TITLE: n-way duel probability problem.
QUESTION [2 upvotes]: Let's start with a simple problem:

A and B enter a paintball duel, where each one has a 50% chance of tagging another. The game will last until there is one player left, who will become the winner. A goes first. Calculate the probability of A being the winner.

Let's call the probability of A being the winner is $x$.
When A goes first, A has a $1/2$ chance of tagging B, and the other $1/2$ B will has a $1/2$ chance of tagging A, and so on. Let's look at how symmetric this problem is: when A misses, B will play the game as if B goes first, in other word, B will take the role of A and have an $x$ chance of winning, which makes A has a $1-x$ chance of winning.
Therefore, we have the equation: $x = 1/2 + 1/2(1-x)\Leftrightarrow x=2/3$.
So the probability of A winning is $2/3$, B winning is $1/3$.
Another way to solve this problem is to think about the fact that when A and B both miss, which happens in $1/4$ of the time, where the game will return to its initial state, where A has an $x$ chance of winning,
We have another equation: $x = 1/2 + (1/4)(x)\Leftrightarrow x=2/3$.
Now, let's move on to the harder one:

A, B and C enter a paintball duel, where each one has a 50% chance of tagging another. Each player will aim at the player of next turn.  The game will last until there is one player left, who will become the winner. A goes first, then B, then C. Calculate the probability of A being the winner.

This problem is harder because A does not automatically win when A tags B, because C will take the next turn and have a 50% chance of tagging A. But if we look carefully, we can see another pattern here, C will play against A where C goes first, which means this problem is reduced to a 2-way duel, where we already have the solution.
Let's call the probability of A being the winner is $x$.
A will have a $1/2$ chance of tagging B, and then C will go first against A, which means A only have $1/3$ chance of winning.
A will have a $1/2$ chance of not tagging B, then B will have a $1/2$ chance of tagging C and then A will go first against B, where A will win with $2/3$ chance; or B will have a $1/2$ chance of not tagging C and then C will have a $1/2$ chance of not tagging A, which makes the game return to its initial state, where A will have an $x$ chance of winning.
Therefore, we have the equation:
$x = (1/2)(1/3 + (1/2)(2/3 + (1/2)(x)))\Leftrightarrow x=8/21$
So the probability of A winning is $8/21$
B will have a 50% chance to take the role of A, only when A misses the first shot. So the probability of B winning is $4/21$
The probability of C winning will be $9/21 $
What's important to take note here is that C will not lose regardless of the result of A aiming at B and have the highest chance of winning despite going last.
I tried to use Python to simulate the n-ways duel, record the winner of a large number of matches (desirably 100000) and the only pattern I got from the results is the probability of the first player winning is double the probability of the second player winning, while I can't figure out the pattern of the remaining ones.
My questions are, how can I figure out the probability of each player winning in an n-way duel without calculating every single case. Is there a pattern or recursive manner hidden in this problem? Why does the denominator 21 exist in a 3-way duel?
Thanks for your time reading through all of this.

REPLY [1 votes]: Assume there are $n$ players. As easy to understand the probability $r(n,k)$ that the $k$-th player will be tagged is:
$$
r(n,k)=\frac1{2^n-1}\times\begin{cases}
1,&k=1;\\
2^{n-k+1},&k\ne1.
\end{cases}\tag1
$$
After the $k$-th player leaves, we have the duel of $n-1$ players with shifted indices so that the $k+1$-th player becomes the first, and $k-1$-th player becomes the last $n-1$-th one.
Thus the probability that the $k$-th player survives all rounds can be recursively computed as ($1\le k\le n$):
$$
p(n,k)=\begin{cases}1,&n=1;\\
\sum\limits_{1\le i<k}r(n,i)p(n-1,k-i)+
\sum\limits_{k<i\le n}r(n,i)p(n-1,k-i+n),&n>1.
\end{cases}\tag2
$$
The solutions of the recurrence (2) are rational numbers $\frac{P_{nk}}{Q_n}$. The numerators $P_{nk}$ for small values $n$ and $k$ are given in the following table:
$$
\begin{array}{l|lrrrrr|r}
n&k\to&1&2&3&4&5&Q_n\\
\hline
1&&1&0&0&0&0&1\\
2&&2&1&0&0&0&3\\
3&&8&4&9&0&0&21\\
4&&104&52&86&73&0&315\\
5&&2272&1136&2180&1896&2281&9765
\end{array}
$$
The denominators (last column of the table) have the form:
$$
Q_n=\prod_{i=1}^n (2^i-1).
$$


TITLE: What are eigenfrequencies of guided electromagnetic waves?
QUESTION [0 upvotes]: I am using a commercial software (COMSOL) to find modes in dielectric waveguides. Each mode is associated with an eigenfrequency. In this context, what is an eigenfrequency?

REPLY [1 votes]: Eigenvalues are the solutions to certain types of partial differential equations, like linear, second order PDEs with bound solutions.
The OP indicated (in the comments) the phenomenon under study is described by the Helmholz equation, which is in essence a wave equation.
Grand examples are the Wave Equation, the Diffusion Equation, the Fourier Heat Equation and the Schrodinger Equation.


text:
\begin{document}

\noindent
Harvey Mudd College	\hfill	Senior Thesis \\
Mathematics		\hfill	Fall, 2002

\begin{center}
  \textbf{\large Revised Problem Statement: \\
    An Eigenspace Approach to Fourier Transforms on Iterated Wreath
  Product Groups}\\
\bigskip
Nate Eldredge \\
\bigskip
Faculty Advisor: Prof. Michael Orrison \\
Second Reader: Prof. Shahriar Shahriari
\end{center}

\section{Abstract}

The discrete Fourier transform (DFT) used throughout science and
mathematics is a special case of a more general Fourier transform,
which can extract components of a signal that are unchanged by a
particular set of operations.  We consider a case where the samples
are leaves of a tree, and are operated on by permuting the tree, a set
of operations which form a so-called iterated wreath product group.
Current methods of computing Fourier transforms rely heavily on
detailed knowledge about the group representations involved; we
propose to develop a technique for computing these transforms by
projecting onto eigenspaces of certain operators on the signal space.

\section{Introduction} \label{intro}

\subsection{Representation Theory} \label{reptheory}

For an introduction to the representation theory of groups, see
\cite{dummit}.  An excellent review of the elements of the subject is
contained in \cite{sagan}.

Let $G$ be a finite group.

\begin{definition}
A \term{representation} of $G$ is a complex vector space $V$ with a
linear action by $G$ (i.e. $g \cdot (a \vec{v} + b \vec{w}) = a (g
\cdot \vec{v}) + b (g \cdot \vec{w})$ for vectors $\vec{v}, \vec{w}$,
scalars $a, b$, and $g \in G$).  In this case, $V$ is also called a
\term{$G$-module} or \term{$\C G$-module}.
\end{definition}

Each element $g$ of $G$ corresponds to an invertible linear mapping
from $V$ to itself, so if a basis for $V$ is chosen, this mapping can
be written as a matrix.  So a representation can also be thought of as
a homomorphism $\phi : G \to GL(V)$, where $GL(V)$ is the general
linear group of $V$.  This is the power of representation theory:
doing computations in a group tends to be difficult, but doing
computations with matrices is often comparatively easy, and
representation theory lets us move between the two.

Note that $V$ may have a subspace $W$ which is closed under the action
of $G$ (i.e., for every $g \in G, \vec{w} \in W$, we have $g \cdot
\vec{w} \in W$).  If $V$ has no such (nontrivial) subspaces, $V$ is
called \term{irreducible}.  Central results in representation theory
state that, up to isomorphism, there are only finitely many
irreducible representations of $G$, and every representation can be
written uniquely as a direct sum of irreducibles.  So when we are
given a particular group, we often want to determine its irreducible
representations.

\subsection{Fourier Transforms} \label{fooya}

A particularly interesting representation of $G$ is its \term{regular
representation} $\C G$.  This can be considered as the vector space of
all complex-valued functions (\term{signals}) $f : G \to \C$, on which
$G$ acts via $(g \cdot f)(s) = f(g^{-1}s)$, for $g, s \in G$.  A natural
basis $\mathcal{B}$ for $\C G$ is given by the set of delta functions
\begin{equation*}
  f_g(h) = \begin{cases}
    1, & h = g \\
    0, & h \ne g.
  \end{cases}
\end{equation*}
Think of this as expressing the signal by its values on all the group
elements.

As we mentioned above, $\C G$ can be written as $\C G = W_1 \oplus
\dots \oplus W_k$, where the $W_i$s are irreducible.  In fact, it can
be shown that \emph{all} the irreducibles appear among them, possibly
with several isomorphic copies of each.  (In some cases we will
combine all isomorphic copies of the same irreducible into a single
\term{isotypic} subspace.  Then $\C G$ decomposes uniquely into
isotypics, even though the decomposition into irreducibles may not be
unique.)  If we take a signal $f \in \C G$ and project it onto these
irreducibles, expressing it as $f = f_{W_1} + \dots + f_{W_k}$, we
extract components of $f$ which are in some sense unchanged by the
action of $G$ (each stays within its own subspace).

As another way to compute this, suppose we start with $f$ in the
natural basis.  If we chose a new basis $\mathcal{B'}$ for $\C G$
which contained bases for these subspaces, we could find these
projections simply by picking out appropriate coordinates (such a
basis is called \term{seminormal}).  In fact, the generalized Fourier
transform can be thought of as just the change of basis $\mathcal{B}
\to \mathcal{B'}$, and is closely related to the irreducible
projections described above.

\begin{example}
  In the case where $G = \Z/n\Z = \{0, 1, \dots, n-1\}$, elements of
  $\C G$ just looks like signals on $n$ points.  It can be shown that
  the irreducible subspaces of $\C G$ are all 1-dimensional, and each
  is spanned by a function of the form $f_i(p) = \omega^{ip}$, where
  $p \in G, 1 \le i \le n$, and $\omega$ is a primitive $n$th root of
  unity.  So a signal $f$ can be written as $f = a_1 f_1 + \dots + a_n
  f_n$.  In fact, this is the usual discrete Fourier transform of $f$.
  So our Fourier transform is a generalization of the usual one.
\end{example}

As we said, the Fourier transform can be accomplished by a change of
basis, but in general this requires $O(\ord{G}^2)$ operations to
compute (since $\dim \C G = \ord{G}$).  In 1965, Cooley and Tukey
\cite{cooley-tukey} described an $O(\ord{G} \log \ord{G})$ algorithm
for $\Z/n\Z$.  Such an efficient algorithm is called a \term{fast
Fourier transform (FFT)}, and FFTs for many other groups have since
been developed.  \cite{clausen} goes into detail about all of these,
while \cite{maslen:cooley-tukey} is a briefer overview.

\subsection{Eigenspace Approaches} \label{eigen}

A problem with most standard FFT algorithms is that they require
substantial understanding of the representation theory of the
underlying group in order to apply.  This is unfortunate, since the
whole beauty of representation theory lies in reducing group problems
to linear algebra.  It would be nice to be able to efficiently compute
Fourier transforms using only tools from linear algebra.

Inspiration comes from the symmetric group $S_n$.  There exists a set
of $n-1$ linear operators on $\C S_n$, called the \term{Jucys-Murphy
elements}, which are diagonal when expressed in the seminormal basis.
In fact, every isotypic subspace can be expressed by intersecting
eigenspaces of these elements (a set of elements with this property is
called a \term{separating set}).  Hence we can project onto isotypics
of $\C S_n$ simply by iteratively projecting onto eigenspaces of the
Jucys-Murphy elements.  This isn't quite a Fourier transform, because
that would require projections onto all the \emph{irreducibles}, but
it's nearly as good for many purposes, and furthermore is well-defined
(in general, the Fourier transform depends on a choice of basis).

For this approach to be efficient, we must be able to compute the
eigenspace projections efficiently.  If the operators can be applied
to vectors inexpensively, we have ways to do this (\cite{summer}).
Fortunately, the Jucys-Murphy elements are block-diagonal when
expressed in the standard basis, so they are very sparse, and cheap to
apply.

Analogues of the Jucys-Murphy elements are known for many other
groups.  \cite{orrison:thesis} considers hyperoctahedral groups,
finite general linear groups, finite symplectic groups, and
automorphism groups of graphs.  \cite{ram} generalizes the
construction to arbitrary Weyl groups.  \cite{diaconis:murphy} extends
them to groups with particular subgroup structures, and proves other
results about them.  \cite{russian} reconstructs the representation
theory of the symmetric group in terms of the Jucys-Murphy elements.

\subsection{Wreath Products} \label{wr}

Consider the following two-level tree:
\begin{center}
\Tree [ [.$a_1$ $b_1$ \dots{} $b_n$ ] [.$a_2$ \quad{} \quad{} \quad{} ]
  \dots{} [.$a_m$ \quad{} \quad{} \quad{} ] ]
\end{center}
 where there are $m$ nodes just below the root, each of which has $n$
children.  Suppose we allow some group of permutations $H \subset S_m$
to permute the upper nodes $a_1, \dots, a_m$, carrying their subtrees
with them.  Suppose also that some other permutation group $G \subset
S_n$ is allowed to permute the nodes $b_1, \dots, b_m$ of a particular
subtree at the bottom.  By taking arbitrary compositions of these, we
obtain the \term{wreath product group} $G \wr H$ of permutations (or
automorphisms) of the tree, a subgroup of all the automorphisms of the
tree.  We can think of these as permutations of the leaves which force
the leaves to stay with their ``siblings.''

We will concern ourselves mainly with iterated wreath products.
Suppose we have a regularly branching tree of height $n$ such that
every non-leaf node has $r$ children.  The group of all permutations
of the tree is given by $S_r \wr \dots \wr S_r$ ($n$ times).  If we
don't allow arbitrary permutations, but only allow nodes to be
cyclically permuted with their siblings, we obtain the group $W_{n,r}
= Z_r \wr \dots \wr Z_r$ ($n$ times), where $Z_r$ is the cyclic group
of order $n$.

\cite{kerber1} and \cite{kerber2} describe the representation theory
and other results about wreath products in great detail.
\cite{rockmore:wreath-fft} presents a Cooley-Tukey style FFT algorithm
on wreath products.  \cite{foote:image1} considers the use of iterated
wreath products and Fourier transforms for image and data analysis,
with special emphasis on the case of $Z_{n,4}$ acting on an image.
\cite{orrison:root} explicitly constructs the representation theory of
$W_{n,r}$.

\section{Proposed Research}

\subsection{Eigenspace Approach to Wreath Product FFTs}

We wish to apply the eigenspace Fourier transform techniques of
section \ref{eigen} to the iterated wreath product groups of section
\ref{wr}.  Specifically:

\begin{problemstatementN} \label{sepset}
  Find a separating set for $W_{n,r} = Z_r \wr \dots \wr Z_r$.  This
  set should consist of matrices which are ``nice'' to compute with;
  possibly sparse or otherwise efficient to apply.  Determine bounds
  on the efficiency of the resulting Fourier transform algorithm.
\end{problemstatementN}

As $W_{n,r}$ itself quickly becomes enormous as $n$ increases
($\ord{W_{n,r}} = r^{(r^n-1)/(r-1)}$), it may not be practical to try
to work on signals in the group algebra.  We may instead want to
consider the representation $T$ given by signals on the leaves of the
tree, whose dimension is only $r^n$.  (This approach was considered by
\cite{foote:image1}).

\begin{problemstatementN}
  Find a separating set with similarly nice computational properties
  for $T$.
\end{problemstatementN}

Similar results may present themselves for other wreath product
groups.

\begin{problemstatementN}
  Find separating sets for more general wreath product groups, such as
  $S_n \wr \dots \wr S_n$, or even just $G \wr H$.
\end{problemstatementN}

\subsection{Other Avenues}

There are other results in the representation theory of wreath
products to be pursued.  \cite{orrison:root} describes a construction
of representations for $W_{n,r}$, but the authors have expressed
interest in seeing it explicitly carried out.

\begin{problemstatementN}
  Explicitly compute all the irreducible matrix representations and
  characters for suitable $W_{n,r}$.  Find a general formula for
  these.
\end{problemstatementN}

To this end, we have already written a computer program to list the
trees with which \cite{orrison:root} indexes the representations.
Computational algebra packages such as GAP and Matlab may be helpful
in actually computing the matrices.

Wreath products are a case of a more general class of \term{semidirect
products}.  Perhaps some of the results from above can generalize.

\begin{problemstatementN}
  Generalize the above problems to arbitrary semidirect products of
  groups.
\end{problemstatementN}

\nocite{*}
\bibliographystyle{plain}
\bibliography{bib}
\end{document}

config: arxiv


TITLE: Question on linear combination of random variables
QUESTION [0 upvotes]: I am looking at this question and I am still confused after reading the accepted answer.
This is the question:
A company ships gift baskets that contain apples and pears. The distributions of weight for the apples, the pears, and the baskets are each approximately normal. The mean and standard deviation for each distribution is shown in the table below. The weights of the items are assumed to be independent.
Apples: Mean=4.72 oz, sigma=0.2 oz
Pears: Mean=5.41 oz, sigma=0.18 oz
Basket: Mean=13.52 oz, sigma=1.88 oz
Let the random variable W represent the total weight of 4 apples, 6 pears, and 1 basket. Which of the following is closest to the standard deviation of W?
(A) 1.90 ounces
(B) 1.97 ounces
(C) 2.26 ounces
(D) 3.76 ounces
(E) 3.83 ounces
Specifically, I was looking at these notes:

It's basically the same question but the method that the questioner used in the hyperlinked question is the same method used in these notes I found.

REPLY [1 votes]: $\newcommand{\Var}{\operatorname{Var}}$Let $X$, $Y$, and $Z$ be the random weights of an apple, pear, and basket respectively.
If I understand correctly, you think that the weight of four apples should be $4X$. That is not the case. This like saying "Randomly generate the value of $X$ and then multiply that by four. That is the weight of the four apples". Do you see how that is not what we want? What we actually want is "Randomly generate four weights independently, $X_1$, $X_2$, $X_3$, and $X_4$. The sum of these is the weight of the four apples (combined)."
Notice that if we take the variance of the $4X$, we get
$$\Var(4X) = 4^2\Var(X) = 16(0.2)^2 = 0.64$$
whereas
$$\Var(X_1+\dotsb+X_4) = \Var(X_1)+\dotsb+\Var(X_4) = 4\Var(X_1) = 4(0.2)^2 = 0.16,$$
where the second equality is true by independence, and we notice that $\Var(X_1) = \dotsb = \Var(X_4)$ for the second equality because the $X_i$ follow the same distribution.
All together, the variance of the weight will be
$$\Var(W) = \Var(X_1+\dotsb + X_4 + Y_1+\dotsb+Y_6+ Z_1),$$
which is not $\Var(4X+6Y+Z)$, and I'll leave you to work out the details.


TITLE: Extending Reeb field from contact submanifold to ambient contact manifold
QUESTION [3 upvotes]: Let $(Y,\lambda)$ be a contact manifold, with a codimension-2 contact submanifold $(S,\lambda|_S)$ (this requires $TS\pitchfork\text{Ker}\lambda$). On $Y$ there is a natural vector field, the Reeb field $R$, determined by $d\lambda(R,\cdot)=0$ and $\lambda(R)=1$.
Does there exist a contact form $\lambda'$ on $Y$ such that its Reeb field $R'$ is parallel to $S$, i.e. $R'|_S\subset TS$?
This is true for $\dim Y=3$, as (in the compact scenario) $S$ will be a transverse knot. Using the Moser trick, there is a tubular neighborhood $N(S)\approx S^1\times\mathbb{D}^2$ with $\text{Ker}\lambda|_{N(S)}=\text{Ker}(dz+r^2d\theta)$. Thus $\lambda|_{N(S)}=e^H(dz+r^2d\theta)$ for some function $H:S^1\times\mathbb{D}^2\to\mathbb{R}$. The Reeb vector field on $(S^1\times\mathbb{D}^2,dz+r^2d\theta)$ is $\partial_z$ with Reeb orbit $S=S^1\times\lbrace 0\rbrace$. So $\lambda'=e^{-H}\lambda$ will do the trick.
In general, I am not sure what obstructions would arise by taking the Reeb field on $S$ and trying to extend it to the ambient $Y$.

REPLY [2 votes]: This is a step of the proof of the neighbourhood theorem for contact submanifolds in Geiges's book.
The idea is find a function $f \colon Y \to \mathbb{R}$ so that $f$ is identically $1$ on $S$ and so that
\[
-df + d\lambda(R', \cdot) = 0
\]
on $TY|_S$. If this holds, then it follows that $d( f\lambda)( R', \cdot) = 0$ on $TY|_S$, so $R = R'$ along $S$.
Note that the only condition on $f$ we have here is that $df(v) =0$ for any $v$ tangent to $S$. Note also that $d\lambda(R', v) = 0$ for any $v$ tangent to $S$ as well, since $d\lambda(R', v) = d\lambda_S(R', v)$ in that case. Thus, such a function exists.


TITLE: Group action on the set of irreducible characters are orthogonal.
QUESTION [3 upvotes]: I am studying representation theory and I came across the following problem:
Let $\mathbb{C}[G]$ be the set of $\mathbb{C}$-valued functions on a finite group $G$ and $S(\subset \mathbb{C}[G])$ be the set of irreducible characters of $G$. Let $\mathbb{C}_{\chi}[G]$ be the subspace of $\mathbb{C}[G]$ generated by the set $\{g* \chi \; : \; g \in G \} $ where $g*\chi$ is the action of $G$ on $\chi$ defined as $(g*\chi)(x) = \chi (g^{-1}x) \; \forall x \in G$.
Let us define the scalar product $<|>: \mathbb{C}[G] \times \mathbb{C}[G] \rightarrow \mathbb{C} $ s.t. $<\phi|\varphi> = \frac{1}{|G|} \sum_{g \in G} \phi(g) \overline{\varphi(g)}$.
Questions:

How do we show that if $\chi_1 \neq \chi_2$, then $\mathbb{C}_{\chi_1}[G]$ and $\mathbb{C}_{\chi_2}[G]$ are othogonal w.r.t. above scalar product?

Show that the subspaces $\mathbb{C}_{\chi}[G]$ are $\mathbb{C}[G \times G]$-submodules of $\mathbb{C}[G]$ irreducible by two non-isomporphs. (the action of $G \times G$ on $\mathbb{C}[G]$ is defined as $((g,h)*f)(x) = f (g^{-1}xh) $).

My attempt:

We have the following:
\begin{align*}\label{key}
\mathbb{C}_{\chi_1}[G] &= \text{span} \{ g*  \chi_1 \; : \; g \in G \} \\[2pt]
\mathbb{C}_{\chi_2}[G] &= \text{span} \{ g*  \chi_2 \; : \; g \in G \}
\end{align*}

The dot product $ \langle \phi,\varphi \rangle $can be written as
\begin{align*}
\langle \phi,\varphi \rangle &= \langle \sum_{g \in G} \lambda_g \left( g* \pi \right) , \sum_{h \in G} \lambda_{h}^{'} \left( h* \pi'  \right)  \rangle \\[2pt]
\text{After simplifying, we get} \\
& =  \sum_{g \in G} \sum_{h \in G} \lambda_g \overline{\lambda_{h}^{'}} \langle  \left( g* \pi \right) , \left( h* \pi'  \right)  \rangle
\end{align*}
If $g=h$, then the product will be $0$ but what if $g \neq h$?
(2) I don't even understand the statement.

REPLY [2 votes]: I am going to use the notation $\mathbb{C}^G$ rather than $\mathbb{C}[G]$.
(1) It suffices to check the basis elements $g\ast\chi$ of $\mathbb{C}^G_\chi$ are orthogonal to the basis elements $h\ast\psi$ of $\mathbb{C}_\psi^G$:
$$ \langle g\ast\chi,h\ast\psi\rangle=\frac{1}{|G|}\sum_{x\in G} \chi(g^{-1}x)\overline{\psi(h^{-1}x)}. \tag{$\circ$}$$
Suppose $\chi$ and $\psi$ are characters of $V$ and $W$. Note $\chi(g^{-1}x)\overline{\psi(h^{-1}x)}$ is then the trace of the linear map $T\mapsto (g^{-1}x)T(h^{-1}x)^{-1}$ on $V\otimes W^\ast \cong\hom(W,V)$. Then $(\circ)$ is the trace of
$$ T\mapsto g^{-1}\left(\frac{1}{|G|}\sum_{x\in G} xTx^{-1}\right)h, $$
on $\hom(W,V)$ but by Schur's lemma the expression in parentheses is zero (it is the projection of $T$ onto $\hom_G(W,V)$), so this is the trace of the zero map, which is zero.
(2) Saying subspaces are irreducible "by two non-isomporphs" does not seem like gramamatical English, so perhaps there is a typo or you transcribed/translated incorrectly.
But we can show $\mathbb{C}_\chi^G$ is an irrep of $G\times G$. The key, at least in the following, is in understanding how the Artin-Wedderburn theorem says the group algebra $\mathbb{C}[G]$ acts on irreps $V$. Recall a rep is an algebra homomorphism $\mathbb{C}[G]\to\mathrm{End}(V)$, and AW says $\mathbb{C}[G]\to\bigoplus\mathrm{End}(V)$ is an isomorphism. Keep in mind, also, that $\mathrm{End}(V)\cong V\otimes V^\ast$ as a $G\times G$-rep (where $(g,h)T:=gTh^{-1}$, which is an irrep since $V$ is. (Recall that if $V$ is a $G$-irrep and $W$ is an $H$-irrep then $V\otimes W$ is a $G\times H$-irrep.)
Suppose $\chi$ is the character of an irrep $\rho_V$. Then any $f\in\mathbb{C}^G_\chi$ is expressible as
$$ f_u(x)=\mathrm{tr}\rho(ux) $$
for some $u\in\mathbb{C}[G]$ (the group algebra). There is some redundancy in representing functions this way, though. Indeed the map $\mathbb{C}[G]\to\mathbb{C}^G_\chi$ given by $u\mapsto f_u$ annihilates every summand of $\mathbb{C}[G]$'s AW decomposition except $\mathrm{End}(V)$. Note also $(g,h)\ast f_u=f_{hug^{-1}}$, so in fact we have an onto $G$-morphism from $\mathrm{End}(V)$ to $\mathbb{C}^G_\chi$. (I am sure there is a better way to explain this, but this is how I think about the problem.)


TITLE: Exterior product of a differential form and its derivative
QUESTION [3 upvotes]: Let $\omega$ be a k-form on a smooth manifold $M$ such that there exists $f\in C^{\infty}(M)$ with $f(x)\ne 0$ for all $x\in M$ and $d(f \cdot \omega)=0$.
I need to show that $\omega \wedge d\omega =0$.
I have only been able to show that $\omega \wedge d\omega =\frac{1}{f} \omega \wedge \omega \wedge df$ but I don't know how to conclude.

REPLY [2 votes]: EDIT: this only works when $k$ is odd. (Thanks Ted Shifrin for pointing this out).
Use the Leibniz rule to expand $d(f\omega)$, then wedge with $\omega$. Finally, use the fact that $f$ is nowhere zero and that $d(f\omega)=0$ (In particular that $\omega \wedge d(f\omega) =0$).
EDIT 2: This is false in general: Take $M=(0,\infty)$, $f$ any nonconstant smooth function on $M$, and $\omega=1/f$.


TITLE: $k[X_1,...,X_{m}]\otimes_kk[X_{m+1},...,X_{n+m}]\simeq k[X_1,...,X_{n+m}]$
QUESTION [1 upvotes]: The ring $k[X_1,...,X_m,X_{m+1},...,X_{n+m}]$, together with the obvious inclusions $$k[X_1,...,X_{m}]\hookrightarrow k[X_1,...,X_{n+m}]\hookleftarrow k[X_{m+1},...,X_{n+m}]$$
is the tensor product of the $k$-algebras $k[X_1,...,X_m]$ and $k[X_{m+1},...,X_{n+m}]$ where $k$ is a commutative ring with unity.

First I want to show this by showing the above two inclusions have a universal property but I don't know how to define the unique $k$-algebra homomorphism $\Phi$ in the following diagram:

where $\phi$ and $\psi$ are the given $k$-algebra homomorphism. How can I define it?
Second, the proof of this statement in Milne's textbook is the following: To verify this we only have to check that, for every $k$-algebra $R$, the map $$\text{Hom}(k[X_1,...,X_{m+n}],R)\to \text{Hom}(k[X_1,...],R)\times\text{Hom}(k[X_{m+1},...],R)\tag{1}$$
induced by the inclusion is a bijection. But this map can be identified with the bijection $$R^{n+m}\to R^m\times R^n.\tag{2}$$
I understand that showing $(1)$ is same as showing universal property but how can $(1)$ be identified with $(2)$? Could you explain this to me?
Milne's book and the statement is in p.42 Example 10.4
Edit: Universal property of polynomial ring in Computational commutative algebra 1 by Martin & Lorenzo

REPLY [1 votes]: I believe that Milne's book wants you to verify that $k[X_1, \dots, X_{n+m}]$ satisfies the universal property of the coproduct of $k[X_1, \dots, X_n]$ and $k[X_{n+1}, \dots, X_{n+m}]$ in the category of commutative unital rings (which is the tensor product of rings).
Note that $\Phi$ is the unique ring morphism satisfying $\Phi i = \phi $ and $\Phi j = \psi$. In particular, $\Phi(X_i) = \phi(X_i)$ for $i=1,\dots, n$ and $\Phi(X_i) = \psi(X_i)$ for $i = n+1, \dots, n+m$. By the universal property of the polynomial algebra, there is a unique morphism $\Phi: k[X_1, \dots, X_{n+m}]\to R$ satisfying this and you are done.
This universal property (of the polynomial algebra) asserts that if $R$ is a commutative algebra, then we have a bijection
$$\text{Hom}_{Alg}(k[X_1, \dots, X_n], R) \cong R^n: f \mapsto (f(X_1), \dots, f(X_n))$$  I.e., an algebra morphism $k[X_1, \dots, X_n]\to R$ is uniquely determined by the choice of $n$ points of the ring $R$. This explains why $(1)$ and $(2)$ are equivalent.


TITLE: Proving following equation is false
QUESTION [0 upvotes]: Problem:

If $f(x)$ is continuous and nonnegative on the interval [a, b], then
  $\int_a^b [f(x)]^{1/3}dx=[\int_a^b f(x)dx]^{1/3}$. Prove that this statement is false.

I dont even know where to start.

REPLY [0 votes]: Let $f(x)=x^3$
$$\int _0^1 (f(x))^{1/3} = \int _0^1 xdx = 1/2=0.5$$
$$(\int _0^1 f(x)dx )^{1/3} =( \int _0^1 x^3dx)^{1/3} = (1/4)^{1/3} \approx 0.6299$$
We have two different answer so this works as a counter example to the statement.


TITLE: Hausdorff space, equivalent statement
QUESTION [0 upvotes]: A topological space $Y$ is a Hausdorff space, iff for every(?) two continuous functions $f,g: X\to Y$ holds, that $\{x\in X| f(x)=g(x)\}$ is closed in $X$.

I am asked to proof this statement. I hope I translated it correctly. I am not sure if it is meant that the condition really has to hold for every two continuous functions. Thats why I marked it with (?).
I tried to proof this, but did not succeed to this point and I wonder if this is even true. There is something that seems odd to me.
I know that a for a topological space it is true, that $X$ is Hausdorff, iff $\Delta=\{(x,x)|x\in X\}\subseteq X\times X$ is closed (with regards to the product topology).
And this seems related to the statement above, but much more clear.
In the question above I am not sure what $X$ is and which topology is used.
Does it matter here?
Can you confirm, that this question is correct?
Thanks in advance.

REPLY [2 votes]: The statement is unfortunately worded.  It makes it too implicit that the intention is for the property to be: for every topological space $X$ and for every pair of continuous functions $f, g : X \to Y$, then $\{ x \in X \mid f(x) = g(x) \}$ is closed.

REPLY [1 votes]: It is a correct statement and in fact the "closed diagonal" characterisation ($\Delta$ is usually called the diagonal of $X$) is the means to prove it, and a special case of it as well:
Defining the equality set $E(f,g) := \{x: \in X: f(x) = g(x)\}$ for two continuous function $f: X \to Y$, we see that it equals $(f \nabla g)^{-1}[\Delta]$, where $f\nabla g: X \to Y \times Y$ defined by $(f\nabla g)(x) = (f(x), g(x))$ is continuous iff $f$ and $g$ are.
So $Y$ Hausdorff implies diagonal closed implies all equality sets closed.
The reverse is easy: if all equality sets are closed (for all domains and all pairs of continuous $f,g$), apply it to the 2 projections from $Y \times Y \to Y$ to get that $\Delta = E(\pi_1, \pi_2)$ is closed and we
get the reverse almost for free:
All equality sets closed implies $E(\pi_1,\pi_2)= \Delta$ closed implies $Y$ Hausdorff.


TITLE: Troubles in Dirac's "Principles of quantum mechanics"
QUESTION [6 upvotes]: I'm reading the Dirac's book about QM and I am finding troubles understanding a proof of a theorem (in my edition at page 32), which says that "there are so many eigenkets of $\xi$ that any ket whatever can be expressed as a sum of such eigenkets". I can't understand the expression
$$ \sum_r {{\chi_r (\xi)} \over {\chi_r (c_r)}} - 1 $$
Then Dirac says that if $c_s$ is substituted for $\xi$, every term vanishes except for the one where $r = s$. My reasoning led me to the expression:
$$ \sum_r {{(\xi - c_1)(\xi - c_2) \cdots (\xi - c_{r-1})(\xi - c_{r+1}) \cdots (\xi - c_{n-1})(\xi - c_n)} \over {(c_r - c_1)(c_r - c_2) \cdots (c_r - c_{r-1})(c_r - c_{r+1}) \cdots (c_r - c_{n-1})(c_r - c_n)}} - 1$$
As I understand, there is no $\xi - c_r$ in $\chi_r (\xi)$ as for definition $\chi_r (\xi) = {\phi(\xi) \over (\xi - c_r)}$. For the same motivations $\chi_r (c_r)$ has no term $(c_r - c_r)$. The author says to substitute $c_s$ to $\xi$ in the expression, so I might have to do:
$$ \sum_r {{(c_s - c_1)(c_s - c_2) \cdots (c_s - c_{r-1})(c_s - c_{r+1}) \cdots (c_s - c_{n-1})(c_s - c_n)} \over {(c_r - c_1)(c_r - c_2) \cdots (c_r - c_{r-1})(c_r - c_{r+1}) \cdots (c_r - c_{n-1})(c_r - c_n)}} - 1 $$
So I tried, if $s \neq r$ then:
$$ \sum_r {{(c_s - c_1)(c_s - c_2) \cdots (c_s - c_s) \cdots (c_s - c_{r-1})(c_s - c_{r+1}) \cdots (c_s - c_{n-1})(c_s - c_n)} \over {(c_r - c_1)(c_r - c_2) \cdots (c_r - c_s) \cdots (c_r - c_{r-1})(c_r - c_{r+1}) \cdots (c_r - c_{n-1})(c_r - c_n)}} - 1 $$
hence $c_s - c_s$ in $\chi_r (c_r)$ is equal to zero, then the sum will be $-r \neq 0$. The case $s = r$ gives me some problem:
$$ \sum_r {{(c_r - c_1)(c_r - c_2) \cdots (c_r - c_{r-1})(c_r - c_{r+1}) \cdots (c_r - c_{n-1})(c_r - c_n)} \over {(c_r - c_1)(c_r - c_2) \cdots (c_r - c_r) \cdots (c_r - c_{r-1})(c_r - c_{r+1}) \cdots (c_r - c_{n-1})(c_r - c_n)}} - 1 $$
There is no $c_r - c_r$ in $\chi_r (\xi)$ term as it has been simplified, so $\chi_r (\xi)$ is never $0$. But the denominator contains the term $c_r - c_r = 0$. How can it be possible to have a null term in the denominator?
Could you please explain this sum, and how Dirac intended it?

REPLY [6 votes]: So, to simplify a bit, we are here given an array of numbers $c_1\dots c_n$ (they are not operators!) that are all distinct, and we are asked to consider the smooth, ordinary function
$$
\phi(x) =(x-c_1)(x-c_2)\dots(x-c_n).$$
This has a root of order 1 at every $c_k$ and thus we can also define the function missing the $k$th root,
$$\chi_k(x)=\phi(x)/(x-c_k)=(x-c_1)\dots(x-c_{k-1})(x-c_{k+1})\dots(x-c_n).$$
This function can be filled in at this point where it is technically undefined at $x=c_k$ with some non-zero value $\chi_k(\c_k),$ known to be non-zero because the original numbers were all distinct.
Your question concerns the value of
$\chi_k(c_\ell)/\chi_k(c_k)$. You appear to expand it out to find 1 or so? It is not 1, it is the Kronecker delta,
$$
\chi_k(c_\ell)/\chi_k(c_k)=\delta_{k\ell}=\{1\text{ if } k =\ell\text{ else } 0\}.$$
As Cosmas says, your problem appears to be getting too abstract too quickly and we may prefer to do this once by example. So consider the numbers [-1,2,3] for our starting list. Then we have three parabolas $\chi_1(x)=(x-2)(x-3)$, $\chi_2(x)=(x+1)(x-3)$, $\chi_3(x)=(x+1)(x-2)$. Working through this expression for $k=2$, say, we would find that
$$
\chi_2(c_1)=(c_1+1)(c_1-3)=({-1}+1)({-1}-3)=(0)({-4})=0,\\
\chi_2(c_2)=(c_2+1)(c_2-3)=(2+1)(2-3)=(3)({-1})={-3},\\
\chi_2(c_3)=(c_3+1)(c_3-3)=(3+1)(3-3)=(4)({0})={0}.
$$
So for these ratios we found $\chi_2(c_1)/\chi_2(c_2)=0/({-3})=0$, and $\chi_2(c_2)/\chi_2(c_2)=1$ more trivially (dividing any number by itself yields 1), and
$\chi_2(c_3)/\chi_2(c_2)=0/({-3})=0$ again.
Learning is pain. Or more precisely, an abstraction provides a sort of pain relief, it organizes a chaos for us, and we can only learn abstractions when we have first been exposed to the pain that demands the organizing. Learning the general rule may require doing a lot of examples. So I have exposed you to a bit of pain, writing all of that stuff out, and now we can start to organize that pain.
Once we see the particular examples, we can now see that by construction $c_\ell$ is defined to be a root for any polynomial $\chi_k$ so long as $k\ne \ell,$ because it is a root of $\phi$ but it is not the root that $\chi_k$ eliminated, unless $k=\ell.$ and this idea that we're going to divide by $\chi_k(c_k)$ produces the Kronecker delta by removing the only piece of information that was left!
In turn you have the expression $$
f(x) ={-1} + \sum_{k=1}^n\frac{\chi_k(x)}{\chi_k(c_k)},
$$you can now appreciate the claim that $$
f(c_\ell)
={-1} + \sum_{k=1}^n\delta_{k\ell}={-1} +1=0.$$


TITLE: Does an increasing and bounded differentiable function have bounded derivative?
QUESTION [0 upvotes]: Suppose that $f:\mathbb{R}\to\mathbb{R}$ is differentiable, non-decreasing and bounded. Does this mean that $f'$ is bounded?
I know for a fact that if we remove the non-decreasing assumption then the answer is no.
What happens when $f$ is non-decreasing? Any hint would be great!

REPLY [3 votes]: No: the graph of $f$ could look like a sequence of increasingly steep but increasingly small "stairs" whose heights are a convergent series.


TITLE: Sections of etale morphisms
QUESTION [5 upvotes]: We all know that smooth morphisms have sections etale locally. However, the following similar statement is not obvious for me:
If X->Y->Z, X is etale over Y, Y is finite and surjective over Z, then a section of X->Y exists etale locally on Z, i.e. there exists an etale cover U of Z such that X_U->Y_U has a section. Where _U means pullback on U.
I think it is supposed to be easy.
Can anyone explain this to me? Thanks.

REPLY [2 votes]: Brian Conrad says: "First reduce to the case when Y is also finitely presented over Z. Then you can use limit arguments, so it becomes an easy application of basic facts about strictly henselian local rings and finite algebras over them. It will be instructive for you to think about it some more for yourself in view of these hints.
I should have also added that you must be assuming $X\to Y$ is surjective, or else it clearly isn't true. That condition enters into the argument when doing analysis of the situation with strictly henselian local rings."


TITLE: Choice in a Particular HOD Type Model
QUESTION [12 upvotes]: Let $V \models ZFC$. Let $P$ be a forcing and $G \subseteq P$ be generic over $V$. Let $x \in V[G]$.
Let $M$ be the class of set that are hereditarily definable (in $V[G]$) using as parameters $x$ and elements of the ground models $V$.
I would like to know if $M$ satisfies Axiom of Choice.
I believe this $M$ is the same as $HOD^{V[G]}(V \cup \{x\})$ as the ordinals of $V[G]$ are already in $V$. The one single elements $x$ as parameter should not affect whether $M$ models choice. (The same reason why $HOD[x]$ always satisfy choice.) So I believe this question is equivalent to whether $HOD^{V[G]}(V)$ satisfies the axiom of choice.
The meaning of parenthesis and bracket above refer to Jech's convention. Thanks for any help.

REPLY [0 votes]: You can do the following: for each $A\in HOD^{V[G]}(V)$, let $\beta\in Ord$ be large enough such that $V_\beta$ contains all parameters in the hereditary definition of $A$. Fix a well ordering of $V_\beta$ in $V$, say $\prec$. Now we attempt to define an injection from $A$ to $ORD$ in $V[G]$ that is ordinal definable with parameter $\prec \in V$. Let $\gamma=type(\prec)$ and consider $A\to \omega\times (\gamma)^{<\omega}$ such that $a\in A$ is mapped to the least (some well ordering of $\omega\times (\gamma)^{<\omega}$ and by standard argument it is OD) $(\varphi, \gamma_0, \cdots,\gamma_n)$ such that $a$ is defined by $\varphi(a, \prec(\gamma_0),\cdots, \prec(\gamma_n))$ in $V[G]$ where $\prec(\gamma_i)$ is the $\gamma_i$-th element in the well ordering $\prec$. Clearly this is injective and $OD^{V[G]}(V)$ but as $A\times (\omega\times (\gamma)^{<\omega})\subset HOD^{V[G]}(V)$ we know this injection is actually in $HOD^{V[G]}(V)$. Done.


TITLE: Infinite summation formula of Bessel functions
QUESTION [4 upvotes]: I would like to find a closed form for the following series involving the Bessel function $J_k(z)$:
$$
\sum_{k=0}^{+\infty}\frac{(\mu)_{k}}{k!(\lambda)_{k}}t^k\left(\frac{z}{2}\right)^{k}J_{k+\nu}(z),
$$
where $(a)_{k}$ is the Pochhammer symbol. Actually, I am trying to compute a probability density function involving such summation. Moreover, The sum
$$
\sum_{k=0}^{+\infty}\frac{(\mu)_{k}}{k!(\lambda)_{k}}\left(\frac{z}{2}\right)^{k}J_{k+\nu}(z)
$$
can be expressed in closed form (see the book by Prudnikov 'Integrals and Series', for instance). Might it be useful? Should I use another approach?
Thank you in advance.

REPLY [1 votes]: Concerning the finite sum, I am not sure you can get it (maybe for particular values of $ \nu $). Here is a possible "closed" expression for your sum. Start by writing $ \lambda := \mu + \alpha $ with $ \alpha > 0 $ and
$$ \frac{ (\mu)_k }{ (\lambda)_k } = \frac{ (\mu)_k }{ (\mu + \alpha)_k } = \mathbb{E}(\beta_{\mu, \alpha}^k) $$
where $ \beta_{\mu, \alpha} $ is a random variable Beta distributed (see wikipedia for the moments and the definition ; all I do is to write your quotient of Pochammers with a Beta integral). Up to this expectation, you are left with computing
$$ A := \sum_{k \geq 0} \frac{x^k }{k! } J_{k + \nu}(z) $$
Now, use the following representation (found on wikipedia) :
$$ J_\alpha(z) = \frac{ (z/2)^\alpha }{ \Gamma(\alpha + \frac{1}{2}) } \int_{ [-1, 1] } e^{i sz } (1 - s^2)^{ \alpha - \frac{1}{2} } \frac{ds}{\sqrt{2\pi } }  $$
valid for $ \alpha > \frac{1}{2} $ and $ z \in \mathbb{C} $ (you must hence suppose that your $ \nu $ is greater than $ \frac{1}{2} $ ; if not, you will have to adapt).
Using dominated convergence and exchange of integral and sum, you are left with
\begin{align*}%$
A & = \int_{ [-1, 1] } (z/2)^\nu \sum_{k \geq 0} \frac{ (x  (1 - s^2) z/2 )^k }{k! \Gamma(\nu + k + \frac{1}{2}) }    e^{i sz } (1 - s^2)^{ \nu - \frac{1}{2} } \frac{ds}{\sqrt{2\pi } } \\
   & = \int_{ [-1, 1] } (2 x (s^2 - 1) )^{-\nu}   J_{\nu + 1/2} (x  (s^2 - 1 ) z  )    e^{i sz } (1 - s^2)^{ \nu - \frac{1}{2} } \frac{ds}{\sqrt{2\pi } }
\end{align*}
where I use the definition of the Bessel function as a series (here again, see wikipedia).
You can then use the previous integral representation of the Bessel function to conclude (remember that $ x = \beta_{\mu, \alpha} t z/2 $ and you have to take the expectation in $ \beta_{\mu, \alpha} $). Your representation is then a triple integral involving classical functions.


TITLE: Why does the Strange Quark have Strangeness -1?
QUESTION [5 upvotes]: I have been trying to find an explanation for the strange quarks negative strangeness value, I understand the term strangeness predates the quark model, but I'm unsure if terminology carry over is the reason for the naming convention.
Apparently, it is also convention to give quantum numbers a positive or negative value depending on the charge of the particle. Does the strange quark's negative charge give it a negative strangeness?

REPLY [1 votes]: The sign of additive quantum numbers is set up to a convention. Take for example the case of the electric current: we use to say in electronics that current is a positive flow of charge, that is a negative flow in the opposite direction.
Having this in mind, a subnuclear physics reason can be that the Gell-Mann & Nishijima formula that relates electric charge $Q$, strong isospin $I_3$, barionic number $B$ and strangeness $S$ (the sum of these two and all other quark numbers is called strong hypercharge, $Y$) states:
$$ Q = \frac{B+S}{2} + I_3 $$
Because of this, reminding that the charge of quark strange is $-\frac{1}{3}$ we obtain the correct solution if we put $I_3 = 0$, $B = \frac{1}{3}$ and $S = -1$.


TITLE: analyse the saddle point of $(x+iy)^n$ at (0,0)
QUESTION [1 upvotes]: I've got the following question.
$f(x,y)=\sum_{k=0}^{\lfloor{n/2}\rfloor}{n\choose k}x^{n-2k}(-y^2)^k$
$(f(x,y) \equiv Re((x+iy)^n)$
I need to explain the function near $(0,0)$.
I've calculated that  $f , \nabla$ and the Hessian of $f$ is zero at $(0,0)$.
Now I want to show this point is a saddle point , meaning I want to find $x,y$ such that $f(x,y)>0$ and $x_1,y_1$ such that $f(x_1,y_1)<0$
I've changed the coordinates to polar coordinates $(x=rcos\theta , y=rsin\theta$ ,  $r>0$ , $0\leq\theta<2 \pi)$ and got the next equation:
$f(x,y)=\sum_{k=0}^{\lfloor{n/2}\rfloor}{n\choose k}({rcos\theta})^{n-2k}(-({rsin\theta})^2)^k$ $\Rightarrow$
$f(x,y)=r^n({cos\theta})^{n}\sum_{k=0}^{\lfloor{n/2}\rfloor}{n\choose k}(-1)^{k}({sin\theta})^{2k}({cos\theta})^{-2k}  $
Now I am looking on $0<\theta<\pi/2$ and I cannot understand is this sum converges to a negative number?
Or maybe there is another method which will help me o determine the directions which the function is increasing/decreasing near this saddle point.
Thank you,
Michael

REPLY [1 votes]: Try
$$
\left(\sqrt{x^2+y^2}e^{i\phi}\right)^n = (x^2+y^2)^{n/2}(\cos(n\phi)+i\sin(n\phi))
$$
with $\phi(x,y) = \arctan(y/x)$
Now, depending on the value of $\phi(x,y)$...


TITLE: Closure of the set $\{f\in C[0,1] : f(x) >0 \ \forall x\in[0,1]\}$
QUESTION [0 upvotes]: Let $C[0,1]$ be defined with the max-norm and $A=\{f\in C[0,1] : f(x) >0 \ \forall x\in[0,1]\}$. Show that the closure of $A$ is the set $\{f \in C[0,1] : f(x) \geqslant 0 \ \forall x\in[0,1] \}$.

The closure of $A$ would be $A \cup \partial A$, but I'm not sure how to find the boundary here? It also seems that there would be alternative ways to show this?

REPLY [0 votes]: If you look in the pointwise topology, $A$ can be seen as $(0,+\infty)^{[0,1]}$ and in this (product, essentially) topology the closure is $[0,+\infty)^{[0,1]}$ by standard facts about products (the continuous functions are dense in these product sets so that transfers to $C([0,1])$ as well). The max-topology is finer so we at least have $\overline{A} \subseteq B:=\{f \in C([0,1]]\mid \forall x: f(x) \ge 0\}$. The reverse follows as $B(f, r) \cap A \neq \emptyset$ whenever $f \in B$ and $r>0$; balls taken in the max-metric, so that $B \subseteq \overline{A}$.


TITLE: Find the last three digits of the $2020^{th}$ fibonacci number.
QUESTION [6 upvotes]: Consider the Fibonacci sequence $1, 1, 2, 3, 5, 8, 13, ...$ What are the last three digits (from left to right) of the $2020^{\text{th}}$ term?

The problem is taken from here. I have no idea how to approach this problem. Maybe the general term of the Fibonacci sequence will be of some use?
$$F_n=\frac{(\frac{1+\sqrt{5}}{2})^n-(\frac{1-\sqrt{5}}{2})^n}{\sqrt{5}}$$
Hints are welcomed. Thanks in advance.

REPLY [4 votes]: By Chinese remainder theorem, it suffices to find the result mod $8$ and mod $125$.
I assume that you know how to do it mod $8$ (by finding period).
For mod $125$, we write $$2^nF_n = \frac 1 {\sqrt 5} \left(\sum_{k = 1}^n\binom n k \sqrt 5^k - \sum_{k = 1}^n\binom n k (-\sqrt 5)^k\right) = 2\sum_{k = 0\\2\nmid k}^n\binom n k 5^\frac{k - 1}2.$$ Now we only have to compute the sum for $k = 1, 3, 5$, as all other terms are obviously multiples of $125$.
This, together with a simple calculation of $2^{2020} \mod 125$, will give you the result.


TITLE: Rewrite hyperbola $Ax^2+Bxy+Dx+Ey+F=0$ into standard form
QUESTION [0 upvotes]: The general conic section is given by $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0.$$
In the case where $B\neq0$, $C=0$, we have $$Ax^2+Bxy+Dx+Ey+F\overset{1}=0,$$ with $B^2-4AC=B^2>0$, so that this is a hyperbola.
How do we rewrite $\overset{1}=$ into the standard (or canonical) form?
(Ideally, please just show the algebra and do not make use of other results from conic sections without also proving them.)

REPLY [1 votes]: Basically, the idea is to rotate your axes so that the given curve is a hyperbola in the usual sense. In order to do this, we let
$$x=X\cos \theta+Y\sin\theta \; \; \; \; \; \; \; \; \; \; \; \; \;  y=-X\sin\theta+Y\cos\theta$$
where $(X,Y)$ are the coordinates in the rotated axes.
For ease of notation, I shall define
$$c=\cos\theta \; \; \; \; \; \; \; \; \; \; \; \; \; s = \sin\theta$$
Plugging this in, we get
$$(Ac^2-Bsc)X^2+(2Asc+Bc^2-Bs^2)XY+(As^2+Bsc)Y^2+(Dc-Es)X+(Ds+Ec)Y=-F$$
With a hyperbola, we want to make sure that the $XY$ term disappears. i.e., we want to choose $\theta$ such that the coefficient of $XY$, which is $2Asc+Bc^2-Bs^2$, is $0$.
\begin{align}
\ & 2Asc+Bc^2-Bs^2=0 \\
\ \implies & A\sin(2\theta)+B\cos(2\theta)=0 \\
\ \implies & \theta=-\frac 12 \tan^{-1} \frac BA
\end{align}
With this particular value of $\theta$, you work out what $\sin \theta$ and $\cos \theta$ are, and plug them back into the equation above to get something in the from of
$$PX^2+QY^2+RX+SY=-F$$
which implies that
$$P \biggl(X+\frac{R}{2P} \biggr)^2-Q \biggl(Y+\frac{S}{2Q} \biggr)^2=\frac{R^2}{4P}+\frac{S^2}{4Q}-F$$
where $P,Q$ should be positive.
Then, via a shift in the axes and some rearrangements, we get the required hyperbola.

Probably easier to understand with an example.
Say we have the equation
$$x^2-\sqrt 3xy+\sqrt 3x+y-\frac 73=0$$
Then as before, we let
$$x=X\cos \theta+Y\sin\theta \; \; \; \; \; \; \; \; \; \; \; \; \;  y=-X\sin\theta+Y\cos\theta$$
$$c=\cos\theta \; \; \; \; \; \; \; \; \; \; \; \; \; s = \sin\theta$$
Plug this in to get
$$(c^2+\sqrt 3sc)X^2+(2sc-\sqrt 3c^2+\sqrt 3s^2)XY+(s^2 - \sqrt 3sc)Y^2+(\sqrt 3c-s)X+(\sqrt 3s+c)Y=\frac 73$$
We want to chose $\theta$ such that $2sc-\sqrt 3c^2+\sqrt 3s^2=0$, so
\begin{align}
\ & 2sc-\sqrt 3c^2+\sqrt 3s^2=0 \\
\ \implies & \sin (2\theta)- \sqrt 3 \cos (2\theta)=0 \\
\ \implies & 2\theta = \frac \pi 3\\
\ \implies & \theta = \frac \pi 6
\end{align}
If $\theta = \frac \pi 6$ then $\cos \theta = \frac{\sqrt 3}{2}$ and $\sin\theta = \frac 12$, so
$$\biggl(\frac 34+\frac 34 \biggr)X^2+\biggl(\frac 14 - \frac 34 \biggr)Y^2+\biggl(\frac 32 -\frac 12 \biggr)X+ \biggl(\frac{\sqrt 3}{2}+\frac{\sqrt 3}{2} \biggr)Y=\frac 73$$
$$\frac 32 X^2- \frac 12 Y^2+X+\sqrt 3 Y=\frac 73$$
$$\frac 32 \biggl(X+\frac 13 \biggr)^2- \frac 12 \bigl(Y-\sqrt 3 \bigr)^2=1$$
And hence
$$\frac{\hat X ^2}{(\sqrt {2/3})^2}- \frac{\hat Y ^2}{(\sqrt 2)^2}=1$$
where $\hat X = X+\frac 13$ and $\hat Y = Y-\sqrt 3$


text:
\section{Normal Space is Regular Space}
Tags: Regular Spaces, Normal Spaces

\begin{theorem}
Let $\struct {S, \tau}$ be a [[Definition:Normal Space|normal space]].
Then $\struct {S, \tau}$ is also a [[Definition:Regular Space|regular space]].
\end{theorem}

\begin{proof}
Let $T = \struct {S, \tau}$ be a [[Definition:Normal Space|normal space]].
From [[Normal Space is T3 Space|Normal Space is $T_3$ Space]], we have that $T$ is a [[Definition:T3 Space|$T_3$ space]].
We also have by definition of [[Definition:Normal Space|normal space]] that $T$ is a [[Definition:Fréchet Space (Topology)|$T_1$ (Fréchet) space]].
From [[T1 Space is T0 Space|$T_1$ Space is $T_0$ Space]] we have that $T$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]]
So $T$ is both a [[Definition:T3 Space|$T_3$ space]] and a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
Hence $T$ is a [[Definition:Regular Space|regular space]] by definition.
{{qed}}
\end{proof}

config: wiki


TITLE: Given $2x+y=180$ is true for a triangle, which quantity is bigger out of 4 choices?
QUESTION [1 upvotes]: Ok so I have the following question:

Each angle in ∆QRS has a degree measurement of either x or y and the angles are expressed by the equation 2x + y = 180
Quantity A
The perimeter of $\triangle QRS$
Quantity B
17
Which of the following statements are true?
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
My initial try was that as the angles are 2x+y=180 the triangle is isosceles and  RS must be $7$, thus the perimeter of $\triangle QRS$ is $7+7+4 = 18 > 17$ and the correct answer is A.
Apparently, the correct answer is D, as someone noted that QS can also be $4$, so $7+4+4=15 < 17$, so it can't be determined. Can anyone explain why is that? Is it because QS is not fixed, so it could also take $4$ as value?

Also isn't the graph misleading in order to make you think only of the first choice?   Thank you!

REPLY [1 votes]: Your reasonement is correct. In fact, without further informations we can't determine the angle adiacent to the base. So there are two different possibilities for the perimetre:
$$P=7+7+4=18>17$$
Or:
$$4+4+7=15<17$$
So the correct answer is $D$ because we don't have much more informations.


TITLE: Find the poles/residues of $f(z)=\frac{\sin(z)}{z^4}$
QUESTION [3 upvotes]: I'm trying to calculate the residua of the following complex function but am encountering problems trying to determine its poles:
$$f(z)=\frac{\sin(z)}{z^4}$$
Expanding the denominator shows that we have a pole at $z=0$ of order four.
We also check the numerator for any removable singularities, and we see that:
$$\sin(z)=0, \forall z=n\pi, n \in \mathbb N$$
Now since the pole we have is $z=0$, we have a removable singularity for $z=0$ in the numerator with $\sin(0)=0$, where, in the above, $n=0$.
Does this mean that, with the one removable singularity, the order of the denominator is reduced by one, to give $z^3$ in the denominator?
If so, what formula or means should I use to calculate the residue at z=0? Since we have a removable singularity, its residue should be zero. I found online that instead of using the formula for a pole of order $m$ People evaluated the limit of f(z) directly i.e:
$$\lim_{z\to0}\frac {\sin(z)}{z^4}$$
Why does this hold instead of having to use the formula for evaluating a pole of order $m$?

REPLY [5 votes]: You have a pole of order $3$. The intuitive idea is that one $z$ is "absorbed" by the $\sin z$ , the other $3$ are untouched.
You do not have a removable singularity (it is a pole!). You would have if there was a $z$ but there is not, so...
How do you make all of this formal? Just write down the Laurent series in $0$..
So you have
$$\sin z = z - \frac{z^3}6 + \frac{z^5}{5!} - \dots $$
So the Laurent expansion of $\frac{\sin z}{z^4} $ is
$$\frac{\sin z}{z^4} = \frac 1{z^3} - \frac1{6z} + \frac{z}{5!} - \dots $$
So the pole is of order $3$, and the residue is $-\frac 16$


TITLE: Nested convex optimization
QUESTION [1 upvotes]: Suppose I have a convex optimization problem of the form $$\min_x f(x) ~~s.t.\\x\in X$$.  Say that $f(x)$ and its (sub)gradient are not given in a closed form, but are determined by solving a convex sub-problem of the form $$\min_y g(x,y)~~s.t.\\y\in Y(x)$$ where $Y(x)$ is a convex feasible region that depends on $x$.  Just to drive the point home, let's say that $g$ and its (sub)gradient are also not given in a closed form, but are determined by solving the convex sub-problem $$\min_z h(x,y,z) ~~s.t.\\z\in Z(x,y)$$, and this last sub-problem does have closed form expressions for the objective function and (sub) gradient. So, our whole problem effectively has three "levels".  What can I say about the complexity of my original problem?  How does the potential for numerical error propagate back up through the inner problems?

REPLY [1 votes]: This is known as bilevel optimization.  My impression is that it is considerably harder than usual convex optimization, even numerically.


TITLE: Can I use energy conservation to find out the new amplitude of SHM?
QUESTION [1 upvotes]: A mass $m_1$ connected to a horizontal spring performs S.H.M. with amplitude $A$. While mass $m_1$ is passing through its mean position another mass $m_2$ is placed on it so that both the masses move together with amplitude $A_1$. The ratio $A_1/A$ is? ($m_2<m_1$).

I know that I can use conservation of momentum to solve this question. I just wanted to know why I can't use conservation of energy to do the same.
I want to equate the energy in SHM for the first SHM at mean position and the energy in SHM for the second SHM with two blocks at the mean position.

REPLY [2 votes]: The question is ambiguous, as the answer depends on how $m_2$ is placed on $m_1$. If it is moving with $m_1$, then you're adding kinetic energy to the system:
$$ \frac 1 2 m_1 v^2 \rightarrow \frac 1 2  (m_1+m_2) v^2$$
and if it's stationary, then you are taking energy out (but conserving momentum):
$$ \frac{p^2}{2m_1} \rightarrow \frac{p^2}{2(m_1+m_2)}$$
The question doesn't specify other than to say it is done such that the amplitude is $A_1$.
Since potential energy goes as $\frac 1 2 kA^2$, and the question was multiple choice, you should be able to pick the correct answer without doing any serious calculation.


TITLE: On Cauchy's Theorem (Proof)
QUESTION [1 upvotes]: I am trying to understand the Cauchy Theorem from this resource. I don't understand why (in 1st paragraph of the proof 1), "If p does not divide |H|, then it divides the order [G:H]". Would someone please explain it?
Thanks a lot

REPLY [0 votes]: Take a look at this theorem;
https://en.wikipedia.org/wiki/Lagrange%27s_theorem_(group_theory)


TITLE: Metropolis Hasting Algorithm. Whats going wrong?
QUESTION [1 upvotes]: Hi I wrote a Langevin Metropolis-Hasting Algorithm on python to evaluate
$$ E[\|X\|^2], $$
where $X\sim \pi$, with
$$\pi(x)= \frac{1}{2\sqrt{2d}}e^{\frac{1}{2}\|x-a\|^2-\frac{1}{2}\|x+a\|^2},~~a_i=\frac{1}{\sqrt{2d}}~\forall i=1,\dots d$$
a Gaussian mixture. I.e I built a Markov Chain for some fixed $h>0$ and $\xi_i\sim N(0,I)$
$$ \tilde{x}_{i+1}:=x_i+ h \nabla \log \pi(x_i) +\sqrt{2 h} \xi_i$$
and setting $x_{i+1}$ according to
\begin{align}
&\mathbb{P}( x_{i+1}=\tilde{x}_{i+1})=\alpha(\tilde{x}_{i+1},x_i)
\\
& \mathbb{P}( x_{i+1}=x_i)=1-\alpha(\tilde{x}_{i+1},x_i).
\end{align}
where the acceptance probability is
$$  \alpha(x,y)=\min(1,\frac{\pi(\tilde{x}_{i+1})\exp\Big( -\frac{1}{4h}\| x_i-\tilde{x}_{i+1}-h\nabla \log \pi(\tilde{x}_{i+1}) \|^2 \Big)}{\pi(x_i)\exp\Big( -\frac{1}{4h}\| \tilde{x}_{i+1}-x_i-h\nabla \log \pi(x_i) \|^2 \Big)}).  $$
I calculate $\nabla \log \pi(x)=-2x$.
$\textbf{My question :}$ Does the above look ok ? when I run my code sometimes it works perfectly other times it gets stuck why could this be?

REPLY [1 votes]: Your proposals and acceptance probability seem correct to me.
The main issue I see here, is that you mention "some" $h>0$. Such parameter usually need to be tuned to make sure that your proposals are efficient. Have you done it ?
Cheers!


TITLE: probability that joint events are infinitely often?
QUESTION [2 upvotes]: If set sequences $A_1, A_2, \cdots$, $B_1, B_2, \cdots$ consist of measurable sets in the sigma-algebra $Q$. Suppose $P(A_k \text{ infinitely often }) = 1$, and $P(B_k^c \text{ infinitely often }) = 0$. What is the probability that infinitely many of joint events $A_k \cap B_k$ will occur?
I can see $P(\cap_{n=1}^{\infty}\cup_{k=n}^{\infty}A_k) = 1$, $\displaystyle P(\cup_{n=1}^{\infty}\cap_{k=n}^{\infty}B_k) = 1$ by De Morgan Law, and it's asking $P(\cap_{n=1}^\infty\cup_{k=n}^\infty (A_k \cap B_k))$, however no idea how to proceed. Can anyone help?

REPLY [2 votes]: It is $1$.  If $A_k$ occurs infinitely often and $B_k^{c}$ occurs at most finitely many times $A_k \cap B_k$ occurs infinitely often.
Note that $P(B_k^{c} \,  \text  {infinitely often })=0$ is same as the statement that all but finitely many $B_k$'s occur with probability $1$.
Let $A$ be the event that $A_k$ occurs infinitely often and $B$ the event that $B_k^{c}$ occurs only finitely many times. It follows from the hypothesis that $P(A)=1$ and $P(B)=1$. Hence $P(A\cap B)=1$. But if $A\cap B$ occurs then $A_k$ occurs infinitely often and $B_k^{c}$ occurs only finitely many times so $A_k\cap B_k$ must occur infinitely often.


TITLE: Convergence of $f_n(x)=nx$
QUESTION [0 upvotes]: Consider the sequence of functions $$f_n(x)=nx$$
As $n$ gets larger, so does the gradient of the line passing through the origin. Graphically, as $n$ goes to infinity, this will converge to the vertical straight line $x=0$, which is not defined since the same value is mapped to (infinitely) many values.
Does this sequence converge to this vertical line? If so, how would one prove it, since you cannot just right this function limit as "$f(x)=...$"?

REPLY [1 votes]: We cannot speak of convergence of functions if we don't know what the domain of the functions is. In your case:

If the domain of the functions is $\{0\}$, then the functions converge to the zero function.
On any set containing a nonzero element, the series of functions does not converge.