text:
\section*{Acknowledgments}
We send many thanks to Josep \`Alvarez Montaner, Manuel Blickle,
Daniel J.\- Hern\'andez, Mel Hochster, Gennady Lyubeznik, Felipe P\'erez, and Uli Walther
for useful mathematical conversations and suggestions related to this work.
We also thank  Josep \`Alvarez Montaner
for computing several examples that appear in this survey.
The first author also thanks No\'e B\'arcenas Torres, Fernando Galaz Garc\'ia, and  M\'onica Moreno Rocha for organizing
the ``Taller de Vinculaci\'on Matem\'aticos Mexicanos J\'ovenes en el Mundo'' in Guanajuato, M\'exico, where this project was initiated,
as well as the National Council of Science and Technology of Mexico (CONACyT) for its support through Grant $210916.$
{\edit{ We thank the referee for a careful reading of this paper, and for several comments and suggestions that improved it.
}}


TITLE: Generating random Bitonic Sequences
QUESTION [0 upvotes]: I'm trying to generate random bitonic sequences to test bitonic merging/sorting networks. However I'm unable to come up with a good way to actually generate different types of these sequences with equal likelihood.
A sequence $x_0, x_1, \ldots, x_{n-1}$ is bitonic if,

$\exists $ $k$ such that  $x_0 \leq x_1 \leq \ldots, x_{k-1} \leq x_k \geq x_{k+1} \geq \ldots \geq x_{n-1}$
$\exists$ $k$ such that  $x_0 \geq x_1 \geq \ldots, x_{k-1} \geq x_k \leq x_{k+1} \leq \ldots \leq x_{n-1}$
a circular shift of any of the above types

My best attempt so far is to randomly permute an input, say $0, 1, \ldots, n-1$, and then pass it through a bitonic sorter and stop before the final merge. However this is a little expensive and I'm trying to find better ways to generate them.

REPLY [2 votes]: Let's say you just want to generate a bitonic permutation of $\{1,2,\dots,n\}$. To do this, we should figure out what bitonic permutations look like.
Given any bitonic permutation, we can cyclic-shift it to get a bitonic permutation beginning with $1$. The result looks like $$1 < x_1 < x_2 < \dots < x_k < n > y_1 > y_2 > \dots > y_{n-k-2}$$
for some $k$, where $x_1, x_2, \dots, x_k$ and $y_1, y_2, \dots, y_k$ are the elements of $\{2, \dots, n-1\}$ in some order. It's possible that either the $x$ sequence or the $y$ sequence is actually empty.
Actually, if I tell you the sets $X =\{x_1, \dots, x_k\}$ and $Y = \{y_1, \dots, y_{n-k-2}\}$, there is only one way to compute the bitonic permutation above, because the elements of $X$ must be in ascending order and the elements of $Y$ must be in descending order.
So we can generate a uniformly random bitonic permutation by randomly choosing each element of $\{2, \dots, n\}$  to be either part of $X$ or part of $Y$. Having done that, we put the elements in their now-uniquely-determined places, and get a sequence of the above form. We finish by applying a random cyclic shift.


TITLE: Is it possible for the number created by ordering $1$ to $n$ where $n > 1$ be a palindrome?
QUESTION [5 upvotes]: Is it possible for the number created by the consecutive numbers $1$ to $n$ where $n > 1$ be a palindrome eg. $1234567\ldots n$?

I believe this is a contest problem, but how would one solve this problem without looking up the hints?

REPLY [3 votes]: A solution can be found on page 43 of Andreescu and Andrica, Number Theory: Structures, Examples, and Problems, which page I was able to access on Google Books.


TITLE: Eliminating square roots in trig identities (Integrating square roots)
QUESTION [1 upvotes]: Evaluate integral:
$$\int_{0}^{2\pi} \sqrt{\frac{1-\cos x}{2}} dx$$
I don't even know where to begin, was absent from class the day of this lecture. I know I'm supposed to use some identity related to $$1+\cos2\theta = 2\cos^2\theta$$ but I'm sort of lost. Any help would be appreciated.

REPLY [3 votes]: Hint. You may use
$$
\sin^2 \frac{x}2=\frac{1-\cos x}2
$$ and $\sin u \geq0, \, u \in [0,\pi].$

REPLY [2 votes]: If $1 + \cos 2\theta = 2 \cos^2 \theta$, then $$1 - \cos 2\theta = 2 \sin^2 \theta,$$ because by adding these two equations together, you would get $$2 = 2 \cos^2 \theta + 2 \sin^2 \theta,$$ where upon dividing by $2$ you recover the circular identity $$1 = \cos^2 \theta + \sin^2 \theta.$$  Consequently, $$\frac{1 - \cos 2\theta}{2} = \sin^2 \theta,$$ and taking square roots gives $$\sin \theta = \sqrt{\frac{1 - \cos 2\theta}{2}}.$$  Now let $\theta = x/2$.  What do you get?


TITLE: Projection on convex set
QUESTION [1 upvotes]: Given a vector $y=(y_k) \in \mathbb{R}^n$, we consider the energy function
$$E:\mathbb{R}^n\rightarrow \mathbb{R}^n,\; x \mapsto \sum_{k=1}^{n-1}(x_{k+1} + x_{k−1} − 2x_k)^2,$$
and the set
$$C = \{x \in \mathbb{R}^n \text{ s.t } |x_k − y_k| ≤ q, \forall k=1,\ldots,n\},$$
where $q$ is a fixed positive real number.

Show that $C$ is convex.
Give an explicit formula for $\pi_C(x)$, the euclidean projection of a vector $x \in \mathbb R^n$ onto $C$.

I have tried to solve this. I proved that the energy function is convex. but somehow i am getting stuck on finding the projection of the vector on set C.

REPLY [0 votes]: $C$ is a  product of $n$ compact intervals $[y_1 - q, y_1 + q] \times \ldots \times [y_n - q, y_n + q]$ and is therefore convex (N.B.: every interval is convex and the Cartesian product of a finite number of convex sets is convex)
For computing the projection onto a hyper-cube, see here.
Finally, for an appropriately chosen sparse $n$-by-$n$ matrix $A$ (exercise), your energy can be written as $E(x) \equiv \|Ax\|_2^2$, and so your problem can be written in the form

$$\underset{x \in C}{\text{minimize }} \|Ax\|_2^2,
$$
which is solvable via a fast projected-gradient scheme like FISTA, when not solvable in closed form...
Hope this helps.


text:
{\bf Problem.} Compute $\arcsin \frac{1}{\sqrt{2}}.$  Express your answer in radians.
{\bf Level.} Level 1
{\bf Type.} Precalculus
{\bf Solution.} Since $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}},$ $\arcsin \frac{1}{\sqrt{2}} = \boxed{\frac{\pi}{4}}.$

set_name: MATH


TITLE: Having confusion in this how can I solve
QUESTION [0 upvotes]: An ellipse is drawn with major axis and minor axis of lengths 10 and 8 respectively. Using one focus as the centre,a circle is drawn that is tangent to the ellipse,with no part being outside the ellipse. Then the radius of the circle is:
Ans : Radius = 2 HOW???

REPLY [0 votes]: Since
$$a = \frac{10}{2} = 5, b = \frac{8}{2} = 4$$
Then, to find the focii
$$c^2 = a^2 - b^2 = 25 - 16 = 9$$
So that $c = 3$. Therefore if we want to draw a circle tangent to the ellipse, with no part being outside the ellipse we still have 2 units inside the ellipse (since $a - c = 5 - 3 = 2$). Hence using one focus as the center, a circle is drawn that is tangent to the ellipse, with no part being outside the ellipse.


TITLE: Is the 'appropriate' constructible hierarchy a model of ZFC?
QUESTION [0 upvotes]: Working in with the constructible hierarchy, let's define $Birth$ and $Rank$ of sets as:
$Birth(x)=\alpha \iff min \ \alpha: x \in L_{\alpha+1}$
$Rank(x)=\alpha \iff min \ \alpha : x  \subseteq L_\alpha  $
Let's call sets $appropriate$, if and only if their birth is their rank, for example $\omega$.
Now if we build a new hierarchy $L'$ that requires sets to be not just constructible from below, but also appropriate. Lets denote the stages of that hierarchy by $L'_\alpha$, so each successor stage $L'_{\alpha+1}$ is the set of all appropriate constructible subsets of the prior stage $L'_\alpha$. Of course $L'$ is the union of all appropriate constructible successor and limit stages.
Question 1: Is $L'$ is an inner model of ZFC?
Question 2: is $L'_\omega = L_\omega $?

REPLY [3 votes]: You cannot possibly expect to have $L'$ as a model of $\sf ZF$, since it is clearly smaller than $L$.
To see why, note that your requirement shows that $\mathcal P(\omega)^{L'}\subseteq L_{\omega+1}$. But this set is countable in $L$, and so certainly not the power set of $\omega$ itself.


TITLE: Inner Product Space and Linear Mapping Theorem
QUESTION [0 upvotes]: I'm having some trouble proving the following theorem:
Let $($$X$,$\langle\cdot | \cdot\rangle$$)$ be an inner product space and $f: X \to \mathbb{R}$ a linear mapping. Prove that there exists a unique $\boldsymbol{\tilde{x}} \in X$ such that $f(\boldsymbol{x})=\langle\boldsymbol{x} | \boldsymbol{\tilde{x}}\rangle$ $\forall \boldsymbol{x} \in X$.
Some tips would be great!

REPLY [0 votes]: I can answer for the finite-dimensional case. I'm not that experienced with infinite dimensional inner product spaces. I'm hoping the argument won't change much for the infinite dimensional case.
For the finite dimensional case with $C^n$ as the underlying field, let $e_1,\ldots ,e_n$ be an orthonormal basis for $C^n$.
Thus, we have $x = \langle x, e_1 \rangle e_1 +\ldots + \langle x, e_n \rangle e_n $
Then from the linearity of inner products and $f$, $$
\begin{align}
f(x) &= f\left(\langle x, e_1 \rangle e_1 +\ldots + \langle x, e_n \rangle e_n \right) \\
&= \langle x, e_1 \rangle f(e_1) +\ldots + \langle x, e_1 \rangle f(e_2) \\
&= \langle x, \overline {f(e_1)} e_1 \rangle + \ldots+\langle x, \overline {f(e_n)} e_n \rangle \\
&= \langle x, \overline {f(e_1)} e_1 +\ldots+\overline {f(e_n)} e_n\rangle
\end{align}
$$
Thus, $f(x) = \langle x, u \rangle$, where $u = \overline {f(e_1)} e_1 +\ldots+\overline {f(e_n)} e_n$ .


TITLE: Counting m-subsets of $\{x_1,\cdots,x_n,y_1,\cdots,y_n\}$
QUESTION [2 upvotes]: I am trying to come up with a combinatorial proof for $$\sum \binom n  k \binom{n-k}{ (m-k)/2}2^k =\binom{ 2n} m$$
where the sum is over k such that $m-k$ is even. I would like to count the left side as m-substets of $\{x_1,...,x_n,y_1,...,y_n\}$ but I can't seem to make a connection between this.  I have tried involving logic from the Delannoy number formula without any success. I have tried to determine what a single term for a fixed k of $\binom n k\binom{n-k}{(m-k)/2}2^k$ counts as well but no insight yet.

REPLY [0 votes]: Permit me to contribute an algebraic proof of the case when $m$ is even.
Suppose we seek to verify that
$$\sum_{q=0}^n {n\choose 2q} {n-2q\choose p-q} 2^{2q} =
{2n\choose 2p}.$$
Observe that the sum is
$$\sum_{q=0}^n {n\choose p-q} {n-p+q\choose n-p-q} 4^q.$$
which is
$$\sum_{q=0}^p {n\choose p-q} {n-p+q\choose n-p-q} 4^q
= 4^p \sum_{q=0}^p {n\choose q} {n-q\choose n+q-2p} 4^{-q}.$$
Introduce the Iverson bracket
$$[[0\le q\le p]]
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{z^q}{z^{p+1}} \frac{1}{1-z}\; dz.$$
This  provides  range  control  so  we  may extend $q$ to $n.$
Introduce furthermore
$${n-q\choose n+q-2p}
= \frac{1}{2\pi i}
\int_{|w|=\epsilon}
\frac{(1+w)^{n-q}}{w^{n+q-2p+1}} \; dw.$$
We thus get for the sum
$$\frac{4^p}{2\pi i}
\int_{|w|=\epsilon}
\frac{(1+w)^{n}}{w^{n-2p+1}}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{p+1}} \frac{1}{1-z}
\sum_{q=0}^n {n\choose q} z^q \frac{1}{w^q (1+w)^q} 4^{-q}
\; dz\; dw
\\ = \frac{4^p}{2\pi i}
\int_{|w|=\epsilon}
\frac{(1+w)^{n}}{w^{n-2p+1}}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{p+1}} \frac{1}{1-z}
\left(1+z\frac{1}{4w(1+w)}\right)^n
\; dz\; dw
\\ = \frac{4^{p-n}}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{2n-2p+1}}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{p+1}} \frac{1}{1-z}
\left(4w(1+w)+z\right)^n
\; dz\; dw.$$
We evaluate  the inner integral using  the negative of the  residue of
the pole at $z=1$ which yields
$$\frac{4^{p-n}}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{2n-2p+1}} (4w+4w^2+1)^n \; dw
\\ = \frac{4^{p-n}}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{2n-2p+1}} (2w+1)^{2n} \; dw
\\ = 4^{p-n} {2n\choose 2n-2p} 2^{2n-2p} = {2n\choose 2p}.$$
If we want to be rigorous we need to verify that the contribution from
the residue at infinity of the last integral in $z$ is zero when $n\ge
p$. We get for the residue
$$-\mathrm{Res}_{z=0} \frac{1}{z^2}
z^{p+1} \frac{1}{1-1/z} (4w(1+w)+1/z)^n
\\= -\mathrm{Res}_{z=0}
z^{p} \frac{1}{z-1} (4w(1+w)+1/z)^n
\\= -\mathrm{Res}_{z=0}
\frac{1}{z^{n-p}} \frac{1}{z-1} (4zw(1+w)+1)^n.$$
This is clearly zero when $n=p.$ For $n\gt p$ we obtain
$$\sum_{q=0}^{n-p-1} {n\choose q} 4^q w^q (1+w)^q.$$
This polynomial has degree $2n-2p-2$  but the integral in $w$ extracts
the coefficient on $2n-2p$ for a zero contribution.
Addendum. We can use the same method to prove the companion identity
$$\sum_{q=0}^n {n\choose 2q+1} {n-2q-1\choose p-q} 2^{2q+1} =
{2n\choose 2p+1}.$$
The sum is
$$\sum_{q=0}^n {n\choose p-q} {n-p+q\choose n-p-q-1} 2^{2q+1}$$
which is
$$\sum_{q=0}^p {n\choose p-q} {n-p+q\choose n-p-q-1} 2^{2q+1}
= 2^{2p+1} \sum_{q=0}^p {n\choose q} {n-q\choose n+q-2p-1} 2^{-2q}.$$
Using exactly the same substitution as before we obtain the integral
$$\frac{2^{2p+1-2n}}{2\pi i}
\int_{|w|=\epsilon}
\frac{1}{w^{2n-2p}}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{p+1}} \frac{1}{1-z}
\left(4w(1+w)+z\right)^n
\; dz\; dw.$$
This time we get from the residue at the pole $z=1$
$$2^{2p+1-2n} {2n\choose 2n-2p-1} 2^{2n-2p-1}
= {2n\choose 2p+1}.$$
For  the residue  at infinity  we  are extracting  the coefficient  on
$w^{2n-2p-1}$ but  the inner  term has degree  $2n-2p-2$, again  for a
contribution of zero.

Addendum  II.  We  can  actually  eliminate  the  Iverson  bracket
starting from
$$4^p \sum_{q=0}^p {n\choose q} {n-q\choose n+q-2p} 4^{-q}.$$
and observing that this is
$$4^p \sum_{q=0}^p {n\choose q} {n-q\choose 2p-2q} 4^{-q}.$$
Now introduce
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2p-2q+1}}
(1+z)^{n-q}
\; dz$$
This is zero when $q\gt p$ so  it provides the range control, which we
have now obtained without the Iverson bracket.
We get for the sum
$$4^p \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2p+1}}
(1+z)^{n}
\sum_{q\ge 0} {n\choose q} 4^{-q} \frac{z^{2q}}{(1+z)^q}
\; dz
\\ = 4^p \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2p+1}}
(1+z)^{n}
\left(1+\frac{1}{4}\frac{z^2}{1+z}\right)^n
\; dz
\\ = 4^p \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{2p+1}}
\left(1+z+\frac{1}{4}z^2\right)^n
\; dz$$
Now put $z=2w$ to get
$$4^p \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{2^{2p+1} w^{2p+1}}
\left(1+2w+w^2\right)^n
\; 2 dw
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{w^{2p+1}}
(1+w)^{2n}
\; dw.$$
This is
$${2n\choose 2p}$$ as claimed.


text:
{\bf Problem.} What is the period of $y = \cos \frac{x}{2}$?
{\bf Level.} Level 1
{\bf Type.} Precalculus
{\bf Solution.} The graph of $y=\cos \frac{x}{2}$ passes through one full period as $\frac{x}{2}$ ranges from $0$ to $2\pi,$ which means $x$ ranges from $0$ to $\boxed{4 \pi}.$

The graph of $y=\cos \frac{x}{2}$ is shown below:

[asy]import TrigMacros;

size(400);

real g(real x)
{
	return cos(x/2);
}

draw(graph(g,-3*pi,3*pi,n=700,join=operator ..),red);
trig_axes(-3*pi,3*pi,-2,2,pi/2,1);
layer();
rm_trig_labels(-5, 5, 2);
[/asy]


TITLE: Lower bound for a probability of 1-dimensional Brownian Motion
QUESTION [1 upvotes]: I'm interested to finding a more analytical solution for one of the problems in Richard Bass's book Stochastic Processes.
Let $B_t$ be an one-dimensional Brownian motion. I'm asked to show that there exists some $\gamma$ such that if $t<\gamma$, then $$P(0\leq B_t \leq \delta/2) \geq \frac{1}{4}$$ and $$P(-\delta/2 \leq B_t \leq 0) \geq \frac{1}{4}.$$
For me it's easy to argue that, due to Gaussian distributions being symmetrical and 1-dimensional Brownian motion having the property $B_t\sim N(0,t)$, the problem is equivalent to estimating $P(B_t \leq |\delta/2|)$. By choosing $\gamma=\delta^2/4$, we are estimating the probability that a normal random variable is within one standard deviation from its mean, and from which we can show that such $\gamma$ exists.
However, I'm left with this lingering feeling that the problem could be solved with more finessé by using BM's theoretical properties. Any thoughts?

REPLY [0 votes]: Here joint distribution of Brownian motion is mentioned nowhere, so this fact would hold for any stochastic process $X$ whose one-dimensional distributions are the same as for the Brownian motion. For example, $X$ can be a white noise with $0$ mean and variance linearly growing in time. As a result, no BM theory is really needed here beyond the fact that $X_t\sim \mathcal N(0, t)$. So your solution is the one that fits the problem the best way


TITLE: Probability of draw the same card I have in my hand
QUESTION [0 upvotes]: The deck is composed of cards from 6 to A:

6 7 8 9 10 J Q K A of Diamonds
6 7 8 9 10 J Q K A of Hearts
6 7 8 9 10 J Q K A of Spades
6 7 8 9 10 J Q K A of Clubs

The deck is shuffled and I draw 05 cards.

Say I drew the cards: 9 9 9 J A
A1) If I exchange the J and A and draw more two cards, what's the probability of one of these new 02 cards being a 9?
A2) If I exchange the J and A and draw more two cards, what's the probability of drawing a pair?
B1) If I exchange only the the J and draw one more card, what's the probability of this new card being a J?
B2) If I exchange only the the J and draw one more card, what's the probability of this new card being a 9?
(I presume you see this is a Poker questions, but the root of the questions is about probability, that's why I'm posting it here.)
Thank you very much.

REPLY [0 votes]: I find the most general and useful approach in solving these types of probability questions is to count the # of desired results and divide by the total # of possibilities. That is, $Probability =\dfrac{\# desired}{\# total}$.
Here's a breakdown of the steps for A1.
Assuming you "discard" (without replacement) the J and A, we had 36 cards total, and drew 5, so we have 31 cards left in the deck, and 1 of those is a 9.
We start by counting how many desired results there are. And represent the two cards by "_ _".
$\textbf{Step 1: Count the positions.}$
_ _
We just need a 9 to be one of the 2 cards, so order doesn't matter. We had 2 spots to choose from, so at this point we have Desired = 2
$\textbf{Step 2: Fix a card.}$
9 _
There's only 1 card that is desired for that spot. Desired = $2 \times 1$
$\textbf{Step 3: Fill rest of positions}$
9 A
We can have any other card for the other position, and there are 30 cards left. Here we have picked "A" arbitrarily to represent any card. Desired = $2 \times 1 \times 30 = 60$
$\textbf{Step 4: Count total possibilities.}$
_ _
We have 31 choices for the first card, and a 30 choices for the second card. Thus, Total = $31 \times 30$.
Thus, $\dfrac{\# desired}{\# total} = \dfrac{2\times 1 \times 30}{31\times 30} = \dfrac{2}{31}$
The method for solving the rest is highly similar, so that should get you started.


TITLE: Why do some factorized states have different probability than others in terms of Clebsch-Gordan coefficients?
QUESTION [3 upvotes]: When adding a spin-1 to a spin-1/2, we have a six-dimensional Hilbert space spanned by the factorized states
$$ \left \{ \left | j_1=1,  m_1 \right \rangle \otimes \left |j_2= \frac{1}{2}, m_2 \right \rangle \right \},$$ where $$m_1= \{-1,0,1 \},$$ and $$m_2=\{-\frac{1}{2}, \frac{1}{2} \}.$$
Equivalently, the same Hilbert space is spanned by the coupled states, resulting from adding the angular momenta, $$ \left \{ \left | j,\ m, \ j_1, \ j_2 \right \rangle \right \},$$ where $$j=|j_1-j_2|,\ |j_1-j_2|+1, \ ...\ ,j_1+j_2, $$ and $$m=-j  , \ ..., \ j.$$
Now, we can write the coupled bases in terms of the factorized ones, since they span the same Hilbert space. As a concrete example, we can write
$$  \left | j=\frac{3}{2},\ m=\frac{1}{2}, \ j_1=1, \ j_2=\frac{1}{2} \right \rangle =
\frac{1}{\sqrt{3}} \left( \left | j_1=1,  m_1=1 \right \rangle \otimes \left |j_2= \frac{1}{2}, m_2=- \frac{1}{2} \right \rangle \right) +\frac{2}{\sqrt{3}} \left( \left | j_1=1,  m_1=0 \right \rangle \otimes \left |j_2= \frac{1}{2}, m_2=\frac{1}{2} \right \rangle \right).$$
The coefficients are just the Clebsch-Gordan coefficients, and they (squared) represent the probability of finding the system in the corresponding factorized state when a measurement occurs. My question is: why are they not equally probable? My understanding is that each of the factorized basis is equally as good the others.
I know how to get the coefficient mathematically using the ladder operators, but I guess I am missing the physical reason why the probabilities are distributed this way.

REPLY [1 votes]: This is a good question which does not have a good answer, except to say that "it's in the math".
The least bad analogy that one can find is to realize that the values of $\hat L_\pm$ depend on $M$ and $J$ when the operators act on $\vert JM\rangle$ states.  Thus, starting from the $\vert J,J\rangle$ state, the process of lowering to $\vert J,J-1\rangle$ will not produce states with the same proportions of each $\vert j_1m_1\rangle$ and $\vert j_2m_2\rangle$ since these proportions are given by the matrix elements of $L_-^{(1)}$ and $L_-^{(2)}$ acting on each of $\vert j_1m_1\rangle$ and $\vert j_2m_2\rangle$ respectively.  This is an entirely mathematical argument.
In Section 27 of his book, Wigner provides a geometrical interpretation to
$\vert\langle j_1m_1;j_2m_2\vert JM\rangle\vert^2$ in terms of the area of triangle with sides $\vec j_1,\vec j_2$ and $\vec J$ with projections $m_1,m_2$ and $M$, assuming large values of the quantities so some sort of classical interpretation can be reached.   The model is based on a precession argument, with different vectors $\vec j_1$ and $\vec j_2$ precessing at different speeds on account of their different lengths and projections along $\vec J$.
Wigner's result is really an average and not a terribly good estimate of $\vert\langle j_1m_1;j_2m_2\vert JM\rangle\vert^2$, as you can see from Fig 2b of this paper:

Schulten K, Gordon RG. Semiclassical approximations to 3/-and 6/-coefficients for quantum-mechanical coupling of angular momenta. Journal of Mathematical Physics. 1975 Oct;16(10)

Schulten and Gordon improve on Wigner by converting a recursion relation to a differential equation, which they then solve using WKB techniques.
It's not that hard to illustrate why Wigner's physical argument is off: the values of the CGs actually oscillates significantly with $m_1$ and $m_2$ for fixed $M$, as illustrated in the graph below for $\langle 12,m; 17,-m\vert 24,0\rangle$.  (The value of $m$ is on the horizontal axis, and the value of the CG on the vertical axis). In particular, for $m=0$, the CG is actually $0$ so there is probability $0$ of finding the factorized state $\vert 12,0\rangle\vert 17,0\rangle$ in the state $\vert 24,0\rangle$.

The oscillation clearly visible in the above figure can also be understood by (approximately) converting a recursion relation for CGs to a differential equation, which this time is recognized to be the same as the quantum harmonic oscillator, shifted from the origin by some amount which depends on the CG.  There's details of this trick in

Rowe DJ, de Guise H. The shifted harmonic approximation and asymptotic SU (2) and SU (1, 1) Clebsch–Gordan coefficients. Journal  of Physics A: Mathematical and Theoretical. 2010 Nov 25;43(50):505307.

and there is an arXiv version of this paper if you cannot access the journal version.
This paper just displaces the question you ask to "why should this be related to a harmonic oscillator" but the authors do not provide an answer to this newer question.


TITLE: Placing points equidistantly along an Archimedean spiral from parametric equations
QUESTION [1 upvotes]: I have an Archimedean spiral determined by the parametric equations $x = r t *cos (t)$ and $y = r t *sin (t)$.
I need to place $n$ points equidistantly along the spiral. The exact definition of equidistant doesn't matter too much - it only has to be approximate.
Using just $r$, $t$ and $n$ as parameters, how can I calculate the coordinates of each equidistant point?
Any algorithm will suffice, as I will eventually be translating this to JavaScript. Thank you very much for your time.
I've already looked at some of answers to similar questions - there's this one, where the formula proposed in the question is already far beyond my understanding; and this one for which the answer seems to use a unit spiral rather than an absolute spiral. A more easy-to-understand solution would be very appreciated.
Thanks!

REPLY [4 votes]: I concur with Aretino's answer; I just wanted to dig in to the details a bit, in the hopes it illustrates some of the options and approaches we can utilise here.
Length of the curve $$\begin{cases} x(t) = r t \cos(t)\\y(t) = r t \sin(t)\end{cases}\tag{1}\label{1}$$from $t_0$ to $t_1$ is
$$s( t_0 ,\, t_1 ) = \int_{t_0}^{t_1} ds(t) \, dt$$
where
$$ds(t) = \sqrt{\left( \frac{d \, x(t)}{d \, t} \right)^2 + \left ( \frac{ d \, y(t) }{d \, t} \right)^2} $$
i.e.
$$\begin{array}{rl} s( t_0 ,\, t_1 ) = & \frac{r \, t_1}{2} \sqrt{ t_1^2 + 1 } - \frac{r \, t_0}{2} \sqrt{ t_0^2 + 1 } \; - \\
\; & \frac{r}{2} \log_e\left(t_0 + \sqrt{t_0^2 + 1}\right) -
 \frac{r}{2} \log_e\left(\sqrt{t_1^2 + 1} - t_1\right)
\end{array}\tag{2}\label{2}$$
In practice, we'd like to know $t_1 = f( d, t_0 )$, i.e. the position $t_1$ on the curve that is distance $d$ from $t_0$ along the curve, fulfilling $s( t_0 , t_1 ) = d$. Unfortunately, there are no algebraic solutions for function $f(d, t_0)$.
Numerically, we can roughly approximate $s'(t_0 ,\, t_1) \approx (t_1^2 -t_0^2) \, r/2$. It is a bit off when $t_0$ is very small (near the center of the spiral), but gets better as $t_0$ and/or $t_1$ increase. (Since the spiral is tightest near the center, $t_0 = 0$, I suspect that humans tend to not perceive that error, which means this approximation should be okay for visual purposes.)
If we need a result to within a specific precision, we can use a binary search to find $t_1$ from $s(t_0 ,\, t_1) = d$, numerically. (This has $O(\log N)$ time complexity, and generally requires $N$ iterations (evaluations of $s(t_0 , t_1)$) to get $N$ bits (binary digits) of precision, so it is quite efficient, too.)
For approximately equally spaced (as measured along the curve) points, we can use $$\tau_n = \sqrt{\frac{2 \, d \, n}{r}}, \qquad 0 \le n \in \mathbb{Z}$$Then,
$$s'( \tau_n ,\, \tau_{n+1} ) = d \tag{3}\label{3}$$
which means that
$$\begin{cases}
x_n = x( \tau_n ) = r \sqrt{\frac{2 \, d \, n}{r}} \cos\left(\sqrt{\frac{2 d \, n}{r}} \right) \\
y_n = y( \tau_n ) = r \sqrt{\frac{2 \, d \, n}{r}} \sin\left(\sqrt{\frac{2 d \, n}{r}} \right) \end{cases}$$
gives us points $(x_n , y_n)$ that are spaced roughly $d$ apart, measuring along the curve.
If we substitute $k = 2 d / r$, Aretino's answer directly follows.


TITLE: Lattice embeddings of a polygon
QUESTION [5 upvotes]: Consider the four lattice polygons below.  Each shape is over the coordinates.

If reflected or flipped on the major axes and diagonals, these four polygons remain distinct.  However, this is the same polygon each time. The last three are rotations of the first, with the rotation matrix built from the arctan of different pythagorean triple based fractions: -12/5, 63/16, and 4/3.
One way to determine if a lattice polygon has a different rotational embedding is to apply all of the arctan pythagorean fraction rotation matrices and to see if the points stay on the lattice.  Is there an easier method?

REPLY [0 votes]: One way to check for congruence is to compute the six squared lengths between vertices for each quadrilateral.  If two quadrilateral have the same set of lengths they are congruent.