TITLE: How do they measure the rotational speed of cricket balls?
QUESTION [18 upvotes]: The 2013 Ashes series (a cricket thing between England and Australia) are underway as of today and it seems they have a new (to me) gimmick. It appears that they are now able to measure almost instantaneously (within a few seconds) the rotational speed of balls bowled by spin bowlers.

On the (golf) driving range, I was able to get similar readings, but they were from a machine only a few feet away from the tee. I don't quite understand that either, but it seems at least easier. Although there, they were even able to determine the rotational axis as well.

But in cricket they also have a rotational axis along the direction of the ball, which is even more difficult to measure (unless they have a third radar somewhere perpendicular-ish to the pitch).
I think I now see that they placed very small black boxes between two of the stumps on either end of the pitch. I guess that must be the first two radars. (One of which will be useless because the batsman is standing in front of it.)

So, essentially: How do they do it? And, how accurate can it be, given that the reflection of the "sides" of the ball may be poor? Or are they using (visible-light) camera images, as suggested by the comments of udiboy and Deer Hunter below? (I'm fairly confident the golf system doesn't do that, but I don't know. Anyways, to make sure, the question is primarily about the cricket.)

Added later.
I found that they measure spin rates in baseball as well: http://sportsillustrated.cnn.com/2011/writers/tom_verducci/04/12/fastballs.trackman/index.html.
The http://trackman.dk site refers to these technologies and (patent/patent application) numbers:

Measuring spin rate of sports balls by radar using multiple harmonic
  spectrum traces. (US2009/0075744, EP1698380,
  DE602006009719.0, GB/EP/1698380, ZL200680006869.0, JP2008/538085A and
  KR10/0947898)
Measuring spin axis orientation of sport balls from trajectory
  measurements by radar. (US2009/0075744, DE602006015036.9,
  GB/EP/1853362, ZL200680006869.0, JP2008/538085A and KR10/0947898)

REPLY [1 votes]: If the balls are kept clean, a high-speed camera coupled to a computer running an image recognition/analysis program ought to be adequate. As an example of the sorts of images you can get, see this cricket match At about 1:35 you can see slow motion with a long lens. The stripe on the ball is very visible.
As a starting point, an upper limit of 3000 rpm (per your first image) gives 50 rps. At normal (at least, as normal as cricket-playing countries get) PAL video rates of 50 fps that's a maximum of 1 revolution per frame, so high speed should easily be able to resolve the rotation. And, of course, there's no guarantee that the 4 figures presented in the image are precise. I suspect the last two significant digits are bogus.
Designing such a system, I expect, would be quite challenging, but the problem would be in designating the period and exact area of interest, and doing it in a few seconds.


text:
\begin{document}
\title{Simultaneous dense and nondense orbits for toral automorphisms}

\author{Jimmy Tseng}
\address{J.T.:  School of Mathematics, University of Bristol, University Walk, Bristol, BS8 1TW UK}
\email{j.tseng@bristol.ac.uk}

\thanks{J.T. acknowledges the research leading to these results has received funding from the European Research Council under the European Union's Seventh Framework Programme (FP/2007-2013) / ERC Grant Agreement n. 291147.}

\begin{abstract} We show that, for pairs of hyperbolic toral automorphisms on the $2$-torus, the points with dense forward orbits under one map and nondense forward orbits under the other is a dense, uncountable set.  The pair of maps can be noncommuting.  Our main tools are a geometric construction and the Baire Category theorem.
\end{abstract}
\maketitle
\section{Introduction}\label{secIntro}  Given a dynamical system $f: X \rightarrow X$ on a set $X$ with a topology, let us say that a point has \textit{dense forward orbit} if its forward orbit closure equals $X$ and \textit{nondense forward orbit} otherwise.  Let the set of points with dense forward orbits be called the \textit{dense set for $f$} and denoted by $D(f)$ and with nondense forward orbits be called the \textit{nondense set for $f$} and denoted by $ND(f)$.  Let $\tilde{f}:X \rightarrow X$ be another dynamical system on the same phase space.  How large is the set $D(f) \cap ND(\tilde{f})$ is a natural question to ask.  It was first asked in print by the author and V.~Bergelson and M.~Einsiedler in~\cite{BET}, in which complete orbits (as well as forward orbits) were considered for commuting pairs of maps on the torus and on certain compact homogeneous spaces.  The technique in the present paper is completely different, but applies to pairs of noncommuting maps as well.  The results in this paper were proven before the results in~\cite{BET} and complement them.  The technique in this paper, which involves building a certain fractal with the help of rational points (i.e. periodic points), could be useful for other considerations.

\subsection{Statement of results}A hyperbolic toral automorphism $T: \TT^d \rightarrow \TT^d$ is induced by a linear mapping $\RR^d \rightarrow \RR^d$, which we also refer to as $T$.  The hyperbolicity of the mapping implies that $\RR^d = E^+_T \oplus E^-_T$ where $T$ expands $E^+_T$, which we refer to as the {\em expanding space} and sometimes denote simply by $E^+$ if the mapping is understood, and where $T$ contracts $E^-_T$, the {\em contracting space} and, likewise, sometimes we denote by $E^-$.  Our main result is the following.

\begin{theo}\label{thmForDenNonDen}Let $T, S:\TT^2 \rightarrow \TT^2$ be hyperbolic toral automorphisms.  If $\dim(E_T^- \cap E_S^-) = 0$, then $ND(T) \cap D(S)$ is a dense, uncountable set.
\end{theo}

\begin{coro}\label{coroForDenNonDen}Let $T:\TT^2 \rightarrow \TT^2$ be an hyperbolic toral automorphism and $\{S_k\}$ be the family of all hyperbolic toral automorphisms such that $\dim(E_T^- \cap E_{S_n}^-) = 0$.  Then $ND(T) \cap \bigcap_k D(S_k)$ is a dense, uncountable set.
\end{coro}

\begin{rema}
We can assert a slightly stronger conclusion---see Remark~\ref{remNonCommResultStronger}.  Note that the theorem is valid for both noncommuting and commuting pairs of maps.  Also, a generalization to (noninvertible) hyperbolic toral endomorphisms should follow with the adjugate (see \cite[Chapter 4]{HoJo} for the definition) in place of the inverse.

\end{rema}

Finally, we note that the nondense sets for the maps we consider are known to be Lebesgue null sets of full Hausdorff dimension~\cite{BFK} (see also \cite{T4, Ur}).

\subsection{Idea of proof}
The fractal whose points miss a certain (open) box under forward iterates of $T$ will be ``geometrically large'':  namely, in any neighborhood of the fractal, there is a point of the fractal that lies in a piece of the hyperplane (or line for $d=2$) which is an affine translate of $E^-_T$.  Moreover, this whole piece is in the fractal, a fact that follows from constructing the box using pieces of the leaves of the expanding and contracting foliations and noting that iterates of the box cannot overlap properly (Section~\ref{secConFracforT}).  The construction of our fractal involves rational points (i.e. periodic points) and could be useful for other considerations.

Next through a given element of a countable dense subset of the torus, there is a piece of $E^-_S$ suitably translated, which under the action of $S^{-1}$ will intersect the piece of the translate of $E^-_T$ alluded to previously (Section~\ref{secGeoFracforT}).  Now an application of the Baire Category theorem gives the result (Section~\ref{secBaireCatArg}).

The reason that this argument does not generalize to both the backwards and forwards orbits of $T$ is that the corresponding fractal would not be ``geometrically large'' because the intersection of the two fractals given by the forward orbits of $T$ and $T^{-1}$ (respectively) would not contain any piece of linear subspace of any dimension $\geq 1$.  The only way to have a piece of linear subspace is for it to be contained in linear subspaces in the two fractals.  If pieces of linear subspace from different fractals intersect, then they must intersect in a point (because $E^- \oplus E^+ = \RR^d$ and the intersection point is an affine translate of the origin), and thus line segments are precluded from the intersection.

Finally, since $d=2$ and the map is hyperbolic, the eigenvalues must be real numbers.  In many instances, we will assume that the eigenvalues are positive.  This assumption does not affect the generality of our theorem, as the last part of the proof of Theorem~\ref{thmForDenNonDen} shows (see Section~\ref{secBaireCatArg}).

\subsubsection*{Acknowledgements}  I thank V.~Bergelson for suggesting the problem and discussions.

\section{Constructing the fractal for $T$}\label{secConFracforT}
Let $d=2$ and $T: \TT^d \rightarrow \TT^d$, $S:\TT^d \rightarrow \TT^d$ be two hyperbolic toral automorphisms.  Given a set $B \in \TT^d$, define \[F:=F_{T^{-1}}(B) := \TT^d \backslash \cup_{n=0}^\infty T^{-n}(B).\]  Note that if $B$ is open, then $F$ is compact.  We refer to the elements of the collection $\{T^{-n}(B)\}_{n=0}^\infty$ as the {\it $T^{-1}$-tubes for $B$}, or simply as \textit{tubes} if the context is clear.  When we are considering two tubes $T^{-m}(B)$ and $T^{-n}(B)$ where $m<n$, we call $T^{-m}(B)$ the \textit{old tube} and $T^{-n}(B)$ the \textit{new tube}.  Also, the notation $E^-(x)$ denotes the affine translate of the contracting space $E^-$ given by $E^- + x$.  For the expanding space $E^+$, we also define similar notation.

\begin{rema} \label{remaEleEmbeddingProp}
Translation on $\TT^d$ is the well-defined map that makes translation on $\RR^d$ commute with the natural projection $\RR^d \rightarrow \TT^d$.  Moreover, the contracting space as embedded naturally in $\TT^d$ lifts to the family of hyperplanes $\{E^-(k)\}_{k \in \ZZ^d}$.  If two elements of this family intersect, we have that $x +k_1 = y + k_2$.  If $k_1 = k_2$, the two elements are the same element.  If not, then $E^-$ contains a non-zero integer vector.  Applying the map on $\RR^d$ corresponding to the toral map $T$, we see that the non-zero integer vector is mapped to integer vectors with smaller and smaller length.  Therefore, some power of this map is not injective, a contradiction of the fact that the map is invertible.  Therefore the elements of this family are pairwise disjoint.  Moreover, a suitable affine translation of an element will be another element.  We may make similar statements for the translates of the contracting space and translates of the expanding space.
\end{rema}

We will construct a small open box (parallelepiped) $B$ with one vertex at the origin $\boldsymbol{0}$ of $\TT^d$.  Our construction takes place inside a small-enough neighborhood of $\boldsymbol{0}$ so that locally we are on $\RR^d$.  Pick an open box (parallelepiped) $B'$ with one vertex at $\boldsymbol{0}$ whose edges are small segments of the eigendirections (i.e. unit eigenvectors) of $T$.  The box $B'$ thus contains a neighborhood of $\TT^d$.  Pick a rational point $y_0 \in B'$ and let $\{y_0, \cdots, y_{N-1}\}$ be the orbit of $y_0$ under $T$.  Also let $B'^- := \overline{B'} \cap E^-$ and $B'^+ := \overline{B'} \cap E^+$ (recall locally we are on $\RR^d$).

To construct the box $B$, we must correctly choose an iterate $y_n$.  Let $v_1, \cdots, v_\ell$ be the contracting eigendirections and let $v_{\ell+1}, \cdots, v_d$ be the expanding eigendirections.  Let $\a>0$.  The vectors $\a v_1, \cdots, \a v_\ell$ form the edges of a closed parallelepiped.  Replacing any subset of the $\a v_i$s with their negation will form $2^\ell -1$ other closed parallelepipeds for a total of $2^\ell$ parallelepipeds.  Denote the union of the $2^\ell$ parallelepipeds by $P(\a, \boldsymbol{0})$ and let $P(\a, y_n):= P(\a, \boldsymbol{0}) + y_n$.  Thus $P(\a, y_n)$ is a compact set with $y_n$ in its interior and whose diameter grows as $\a$ grows.  Moreover, $P(\a, y_n)$ lies in the contracting space through $y_n$ (this space is just $E^-$ with the origin translated to $y_n$).  Therefore, as $\a$ grows, $P(\a, y_n)$ becomes larger and it may even wrap around $\TT^d$, but, nevertheless, there is a unique least $\a$ for which $P(\a, y_n)$ meets $B'^+$.  As alluded to above, this intersection is exactly one point.  Denote this unique $\a$ by $\a_n$.  (By the ergodicity of $T^{-N}$, $\a_n$ is finite.)

Let $0 \leq m<N$ be an index such that $\a_m$ is minimal.  And consider the following cases.

\subsection{The index $m$ is unique}  Thus $P(\a_m, y_m)$ meets $B'^+$ and is the only $P$ to do so.  Call this intersection point $x$.

\subsubsection{The parallelepiped $P(\a_m, y_m)$ meets $B'$.}  Recall that $B'$ is open and thus does not contain $B'^+$.  Since $x \in E^+$, $x$ can be uniquely written as a linear combination of $v_{\ell+1}, \cdots, v_d$ and, by the construction of $B'$, the coefficients are all nonnegative.  If one of the coefficients is zero, then it follows that the corresponding coordinate for $y_m$ is zero since we have only added vectors in $E^-$.  However, the action of $T$ preserves this zero coordinate and thus $y_0$ must have this same coordinate be zero, a conclusion that contradicts the fact that $y_0$ was chosen in $B'$.  These unique coordinates of $x$ determine a unique closed $d-\ell$-parallelepiped $S$ with $x$ as the vertex (note that the interior of this parallelepiped is an open set in the subspace topology of $E^+$).

For every $n \neq m$, there is a positive minimum distance between $P(\a_m, y_n)$ and  $B'^+$.  Consequently, there exists an open $d$-parallelepiped $B$ with $S$ as a face and with segments of $v_1, \cdots, v_\ell$ as edges of $\boldsymbol{0}$ and such that $B$ does not meet any $P(\a_m, y_n)$ (including $n = m$ since $B$ is open).  Finally, let us distinguish the contacting face $B_{\boldsymbol{0}}:= \overline{B} \cap E^-$ and its parallel face $B_x := \overline{B} \cap P(\a_m, y_m)$.  Note that we can and do choose $B$ so that $B_x$ is an (arbitrarily small) $\ell$-parallelepiped.

\subsubsection{The parallelepiped $P(\a_m, y_m)$ does not meet $B'$.}  Picking an $\a$ slightly bigger than $\a_m$ will result in $P(\a, y_m)$ meeting $B'$, but $P(\a, y_n)$ not meeting $B'^+$ for any $n \neq m$.  Choose $B$ and denote $B_{\boldsymbol{0}}$ and $B_x$ as above.

\subsection{The index $m$ is not unique}  Let $m_0, \cdots, m_k$ be such that $\a_m:= \a_{m_0} = \cdots = \a_{m_k}$.  Thus every element of $\{P(\a_m, y_{m_i})\}_{i=0}^k$ meets $B'^+$ and are the only $P$ to do so.

\subsubsection{At least one of the $P(\a_m, y_{m_i})$ meets $B'$}  Consider only those elements of $\{P(\a_m, y_{m_i})\}_{i=0}^k$ that meet $B'$ and consider their intersection points with $B'^+$.  By renaming the elements of the orbit of $y_0$ if necessary, we may assume that these intersection points $x_0, \cdots, x_j$ correspond to $y_0, \cdots, y_j$ where $0\leq j \leq k<N$.

If $x_h = x_i$, then $y_h - y_i$ lies in $E^-$.  Thus there is some $n$ such that $T^n(y_h - y_i)$ is shorter than all the distances between any distinct pairs in the orbit of $y_0$.  Consequently, $y_h = y_i$; thus $h=i$; and all the $x_i$s are distinct.

As before, each $x_i$ determines a closed $(d-\ell)$-parallelepiped $S_i$.  And moreover, there exists at least one of the $S_i$s which contains exactly one of the elements of $\{x_0, \cdots, x_j\}$ (namely the element that determines it).  Let us show why.  Let $S_h$ contain more than one element:  so it contains $x_h$ by construction and an $x_i$ by supposition.  In order for $x_i$ to be in $S_h$, at least one coordinate of $x_i$ must be strictly less than the corresponding coordinate for $x_h$.  Hence, it follows that $S_i \subset S_h$.  Since there is only a finite number of such $S$s, this process must terminate.

Let us pick $S_i$ such that it only contains $x_i$ from the set $\{x_0, \cdots, x_j\}$.  Disregarding the $P$s that do meet both $B'$ and $B'^+$, form $B$ as before and thus $B$ does not meet this pruned subset of $P$s.  Denote $B_{\boldsymbol{0}}$ and $B_x:=B_{x_i}$ as before.  And note that $B$ will not meet $B_x$ as before.

What is left to show is that $B$ will not meet any $P$ from the disregarded subset.  Assume that $B$ does meet a point $z$ in one of the disregarded $P$s.  Since $B$ is a parallelepiped with vertex $\boldsymbol{0}$ whose edges are segments of the eigendirections, then the point whose expanding eigendirection coordinates are the same as that of $z$ but whose contracting eigendirection coordinates are all zero lies on $\overline{B} \cap E^+
\subset B'^+$.  Therefore it must be one of the $\{x_0, \cdots, x_j\}$ and hence it must be $x_i$.  Consequently, $y_i -z \in E^-$ and $z \in B_x$, which we know does not meet $B$, a contradiction.

\subsubsection{None of the $P(\a_m, y_{m_i})$ meet $B'$}  Picking an $\a$ slightly bigger than $\a_m$ will result in every element of $\{P(\a, y_{m_i})\}_{i=0}^k$ meeting $B'$, but $P(\a, y_n)$ not meeting $B'^+$ for any $n \notin \{m_0, \cdots, m_k\}$.  Choose $B$ and denote $B_{\boldsymbol{0}}$ and $B_x$ as above.  This concludes the construction of the open parallelepiped $B$.

\begin{rema}\label{remaNoSelfIntersectTubes}
Having constructed $B$, let us consider the family of $T^{-1}$-tubes for $B$.  Hyperbolic toral automorphisms are bijections and hence no tube can intersect itself.  (By placing $B$ into a bigger parallelepiped, one sees that not only does the tube not intersect itself, but also, restricted to any small-enough neighborhood where the tube has path components, there is a positive minimum distance between path components---moreover, this distance depends only on the tube.)
\end{rema}

\subsection{Proper overlaps}  The notion of proper overlaps, to be defined in this section, is local.  However, we show below that proper overlaps cannot occur anywhere; thus a global condition on the family of tubes is obtained.

Two distinct tubes can intersect.  We must distinguish a special type of intersection, which we now define.  At every vertex of the box $B$, we can erect a local coordinate system given by the eigendirections.  This allows us to identify for each vertex $x$ a contracting face, which is spanned by the contracting eigendirections at $x$, and an expanding face, which is spanned by the expanding eigendirections at $x$.  Since we are considering tubes, we are considering iterates of the inverse map and the contracting face expands and the expanding face contracts.

Let $\Cc$ denote the contracting face of $\boldsymbol{0}$ and $\Ee$, the expanding face.  When the edges of the vertex $\boldsymbol{0}$ of $B$ are regarded as vectors, there is a unique vertex of $B$ that is the sum of these vectors; let $\Cc'$ denote the contracting face through this vertex and $\Ee'$, the expanding face.  (Informally, $\Cc$ and $\Cc'$ are a pair of parallel faces of $B$, as are $\Ee$ and $\Ee'$.) Also, define $\Cc_m := T^{-m}(\Cc)$; likewise for $\Cc'_m$, $\Ee_m$, and $\Ee_m'$.

Let $y \in T^{-m}(B)$.  Since the tube is open, there exists around $y$ an open ball which is contained in the tube.  Borrowing the terminology for $P(\a, y)$ above, form the analogous compact set $Q(\a, y)$ with the expanding eigendirections.  Note that there is an $\a>0$ small enough so that $Q(\a,y)$ is contained in the tube.  As $\a$ grows, $Q(\a,y)$ must meet $\Cc_m$ and $\Cc'_m$ in two distinct points because the tube is just a long $d$-parallelepiped (which is embedded in $\TT^d$ in the usual way, a fact that is immaterial to us because our construction is local) and thus convex.  Note that we do not grow $\a$ so that it meets $\Cc_m$ twice because, locally, we want to stay in the same path component as $y$ (see Remark~\ref{remaNoSelfIntersectTubes}).  The same applies for $\Cc'_m$.

Consider an old tube $T^{-m}(B)$ and a new tube $T^{-n}(B)$.  Let $y \in T^{-m}(B) \cap T^{-n}(B)$.  Since the intersection of the two tubes is an open set, there exists an $\a > 0$ such that $Q(\a,y)$ is contained in this intersection.  As $\a$ grows, let us say that $Q(\a, y)$ meets each of the faces $\Cc_m$, $\Cc'_m$, $\Cc_n$, and $\Cc'_n$ at $\a$ equals $\a_m$, $\a'_m$, $\a_n$ and $\a'_n$, respectively.  Denote the intersection points $x_m$, $x'_m$, $x_n$, and $x'_n$, respectively.  Some of the intersection points (and thus some of the $\a$s) may coincide.  However, since the primed contracting faces lie in a contracting space translated to an iterate of the rational point $y_0$, they cannot coincide with the non-primed contracting faces (both of which lie in the contracting space).  (Also, all of these contracting spaces are embedded in the usual way in $\TT^d$, which is immaterial because we are working locally.)

The notion of a slice is convenient.  A \textit{slice of a tube $T^{-k}(B)$ through
$z$} is the connected component, containing $z$, of the intersection of the tube with $E^+(z)$ where $z$ is in the tube.  A \textit{closed slice} is the same definition, except that the tube is replaced by a closed tube.  Let $\Cc_k$, $\Cc'_k$, $x_k$, and $x'_k$ mean the same as above.

\begin{lemm}\label{lemmSlicesAreTranslates} The slice is a $(d-\ell)$-parallelepiped with vertices $x_k$ and $x'_k$.  Moreover, the slice is just the expanding face of the tube $E^+ \cap T^{-k}(B)$ translated by adding $x_k$.
\end{lemm}
\begin{proof}
Lift, via the natural projection $\RR^d \rightarrow \TT^d$, the tube to $\RR^d$.  We obtain an infinite family of parallelepipeds (see Remark~\ref{remaEleEmbeddingProp}).  It is clear that the family of parallelepipeds are pairwise disjoint.  Exactly one of these $\Tt$ has the origin of $\RR^d$ as vertex.  Let $\tilde{x}_k$ and $\tilde{x'}_k$ denote the lifts of $x_k$ and $x'_k$, respectively, to $\Tt$---these lifts are unique (if they are not, then $E^-$ contains a non-zero integer vector and now follow Remark~\ref{remaEleEmbeddingProp}.)

The space $E^+(y)$ lifts to a unique translate of the expanding space (which by abuse of language we also call $E^+(y)$) containing $\tilde{x}_k$ and $\tilde{x'}_k$.

The point $\tilde{x}_k$ lies in the contracting space $E^-$. Note that $E^+(y)- \tilde{x}_k = E^+$.  Thus the point $\tilde{x'}_k-\tilde{x}_k$ lies in $E^+$ and has the same expanding coordinates as $\tilde{x'}_k$.  Now $\tilde{x'}_k$ lies in the unique lift of $\Cc'_k$ to $\Tt$ and therefore has the same expanding coordinates as the unique vertex of $\Tt$ which is the sum of the edges of $\Tt$ with the origin as vertex.  Consequently, the parallelepiped $E^+ \cap \Tt$ has one vertex at the origin and is uniquely determined by the vertex at $\tilde{x'}_k-\tilde{x}_k$. By the definition of a parallelepiped, every point in the lifted slice (which has the same contracting coordinates as $\tilde{x}_k$) is just the translate of $E^+ \cap \Tt$ by adding $\tilde{x}_k$.  The lemma follows.

\end{proof}

Let the eigenvalues of $T$ be positive real numbers.  Let the old and new tubes intersect at a point $y$.  Let $S_m$ be the slice of the old tube through $y$ and $S_n$, the new tube.   We say that the two tubes \textit{overlap properly at $y$} if $S_n \cap \{x_m, x_m'\} \neq \emptyset$.  Note that by the dynamics of the system and the fact that the tubes are open sets, once the overlap property holds (or, respectively, does not hold) at $y$, then there is a small neighborhood of $y$ in the intersection of the two tubes for which the overlap property holds (or, respectively, does not hold). By (the analog of) Lemma~\ref{lemmContainsSlice}, we note that, if $S_n$ contains both $x_m$ and $x_m'$, then $\overline{S}_m \subset S_n$, a contradiction of Lemma~\ref{lemmNewTubesAreSmaller}.  Thus, when the tubes overlap properly, $S_n$ contains either $x_m$ or $x_m'$, but not both.

\begin{lemm}\label{lemmSliceofOldandNewTubes} Let the eigenvalues of $T$ be positive real numbers.  For any $j \leq m$ and any slice $S_m$ of a tube $T^{-m}(B)$, $T^j (S_m)$ is the slice $S_{j-m}$ of the tube $T^{j-m}(B)$, where $T^j(x_m) = x_{j-m}$ and $T^j(x_m') = x_{j-m}'$.
\end{lemm}
\begin{proof}

Applying $T^j$ to $x_m$ yields a point in the contracting space; applying it to the edges of $S_m$ with vertex $x_m$ preserves the orientation of every edge because these edges point in the eigendirections.  Since every point in $S_m$ is a (appropriate) linear combination of edges, every point of $T^j (S_m)$ is a (appropriate) linear combination of $T^j$ applied to the edges.  In particular, $T^{j}(x_m) = x_{j-m}$ and $T^{j}(x_m') =x_{j-m}'$.  By Lemma~\ref{lemmSlicesAreTranslates}, $T^j (S_m)$ is a slice $S_{j-m}$ of the new tube.
\end{proof}

Let $S_m$ denote the slice of the old tube through $y$ and $\overline{S}_n$ denote the closed slice of the new tube through $y$.
\begin{lemm}\label{lemmContainsSlice} Let the eigenvalues of $T$ be positive real numbers.  If both $x_n$ and $x_n'$ are in $S_m$, then $\overline{S}_n \subset S_m$.
\end{lemm}

\begin{proof}

By Lemma~\ref{lemmSlicesAreTranslates}, both slices are parallelepipeds with the unprimed vertex in the contracting space (as embedded in the torus) and the edges of these vertices pointing in the eigendirections.  Since $x_n, x_n' \in S_m$, these vertices can be uniquely expressed in coordinates with respect to $x_m$. Also every point in $\overline{S}_n$ can be uniquely expressed in coordinates with respect to $x_n$.  Let us write both coordinate systems explicitly.  Let $e_{\ell +1}, \cdots, e_d$ be the edges of $x_m$ corresponding to their respective eigendirections. Then the coordinates of $x_m'$ is all ones.  Let $(c_{\ell +1}, \cdots, c_d)$ be the coordinates of $x_n$ and $(c_{\ell +1}', \cdots, c_d')$ the coordinates of $x_n'$.  Then $0 < c_i, c_j'< 1$.  Since the vector $x_n'$ is equal to the sum of the vector $x_n$ with the sum of positive multiples of the expanding eigendirections, we have that $c_j' > c_j$.  Moreover, $(c_j'-c_j)|e_j|$ is the Euclidean length of the edge $\tilde{e}_j$ in the direction $v_j$ of the vertex $x_n$ for $S_n$.  Let $\a_{\ell+1}\tilde{e}_{\ell+1} + \cdots + \a_d \tilde{e}_d$ be a point in $\overline{S}_n$.  Then $0 \leq \a_j \leq 1$ and this same point is uniquely expressed as \begin{align*}& \a_{\ell+1}(c_{\ell+1}'-c_{\ell+1})|e_{\ell+1}| e_{\ell+1}/|e_{\ell+1}| + \cdots + \a_d (c_d'-c_d)|e_d| e_d/|e_d| + x_m \\ =  \ &  [\a_{\ell+1}(c_{\ell+1}'-c_{\ell+1}) + c_{\ell+1}] e_{\ell+1} + \cdots + [\a_d (c_d'-c_d)+c_d] e_d.\end{align*}  As $0 <\a_j (c_j'-c_j) +c_j \leq c_j' <1$, this point is in $S_m$.
\end{proof}

\begin{lemm}\label{lemmNewTubesAreSmaller} Let the eigenvalues of $T$ be positive real numbers.  Some vertex of the slice of the old closed tube through $y$ is not contained in the slice of the new closed tube through $y$.
\end{lemm}

\begin{proof}
Assume not.  Let $S_m$ denote the slice through the old closed tube containing $y$.  By Lemma~\ref{lemmSliceofOldandNewTubes}, $T^{-(n-m)} (S_m)$ is a slice through the new closed tube.  By Lemma~\ref{lemmSlicesAreTranslates}, the slice of the new closed tube through $y$ is an affine translate of $T^{-(n-m)} (S_m)$. By assumption, every vertex of the old tube's slice is in the new tube's slice. Therefore all points of the old tube's slice can be uniquely expressed in coordinates (that is a tuple of real numbers between 0 and 1) given by $x_n$ and its edges.  Translate the old tube's slice so that the vertex $x_m$ coincides with $x_n$. In coordinates, this translation is equivalent to coordinate-wise subtraction of the coordinates of $x_m$.  The resulting coordinates of any point in the translation of the old tube's slice are tuples of real numbers between 0 and 1; thereby showing that the translated slice lies completely inside the new tube's slice.  Hence, \[\dist\big(T^{-(n-m)}(x_m),  T^{-(n-m)}(x_m')\big) = \dist(x_n, x_n') \geq \dist(x_m, x_m'),\] a contradiction of the fact that $x_m$ and $x_m'$ lie on an affine translate of the expanding space and the inverse map contracts on this space.
\end{proof}

\begin{prop}\label{propNoProperOverlaps} Let the eigenvalues of $T$ be positive real numbers.  For the constructed box $B$, two distinct tubes do not properly overlap anywhere.
\end{prop}

\begin{proof}

Let the old tube be $T^{-m}(B)$ and the new tube be $T^{-n}(B)$.  Let $y \in \TT^d$.  If $y$ is not in the intersection of the two tubes, then the two tubes do not properly overlap at $y$.

Thus we need only consider when $y$ is in the intersection of the two tubes.  Assume that the two tubes overlap properly at $y$. Let $S_m$ be the slice of the old tube through $y$ and $S_n$, the new tube.  Since the overlap is proper, $S_n$ contains either $x_m$ or $x_m'$.

\textit{Case:  $x_m \in S_n$.}  Applying $T^m$ to the old tube will map it into the original box $B$.  Also, the point $x_m$, which lies in the contracting space as embedded in the torus, will return to the contracting face of $B$, namely the set $B_{\boldsymbol{0}}$.  For all $k \geq m$, $T^k(x_m) \in B_{\boldsymbol{0}}$.  Applying $T^n$ to the new tube will map it into the original box $B$.  Since $B$ is open, it does not contain any point in any of its faces and therefore $T^n(x_m) \notin B_{\boldsymbol{0}}$, a contradiction.

\textit{Case:  $x_m' \in S_n$.}  Similar to the previous case, $T^m(x_m') \in B_x$.  As before, applying $T^n$ to the new tube will map it into $B$.  Therefore $T^n(x_m') \in B$.  Let $P(\a, y_m)$ be the $P$ chosen in the construction of $B$.  (To prevent notation clash between the two different meanings of $m$, let us call this $P$, $P(\a, y_k)$.)

Now, by construction, $P(\a, y_k)$ contains $B_x$ and no element of $\{P(\a, y_\ell)\}_{\ell=0}^N$ meets $B$.  Since $T$ permutes the $y_\ell$s and is contracting on any $P(\a, y_\ell)$, we have that for each $P(\a, y_\ell)$ there exists a $P(\a, y_j)$ such that $T(P(\a, y_\ell)) \subsetneq P(\a, y_j)$.  Thus, $T^i(x_m') \notin B$ for every $i \geq m$, which contradicts the fact that $T^n$ maps the new tube onto $B$.

 \end{proof}

\section{The geometry of the fractal for $T$}\label{secGeoFracforT}

The fractal is the set $F:=F_T :=F_T(B):=\TT^d \backslash \bigcup_{n=0}^\infty T^{-n} B$.

\begin{lemm}\label{lemmDisjointpieces} The sets $\Cc_M, \Cc_M'$ and $\bigcup_{n=M+1}^\infty T^{-n} B$ are pairwise disjoint.
\end{lemm}
\begin{proof}

The contracting space and its translate are leaves in the stable foliation for $T$.  As distinct leaves of the same foliation are disjoint, the only way for the contracting space and its translate to intersect is if they coincide.  Recall that $\Cc_M$ is contained in the contracting space and $\Cc_M'$ is in the translate of the contracting space through the rational point $y_m$. Thus, if we assume that $\Cc_M, \Cc_M'$ are not disjoint, then $y_m$ is in the contracting space.  Since $y_0$ is not an integral point, no point in its orbit is either.  Thus, there exists a positive minimal distance between the orbit of $y_0$ and $\boldsymbol{0}$, which contradicts the fact that $T^{i}(y_m)$ is shrinking towards $\boldsymbol{0}$ as $i \rightarrow \infty$.

Assume that $\Cc_M$ meets $\bigcup_{n=M+1}^\infty T^{-n} B$.  Then there exists $y \in \Cc_M \cap T^{-n} B$ for some $n > M$.  Call the tube corresponding to $M$ the old tube and $n$, the new tube.  Taking slices through the two tubes at $y$ shows that they overlap properly at $y'$, a point in the intersection of the two tubes.  This contradicts Proposition~\ref{propNoProperOverlaps}.  The same proof suffices for $\Cc_M'$.
\end{proof}

Let $y \in F$.  A \textit{piece of space parallel to the contracting space} is a translate of an open subset of the contracting space.

We must now distinguish two types of slices:  interior and boundary. Given a tube $T^{-k}B$, the two distinct \textit{boundary slices} are $\Ee_k$ and $\Ee_k'$.  They cannot coincide.  All other slices of the tube are \textit{interior slices.}

The following is the key lemma.\footnote{Case~2 of the proof of this lemma requires $d=2$.  All of the proceeding (Section~\ref{secConFracforT} and Section~\ref{secGeoFracforT} up to Case 2 of this lemma) is also valid for $d \geq 2$ with $T$ diagonalizable with all positive real eigenvalues.}

\begin{lemm}\label{lemmForwardFracContainsContrSpace} Any open subset of $F$ contains a piece of space parallel to the contracting space.
\end{lemm}
\begin{proof}
Let $U$ be an open ball of $\TT^d$ around $y \in F$.  By ergodicity, the orbit of a point in $B$ meets $U$.  Therefore, some tube meets $U$.  Since $y \in F$, no tube can completely contain $U$.  Let $M \geq 0$ correspond to the first tube that meets $U$.  Since $U$ is not completely contained in this tube, $U$ meets the boundary of the tube.  There are two main cases.

\textit{Case 1:  $U$ meets $\Cc_M \cup \Cc_M'$.}  Let $z$ be a point in this intersection and assume that it is in $\Cc_M$.  Since $U$ is open, $U$ meets the interior of $\Cc_M$.  The intersection of $U$ and this interior is a piece of space parallel to the contracting space. By Lemma~\ref{lemmDisjointpieces}, this interior will be in $F$.  The same proof suffices for $\Cc_M'$.

\textit{Case 2: $U$ does not meet $\Cc_M \cup \Cc_M'$.}

In this case, we finally use the assumption of $d=2$.  Tubes are parallelograms which may wrap around the torus; slices are intervals.  Since $U$ does not meet $\Cc_M \cup \Cc_M'$ and it cannot be contained inside the tube, it must meet the interior of $\Ee_M$ or the interior of $\Ee_M'$.  Without loss of generality, we may assume that $U$ is a (open) parallelogram containing $y$ with sides parallel to the expanding and contracting directions.  Also, since all eigenvalues are assumed to be positive, at any point in any $T^{-k}$-tube near the interior of $\Ee_k'$, we may move in the negative contracting eigendirection and still remain inside the tube.  Likewise, at any point in any $T^{-k}$-tube near the interior of $\Ee_k$, we may move in the positive contracting eigendirection and still remain inside the tube. Clearly, we may move almost as far as the length of $\Cc_k$.  Without loss of generality, we may assume that diam$(U)$ is much smaller than the length of $\Cc_0$.

Since $T$ is an automorphism, $\Ee_M$ and $\Ee_M'$ cannot intersect.  Thus, if $U$ meets both the interior of $\Ee_M$ and the interior of $\Ee_M'$, then, by shrinking $U$ only in the contracting direction, we can have $U$ contain $y$ and the interior of either $\Ee_M$ or $\Ee_M'$. Since $y$ is outside the tube and since the tubes and $U$ are open parallelograms with the sides along the eigendirections, the slice $S_U$ of $U$ through $y$ is outside of the tube.  Moreover, since $U$ only meets the interior of $\Ee_M$ or $\Ee_M'$, the slice $S_U$ divides the parallelogram $U$ into two open halves:  one half $W$ intersecting the tube and the other $V$ not intersecting the tube.

\textit{Case 2A:  $U$ meets $\Ee_M$.}  First note that $V$ is in the negative contracting direction with respect to $W$.  Let $n > M$ correspond to the tube that next intersects the open parallelogram $V$.  If $U$ meets $\Cc_n \cup \Cc_n'$, then we are in Case 1 and thus finished.  Otherwise, $U$ must meet the new tube in the interior of $\Ee_n$ or the interior of $\Ee_n'$.  Now $\Ee_n \subset \Ee_M$ and thus $\Ee_n \cap U \subset W \cup S_U$.  Since $V$ does not meet $\Cc_n \cup \Cc_n'$, then either $V$ meets the interior of $\Ee_n'$ or $V$ is completely contained in the new tube.  If the latter, we claim that the interior of $\Ee_n'$ also meets $U$.  If not, then starting at $\Ee_n$ (which meets $W\cup S_U$) we may move in the positive contracting direction and stay in the new tube until we meet $\Ee_n'$.  As we move along, we meet $V$ because the new tube meets $V$.  Since $U$ does not meet $\Ee_n'$ we move across $U$.  Thus the new tube completely contains $U$, a contradiction.    Therefore, in either case, $U$ must meet the interior of $\Ee_n'$.   Now the closed new tube restricted to $U$ is the union of (two) disjoint path-components.  One of these path-components meets $\Ee_n'$.  Let $C'$ denote the interior of this path component.  Note that $C'$ meets $V$.  Since we are allowed to move in the negative contracting direction starting at $\Ee_n'$ and remain in the new tube, $C'$ only meets $V$.  Thus, either the boundary slices $\Ee_M$ and $\Ee_n'$ intersect in $U$ and therefore both contain the slice $S_U$ or they are distinct slices and therefore contain distinct slices of $U$ with only one of these slices allowed to possibly be $S_U$.  But since these boundary slices cannot intersect, there exists an open parallelogram $\tilde{U}$ contained in $U$ and containing $S_U$ and meeting either $\Ee_M$ or $\Ee_n'$, but not both.\footnote{If there exists $M < \ell < n$ such that the $T^{-\ell}$-tube meets $W \cup S_U$, then $T^{-\ell}(B) \cap U \subset T^{-M}(B) \cap U$.}  The slice $S_U$ divides $\tilde{U}$ into two halves:  one half $\tilde{W}$ intersecting one of the tubes and the other $\tilde{V}$ not intersecting either tube.

\textit{Case 2B:  $U$ meets $\Ee_M'$.}  First note that $V$ is in the positive contracting direction with respect to $W$.  Let $n > M$ correspond to the tube that next intersects the open parallelogram $V$.  If $U$ meets $\Cc_n \cup \Cc_n'$, then we are finished.  Otherwise, $U$ must meet the new tube in the interior of $\Ee_n$ or the interior of $\Ee_n'$.  Now $\Ee_n \subset \Ee_M$.  If $U$ meets the interior of $\Ee_n$, then $U$ also meets the interior of $\Ee_M$, a contradiction.  Therefore, $U$ must meet the interior of $\Ee_n'$.   Since we are allowed to move in the negative contracting direction and remain in the new tube, the new tube only meets $W$, a contradiction of the assumption that the new tube meets $V$.  Consequently, for Case 2B, $U$ must meet $\Cc_n \cup \Cc_n'$.

Cases 2A and 2B now provide a recursive algorithm, which must terminate in a finite number of steps by entering Case 1.  If not, then the only way for the algorithm to continue is for Case 2A to be repeatedly used.  Now at each step, the constructed $\tilde{U}$ and $\tilde{V}$ are open parallelograms both having sides in the expanding direction the same length $L_U$ as the sides of the expanding direction of $U$.  Consequently, there exist an increasing sequence of times corresponding to tubes which are the first to intersect the sequence of $\tilde{V}$ and the intersection will not contain the parts of the tubes that are the iterates of $\Cc$ or $\Cc'$.  Therefore, a slice from any tube in this sequence will be longer than or equal to $L_U$.  But since this sequence is comprised of inverse iterates of $B$, the expanding direction is contracting and thus the length of the slices must become smaller than $L_U$, a contradiction. This proves the lemma.

\end{proof}

\section{A Baire category argument}\label{secBaireCatArg}

In this section, we prove Theorem~\ref{thmForDenNonDen}.  First, we require two lemmas:

\begin{lemm}\label{lemmNonDenseisNowhereDense} A nondense orbit is a nowhere dense orbit.
\end{lemm}
\begin{proof}
Assume not.  Let $x \in \TT^d$ have a nondense orbit that is not nowhere dense.  Thus $\overline{\Or_T^+(x)}$ contains a nonempty open set $U$.  By ergodicity, there is some point $y \in U$ with dense orbit.  Thus, \[\TT^d = \overline{\cup_{n=0}^\infty T^n(U)} \subset \overline{\Or_T^+(x)},\] a contradiction.   \end{proof}

\begin{lemm}\label{lemmNondenseOrbitisNowhereDense} Let $n \in \NN$.  The point $x \in \TT^d$ has nondense orbit under $T \iff$ the point $x \in \TT^d$ has nondense orbit under $T^n$. \end{lemm}

\begin{proof}
The forward implication is obvious.  We prove the reverse implication. Assume it is false.  Note that
\[\overline{\Or_T^+ (x)} = \cup_{i=0}^{n-1} T^i(\overline{\Or_{T^n}^+(x)}).\]  By Lemma~\ref{lemmNonDenseisNowhereDense}, the right-hand side is a finite union of closed sets with empty interior and hence has an empty interior itself, a contradiction.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thmForDenNonDen}]  Since $d=2$, the eigenvalues of $T$ are real.  There are two cases. The first is that all the eigenvalues are positive.
By Lemma~\ref{lemmForwardFracContainsContrSpace}, we have that an open ball $U$ containing a point $y \in F_T$ also contains an open interval $I$ completely contained in $F_T$ and parallel to $E_T^-$.  Thicken this interval slightly to obtain an open set $V \subset U$.

Consider an open parallelogram $P:=P(r, \ell^{-1})$, with edges parallel to the contracting and expanding directions of $S$, centered around a point $r \in \QQ^d \cap \TT^d$ of diameter $\ell^{-1}$ where $\ell$ is a natural number.  Since $r$ is rational, it is periodic under $S$.  Let $N$ be its period.  Let $\Cc_P$ denote the line segment through $r$ in $P$ in the $\pm E_S^-$ directions.  Since $S^{-N}$ fixes $r$, it fixes the leaf of the contracting foliation through $r$.  Moreover, an interval $D$ of this leaf containing $r$ will be expanded by $S^{-N}$ and also contain $D$.  In particular, $\Cc_P \subset S^{-N} \Cc_P$.  Since $S^{-N}$ is ergodic, there exists $m \geq 0$ such that the $S^{-mN}$ tube for $P$ will meet $V$ and $V$ completely contains a slice of this tube (because slices under $S^{-1}$ contract).  Therefore, $S^{-mN}\Cc_P$ meets $V$.  Now this iterate of $\Cc_P$ is, locally in $V$, a line segment in the $\pm E_S^-$ direction, which by supposition is different from the $\pm E_T^-$ direction.  Applying $S^{-N}$ more times will expand $\Cc_P$ (which stays on the same leaf) so that it cuts all of $V$ in the $\pm E^-_S$ directions.  This iterate of $\Cc_P$ will intersect $I$ by incidence geometry.  And this intersection point, which belongs to $F_T$, will return to $P$ under enough iterates of $S$.

Thus, for all $r$ and $\ell$, the sets \[\bigcup_{n=0}^\infty S^{-n}P(r, \ell^{-1}) \cap F_T\] are open dense subsets of the Baire space $F_T$.  Consequently, the Baire Category Theorem implies that the set \[A: = \bigcap_{r \in \QQ^d \cap  \TT^d} \bigcap_{\ell = 1}^\infty \bigcup_{n=0}^\infty S^{-n} P(r, \ell^{-1})\] restricted to $F_T$ is a dense $G_\delta$ subset of $F_T$.

Now consider the box $B$ used in the construction of $F_T$.  Take a shrinking sequence of such boxes $\{B_j\}_{j=0}^\infty$ ($\boldsymbol{0}$ is a vertex in all of the boxes) and form the corresponding fractals $F_j := F_T(B_j)$.  Let $F := \cup_{j=0}^\infty F_j$; we note that $F$ is a subset of the set $ND(T)$ of points in the torus which have nondense forward orbits under $T$.  The set $F$ is also winning and hence dense. (The set $F$ is a superset of the winning set of points whose forward orbits miss a neighborhood of $\boldsymbol{0}$).  The set $A$ is dense in all $F_j$.

We now show that $A \cap F$ is dense.  Let $W$ be an open set of $\TT^d$.  Since $F$ is dense, $W$ contains a point $z \in F$.  But, $z \in F_j$ for some $j$.  Since $A$ restricted to $F_j$ is dense in $F_j$, there exists some $a \in A \cap F_j$ such that $a \in W$. Consequently, $A \cap F$ is a dense uncountable subset of the torus.

Finally, it is clear that $A$ is a subset of the set of points in the torus which have dense forward orbits under $S$.  This proves the theorem for $T$ with only positive eigenvalues.

For the general case of $T$ with real eigenvalues, we have that $T^2$ has positive eigenvalues.  By Lemma~\ref{lemmNondenseOrbitisNowhereDense}, the sets $ND(T)$ and $ND(T^2)$ are the same.  Consequently, $A \cap ND(T)$ is dense and uncountable.  This proves the theorem.

\end{proof}

\begin{proof}[Proof of Corollary~\ref{coroForDenNonDen}]  The proof is immediate from the proof of the theorem.\end{proof}

\begin{rema}\label{remNonCommResultStronger}
We may assert a little more (still assuming that $\dim(E_T^- \cap E_{S_n}^-) = 0$ and also adding in the additional assumption that all eigenvalues of $T$ are positive) than the conclusion of Theorem~\ref{thmForDenNonDen} or Corollary~\ref{coroForDenNonDen}.  Fix some open box $B$ as in the construction of the fractal $F_T$.  Then the set of points that miss $B$ under forward orbits of $T$ and that have dense forward orbits under all the $S_n$ is a dense $G_\delta$ subset of $F_T \subset \TT^d$.  Also, note that we may replace the role of $\boldsymbol{0}$ in this proof of the theorem with any rational point on the torus.  This is done by replacing $T$ with the power of $T$ that fixes the rational point and using Lemma~\ref{lemmNondenseOrbitisNowhereDense}.
\end{rema}

config: arxiv


TITLE: Difficult Functions Evaluation Problem
QUESTION [2 upvotes]: I have a question about finding the value of a certain function that I cannot wrap my head around.
The question is:  Given a function $f(x)$ satisfying $$f(x) + 2f\left(\frac{1}{1-x}\right) = x,$$ Then find $f(2).$
So far, I have tried plugging 2 into the original equation to yield $f(2) + 2f(-1) = 2.$
Next, I plugged $\displaystyle \frac{1}{2}$ into the original equation to yield $\displaystyle f\left(\frac{1}{2}\right) + 2f(2) = \frac{1}{2}$.
However, I do not know how to solve this system of equations for $f(2).$
Please let me know of any hints you may have. Many thanks.

REPLY [2 votes]: Given $$f(x)+2f\left(\frac{1}{1-x}\right) = x...................(1)$$
Replace $\displaystyle x\rightarrow \frac{1}{1-x}\;,$ We get
$$f\left(\frac{1}{1-x}\right)+2f\left(\frac{x-1}{x}\right) = \frac{1}{1-x}.......(2)$$
Again replace $\displaystyle x\rightarrow \frac{1}{1-x}\;,$ We get
$$f\left(\frac{1}{1-x}\right)+2f(x) = \frac{x-1}{x}..............(3)$$
Now from $(1)$ and $(2)$
$$f(x)-4f\left(\frac{x-1}{x}\right)=x-\frac{2}{1-x}..........(4)$$
Multiply $(3)$ by $4\;,$ We get
$$4f\left(\frac{1}{1-x}\right)+8f(x) = \frac{4(x-1)}{x}..........(5)$$
Add $(4)$ and $(5)$
$$9f(x) = x-\frac{2}{1-x}+\frac{4(x-1)}{x}$$
So $$9f(2)=2+2+2=6\Rightarrow f(2) = \frac{2}{3}$$


TITLE: Area under the parabola is always 1. But why?
QUESTION [1 upvotes]: I have the following matlab function:
function y=par(x,s)
%parabolic density function with basis = 2s
y=0*x;
ind=-s<x & x< s;
y(-s<x & x< s)=(3/(4*(s^3)))*(s*s-x(ind).*x(ind));

This function implements
$$y_s(x) = \begin{cases}
\frac{3}{4s^3}(s^2-x^2) & \text{if } -s<x<s, \\
0 & \text{otherwise.}
\end{cases}$$
The area under the parabola is always one. I tested it with the following matlab procedure:
s = 0.5:0.1:100.5;
for i=1:length(s)
    q =  quad(@(x)par(x,s(i)),-s(i),s(i));
end

q is always 1 here.
btw.
quad is a matlab function to evaluate the integral. Here it calculates $q = \int_{-s}^s y_s(x)\,dx$.
Can somebody explain to me why it is always one ?

REPLY [1 votes]: It may help to rewrite your integrand as
$$ y_s(x) = \frac{3}{4s^3}(s^2 - x^2) = \frac 1s \frac 34 \left( 1 - \left( \frac xs \right)^2 \right). $$
Now, let's first consider the case $s = 1$, which simplifies the integrand to
$$ y_1(x) = \frac 34 (1 - x^2), $$
which is to be integrated from $-1$ to $1$.  From high school calculus, I'm sure you'll remember that
$$ \int_0^1 x^2 \,dx = \frac 13 $$
which implies that
$$ \int_0^1 (1 - x^2) \,dx = 1 - \frac 13 = \frac 23 $$
and thus, since $(-x)^2 = x^2$,
$$ \int_{-1}^1 (1 - x^2) \,dx = 2 \cdot \frac 23 = \frac 43. $$
Multiply that by $\frac 34$, and you get $1$.
Now, what about $s \ne 1$?  The rewritten integrand has a factor of $\frac 1s$ in front of it, so the height of the parabola is divided by a factor of $s$.  On the other hand, the argument $x$ is also divided by $x$, which effectively multiplies the width of the parabola (and, not coincidentally, the range over which it is integrated) by $s$.  Those two effects cancel out, leaving the area unchanged.
In fact, we can generalize this result: for any function $f$ integrable on $[a,b]$ and any non-zero constant $s$,
$$\begin{aligned}
\int_a^b f(x) dx
&= \frac 1s \int_{sa}^{sb} f\left(\frac xs\right) \,dx \\
&= \int_{sa}^{sb} \frac 1s f\left(\frac xs\right) \,dx .
\end{aligned}$$


TITLE: Why gravitational constant $G$ is used to calculate gravity between objects?
QUESTION [1 upvotes]: I have recently read about newtons law and i dont seem to understand why G is used. what is the logic behind G?

REPLY [6 votes]: Human beings invented the meter, the kilogram, and the newton.   Mother nature did not take those into account when she invented gravity and the magnitude of its force.  So humans had to add $G$ in order to get the correct size of the force when using meters, kilograms, and newtons to measure things.


TITLE: $f(x)=x\cdot \ln(x)$ uniformly continuous in $(0, \infty)$
QUESTION [0 upvotes]: I have to decide if the function $f(x)=x\cdot \ln(x)$  in the interval $(0,+\infty)$ is uniformly cotinuous .

REPLY [0 votes]: $$f(x)=x\cdot \ln(x)$$ is not uniformly continuous on $(0,\infty)$
Note that $$f'(x)=\ln(x)+1$$ grows without bound so you can find two points $x$ and $y$  as close as you wish, where $|f(x)-f(y)|>1$
Let $\epsilon>0$ be given. Choose an $x> e^{2/{\epsilon}}$ and let $y=x+\epsilon /2$
According to the Mean Value Theorem there exist a $c$ between $x$and $y$ such that $$|f(x)-f(y)|=|f'(c)||x-y|= |1+\ln c|(\epsilon /2) =(1+2/{\epsilon})(\epsilon /2)>1$$
That contradicts the uniform continuity.
The trouble is infinity not zero.


text:
\begin{document}
\title{Conjugacy Growth in Polycyclic Groups}
\author{M. Hull}
\maketitle
\begin{abstract}
In this paper, we consider the conjugacy growth function of a group, which counts the number of conjugacy classes which intersect a ball of radius $n$ centered at the identity. We prove that in the case of virtually polycyclic groups, this function is either exponential or polynomially bounded, and is polynomially bounded exactly when the group is virtually nilpotent. The proof is fairly short, and makes use of the fact that any polycyclic group has a subgroup of finite index which can be embedded as a lattice in a Lie group, as well as  exponential radical of Lie groups and Dirichlet's approximation theorem.
\end{abstract}

\section{Introduction}
Given a finitely generated group $G$, along with some fixed finite generating set $S$, let $B(n)$ denote the ball of radius $n$ centered at the identity with respect to the standard word metric induced by $S$, and denote the corresponding norm function by $\vert\cdot\vert_G$.  We denote the {\em conjugacy growth function} of $G$ by $\gamma_G^c(n)$, defined as
\[
\gamma_G^c(n)=\#\text{conjugacy classes } C \text{ such that } C\cap B(n)\neq\emptyset
\]

This function was introduced  to count closed geodesics up to free homotopy on complete Riemannian manifolds \cite{IB}; a more complete history of this function can be found in \cite{GS}. We will also use $\gamma_G(n)$ to denote the standard growth function, that is $\gamma_G(n)=\#B(n)$. We consider these functions up to the following equivalence; $f\preceq g$ if there exist constants $C$ and $D$ such that $f(n)\leq Cg(Dn)$  $\forall n\in\N$, and $f\sim g$ if $f\preceq g$ and $g\preceq f$. We say that $f$ is {\em polynomially bounded} if $f\preceq n^d$ for some $d\in\N$ and $f$ is {\em exponential} if $f\sim 2^n$. Note that unlike the standard growth function, polynomially bounded does not imply polynomial in the case of conjugacy growth functions \cite[Example 2.4]{GS}. In \cite{GS}, it was conjectured that every amenable group of exponential growth has exponential conjugacy growth, and an amenable growth of polynomial conjugacy growth is virtually nilpotent. Our main result gives a proof of this conjecture in the case of virtually polycyclic groups.
\begin{theorem}
Let $P$ be a virtually polycyclic group. Then the conjugacy growth function of $P$ is either polynomially bounded or exponential. Furthermore, the conjugacy growth function of $P$ is polynomially bounded if and only if $P$ is virtually nilpotent.
\end{theorem}
This theorem is an extension of a well known result for the ordinary growth function proved by Wolf in \cite{W}.
The author would like to mention that he is aware of an independent proof  of this result by Emmanuel Breuillard and Yves De Cornulier for the more general case of virtually solvable groups. The author would also like to thank Denis Osin for providing some important ideas in the proof, and Curt Kent for many helpful conversations.

\section{Proofs}
Recall that if $P$ is virtually nilpotent, then $\gamma_P$ is polynomial \cite{B}. Since clearly $\gamma_P^c\leq\gamma_P$, we have that $\gamma_P^c$ is polynomially bounded. Thus, it remains to show that if $P$ is not virtually nilpotent, then $\gamma_P^c$ is exponential. Since for any group $G$, $\gamma_G^c\leq\gamma_G\preceq2^n$, we only need show that $2^n\preceq\gamma_P^c$. We begin with the following straightforward lemma.
\begin{lemma}
Let $H$ be a subgroup of finite index in $G$. Then $\gamma_H^c\preceq\gamma_G^c$.
\end{lemma}
\begin{proof}
Let $S$ and $T$ be finite generating sets for $G$ and $H$ respectively. Let $k=\max\{\vert t\vert_G: t\in T\}$. Let $g_1,...g_m$ be a set of left coset representatives for $H$. Suppose $h_1$, and $h_2$ are conjugate in $G$ but not in $H$. Then for some $g_i$ and $h\in H$, $g_ihh_1h^{-1}g_i^{-1}=h_2$. Thus, $g_i$ combines the conjugacy classes of $h_1$ and $h_2$;  it follows that if $h\in H$ and $\vert h\vert_H\leq n$, then $\vert h\vert_G\leq kn$ and the  conjugacy class of $h$ in $G$ splits into at most $m+1$ conjugacy classes in $H$. Thus, $\gamma_H^c(n)\leq(m+1)\gamma_G^c(kn)$.
\end{proof}
Given a connected Lie group $\Gamma$ generated by a compact, symmetric neighborhood of the identity $\Omega$, we define $\vert g\vert_{\Gamma}=\inf\{n:g\in\Omega^n=\Omega . . . \Omega\}$. Since this is independent (up to quasi-isometry) of the choice of $\Omega$, we refer to this as the {\em canonical norm} on $\Gamma$. This norm induces a metric in the usual way, that is $dist(g,h)=\vert g^{-1}h\vert_{\Gamma}$. Note that this metric is quasi-isometric to the metric induced by any left-invariant Riemannian structure on $\Gamma$. Also, if $\Delta$ is a connected Lie subgroup of $\Gamma$ with canonical norm $\vert\cdot\vert_{\Delta}$, then there exists a universal constant $D$ such that $\vert h\vert_{\Gamma}\leq D\vert h\vert_{\Delta}$ for all $h\in\Delta$.

We define the {\em exponential radical} of $\Gamma$, denoted $Exp(\Gamma)$ to be the set of all $g\in\Gamma$ satisfying
\[
\frac{1}{c}\log(\vert n\vert+1)-\epsilon\leq \vert g^n\vert_{\Gamma}\leq c\log(\vert n\vert+1)+\epsilon
\]

for some constants $c, \epsilon$ and for all $n\in\Z$. Such $g$ are called {\em strictly exponentially distorted}. For convenience, we consider $1$ to be strictly exponentially distorted. A construction of the exponential radical is given in \cite{O}.

We will also make use of Dirichlet's approximation theorem, which can be found in \cite[Theorem 200]{HW}.
\begin{theorem}
For any real numbers $c_1,..., c_k$, and any integer $m>0$, there exist integers $q, p_1,...,p_k$ such that $\vert q\vert\le m^k$ and $\vert qc_i-p_i\vert<\frac{1}{m}$.
\end{theorem}

Now we are ready to prove the main theorem.

\begin{proof}
Let $P$ be a group which is virtually polycyclic but not virtually nilpotent. Then $P$ has a subgroup $P_0$ of finite index which can be embedded as a lattice into a connected, simply-connected, solvable Lie group $\Gamma$ \cite[Theorem 4.28]{R}. Furthermore, since every lattice in a connected, solvable Lie group is uniform \cite[Theorem 3.1]{R}, this embedding is a quasi-isometry; this is a consequence of the Milnor-\v{S}varc Lemma (see, for example, \cite{S}). In view of Lemma 1, it suffices to show that $\gamma_{P_0}^c$ is at least exponential.

Let $\Delta$ be the maximal connected nilpotent normal subgroup of $\Gamma$, and $Z(\Delta)$ the center of $\Delta$. We identify $P_0$ with its image in $\Gamma$. Then $P_0\cap\Delta$ is a lattice in $\Delta$ \cite[Corollary 3.5]{R}, and furthermore $P_0\cap Z(\Delta)$ is a lattice in $Z(\Delta)$ \cite[Proposition 2.17]{R}. Since $Z(\Delta)$ is invariant under any automorphism of $\Delta$, $Z(\Delta)\lhd\Gamma$, and thus $P_0\cap Z(\Delta)\lhd P_0$. Note that since $P$ is not virtually nilpotent, $P_0$ is also not virtually nilpotent.  It is proved in \cite{O} that if $\Gamma$ contains a lattice which is not virtually nilpotent, then $Exp(\Gamma)\neq\{1\}$, and in particular $Exp(\Gamma)\cap Z(\Delta)\neq\{1\}$. Let $v\in Exp(\Gamma)\cap Z(\Delta)$, $v\ne 1$. Since $Z(\Delta)$ is a connected, simply-connected, abelian Lie subgroup \cite[Chap. III, Sec. 9, Proposition 15]{NB}, we have $Z(\Delta)\cong\R^k$ for some $k$. Thus, we can identify  $P_0\cap Z(\Delta)$ with $\Z^k$ inside of $\R^k$.

Fix $n\in\N$ sufficiently large. Applying Dirichlet's approximation theorem to the coordinates of $v$ with $m=2^{\frac{n}{k+1}}$,  we can find $q\in\Z$ and $u\in\Z^k$ such that $\vert\vert qv-u\vert\vert\leq\frac{\sqrt{k}}{m}$ where $\vert\vert\cdot\vert\vert$ denotes the standard Euclidean norm. Now, let $t\in\Z$, $1\leq t\leq m$. Then $\vert\vert tqv-tu\vert\vert\leq\sqrt{k}$. Since  the Euclidean norm on $\R^k$ is quasi-isometric to a canonical norm on $Z(\Delta)$ which is a connected Lie subgroup of $\Gamma$, there is a constant $C_0$ such that $\vert tqv-tu\vert_{\Gamma}\leq C_0\vert\vert tqv-tu\vert\vert\leq C_0\sqrt{k}$. Since $v$ is strictly exponentially distorted in $\Gamma$ and $\vert tq\vert\leq m^{k+1}=2^n$, $\vert tqv\vert_{\Gamma}\leq C_1n$ for some constant $C_1$. Thus, there exists a constant $C_2$ such that $\vert tu\vert_{\Gamma}\leq C_2n$. Finally, $P_0$ is embedded quasi-isometrically in $\Gamma$, so there is a constant $C$ such that $\vert tu\vert_{P_0}\leq Cn$.

Now suppose $w, tw\in\Z^k$ and $gwg^{-1}=tw$ for some $g\in P_0$. Since $\Z^k\lhd P_0$, conjugation by $g$ is an automorphism of $\Z^k$; thus, we can identify this automorphism with a matrix $\phi_g\in GL(k, \Z)$. By the rational root theorem, the only rational eigenvalues of $\phi_g$ are $1$ and $-1$; it follows that $t=\pm 1$. Therefore, $\{tu: 1\leq t\leq m\}$ are all representatives of distinct conjugacy classes in $P_0$. Thus, we have found $2^{\frac{n}{k+1}}$ non-conjugate elements in the ball of radius $Cn$ in $P_0$, so $2^n\preceq\gamma_{P_0}^c$.
\end{proof}


TITLE: Multiplying sparse matrices
QUESTION [5 upvotes]: If I have two sparse matrices, $A$ and $B$. Let's say $A$ has $k$ non-zero entries and $B$ has $j$ non-zero entries. Let's assume all I know is the amount of non-zero entries each matrix has, I don't know where they are or what their value is. The dimensions of the matrices are known and are compatible, so it cannot be assumed that they are always square matrices (although they could be in some cases). So, let's assume A is a MxN matrix and B is an NxP matrix - making AB an MxP matrix.
If I multiply these two sparse matrices together (A*B), what is the maximum amount of non-zero values the product of the matrices could have? I have a feeling it must just be $k+j$ but I can't define it mathematically. I'm not after a completely formal proof.

REPLY [3 votes]: It cannot be k+j as it can be seen in the next counterexample:
We define k = 3 and j = 2 and create the next two matrices, with dimension that agree for the multiplication as you stated:
$A=\left( \begin{array}{ccc}
1 & 0 & \cdots & 0 \\
1 & 0 & \cdots & 0\\
1 & 0 & \cdots & 0 \\
0 & 0 & \cdots & 0 \\
\vdots & \vdots &\ddots &\vdots\\
0 & 0 & 0 & 0\end{array} \right),B=\left( \begin{array}{ccc}
1 & 1 & 0 &\cdots & 0 \\
0 & 0 & 0 &\cdots & 0\\
0 & 0 & 0 &\cdots & 0 \\
\vdots & \vdots& \vdots &\ddots &\vdots\\
0 & 0 & 0 & 0 & 0\end{array} \right)$
Which multiplying them will give us the next matrix:
$AB=\left( \begin{array}{ccc}
1 & 1 & 0 &\cdots & 0 \\
1 & 1 & 0 &\cdots & 0\\
1 & 1 & 0 &\cdots & 0\\
0 & 0 & 0 &\cdots & 0 \\
\vdots & \vdots& \vdots &\ddots &\vdots\\
0 & 0 & 0 & 0 & 0\end{array} \right) $
As it can be seen, the result gives us a matirx with six nonzero entries, which is a higher value than $k+j=5$. It gives the intuition that the maximun amount of nonzero entries will be $k*j$.


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  	\rput(-2,-2){\pscircle[fillcolor=white,fillstyle=solid](0,0){0.75}\rput(0,0){\(\tilde{\texttt{4}}\)}}
  	\rput(6,6){\pscircle[fillcolor=white,fillstyle=solid](0,0){0.75}\rput(0,0){\(\tilde{\texttt{4}}\)}}
  	\rput(6,-2){\pscircle[fillcolor=white,fillstyle=solid](0,0){0.75}\rput(0,0){\(\tilde{\texttt{4}}\)}}
  }
  \rput(4,0){
  	\rput(-10,6){\pscircle[fillcolor=white,fillstyle=solid](0,0){0.75}\rput(0,0){\(\tilde{\texttt{3}}\)}}
  	\rput(-10,-2){\pscircle[fillcolor=white,fillstyle=solid](0,0){0.75}\rput(0,0){\(\tilde{\texttt{3}}\)}}
  	\rput(-2,6){\pscircle[fillcolor=white,fillstyle=solid](0,0){0.75}\rput(0,0){\(\tilde{\texttt{3}}\)}}
  	\rput(-2,-2){\pscircle[fillcolor=white,fillstyle=solid](0,0){0.75}\rput(0,0){\(\tilde{\texttt{3}}\)}}
  	\rput(6,6){\pscircle[fillcolor=white,fillstyle=solid](0,0){0.75}\rput(0,0){\(\tilde{\texttt{3}}\)}}
  	\rput(6,-2){\pscircle[fillcolor=white,fillstyle=solid](0,0){0.75}\rput(0,0){\(\tilde{\texttt{3}}\)}}
  }

  \psline[linestyle=dashed,style=curveblue](-8,-4)(8,4)

  \rput(0,2){
  	\psline[style=curveblue](-6,-2)(-2,-2)
  	\psarc[style=curveblue](-2,0){2}{-90}{0}
  	\psarc[style=curveblue](2,0){2}{90}{180}
  	\psline[style=curveblue](6,2)(2,2)

  	\psarc[style=curveblue](6,0){2}{0}{90}
  	\psarc[style=curveblue](-6,0){2}{180}{-90}
  }
  \rput(0,-2){
  	\psline[style=curveblue](-6,-2)(-2,-2)
  	\psarc[style=curveblue](-2,0){2}{-90}{0}
  	\psarc[style=curveblue](2,0){2}{90}{180}
  	\psline[style=curveblue](6,2)(2,2)

  	\psarc[style=curveblue](6,0){2}{0}{90}
  	\psarc[style=curveblue](-6,0){2}{180}{-90}
  }
  \end{pspicture}
}

{
  \psset{unit=0.25}

  \section{\textbackslash pretzeltangle}

  \begin{pspicture}[showgrid=false](-8,-8)(8,8)

  \psellipticarc[linestyle=dotted,linecolor=gray](0,0)(7.5,1.1){0}{180}

  \rput{-45}(0,0){
    \rput(-4,-4){
      \psset{linewidth=\stringwidth}

      \psecurve[linewidth=\stringwhite,linecolor=white]
      (-5,3)
      (-1,4)(3,3)(4,-1)
      (3,-5)
      \psecurve[linecolor=black]
      (-5,3)
      (-1,4)(3,3)(4,-1)
      (3,-5)
      \psecurve[linewidth=\stringwhite,linecolor=white](5,7)
      (2,6)
      (2,2)
      (6,2)
      (7,5)
      \psecurve[linecolor=black](5,7)
      (2,6)
      (2,2)
      (6,2)
      (7,5)
      \psecurve[linewidth=\stringwhite,linecolor=white]
      (-1,4)(3,3)(4,-1)
      (3,-5)
      \psecurve[linecolor=black]
      (-1,4)(3,3)(4,-1)
      (3,-5)

      \psecurve[linecolor=black]
      (2,2)
      (6,2)
      (7,5)(3,5)(4,9)(3,13)
      \psecurve[linewidth=\stringwhite,linecolor=white]
      (13,3)(9,4)(5,3)(5,7)
      (2,6)
      (2,2)
      \psecurve[linecolor=black]
      (13,3)(9,4)(5,3)(5,7)
      (2,6)
      (2,2)
      \psecurve[linewidth=\stringwhite,linecolor=white]
      (6,2)
      (7,5)(3,5)(4,9)
      \psecurve[linecolor=black]
      (6,2)
      (7,5)(3,5)(4,9)
    }
  }

  \psellipticarc[linewidth=3pt,linecolor=white](0,-0.13)(7.5,1.1){180}{0}

  \psellipticarc[linecolor=gray](0,0)(7.5,1.1){180}{0}

  \rput{-45}(0,0){
    \psarc[style=parawide](0,0){5}{0}{90}
    \psarc[style=parawide](0,0){5}{90}{180}
    \psarc[style=parawide](0,0){5}{180}{270}
    \psarc[style=parawide](0,0){5}{270}{360}
    \psdots(5;0)(5;90)(5;180)(5;270)
  }

  \uput{5.5}[135](0,0){\texttt{1}}
  \uput{5.5}[-135](0,0){\texttt{2}}
  \uput{5.5}[-45](0,0){\texttt{3}}
  \uput{5.5}[45](0,0){\texttt{4}}

  \pscircle(0,0){7.5}

  \end{pspicture}
}

{
  \psset{unit=0.25\baselineskip}

  \section{\textbackslash InlineTrivialTangle}

  \begin{pspicture}[showgrid=false](-2,-2)(2,2)
  \psrotate(0,0){45}{
    \dotcircle
    \pscustom{
      \psline(0,2)(0,1)
      \psarcn(-1,1){1}{0}{-90}
      \psline(-1,0)(-2,0)
    }
    \pscustom{
      \psline(0,-2)(0,-1)
      \psarcn(1,-1){1}{180}{90}
      \psline(1,0)(2,0)
    }
  }
  \end{pspicture}
}

{
  \psset{unit=0.25}

  \section{\textbackslash xI}

  \begin{pspicture}[showgrid=false](-5.5,-5.5)(5.5,5.5)

  \psline*[linecolor=lightgray](4,0)(0,-4)(4,-4)(4,0)

  \mysquare

  \psline[style=curveblue]{->}(4,0)(0,-4)
  \psline[style=curveblue](4,0)(5,1)
  \psline[style=curveblue](0,-4)(-1,-5)

  \uput{1.25}[-45](0,0){\textcolor{blue}{\(\psi\)}}
  \rput(4,0){\psdot(0,0)\uput{0.5}[157.5](0,0){\(x\)}}
  \rput(0,-4){\psdot(0,0)\uput{0.5}[112.5](0,0){\(x'\)}}
  \end{pspicture}

  \section{\textbackslash xII}

  \begin{pspicture}[showgrid=false](-5.5,-5.5)(5.5,5.5)

  \psline*[linecolor=lightgray](4,0)(-4,0)(-4,-4)(4,-4)(4,0)

  \mysquare

  \psline[style=curveblue]{->}(4,0)(-4,0)
  \psline[style=curveblue](4,0)(5,0)
  \psline[style=curveblue](-4,0)(-5,0)

  \uput{0.25}[90](0,0){\textcolor{blue}{\(\psi\)}}
  \rput(4,0){\psdot(0,0)\uput{0.5}[45](0,0){\(x\)}}
  \rput(-4,0){\psdot(0,0)\uput{0.5}[135](0,0){\(x'\)}}
  \end{pspicture}

  \section{\textbackslash xIII}

  \begin{pspicture}[showgrid=false](-5.5,-5.5)(5.5,5.5)

  \psline*[linecolor=lightgray](4,0)(0,4)(-4,4)(-4,-4)(4,-4)(4,0)

  \mysquare

  \psline[style=curveblue]{->}(4,0)(0,4)
  \psline[style=curveblue](4,0)(5,-1)
  \psline[style=curveblue](0,4)(-1,5)

  \uput{3}[45](0,0){\textcolor{blue}{\(\psi\)}}
  \rput(4,0){\psdot(0,0)\uput{0.5}[22.5](0,0){\(x\)}}
  \rput(0,4){\psdot(0,0)\uput{0.5}[67.5](0,0){\(x'\)}}
  \end{pspicture}

  \section{\textbackslash xyI}

  \begin{pspicture}[showgrid=false](-5.5,-5.5)(5.5,5.5)

  \psline*[linecolor=lightgray](4,2)(4,-2)(0,0)(4,2)

  \mysquare

  \psline[style=curvered](4,2)(0,0)
  \psline[style=curvered](4,2)(5,2.5)
  \psline[style=curvered](0,0)(-1,-0.5)
  \psline[style=curveblue](4,-2)(0,0)
  \psline[style=curveblue](4,-2)(5,-2.5)
  \psline[style=curveblue](0,0)(-1,0.5)

  \rput(4,2){\psdot(0,0)\uput{0.5}[-35](0,0){\(\tilde{x}\)}}
  \rput(4,-2){\psdot(0,0)\uput{0.5}[35](0,0){\(\tilde{y}\)}}
  \psdot(0,0)
  \end{pspicture}

  \section{\textbackslash xyII}

  \begin{pspicture}[showgrid=false](-5.5,-5.5)(5.5,5.5)

  \psline*[linecolor=lightgray](4,0)(4,-4)(0,-4)(0,0)(4,0)

  \mysquare

  \psline[style=curvered](4,0)(0,0)
  \psline[style=curvered](4,0)(5,0)
  \psline[style=curvered](0,0)(-1,0)
  \psline[style=curveblue](0,-4)(0,0)
  \psline[style=curveblue](0,-4)(0,-5)
  \psline[style=curveblue](0,0)(0,1)

  \rput(4,0){\psdot(0,0)\uput{0.5}[45](0,0){\(\tilde{x}\)}}
  \rput(0,-4){\psdot(0,0)\uput{0.5}[-135](0,0){\(\tilde{y}\)}}
  \psdot(0,0)
  \end{pspicture}

  \section{\textbackslash xyIII}

  \begin{pspicture}[showgrid=false](-5.5,-5.5)(5.5,5.5)

  \psline*[linecolor=lightgray](4,0)(4,-4)(-4,-4)(-4,0)(0,2)(4,0)

  \mysquare

  \psline[style=curvered](4,0)(0,2)
  \psline[style=curvered](4,0)(5,-0.5)
  \psline[style=curvered](0,2)(-1,2.5)
  \psline[style=curveblue](-4,0)(0,2)
  \psline[style=curveblue](-4,0)(-5,-0.5)
  \psline[style=curveblue](0,2)(1,2.5)

  \rput(4,0){\psdot(0,0)\uput{0.5}[45](0,0){\(\tilde{x}\)}}
  \rput(-4,0){\psdot(0,0)\uput{0.5}[135](0,0){\(\tilde{y}\)}}
  \psdot(0,2)
  \end{pspicture}

  \section{\textbackslash xyIV}

  \begin{pspicture}[showgrid=false](-5.5,-5.5)(5.5,5.5)

  \psline*[linecolor=lightgray](4,-2)(4,-4)(-4,-4)(-4,4)(-2,4)(2,2)(4,-2)

  \mysquare

  \psline[style=curvered](4,-2)(2,2)
  \psline[style=curvered](4,-2)(4.5,-3)
  \psline[style=curvered](2,2)(1.5,3)
  \psline[style=curveblue](-2,4)(2,2)
  \psline[style=curveblue](-2,4)(-3,4.5)
  \psline[style=curveblue](2,2)(3,1.5)

  \rput(4,-2){\psdot(0,0)\uput{0.5}[25](0,0){\(\tilde{x}\)}}
  \rput(-2,4){\psdot(0,0)\uput{0.5}[45](0,0){\(\tilde{y}\)}}
  \psdot(2,2)
  \end{pspicture}

}

{
  \psset{unit=0.15}

  \section{\textbackslash PairingIrrRat}

  \begin{pspicture}[showgrid=false](-28,-24)(28,24)
  \rput{90}(0,0){
	  \pscircle(0,-15){0.5}
	  \pscircle(0,15){0.5}

	  \uput{7pt}[-90](0,15){\rput{-90}(0,0){1}}
	  \uput{7pt}[90](0,-15){\rput{-90}(0,0){4}}
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	    \psbezier(2,0)(2,17)(1.25,17)(0,17)
	    \psbezier(-2,0)(-2,17)(-1.25,17)(0,17)

	    \psbezier(6,0)(6,18)(3,19)(0,19)
	    \psbezier(-6,0)(-6,18)(-3,19)(0,19)

	    \psbezier(18,0)(18,19)(9,25)(0,25)
	    \psbezier(-18,0)(-18,19)(-9,25)(0,25)

	    \psbezier(22,0)(22,20)(11,27)(0,27)
	    \psbezier(-22,0)(-22,20)(-11,27)(0,27)

	    \psbezier(2,0)(2,-8)(-2,-8)(-2,-15)
	    \psbezier(6,0)(6,-8)(2,-8)(2,-15)
	    \psbezier(-2,-15)(-2,-19)(2,-19)(2,-15)

	    \psbezier(-22,0)(-22,-8)(-11,-9)(0,-9)
	    \psbezier(-18,0)(-18,-12)(-11,-25)(0,-25)
	    \psbezier(0,-9)(10,-9)(10,-25)(0,-25)

	    \psset{linestyle=dotted}

	    \psbezier(-10,0)(-10,2)(-10,4)(-9.5,8)
	    \psbezier(-14,0)(-14,2)(-14,4)(-13,9)

	    \psbezier(10,0)(10,2)(10,4)(9.5,8)
	    \psbezier(14,0)(14,2)(14,4)(13,9)

	    \psbezier(18,0)(18,-10)(10,-20.5)(0,-20.5)
	    \psbezier(22,0)(22,-11)(11,-22.5)(0,-22.5)

	    \psbezier(-2,0)(-2,-2)(-3.5,-4)(-3.5,-8)
	    \psbezier(-6,0)(-6,-2)(-6.5,-4)(-6.5,-8)

	    \psbezier(-10,0)(-10,-2)(-10,-4)(-9.5,-8)
	    \psbezier(-14,0)(-14,-2)(-14,-4)(-13,-9)

	    \psbezier(10,0)(10,-2)(10,-4)(9.5,-8)
	    \psbezier(14,0)(14,-2)(14,-4)(13,-9)
	  }

	  \psline[linecolor=blue](-24,0)(24,0)
	  \multido{\i=-22+4}{12}{
	    \psdot(\i,0)
	  }
	  \psdot[dotsize=7pt,dotstyle=|,dotangle=135](-22,0)
	  \rput[r]{-90}(-20,1){\(x_1\)}
	  \psdot[dotsize=7pt,dotstyle=|,dotangle=135](-18,0)
	  \rput[r]{-90}(-16,1){\(x_2\)}
	  \psdot[dotsize=7pt,dotstyle=|,dotangle=135](-6,0)
	  \rput[r]{-90}(-4,1){\(x_{2\alpha-1}\)}
	  \psdot[dotsize=7pt,dotstyle=|,dotangle=135](-2,0)
	  \rput[r]{-90}(0,1){\(x_{2\alpha}\)}

	  \psdot[dotsize=7pt,dotstyle=|,dotangle=135](2,0)
	  \rput[r]{-90}(4,1){\(x_{2\alpha+1}\)}
	  \psdot[dotsize=7pt,dotstyle=|,dotangle=135](6,0)
	  \rput[r]{-90}(8,1){\(x_{2\alpha+2}\)}
	  \psdot[dotsize=7pt,dotstyle=|,dotangle=-45](18,0)
	  \rput[l]{-90}(16,-1){\(x_{4\alpha-1}\)}
	  \psdot[dotsize=7pt,dotstyle=|,dotangle=-45](22,0)
	  \rput[l]{-90}(20,-1){\(x_{4\alpha}\)}
	}
  \rput[t](-17,-6){\(\textcolor{red}{\Special_{\alpha}(0;\TEIV,\TEI)}\)}
  \rput[l](1,12){\(\textcolor{blue}{\Rational(\tfrac{1}{n})}\)}
  \end{pspicture}

  \section{\textbackslash PairingIrrIrrParallel}

  \begin{pspicture}[showgrid=false](-49,-28)(49,28)
  \psset{arrowsize=4pt 1}
  \pscircle(-15,0){0.5}
  \pscircle(15,0){0.5}

  \uput{5pt}[180](15,0){4}
  \uput{5pt}[0](-15,0){1}

  {
    \psset{linecolor=red}

    \psbezier[linestyle=dotted](22,-9)(19,-13)(15,-13)(13,-13)
    \psbezier(13,-13)(5,-13)(-5,-9)(-14,-9)
    \psbezier(-14,-9)(-22,-9)(-22,5)(-14,5)
    \psbezier(-14,5)(0,5)(12,-5)(17,0)
    \psbezier(17,0)(20,3)(0,7)(-14,7)
    \psbezier(-14,7)(-25,7)(-25,-11)(-14,-11)
    \psbezier(-14,-11)(-5,-11)(5,-15)(10,-15)
    \psbezier[linestyle=dotted](10,-15)(15,-15)(19.5,-15)(22,-11.5)

    \psbezier[linestyle=dotted](33,-8)(30,-22)(15,-22)(13,-22)
    \psbezier(13,-22)(5,-22)(-5,-18)(-10,-18)
    \psbezier[linestyle=dotted](-10,-18)(-15,-18)(-19.5,-18)(-24,-13.5)

    \psbezier[linestyle=dotted](35,-8)(32,-24)(15,-24)(10,-24)
    \psbezier(10,-24)(5,-24)(-5,-20)(-13,-20)
    \psbezier[linestyle=dotted](-13,-20)(-18,-20)(-24,-18)(-28,-13)

    \psbezier[arrowsize=4pt 1]{->}(34,-8)(35,0)(38,0)(37,8)

    {\psset{yunit=-1}
    	\psbezier[linestyle=dotted](22,-9)(19,-13)(15,-13)(13,-13)
    	\psbezier(13,-13)(5,-13)(-5,-9)(-14,-9)
    	\psbezier(-14,-9)(-22,-9)(-22,5)(-14,5)
    	\psbezier(-14,5)(0,5)(12,-5)(17,0)
    	\psbezier(17,0)(20,3)(0,7)(-14,7)
    	\psbezier(-14,7)(-25,7)(-25,-11)(-14,-11)
    	\psbezier(-14,-11)(-5,-11)(5,-15)(10,-15)
    	\psbezier[linestyle=dotted](10,-15)(15,-15)(19.5,-15)(22,-11.5)

    	\psbezier[linestyle=dotted](33,-8)(30,-22)(15,-22)(13,-22)
    	\psbezier(13,-22)(5,-22)(-5,-18)(-10,-18)
    	\psbezier[linestyle=dotted](-10,-18)(-15,-18)(-19.5,-18)(-24,-13.5)

    	\psbezier[linestyle=dotted](35,-8)(32,-24)(15,-24)(10,-24)
    	\psbezier(10,-24)(5,-24)(-5,-20)(-13,-20)
    	\psbezier[linestyle=dotted](-13,-20)(-18,-20)(-24,-18)(-28,-13)

    	\psbezier[arrowsize=4pt 1]{->}(34,-8)(35,0)(38,0)(37,8)
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  }
  {
    \psset{xunit=-1,linecolor=blue}

    \psbezier[linestyle=dotted](22,-9)(19,-13)(15,-13)(13,-13)
    \psbezier(13,-13)(5,-13)(-5,-9)(-14,-9)
    \psbezier(-14,-9)(-22,-9)(-22,5)(-14,5)
    \psbezier(-14,5)(0,5)(12,-5)(17,0)
    \psbezier(17,0)(20,3)(0,7)(-14,7)
    \psbezier(-14,7)(-25,7)(-25,-11)(-14,-11)
    \psbezier(-14,-11)(-5,-11)(5,-15)(10,-15)
    \psbezier[linestyle=dotted](10,-15)(15,-15)(19.5,-15)(22,-11.5)

    \psbezier[linestyle=dotted](-24,-13.5)(-19.5,-18)(-15,-18)(-10,-18)
    \psbezier(-10,-18)(-5,-18)(5,-22)(13,-22)
    \psbezier(13,-22)(23,-22)(33,-15)(33,-5)
    \psbezier(33,-5)(33,0)(35,0)(35,5)
    \psbezier(35,5)(35,16)(24,24)(13,24)
    \psbezier(13,24)(5,24)(-5,20)(-13,20)
    \psbezier[linestyle=dotted](-13,-20)(-18,-20)(-24,-18)(-28,-13)

    {\psset{yunit=-1}
    	\psbezier[linestyle=dotted](22,-9)(19,-13)(15,-13)(13,-13)
    	\psbezier(13,-13)(5,-13)(-5,-9)(-14,-9)
    	\psbezier(-14,-9)(-22,-9)(-22,5)(-14,5)
    	\psbezier(-14,5)(0,5)(12,-5)(17,0)
    	\psbezier(17,0)(20,3)(0,7)(-14,7)
    	\psbezier(-14,7)(-25,7)(-25,-11)(-14,-11)
    	\psbezier(-14,-11)(-5,-11)(5,-15)(10,-15)
    	\psbezier[linestyle=dotted](10,-15)(15,-15)(19.5,-15)(22,-11.5)

    	\psbezier[linestyle=dotted](-24,-13.5)(-19.5,-18)(-15,-18)(-10,-18)
    	\psbezier(-10,-18)(-5,-18)(5,-22)(13,-22)
    	\psbezier(13,-22)(23,-22)(33,-15)(33,-5)
    	\psbezier(33,-5)(33,0)(35,0)(35,5)
    	\psbezier(35,5)(35,16)(24,24)(13,24)
    	\psbezier(13,24)(5,24)(-5,20)(-13,20)
    	\psbezier[linestyle=dotted](-13,-20)(-18,-20)(-24,-18)(-28,-13)
    }
	  \psline{->}(12.8,-1.4)(13.6,-1.4)
	  \psline{->}(12.8,1.4)(13.6,1.4)
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  {
  	\setlength{\fboxsep}{1pt}
	  \rput(0,2.2){\psdot(0,0)\uput{3pt}[180](0,0){\colorbox{white}{\(y_{22}^1\)}}}
	  \rput(0,5.95){\psdot(0,0)\uput{3pt}[0](0,0){\colorbox{white}{\(y_{11}^1\)}}}
	  \rput(8.5,4.45){\psdot(0,0)\uput{3pt}[-90](0,0){\(y_{12}^1\)}}
	  \rput(-8.5,4.45){\psdot(0,0)\uput{3pt}[-90](0,0){\(y_{21}^1\)}}

	  \rput(0,11){\psdot(0,0)\uput{3pt}[-90](0,0){\(x_{11}^2\)}}
	  \rput(0,13.05){\psdot(0,0)\uput{3pt}[90](0,0){\(x_{22}^2\)}}
	  \rput(4.8,12){\psdot(0,0)\uput{5pt}[0](0,0){\colorbox{white}{\(x_{12}^2\)}}}
	  \rput(-4.8,12){\psdot(0,0)\uput{5pt}[180](0,0){\colorbox{white}{\(x_{21}^2\)}}}

	  \rput(0,19.9){\psdot(0,0)\uput{2pt}[-90](0,0){\(x_{11}^\beta\)}}
	  \rput(0,22.05){\psdot(0,0)\uput{2pt}[90](0,0){\(x_{22}^\beta\)}}
	  \rput(4.8,21){\psdot(0,0)\uput{5pt}[0](0,0){\colorbox{white}{\(x_{12}^\beta\)}}}
	  \rput(-4.8,21){\psdot(0,0)\uput{5pt}[180](0,0){\colorbox{white}{\(x_{21}^\beta\)}}}
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	{
		\psset{yunit=-1}
		\setlength{\fboxsep}{1pt}
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TITLE: Prove for some $n\in\mathbb{N}$ there are infinitely many primes $p$ , S.T the numbers $p-1,p+1,p+2$ have $n$ different prime factors .
QUESTION [6 upvotes]: Prove for some $n\in\mathbb{N}$ there are infinitely many primes $p$  , S.T the numbers $p-1,p+1,p+2$ have $n$ different prime factors .
Attempt:
by the fundamental theorem of arithmetic's
we know that
$$p-1=p_1^{a_1}...p_k^{a_n}$$
also $$p+1=q_1^{a_1}...q_k^{a_n}$$
$$p+2=w_1^{a_1}...w_k^{a_n}$$
Noticing that all the factorizations consists of $n$ different prime factors.

I tried proving a stronger case with adding $p$ and its factorization to achieve 4 consecutive numbers. However that led me no where.
I tried to solve it with the Chinese remainder theorem using the abstract idea that there is only one unique $x_0$ solution. Not sure how to continue with this idea.

REPLY [4 votes]: As pointed out in the comments:
If the question is asking for exactly $n$ factors, then the claim is false for every $n$, for the following simple reason:
If $p \neq 2,$ then $p \equiv 1 \pmod{2},$ thus $p-1$ and $p+1$ are both divisible by $2$, thus can not have different prime factorizations.
However, the statement is true for at least $n$ different prime factors:
Let $p_1,...,p_n, q_1,...,q_n, r_1,...,r_n$ be different odd prime numbers.
Let $P = p_1p_2 \cdots p_n, Q =q_1q_2\cdots q_n$ and $ R= r_1r_2\cdots r_n.$
(1) By the  Chinese Remainder Theorem,
\begin{align*}
x-1 &\equiv 0 &\pmod{P} \\
x+1 &\equiv 0 &\pmod{Q}\\
x+2 &\equiv 0 &\pmod{R} \\
\end{align*}
has a solution.
(2) Let $x\in \mathbb{N}$ be a solution of the congruence equations. Note that  $x_k = x+ k \cdot PQR $ is also a solution for all $k\in \mathbb{N}$.
(3) Now $(x,PQR) =1$, so by Dirichlet's theorem of primes in arithemetic progression you get the result.


TITLE: Two General Relativity questions
QUESTION [0 upvotes]: Hi When contracting $T^{\mu \nu}$ with $ g_{\mu \nu}$ does one get $T^{\mu \nu}_{\mu \nu} = T$?
is the metric tensor already a sum over its component, so it is effectively a trace of a matrix with its components. e.g $$g^{\mu\nu}=Tr A $$ if A is a matrix with the same components as $g^{\mu\nu}$.

REPLY [1 votes]: Contraction implies a sum over indices, i.e.
$T^{\mu\nu}g_{\mu\nu}=\sum_{\mu=0}^3\sum_{\nu=0}^3T^{\mu\nu}g_{\mu\nu}=T.$
An expression like $T^{\mu\nu}_{\mu\nu}$ makes no sense, since the amount of indices of $T^{\mu\nu}$ does not change.
Furthermore, it does not make a statement about the trace of an object before it is summed.


TITLE: The quotient of $\Bbb R^3$ by a finite group.
QUESTION [3 upvotes]: Let $\Gamma$ be a finite subgroup of $SO(3)$ acting on $\Bbb R^3$. What sort of space do we get by taking the quotient $\Bbb R^3/\Gamma$? Is that a manifold?
The group $\Gamma$ is compact since it is finite, so $\Bbb R^3/\Gamma$ is a manifold whenever the action is free. But what if the action is not free? Do we have examples where the action is not free?

REPLY [3 votes]: The quotient is always a 3-dimensional manifold, actually it is always homeomorphic to $R^3$. One way to see this is to note that the quotient is a cone over $S^2/\Gamma$. Now, you have to check that $S^2/\Gamma$ is homeomorphic to $S^2$, whenever $\Gamma$ is a finite subgroup of $SO(3)$. To see this, first, observe that it is a compact 2-dimensional manifold $S$ (I will leave it to you as an exercise).  Second, it is a general theorem due to Armstrong that if $G$ is a group acting properly discontinuously on a simply connected simplicial complex $X$, then $\pi_1(X/G)$ is isomorphic to the quotient group of $G$ by the normal subgroup generated by all elements having nonempty fixed point sets in $X$. In your case, every element of $\Gamma$ has a fixed point in $S^2$, hence,
$\pi_1(S)= 1$. Therefore, $S\cong S^2$ by the classification of surfaces.
Edit. Just for completeness:  If you allow for finite subgroups of $O(3)$, then the quotients are  not necessarily manifolds. First of all, they could be manifolds with boundary, in the case when the subgroup $\Gamma$ contains reflections. If $\Gamma=\{I, -I\}$, the quotient is the cone over $RP^2$, and, hence, is not even a manifold with boundary.


TITLE: Asymptotic equivalent of expected value
QUESTION [3 upvotes]: Let $X_1,...,X_n$ i.i.d random variables, square integrable, and with $E[X_1]=0$.
I am trying to find a simple asymptotic equivalent when $n \to +\infty$ of :
$E[|X_1+...+X_n|]$
I think the result is $\sqrt{\frac{2n}{\pi}E[X_1^2]}$ (derived easily when $X_i$ are $\mathcal{N}(0,\sigma^2)$ )
Furthermore, if I use the CLT, I get the same result assuming I can pass the limit into the expectation (which is a big assumption).
Thanks for your help !

REPLY [3 votes]: We indeed have $|X_1+\dots+X_n|/\sqrt n\to |N|$ in distribution where $N$ is Gaussian, centered and $\mathbb EN^2=\mathbb E[X_1^2]$.
To conclude the convergence of the expectations, we need uniform integrability. Indeed, if $(Y_n)_{n\geqslant 1} $ is a uniformly integrable sequence of non-negative random variables which converges in distribution to $Y$, then $\mathbb E[Y_n] \to\mathbb E[Y]$. To see this, note that
$$|\mathbb E[Y_n]-\mathbb E[Y]|\leqslant\int_0^R
|\mathbb P\{Y_n\gt t  \}-\mathbb P\{Y\gt t  \}|\mathrm dt +
\int_R^{\infty}\mathbb P\{Y_n\gt t  \}\mathrm dt +\int_R^{\infty}\mathbb P\{Y\gt t  \}\mathrm dt.$$
The second term is controlled by $\mathbb E[Y_n\mathbb 1\{Y_n\gt R\} ]$, hence using uniform integrability we can make this term arbitrarily small and uniformly in $n$. To handle the first term, use pointwise convergence of the cumulative distribution functions.