents the entire set $\mathbb{R}\setminus\{0\}$.  On the other hand, the point $b$ really is just a single skinny point, and any open set around it will have to touch the fat point $a$ which surrounds it.


TITLE: Every monic is a kernel
QUESTION [4 upvotes]: This is part of Weibel's Exercise 1.2.2, where I have to show that in the category R-Mod, every monic is a kernel.
A monic morphism is defined to be a map $i \colon A \to B$ such that if $g \colon A \to A$ is such that $i \circ g = 0$, then $g = 0$, and a kernel of a homomorphism $f \colon B \to C$ is a map $i \colon A \to B$ such that $f \circ i = 0$ and $i$ is universal.
Attempt
Take a monic map $i \colon A \to B$ and $f \colon B \to 0$ be the zero homomorphism. Then obviously $f \circ i = 0$ and also if $i' \colon A \to B$ is such that $f \circ i' = 0$ there is a unique map $u \colon A' \to A$ defined to be the composition
$$u \colon A' \to 0 \to A$$
It is unique since it is the composition of the zero map $A' \to 0$ and the zero map $0 \to A$.
It seems to be a problem with my attempt since I didn't use the fact that $i$ is monic, but I can't find any flaws in the argument, so I'd like if anyone could point them out and maybe a hint to complete the exercise. Thank you.

REPLY [2 votes]: The map $u: A'\to A$ is not unique such that the composition $A' \to A \to B$ is zero. Indeed, any map $A'\to A$ will have the property that the composition is zero.


TITLE: scheduling a tournament with constraints
QUESTION [1 upvotes]: I want to schedule a wrestling tournament with $7$ teams, where the teams meet once a week for $4$ weeks,
and each week there are $4$ teams at school $M$ and $3$ teams at school $N$.
I would like to find out if it is possible to schedule the teams so that
1) every pair of teams is at the same school for at least one week and
2) each team is at school $M$ at least twice.
I haven't been able to produce a schedule satisfying both conditions, so is this  possible?

REPLY [1 votes]: There are 4 meetings with partition $4+3$ so that we will have at most $$4\cdot \left(\frac{4\cdot 3}{2} + \frac{3 \cdot 2}{2}\right) = 36$$ pairs possible. On the other hand, we have to accommodate $\frac{7\cdot 6}{2} = 21$ pairs, meaning, that we can have at most $36-21 = 15$ instances of previously used pairs.
On the other hand, we have to satisfy requirement "each team at school $M$ at least twice" and that "each pair meets at least once". There are $16$ slots for $M$ and $14$ double visits, at least 5 teams have to visit $M$ exactly twice.
Observe, that for each pair of teams which visit $M$ exactly twice, they have to meet at least twice, once in $M$ and once in $N$. That means, every such pair produces one redundant pair as well, thus 10 redundant pairs.
Now we have two cases, either there is a team $G$ which never visits $N$, or there are two teams $F,G$ who visit $M$ three times. In the first case that team it produces $6$ additional redundant pairs (one with every other team), for a total of at least $16$, so that's impossible.
In the second case if teams $F,G$ doesn't visit $M$ on the same days (thus produce only 1 redundant pair), we have only 1 non-redundant-pair schedule (the one that visits $M$ when $F$ and $G$ are non visiting $M$). Any other schedule gives two additional redundant pairs with at least one of $F$ or $G$, and repeated non-redundant-pair schedule also gives two additional redundant pairs.
In total that is at least $10+1+2\cdot 4 = 19$, contradiction.
Yet, if $F$ and $G$ visit $M$ on the same days (3 more redundant pairs), then there are only 3 non-redundant-pair schedules (for 5 teams) – either two schedules are the same (2 additional redundant pairs each) or some teams use other schedules (2 additional redundant pairs each), for a total of at least $10+3+2\cdot 2 = 17$, also impossible.
Therefore, all the cases imply we would need at least $16$ redundant pairs, which means there is not enough time/space for non-redundant pairs, and there is no assignment that satisfies given conditions.
I hope this helps $\ddot\smile$


TITLE: How do I calculate with tensors on modules?
QUESTION [0 upvotes]: In a homework, I have to work with tensor products of chain complexes and have to show the following:
Let $\mathbb{K}$ be a principal ideal domain. For two chain complexes $A_{\bf{.}}$, $B_{\bf{.}}$ over $\mathbb{K}$ with boundary operators $\partial^{A}$, $\partial^{B}$, we define $C_{\bf{.}}$ by setting $C_n := \sum_{k+l=n} A_k \otimes B_l$ and $\partial^{A \otimes B} (a \otimes b) := \partial^{A} (a) \otimes b + (-1)^k a \otimes \partial^{B} b $ for two generators $a \in A_k , b \in B_l $.
Show that this is a chain complex.
I have problems computing the following :
$\partial^{A \otimes B} (a \otimes b + c \otimes d) $
$\partial^{A \otimes B} (a \otimes b) \circ \partial^{A \otimes B} (c \otimes d) $
, because if one has $(a \otimes b + c \otimes d  = e \otimes f) $i can't write down simple e,f.

REPLY [2 votes]: Since $\partial^{A+B}$ is a homomorphism, it must be linear. When we define $\partial^{A+B}$ only on elements of the form $a \otimes b$, we get the full homomorphism $\partial^{A+B}$ by extending it linearly to arbitrary elements of $A_k \otimes B_l$. So this means that $\partial^{A+B}(a \otimes b + c \otimes d) = \partial^{A+B}(a \otimes b) + \partial^{A+B}(c \otimes d)$.
See if you can show it is a chain complex now.


TITLE: Is $(u_\lambda^2)$ also an approximate unit when $(u_\lambda) $ is an approximate unit?
QUESTION [3 upvotes]: On P145 of  Murphy's book C$^*$-Algebras and Operator Theory, when he proves a character $\tau$ of an abelian C* algebra has norm one, he uses that $(u_\lambda^2)$ is also an approximate unit when $(u_\lambda) $ is an approximate unit.
But why? An approximate unit of a C* algebra $A$ is an increasing net of positive elements in the closed unit ball such that $u_\lambda a\to a\forall a \in A$. Note that $0\leq u\leq v$ dose not imply that $u^2\leq v^2$. Why is $(u_\lambda^2)$ increasing?

REPLY [2 votes]: In a general $C^*$-algebra it is correct that $0\leq u\leq v$ does not imply $0\leq u^2\leq v^2$.  But we are considering an abelian $C^*$-algebra. Then the algebra is (up to $*$-isomorphism) $C_0(X)$ for some space $X$, and the implication $0\leq f\leq g$ implies $0\leq f^2\leq g^2$ holds for $f,g\in C_0(X)$.


TITLE: Show that every finite-dimensional topological vector subspace is closed.
QUESTION [1 upvotes]: Let $X$ be a normed topological vector space. Show the following:
(i) If $0\neq v \in X$, then $\{\alpha v:\alpha\in \mathbb{R}\}$ is closed.
(ii) If $Y$ is a closed vector subspace of $X$ and $w\in X-Y$, then $Y\cup \{\beta w:\beta \in \mathbb{R}\}$ is closed.
Please help me. I am stucked with this question in the book.

REPLY [3 votes]: ii) follows from i) as union of 2 closed sets is closed.
Hint for i) note that {x} is a closed set. Then try to use that scalar mult. is continuous. (You need the fact that $\mathbb{R}$ is complete).


TITLE: question on finding limit
QUESTION [0 upvotes]: Let
$x_n=\left(1-\frac13\right)^2\left(1-\frac16\right)^2\left(1-\frac1{10}\right)^2.....\left(1-\frac1{\frac{n(n+1)}2}\right)^2,n\ge2$
then find $\lim_{n\rightarrow \infty}~ x_n$  ?
while answering please mention the underlying concept too.

REPLY [3 votes]: Hint:
Try rewriting the expression
$$1-\frac1{\frac{n(n+1)}2}$$
in the form
$$\frac{(n+a)(n+b)}{(n+c)(n+d)}$$
for constants $a,b,c$ and $d$.


TITLE: Can I relate the L1 norm of a function to its Fourier expansion?
QUESTION [5 upvotes]: I would like to express the integral of the absolute value of a real-valued function $f$ (over a finite interval) in terms of the Fourier coefficients of $f$. Failing that, I would like to know of any constraints or statistical correlations (in a sense explained in the motivation) relating these quantities.
Motivation: This comes from a biophysics application, but is perhaps best explained as follows.  If a rubber band of tension $t$ is stretched along the $x$ axis from $0$ to $L$, then it is easy to calculate the thermal fluctuations of its arc-length by letting $z(x)$ be the (small) deviation from the $x$ axis, and then writing the energy (arc-length times tension) in terms of the Fourier coefficients of $z(x)$.  The Boltzmann weight turns out to be a Gaussian since in the limit of small deviations the arc-length becomes a sum of squares of the Fourier coefficients.  My problem is more complicated: We have two rubber bands stretched over the same interval, with deviations $z_{1}(x)$ and $z_{2}(x)$.  The energy includes not only the stretching of the rubber bands, but also a term proportional to the (positive) area enclosed between them, which is
$\int_{0}^{L}|z_{1}(x) - z_{2}(x)|dx$
Hence my question.  So it would be nice to know how this area can be related to the Fourier coefficients of $z_{1}$ and $z_{2}$ or perhaps just to the arc-lengths of the rubber bands.  By "statistical correlations" I am referring to the Boltzmann probability distribution with energy equal to the stretching energy plus the area-energy.
Edit: Specifics on the Boltzmann probability distribution, more motivation.
The state of the system is the pair of functions $z_{1}(x)$ and $z_{2}(x)$ describing the deviation of the two rubber bands from the x axis.  Let's say it's the set of pairs of functions defined on [0,L] and that these functions are identified with a finite number of Fourier coefficients - I am a physicist and would like to avoid nasty functions or mathematically honest discussions of path integrals.
The probability of occurrence of a state (z_{1}, z_{2}) is (before normalization)
$\exp(-\beta E\left[z_{1},z_{2}\right])$
where $E\left[z_{1},z_{2}\right]$ is the energy of the system, which in this case is the functional
$E\left[z_{1},z_{2}\right] = \frac{t}{2}\int_{0}^{L}\left[(\frac{dz_{1}}{dx})^{2}+(\frac{dz_{2}}{dx})^{2}\right]dx + \kappa \int_{0}^{L}|z_{1}(x)-z_{2}(x)|dx$
Where the tension t and the "surface tension" $\kappa$ are just numbers; set them equal to 1 if you wish.  The first integral is the energy cost of stretching the rubber bands (in a linearized regime) and the second is the strange term proportional to the area enclosed between them.  Without the second term, it is easy to diagonalize this functional in terms of the Fourier series of the two functions. That is why I was interested in writing the second term in terms of Fourier coefficients.  That may be too much to ask, but perhaps it is still possible to calculate some quantities such as the statistical average of $z_1(x)^{2}$ - that is the kind of thing I ultimately want to know.
I realize this is an unnatural-looking problem, so I will just mention that it's not really about rubber bands, but rather about fluctuating interfaces which occur in lipid bilayers with coexisting phases.  There are two phase boundaries (one for each monolayer) with their respective "tensions" but there is also a term proportional to the area between them.

REPLY [5 votes]: This answer really has to do with the physics of it: are you sure about the area energy term?
Let me simplify a little bit: Consider a case where $z_1(x)=0$ and $z_2(x)=w(x)$. According to your formula I should get a term proportional to $w'(x)^2$ for the kinetic  term  which is typical and no one will object to you for that. You potential energy would then be proportional to $\int_0 ^L |w(x)| dx$. That does raise an alarm: nonlinear problem.
Unless you really have nonlinear physics going on (which is likely, see below) you should have had $w(x)^2$. That would bring back the $L_2$ norm and everything will be very simple. Just transform $z_2(x)=z_1(x)+w(x)$, get rid of $z_2$ and you will get two uncoupled (continuum of) modes, one (corresponding to $z_1$) is a free particle and the other a Harmonic oscillator. Your Boltzmann statistic will determine how these modes are filled up as you know.
If you think of your rubber band as a collection of springs and masses, $w(x)^2$ is the actual term but collections of springs and masses hardly exist outside textbooks and problem sets. As a physicist you know: molecules interact. The actual expansion (if you are still interested in writing an effective field theory) will involve (interacting) terms of the type $w(x)^2+O[w(x)^4]$. Quantitative results in this model might involve renormalization. See the wikipedia page on quartic interaction that even describes how you should quantize it (Fourier transform). My field theory is rusty so you might already know more than I do.
What if $|w(x)|$ is what you have: If $|w(x)|$ is not too small or too large, just approximate it with $[w(x)]^2$ to linearize the problem. People might point out that this is not a mathematically good approximation but it will physically make sense: very small stretchings are not physically possible and very large stretchings would be ruled out by the Boltzmann statistics as they would correspond to exponentially rare high energy modes. So I would just introduce a factor so that $w(x)$ and $w(x)^2$ coincide where $\kappa w(x)^2 L\approx kT$.
There are a famous nonlinear equations that come cheerfully close to your problem but miss it. Example: Sine-Gordon Equation where instead of $|w(x)|$ you would have $1-\cos [w(x)]$. Actually  more up to the point would be the Sinh-Gordon... which reminds me of the Toda field theory which describes a Toda lattice. A Toda lattice is a nonlinear set of coupled equations that describe a set of nonlinearly coupled particles. If you are working in the liquid state, I doubt that they would be relevant. $w(x)^2$ should be good enough but the dissipative terms will be more troublesome.
Edit: Couldn't resist the pun.  Seems that the relevant equation is a "SIGN-Gordon equation":
$$\varphi_{tt}- \varphi_{xx} + \text{sgn}(\varphi) = 0.$$
Not sure if it is really simple, messy, or plain difficult to solve in your case. An option is to try solve it like a wave equation with a sign changing external force term and then parametrize the solutions based on their energy and apply the Boltzmann statistics. Forgo the Hamiltonian altogether. I think it will be messy.
Another Edit: as the commenters mentioned before, the solution to the absolute value potential Schrodinger eqn. involves Airy functions. So you could in principle use the Airy function eigenstates to find the time-independent modes and fix their energy. This could quantize your problem. The solution in terms of the Airy functions can be found for example here, [PRL, 94, 176805 (2005)].
Yet Another Update: Apparently you have the "Signum-Gordon" equation. Thanks to this answer.


TITLE: Is the evolute always a regular curve?
QUESTION [2 upvotes]: Is the evolute always a regular curve? (that is, the tangent vector of this curve is nonzero at any time)

REPLY [0 votes]: Evolutes in plane (equation  y= x + R  n where n is normal vector) have a cusp at point of symmetry in smooth continuous curves as radius of curvature R is a maximum or minimum with third order derivative vanishing.


TITLE: Proving that a ratio between two functions is decreasing
QUESTION [2 upvotes]: I have a question that seems to be easy, but I haven't been able to prove it. Any help would be appreciated:

Let $f,g\rightarrow[0,1]$ such that $f(0)=0$, $f'(0)>0$, $f(x)\geq0$, $f''(x)\leq0$, $\forall x$ and $g(0)=0$, $g(x)\geq0$, $g'(x)\geq0$, $g''(x)\geq0$, $\forall x$.
  Prove that $\frac{f(x)}{g(x)}$ is decreasing.

I tried to use the property that say: if $\frac{a}{b}<\frac{c}{d}$ then $\frac{a+c}{b+d}<\frac{c}{d}$. I used it with the fact that $f(x)=f(0)+\lim_{n\to \infty} \sum_{i=0}^{n-1} f'(\frac{ix}{n})\frac{x}{n}$ . With this two elements I can prove that if $\lim_{x\to 0} \frac{f(0)}{g(0)}$ exists,, then the solution is true. However, I haven't been able to prove it for the general case.
Any help is greatly appreciated.

REPLY [2 votes]: From the given information, we know that $f$ is positive and concave and $g$ is positive and convex. For any $0 < y < x$, let $y=tx$.  Then $0<t<1$. From the concavity of $f$, we have
$$tf(x)+(1-t)f(0) \le f(tx) = f(y)$$ Since $f(0)=0$, that means $$tf(x) \le f(y)$$ Likewise, the convexity of $g$ gives  $$tg(x) \ge g(y)$$ Dividing the two inequalities above and defining $r(x) \equiv \frac{f(x)}{g(x)}$, we get $$r(x) \le r(y)$$ That is, $r$ is decreasing.


TITLE: Can QED explain this or do I have revert to the classical model of light?
QUESTION [0 upvotes]: I want to know can QED can explain this image,like why there are someplaces with low light (shadows) like behind the chair, and why there are some places that are bright(rest of the floor).
I know that this depends on the scene that we are investigating but given a scene like the one in the image below,can QED explain why some places are bright and dark (shadows), or I have to use the classical model of light (Light travels in straight lines) to explain why some places are dark (shadows) and bright.
Like classical model of light can tell, light coming form the light source travels in straight lines and reaches the floor with no obstacles in between so they are more bright, and light behind the chair reaches after bouncing many times, so it has less energy and so the place behind the chair is not so bright.
Can QED explain this, that why some places are dark (behind the chair) and some bright (rest of the floor)?

REPLY [11 votes]: Would you dig a ditch with a surgeons scalpel?
Yes, quantum mechanics ultimately underlies all physical observations but the mathematical expressions for large dimensions with respect to $\hbar$ become cumbersome and are replaced by the simplest ones for the appropriate study.
Thermodynamics, for the study of bulk matter, blends smoothly with quantum statistical mechanics, and for most engineering uses thermodynamics is by far the simplest and more useful mathematical tool.
Classical electrodynamics for the study of light, blends mathematically with quantum field theory, but its mathematical formulation is appropriate for the large numbers and dimensions we live and observe in.
See this link on how classical fields emerge from the underlying quantum mechanical framework.


TITLE: Expected area of a random $n$-gon
QUESTION [7 upvotes]: Choose $n$ points $\{z_1, \ldots, z_n\}$ from the unit circle $\partial \mathbb{D} = \{z \in \mathbb{C}: |z| = 1\}$ uniformly at random, and let $P_n$ be the convex hull of the $z_i$'s. Let $X_n = area(P_n)$. What can be said about $X_n$? Can $\mathbb{E}X_n$ be computed? It's easy to see that $P_n$ converges to $\mathbb{D}$ pointwise almost surely, so $X_n \to \pi$ a.s. as $n \to \infty$. What is the rate of convergence?
It is 'easy' to get a direct computation for $\mathbb{E}X_3$, by writing down a formula for the area of the triangle and integrating directly: for example, you can assume $z_1 = 1$, and that $z_2 = e^{i \theta}, z_3 = e^{i \phi}$ with $\theta < \phi$, and integrate directly. But this is a bit tedious, and especially so for large $n$! Is there an easier way to do this computation?
One idea is to rely on order statistics: write $z_k = \exp(i \theta_k)$ for iid uniform variables $\theta_k \in [0,2\pi)$, and let $\{\phi_k\}_{k=1}^n$ be the ordered sequence of the $\theta_k$, i.e.
$\phi_1 = \min\{\theta_1, \ldots, \theta_n\}$,
$\phi_2 = $ 2nd largest element of $\{\theta_1, \ldots, \theta_n\}$,
...
$\phi_n = \max\{\theta_1, \ldots, \theta_n\}$.
The distributions of the $\phi$'s can be written down explicitly - though it's a bit messy - and $P_n$ can be divided into $n$ triangles, yielding
$X_n = \frac{1}{2} \sum_{k=1}^n \sin(\phi_k - \phi_{k-1})$,
with the convention $\phi_0 = \phi_n - 2\pi$. I haven't tried writing down this integral out of fear, but it may be do-able. (I suspect there is a clever symmetry argument to avoid needing the order statistics.)
Heuristically, if the $z_n$ are 'well-distributed,' the angle distance between adjacent $z_i$'s is roughly $\frac{2\pi}{n}$, so the error between the areas of the disk and $P$ is roughly
$n(\frac{\pi}{n} - \frac{1}{2} \sin(\frac{2\pi}{n})) = \frac{4\pi^3}{3} n^{-2} + O(n^{-4})$.
Thus we should have
$n^2(\pi - \mathbb{E}X_n) \to L \in (0,\infty)$ $\hspace{.5cm} (\star)$
Perhaps this argument can be formalized, and the constant $L$ can be determined.
After that, the usual next step is to hope for something like
$n^2(\pi - X_n) \to_d Z$ for some random variable $Z$,
or a CLT like
$\gamma_n (X_n - \mathbb{E}X_n) \to_d N(0,1)$ for some constants $\gamma_n$.
Also: what if the $z_k$ are selected uniformly from the unit sphere in $d$-dimensions? I haven't thought much about this, but maybe a statement analogous to $\star$ can be proved.
Edit 1: A quick google search revealed this site, which has some additional interesting calculations. In particular, it claims to compute $L$ via the method I suggested, which would answer $\star$.

REPLY [1 votes]: Both the site you linked to and the thesis linked to by zhoraster yield $L=4\pi^3$.
From the site, we have
$$
\lim_{n\to\infty}\frac{\pi-\mathbb EX_n}{\pi-\frac n2\sin\left(\frac{2\pi}n\right)}=6\;.
$$
With $\pi-\frac n2\sin\left(\frac{2\pi}n\right)=\frac23\pi^3n^{-2}+O\left(n^{-4}\right)$, it follows that $\lim_{n\to\infty}n^2\left(\pi-\mathbb EX_n\right)=4\pi^3$.
From the thesis, we have (p. $63$)
$$
\mathbb EX_n=\frac3{\pi^2}\binom n3\int_0^\pi\sin^2h\left(\frac h\pi\right)^{n-3}\mathrm dh\;.
$$
This is convenient for calculating $\mathbb EX_n$ for small $n$, but we can obtain a simpler form that's more convenient for large $n$ by repeatedly integrating by parts and substituting $x=2h$, yielding
\begin{eqnarray*}
\mathbb EX_n&=&\frac n2\int_0^{2\pi}\cos x\left(\frac x{2\pi}\right)^{n-1}\mathrm dx\\
&=&\pi+\pi\int_0^{2\pi}\sin x\left(\frac x{2\pi}\right)^n\mathrm dx\;.
\end{eqnarray*}
Integrating by parts twice more yields
\begin{eqnarray*}
\mathbb EX_n&=&\pi+\frac{4\pi^3}{(n+1)(n+2)}+O\left(n^{-4}\right)\\
&=&\pi+4\pi^3n^{-2}+O\left(n^{-3}\right)\;,
\end{eqnarray*}
again leading to $L=4\pi^3$.


text:
\section{Results}
\label{sec:results}
\noindent In this section the numerical method suggested in this paper is tested against the validation cases in \S\ref{sec:valid}. Each test case is described in detail below. The numerical method described in \S\ref{sec:nummethod} will be denoted the BIE method for the remainder of the paper.

\input{srcfiles/steady}

\input{srcfiles/cleanvalid}

\input{srcfiles/surfvalid}

\subsection{The Swiss roll}
\noindent To further assess the BIE method, a more complicated simulation is set up to test the robustness of the method. The setup is inspired by that in \cite{Ojala2015}. The drop configuration is shown in Figure~\ref{fig:swissroll_domain}. No far-field flow is imposed, instead the drops will be allowed to deform until circular under surface tension. A drop is deemed circular when
\begin{align*}
  \left| 1 - \dfrac{\max{(|z-c|)}}{\text{mean}(|z-c|)}\right| < 10^{-4},
\end{align*}
where $c$ is the centre of each drop and $z$ its interface discretisation.

\begin{figure}[h!]
\centering
\includegraphics[width=0.6\textwidth]{graphics/sr_t0.png}
\caption{Drop configuration for the swiss roll simulation.}
\label{fig:swissroll_domain}
\end{figure}

The largest drop (i.e. the roll) will be covered with surfactants with an initial uniform concentration $\rho_0=1$. Other parameters are $Pe=10$ and $E=0.1$. The authors note that the standard nondimensionalisation of length by initial radius is not appropriate in this case; instead the characteristic length of half of the length of a square box containing the initial drop configuration is used. Initially, the roll is covered with $375$ panels and the ellipses $37$ panels each. For this simulation $\lambda_k=1$ for all drops $k$ is considered, in order to make the simulation run in a reasonable time on a standard workstation. The simulation has also been run with 50 \% more points without any visible difference. Comparing the final circle radius and centre point between the two discretisations, they differ on the level of the circular tolerance imposed above. The simulation takes approximately $10^5$ time steps.

How the drops deform in time is shown in Figure~\ref{fig:swissroll_evolution}. The drops reach their circular form at time $t=70$. The minimal distance between drops measured during the simulation is $9\cdot10^{-5}$. This can be compared to a corresponding simulation without surfactants, where the drops are circular at time $t=32$, with minimal distance $3\cdot10^{-4}$.

\begin{figure}
\centering
\subfigure[][$t=2.5$]{
\label{fig:ex3-a}
\frame{\includegraphics[trim={4cm 2.75cm 5cm 2cm},clip,height=2in]{graphics/sr_t2p5.png}}}
\hspace{0.5cm}
\subfigure[][$t=5$]{
\label{fig:ex3-b}
\frame{\includegraphics[trim={4cm 2.75cm 5cm 2cm},clip,height=2in]{graphics/sr_t5.png}}} \\
\subfigure[][$t=10$]{
\label{fig:ex3-c}
\frame{\includegraphics[trim={4cm 2.75cm 5cm 2cm},clip,height=2in]{graphics/sr_t10.png}}}
\hspace{0.5cm}
\subfigure[][$t=20$]{
\label{fig:ex3-d}
\frame{\includegraphics[trim={4cm 2.75cm 5cm 2cm},clip,height=2in]{graphics/sr_t20.png}}} \\
\subfigure[][$t=30$]{
\label{fig:ex3-e}
\frame{\includegraphics[trim={4cm 2.75cm 5cm 2cm},clip,height=2in]{graphics/sr_t30.png}}}
\hspace{0.5cm}
\subfigure[][$t=50$]{
\label{fig:ex3-f}
\frame{\includegraphics[trim={4cm 2.75cm 5cm 2cm},clip,height=2in]{graphics/sr_t50.png}}}
\caption{Deformation over time for the swiss roll domain. Surfactant concentration in colour of the same scale as in Figure~\ref{fig:swissroll_domain}.

}
\label{fig:swissroll_evolution}
\end{figure}


TITLE: are linear functionals on C[0, 1] bounded and thus continuous
QUESTION [1 upvotes]: I'm currently on a theorem showing a general representation of linear functionals on the space of continuous functions on the interval $[0,1]$.
My problem is on the beginning of the proof.
First we define a linear functional $f(x)$ on the space $C[0,1]$.
After that using the Cantor theorem we show that every continuous function on a compact interval is actually bounded and thus $C[0,1]$ is a subspace of $M[0,1]$ - bounded functions on the interval $[0,1]$.
Using the Hahn-Banach theorem we show  that our functional $f$ can be extended to a functional $F$ on the whole space $M[0, 1]$ with the same norm.
Maybe I've missed something foundamental, but how do we know that the functional $f$ is bounded, so that it has a norm and it's norm is the same with the extension $F$?
Thanks in advance!

REPLY [3 votes]: No, there exist linear functionals on $C[0,1]$ that are not bounded.
In fact, for every infinite dimensional space there exists a discontinuous (and therefore unbounded) linear functional, see here.
I would suspect that there is an additional assumption somewhere that the linear functional is bounded.


TITLE: what would be a euclidean argument for why the characteristic axiom always holds in hyperbolic geometry?
QUESTION [0 upvotes]: what would be a euclidean argument for why the characteristic axiom always holds in hyperbolic geometry?
Characteristic axiom states Given a line k and a point p not on k, there are at least two lines on p that do not intersect k.

REPLY [0 votes]: First of all, the axiom is, after all, just presumed to hold -- that is after all the point of axioms.
However, you can get some intuitive grasp on why geometry on a hyperboloid would have this property.  Consider a line $\ell$ on the hyperboloid and a nearby exterior point $P$.  Now try to make a parallel line through $P$ in a naive way, by taking two points $(A,B)$ very near to $P$ and
on opposite sides to the shortest line from $P$ to $\ell$, and choosing then so that the minimum distances of $A$ and $B$ to $\ell$ are equal.  Line $AB$ comes very close to $P$; tweak it by increasing the equal distances from $A (B)$ to $\ell$ by equal amounts until the new line passes through $P$.
Now this line $APB$ is actually a geodesic on the hyperboloid, and you will find that as you go further away from $P$, the shortest distance of a point $C$ on $APB$ to the line $\ell$ becomes greater and greater.
Finally, imagine "tilting" this line by an angle at $P$ which is small enough that the rate of approach to $\ell$ due to the tilt (on one side) is less than the rate of increase in the distance due to the concave shape of the hyperboloid.  This tilted line will also never meet $\ell$.  So you can make at least two lines through $p$ that never meet $\ell$.
It is in fact possible to work out explict equations for these lines, but that doesn't add much to the intuition.
You can prove, without introducing new axioms, that if there are at least $2$ such lines through any point, then there are infinite such lines.