TITLE: If a ball spinning on a rod hits another ball, what is conserved linear or angular momentum?
QUESTION [1 upvotes]: Suppose a 1-kg ball A is fixed to a spoke 0.2 m long, which is attached to an axle so that the ball can rotate (v=10m/s, KE=50J, $\omega$=50 rps, L=2, p=0)
Now, there is a second ball B (m=1kg), attached to a 0.1 m stick that can likewise rotate, but at rest. An elastic collision takes place at the point where the two circular paths intersect, ball A stops dead and ball B starts moving at 10 m/s (KE = 50J v, KE is always conserved),
case a) - if ball B is not attached to a stick it acquires linear momentum (v=p = 10 m/s), but ball A had no linear momentum (p = 0),
case b) - if it is, it acquires angular momentum L = 1, ($\omega$= 100 rps), but ball A had L = 2. In this second case is angular momentum  conserved?
@JohnRennie,
your answer covers the angular momentum of B in case a), what about its linear momentum (from 0 to 10)?
can you also address the angular momentum of B in case b) ?

REPLY [4 votes]: Consider the following diagram:

This shows a mass $m$ moving past a point $P$ in a straight line. Note that the mass isn't connected to $P$ in any way - it's just moving past in a straight line.
The angular momentum of $m$ about $P$ is given by:
$$ \vec{L} = \vec{r} \times m\vec{v} $$
So the direction of $\vec{L}$ is normal to the screen and the magnitude is:
$$ L = rmv \sin\theta \tag{1} $$
and since:
$$ \sin\theta = \frac{d}{r} $$
Substituting this into equation (1) gives:
$$ L = mvd \tag{2} $$
Note that everything in equation (2) is a constant, so that tells us that $L$ is a constant as well, so angular momentum is conserved even though at first glance this isn't a system that is rotating.
And this provides the answer to your question. Even though in your second example ball $B$ isn't attached to anything it still has an angular momentum and this angular momentum is still conserved.
Response to comment:
Your distinction between angular and linear momentum is an artificial one. If you look at my working above the particle obviously has a linear momentum, but also an angular momentum. What's more the value of the angular momentum depends on where you fix the point $P$ so there is no unique value of angular momentum.
When you have a ball on a rotating rod the direction of linear momentum is not conserved because there is a force acting (through the rod) and Newton's laws tells that force is the rate of change of momentum:
$$ \vec{F} = \frac{d\vec{p}}{dt} $$
However there is a theorem (Noether's theorem) thats tell us angular momentum is conserved if the force is independant of angle. So if we calculate the angular momentum about the pivot point we'll find that this quantity is a constant. That's why it's useful for calculating trajectories.
But it is incorrect to say that the linear momentum of the ball attached to the rod is zero. At any time the linear momentum of the ball $A$ is $\vec{p} = m\vec{v}$, but the direction of $\vec{p}$ (though not its magnitude) is changing continuously with time due to the force applied by the connecting rod.
Back to your problem: before the collision the linear momentum of $A$ is mv, but it's direction is continually changing with time. However the angular momentum is constant because the force on $A$ is central.
During the collision a force acts between $A$ and $B$. This force acts normally to the connecting rods, i.e. it's not a central force, so the angular momenta will not be constant. If we consider the collision to take an instant then the only force acting is the one between the two balls so the total linear momentum will be conserved. Before the collision $p_A = mv$ and $L_B = 0$. After the collision $p_A = 0$ and $p_B = mv$, so the total linear momentum $p_A + p_B$ is conserved.
Immediately after the collision $B$ starts rotating about its pivot due to the force applied by its connecting rod. The force applied by its rod means the direction of $p_B$ now changes with time, though its magnitude doesn't. The angular momentum $L_B$ is constant because the force applied by the rod is central around $B$.
Note that it still makes perfect physical sense to calculate $L_A$ - it's just:
$$ L_A = \vec{r}_A \times m\vec{v_B} $$
but the force applied to $B$ by its rod isn't centrally symmetric about $A$, so $L_A$ is not constant (and therefore not terribly useful).
If $B$ isn't connected to a pivot, so it moves off in a straight line, then no forces act on $B$. That means both $\vec{p}_B$ and $\vec{L}_B$ calculated about any point are constant.

REPLY [0 votes]: Ball A has linear momentum pointing in the direction of its flight path. But it is not conserved, because you have a force acting on it (centripetal force), that as no counter part (unless you specify where the axle is mounted and allow for the mount to move as well).
Observe, that you can ascribe an angular momentum to B as well even when it does not move on a circular trajectory (As explained by John Rennie).


TITLE: Factor a polynomial over a finite field
QUESTION [0 upvotes]: I have a polynomial: $x^6 + x^4 + x^3 + x + 1$, and I need to factor it over $\mathbb{F}_2$, $\mathbb{F}_4$ and $\mathbb{F}_{16}$. I need to do it manually.
I wasn't even able to factor it over $\mathbb{F}_2$. I found a list of all irreducible polynomials of degree $\leqslant 5$ (here) and tried to divide the original polynomial by each of those, but with each one of them I got a remainder not equal to 0. So I either made a mistake with division or with the way I tried to solved this problem.
I have tried to read about various factorization algorithms, such as Berlekamp's algorithm, but they seem overcomplicated for this particular problem. I think there should be an easier approach to this problem.
In general, how to I try to come up with the factors? How do I know which polynomials are irreducible and which irreducible polynomials should I "look for" in my given polynomial?

REPLY [0 votes]: One very useful fact when dealing with polynomials over finite fields, is that all irreducible polynomials over $F_p$ of degree $r$ divide the polynomial $x^{p^r} - x$; and all other irreducible factors of that polynomial have a degree dividing $r$.
While this takes some time to apply when $r$ is large, it gives you a way to know how many irreducible polynomials of degree $r$ there are, which is useful since most of the times you can find some irreducile polynomials by hand. This is particularly easy when $r$ is prime: in this case, all factors of $x^{p^r} - x$ of degree different than $r$ are those of degree one, and there are exactly $p$ of them: one for each root (every element of $F_p$ is a root). So the polynomial has $\frac{p^r - p}{r}$ irreducible factors of degree $r$, and these are actually all the polynomials of degree $r$ over $F_p$.
I hope this helps!


TITLE: Does the LHC Beam Dump experience deceleration forces, or only heat?
QUESTION [1 upvotes]: When the proton beam is dumped into the beam dump, will the dump actually experience the deceleration force of a 200 mph freight train that is being stopped to a standstill in about a microsecond?
If yes, that must be millions of tons of force.
Or is the kinetic energy of the beam somehow converted into heat without generating deceleration forces?

REPLY [3 votes]: Force is the rate of momentum transfer, not the rate of energy transfer.
For slow objects like a 200mph train, momentum is $p\approx mv$ and kinetic energy is $T\approx\frac12mv^2,$ and so $\frac pT\approx\frac2v.$ In the case of the train, $\frac pT\approx0.01\,\mathrm{mph}^{-1}.$
For fast objects like the LHC protons, momentum is $p=\gamma mv$ and the kinetic energy is $T=(\gamma-1)mc^2\approx\gamma mc^2$ (where $\gamma=1/\sqrt{1-\frac{v2}{c^2}}$ shows up due to special relativity), and so $\frac pT\approx\frac v{c^2}.$ Approximating further $v\approx c,$ $\frac pT\approx\frac1c.$
Comparing these formulae, the LHC proton beam has only $\frac{1/c}{0.01\,\mathrm{mph}^{-1}}\approx0.00000015$ times the momentum as the "freight train of equivalent energy". The beam might have the energy of a train going 200mph, but it has the momentum of one going $0.00000015$ times as fast, or less than 2 inches per hour. Thus, dumping the beam requires handling a large amount of energy but doesn't produce nearly as much recoil as you are expecting. This is done (per the article) by heating a large mass that doesn't move significantly.
Note that SR is not required to get this qualitative result; without SR the $1/c$ would simply be $2/c$ and the momentum estimate would be (incorrectly) doubled. The fundamental intuition that is valid both relativistically and nonrelativistically is that as something speeds up, its momentum grows slower than its kinetic energy. For something going as ludicrously fast as an LHC proton, you can have a very large amount of energy without much momentum.


TITLE: Can a bounded analytic function have unbounded derivative?
QUESTION [0 upvotes]: Inspired by a physics problem, can a bounded analytic function have unbounded derivative?
I have seen, for instance, this question, but that one and the others all seem to take advantage of the fact that the function is not differentiable around the point where it has an infinite derivative, but in physics we usually deal with analytic function that ¨behave nicely¨ on ¨nice looking domains¨.
Addendum:
Please consider functions that are analytic on either $\mathbb{R}$ or $[0,1]$.

REPLY [0 votes]: When a function $f$ is analytic on an open interval I, then the derivative of $f$ is bounded on every closed sub-interval of $I$.
But notice the following examples, where an analytic function is bounded but its derivative is not bounded like:

$\sqrt x$ at $x=0$.
$\arcsin x$ at $x=\pm1$.

In these specific cases, the $x$'s are called branch points, and the derivatives are $+\infty$ in either case.
Notice however, that the functions are not analytic at these branch points, even though they are continuous there.  The functions are only analytic on $\Bbb R^+$ resp. $(-1,1)$ Theses examples carry over to $\Bbb C$ by the way; and even though the square-root extends to negative numbers, the square-root is still not analytic in $0$.


TITLE: is ∀x∃y(Lxy) and ∃x∀yLyx provable from each other?
QUESTION [1 upvotes]: The proof I have is :
$\begin{aligned}1&\ &\forall x\exists yLxy\\
2&\ &\exists y\ \text{(Lay)}\ \forall E_1\\
3&\ &\text{Assume Lab}\\
4&\ & \forall y Lyb\ \forall I_3∀yLyb\\
5&\ &\exists x\forall yLyx \exists I_4\\
6&\ & \exists x\forall yLyx ∃x∀yLyx\ \exists E_2 ∃, 3–5 \\\end{aligned}$
But I don't know if steps 4 and 5 are proper application of the universal and existential introductions.

REPLY [1 votes]: It is not true.  Let $Lxy$ be $x=y$ and the universe be $\{1,2\}$  Then $\forall x \exists y Lxy$ is true because for whatever $x$ you pick you can make $y$ be the same.  $\exists x \forall y Lyx$ is false because whatever $x$ you pick there is an unequal $y$.


TITLE: Why do we choose $\delta := \min\{\delta_1,\delta_2,...,\delta_k\}$?
QUESTION [1 upvotes]: I have a question. If I want to prove some limits, I use the $\delta-\epsilon$ definition. And I have to choose my $\delta := \min\{\delta_1,\delta_2,...,\delta_k\}$ to be sufficient. Why does it work? I have found one usefull answer, but I don't understand a very important place. So if we have the precise definition of limit.
$\lvert x-a \rvert<\delta$ then $\lvert f(x)-L \rvert< \epsilon$
which we might denote $P(\delta)$, regarding $f,a,L$ and $\epsilon$ as given/known.
If the condition $P(\delta)$ is true for some $\delta>0$ and if $0<\delta'<\delta$, then $P(\delta')$ is also true, $\textbf{because its hypothesis is logically more strict}$. I don't understand, why does it work? For example if $\lvert x-a \rvert<\delta$ and $0<\delta<\delta'$ its clearly that it will work for $\delta'$ because its greater, but why will it work for $\delta'$ if $0<\delta'<\delta$?
Thank you for help!

REPLY [2 votes]: If we have $\delta:=min\{\delta_1,...,\delta_k\}$, then by definition, $\delta \leq \delta_i$ for each $i$. Thus if $|x-a| \leq \delta$, we have $|x-a| \leq \delta_i$ for each $i$. So this definition allows us to take the most stringent of conditions.


TITLE: Hilbert space of a quantum system
QUESTION [8 upvotes]: The first postulate of Quantum Mechanics as I've learned can be stated as:

The states of a quantum system can be described by vectors in a Hilbert space.

I've seem also some people also requiring the Hilbert space be separable. Anyway, the point here is: we are associating a Hilbert space to quantum systems in order to describe their states. Usually, in basic treatments of Quantum Mechanics, when dealing with a spinless particle, the Hilbert space is usually taken to be $L^2(\mathbb{R}^n, d^nx)$, and states are taken to be equivalent to wavefunctions.
On the other hand, there is the more general situation on which the wavefunction is just the position representation of the state. And in truth there are situation on which there is no wavefunction. This led me to wonder how does one really choose the Hilbert space.
There is, however, the following theorem:

Let $\mathcal{H}_1$ and $\mathcal{H}_2$ be two Hilbert spaces, then they are isometrically isomorphic if and only if they have the same dimension.

In that case, the question of "which Hilbert space to choose" seems to be nonsensical. After all, choosing one with a certain dimension, any other would work as well. This makes it quite strange also to associate a Hilbert space to a system, if this association is not unique.
Still, there is the question of the dimension. I believe that one could argue that this could be determined by the observables. But still, one would need to know the spectrum of the operator before even defining the space upon which it is defined!
Anyway, I'm quite confused with these questions. How does one choose a Hilbert space to associate with a system? How does one deal with this ambiguity given by that theorem? If the observables determine the space, how does this happen if one can only define the operators after knowing the space?

REPLY [0 votes]: Choosing a Hilbert space is easy. It just needs to be mathematically convenient, while still completely characterising the system. (You could use $l_2$ to represent any quantum state, but you often don't because that is often mathematically very inconvenient.)
As for defining operators, it is true one cannot explicitly define an operator without defining its domain (i.e. the Hilbert space), but most operators have properties, like commutation relations, that must be true no matter what Hilbert space you choose. These properties greatly inform your choice of Hilbert space because, again, certain Hilbert spaces are mathematically far more convenient than others.
For example, if you require the use of two operators $A$ and $B$ with the property
$-i[A, B] = 1$
then the Hilbert space you choose cannot possibly be finite. This happens in quantum mechanics all the time, for example when $A$ is position and $B$ is momentum.


TITLE: Alternating sequence of ascending power of 2
QUESTION [1 upvotes]: We know that any whole number can be expressed as a binary number. But what if we change this a little bit?
Any positive whole number can be expressed as a finite series of ascending powers of $2$ with alternating sign. For example, $3=-1+4=-2^0+2^2$, $6=2-4+8=2^1-2^2+2^3$.
Question: Giving a positive number $n$, in how many ways can $n$ be expressed in such a fasion given above?
I would be pleased if anyone caould show me if there's any article or books on this topic.
I am not clear if a use the wrong tags. If that happens, pls point out.

REPLY [3 votes]: There is only one way to do this. Notice that $2^n -  2^m = \sum\limits_{i=m}^{n-1} 2^i$.
For a positive integer $N$ consider a set $S = \{i \in \mathbb{N}\cup\{0\} \colon i\text{-th bit in the binary representation of } N \text{ is }1\}$ Then $S$ can be represented as unite of maximal-inclusive integer intervals $S = \bigcup\limits_{i=1}^k \{\alpha_i, \alpha_i + 1, \ldots, \beta_i\}$ such that $\alpha_{i+1} > \beta_{i} + 1$ in the unique way. For example $S = \{0,1,3,5,6,7\} = \{0,1\} \cup \{3\} \cup \{5,6,7\}$. If $S$ is represented in a such way then $N = \sum\limits_{x\in S} 2^x = \sum\limits_{i=1}^k \sum\limits_{j=\alpha_i}^{\beta_i} 2^j=\sum\limits_{i=1}^k (2^{\beta_i + 1} - 2^{\alpha_i})$ by the first observation. $\alpha_1 < \beta_1 + 1 < \alpha_2 < \ldots < \alpha_k < \beta_k + 1$. On the other hand if $N$ is represented in that way then $S = \bigcup\limits_{i=1}^k \{\alpha_i, \ldots, \beta_i\}$. Therefore there is only one way if the signs alternating like $-1, +1, -1, \ldots$.


TITLE: Surface Parameterization and Coordinate Systems -- how do they mesh?
QUESTION [0 upvotes]: I am currently thinking about computing the surface area of surfaces in $\mathbb{R}^3$ through the lens of a traditional multivariable course, and I'm a bit confused about how parameterizations are meant to work. As a toy example, let's say I want to find the surface area of the sphere $x^2 + y^2 + z^2 = 1$. I can use the parameterization given by
$$r(u,v) = \langle \cos(u)\sin(v), \sin(u)\sin(v), \cos(v)\rangle$$
and set up the corresponding integral
$$\int_0^{\pi}\int_0^{2\pi} \sin(v)~dudv$$
where $\sin(v) = \|r_u \times r_v\|$. This is all fine and good. However, if I want to think about my parameterization landing in $\mathbb{R}^3$ from the perspective of spherical coordinates, I can use the parameterization
$$r(u,v) = \langle 1, u, v \rangle.$$
In this case, the function $\|r_u\times r_v\|$ is just equal to 1. If I take this at face value, I should think that I've done something wrong since this would give me $4\pi^2$ as the surface area, not $4\pi$. However, If I close my eyes and throw in the Jacobian for spherical coordinates as it is along the sphere, then I get the "correct" integral.
Why should I use the Jacobian here? I'm not doing a triple integral, and there's no volume form running around here. I'm strictly integrating a function $f(u,v)$ over a region in the $uv$-plane. What gives?

REPLY [0 votes]: The reason is because the term $r^2sin(\theta)d\theta d\phi$ is literally what you call $r_u \wedge r_v$$dudv$. Namely, the term $r\sin(\theta)d\phi$ is the increment of arclength along the $\phi$ direction and $rd\theta$ is the arclength increment along the $\theta$ direction, and to find the incremental area you just multiply them (or wedge them) together resulting in $r^2sin(\theta)d\theta d\phi$ (you can do that, since they form a pointwise orthogonal frame/basis). That makes it clear why the Jacobian appears in computing the area and it should also give you an idea what the Jacobian factor means geometrically. To find the incremental volume in spherical coordinates you just wedge along the increment in the $r$ direction resulting in $r^2sin(\theta)d\theta d\phi dr$, hence why it appears in computing the area, because computing the volume just adds a factor of $dr$ to it. My advice to you: Revise the spherical coordinates from a geometric perspective, it will make all these formulas more clear.


TITLE: About definition of transitive set (with sets or ur-elements)!!
QUESTION [0 upvotes]: I redeaded the following definitions:
--1) let $A$ a set, $A$ is transitive set if $\forall B \in A (B \subseteq A )$
--2) let $A$ a set, $A$ is transitive set if $\forall x \in A (x \subseteq A )$
I think that 2) is correct, in fact if $A:=\{\emptyset, \{\emptyset\},a,f\}$ then in case 1) $A$ is transitive set but in case 2) $A$ is not transitive set because  $a,f \nsubseteq A$... is correct?
Thanks in advance!

REPLY [1 votes]: The two definitions are equivalent. They are also equivalent to this one:

A set $A$ is transitive if $\forall\heartsuit\in A(\heartsuit\subseteq A)$.

The only difference is what we're calling the variable ranging over $A$.
As for your particular example, it depends on what $a$ and $f$ are. In the case that $a=\bigl\{\emptyset,\{\emptyset\}\bigr\}$ and $f=a\cup\{a\},$ then $A$ is indeed transitive. It also works if $f=\bigl\{\emptyset,\{\emptyset\}\bigr\}$ and $a=f\cup\{f\}.$ It also works if $a$ and $f$ are elements of $$\Bigl\{\emptyset,\{\emptyset\},\bigl\{\emptyset,\{\emptyset\}\bigr\}\Bigr\}.$$ Otherwise, you're correct that $A$ is not transitive, regardless of what $a$ and $f$ might be (sets or ur-elements).


TITLE: Find the limit of this function
QUESTION [1 upvotes]: Find
$$\lim_{n\to\infty}\left(1+x^n+\left(\frac {x^2}2\right)^n\right)^{\frac{1}{n}}$$
$(x\ge0)$.
Actually I could not find it when $x\ge1$, but when $0\leq x\lt1$ I found that the limit is $1$. Help me please.

REPLY [3 votes]: $$\begin{align} \lim_{n\to\infty} \left(1+x^n+2^{-n}x^{2n}\right)^{1/n} \end{align}$$
Let's consider the roots of the quadratic: $x^n=\frac{-1\pm\sqrt{1-2^{2-n}}}{2^{1-n}}$
Thus, we have
$$\begin{align} &\lim_{n\to\infty} \left(1+x^n+2^{-n}x^{2n}\right)^{1/n} \\
=&\lim_{n\to\infty} (2^{-n})^{1/n}\left(x^n+\frac{1+\sqrt{1-2^{2-n}}}{2^{1-n}}\right)^{1/n}\left(x^n+\frac{1-\sqrt{1-2^{2-n}}}{2^{1-n}}\right)^{1/n} \\
=&\lim_{n\to\infty} \frac{1}{2}\left(x^n+\frac{2}{2^{1-n}}\right)^{1/n}\left(x^n+\frac{0}{2^{1-n}}\right)^{1/n} \\
=&\lim_{n\to\infty} \frac{x}{2}\left(x^n+2^n\right)^{1/n} \\
=&\frac{x}{2}\max(x,2)  \end{align}$$


TITLE: Similar method, different answers
QUESTION [0 upvotes]: Ryan goes to church 2 days a week and college 3 days a week. On exactly one day he goes to church and college both. He went to college on Friday what is the probability he went to church the next day.
Textbook approach:

My approach :
CASE : 1 HE WENT TO CHURCH ON FRIDAY

Mo Tu Wed Thu Fri(COLLEGE,CHURCH) Sat Su
All cases

There are 6 choices to select the next day to visit church : 6C1 = 6
For each of the above choices, there are 5 choices to select 2 college dates : 5C2 = 10

Favorable cases

Church on Saturday
Select 2 days of college from the rest of 5 days : 5C2 = 10

Probability due to CASE 1 : (10/60)
CASE : 2 HE DID NOT GO TO CHURCH ON FRIDAY
Mo Tu Wed Thu Fri(COLLEGE) Sat Su
All cases

6 days to select 2 college dates : 6C2 = 15
Select 1 out of those 2 dates for church, select 1 more date out of the left 4 days : 2C1 x 4C1 = 8

Favorable cases

Sat is when he visits church
Out of the remaining 5 days(Fri is not church) select 1 more day for church : 5C1 = 5
Out of the 2 days of church select 1 day for college : 2C1 = 2
Out of the remaining 4 days select 2 for college : 4C2 = 6

Probability due to CASE 2 : (60/120)
Total probability : (10/60)+(60/120) = 2/3

REPLY [0 votes]: Note that you get the same probability as book author for the first case: $\frac{10}{60} = \frac{1}{6}$.
The first problem is the probability in the second case. It is already known that Friday is day for college, so in the last sentence you should select only 1 day (not 2) for college out of the remaining 4 days: $\binom{4}{1} = 4$. Then you get probability in the second case $\frac{40}{120} = \frac{1}{3} = \frac{2}{6}$.
The second and the main problem is that two cases described in both solutions are neither equiprobable nor summable. The probability in the first case can't be just added to the probability in the second case. An easy argument is that you could potentially get a result greater than 1, that is not a good idea. More rigorous argument is Bayes' theorem.
The probability that Ryan went to church on Friday subject to he went to college on Friday is $\frac{1}{3}$, because one of 3 days he goes to college he also goes to church. For the same reason the probability that Ryan didn't go to church on Friday subject to he went to college on Friday is $\frac{2}{3}$.
A rigorous solution is the following.
$$\mathrm P\{\text{Ryan went to church on Saturday} \mid \text{Ryan went to college on Friday}\} = \mathrm P\{\text{Ryan went to church on Saturday} \mid \text{Ryan went to college and church on Friday}\} \cdot \mathrm P\{\text{Ryan went to church on Friday} \mid \text{Ryan went to college on Friday}\} + \mathrm P\{\text{Ryan went to church on Saturday} \mid \text{Ryan went to church and didn't go to church on Friday}\} \cdot \mathrm P\{\text{Ryan didn't to to church on Friday} \mid \text{Ryan went to college on Friday}\} =
\frac{1}{6}\cdot \frac{1}{3} + \frac{1}{3} + \frac{2}{6} = \frac{5}{18}.$$


TITLE: Two problems on combinatorics
QUESTION [2 upvotes]: Suppose we have a bag which has chips of four colors numbered $1$ to $13$, i.e. in total $52$ balls. Now what is the difference between these two problems.

Problem-$1$- In how many ways can you choose $3$ chips with one number and $2$ chips with a  different number?

My approach- First we have to calculate no. of ways to select two different numbers from $\{1,2,\dots, 13\}$ , which are $13\times 12$. Now suppose we chose number $2$ and $4$, then I have $^4C_3$ choices for chip number $2$ and $^4C_2$ choices for chip number $4$. So the answer is $12\times 13\times 4\times 6=3744$. The book also worked like this, thus same answer.

Problem-$2$- In how many ways can you choose $2$ chips with one number and $2$ chips with a different number, and one chip of an other number?

My doubt- Now should not be the approach here same like in first question, and thus we should get answer $13\times 12\times 11\times ^4C_2 \times ^4C_2 \times ^4C_1= 13\times 12\times 11\times 6 \times 6 \times 4 =247104$.
But the book says (on page $86$) that "in this problem there is no way to distinguish the difference between a pair of chips numbered C  and a pair of chips numbered D. In this problem you choose your chips at once. The chips comes to you at same time."
What is different in this problem, why do now suddenly chips come at same time. It is word to word same as previous problem just one more slot. The book calculates the answer as $^4C_2 \times ^4C_2 \times ^{13}C_2 \times 44=123552$ i.e. half of what I calculated.
Aren't both problems same?

REPLY [0 votes]: The book answer is right, and it has already been explained why,
but I would prefer to arrange the terms differently for greater conceptual clarity
$${13\choose 2}{4\choose2}^2 {11\choose 1}{4\choose 1}$$


TITLE: Find a formula in $n, a, m, d$ for the sum: $(a+md)+(a+(m+1)d+(a+(m+2)d)+ ...+ (a+(m+n)d$
QUESTION [0 upvotes]: where m and n are integers, $n$ greater than or equal to $0$ and $a$ and $d$ are real numbers. Justify your result.
By distributing and factoring, I have gotten up to $(a+md)n+d(1+...+n)$ but I am stuck on what to do next.

REPLY [0 votes]: Hint:
Change your variables!
Set $A:= a+md$ , $D:=d$ and $N:=n$. Your sequence turns out to be
$A,A+D,A+2D,...,A+ND$.
Now what is the sum of an arithmetic progression?


TITLE: $\bigwedge^{2}T\cong\mathbb{Z}/2\mathbb{Z}$
QUESTION [1 upvotes]: Hello I have the next issue
Consider $T$ the subgroup of diagonal matrices of $GL(2, 3) $ (group of invertible matrices of $2$ by $2$ over a field of three elements)
It is well known that $T$ is isomorphic to the Klein four group, therefore
$\mathbb{Z}/2\mathbb{Z} \cong H_{2}(T, \mathbb{Z})\cong\bigwedge^{2}T $
I would like to give isomorphisms from $\mathbb{Z}/2\mathbb{Z}$ to $\bigwedge^{2}T$
Any hint would be helpful, thank you!

REPLY [1 votes]: Fix your well known iso $T \simeq \mathbb{Z}_2 \oplus \mathbb{Z}_2$. You get for free that this is a $\mathbb{Z}_2$-vector space isomorphism; hence you obtain a basis $\{\alpha, \beta\}$ for $T$.
Now map
$$
1 \in \mathbb{Z}_2 \mapsto \alpha \wedge \beta.
$$
This map is a $\mathbb{Z}_2$-linear monomorphism (the wedge on the RHS is a basis element of the exterior algebra) and by a dimension argument it must be an iso.


TITLE: What is the difference between a multivariate function and a vector-valued function?
QUESTION [2 upvotes]: I am not sure how to distinguish between a multivariate function and a vector-valued function when they are defined. For example, if I define a function as $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$, how do I know whether it's defining something like $f(x, y) = x + 2y$ or $\vec{v}(x, y) = (x + y)\hat{i} + (x - y)\hat{j}$?

REPLY [5 votes]: The difference is in if the domain or the target space has multiple dimensions.  We say that a function is multivariate if it has multiple variables as an input (another way to view this is that it has a vector as an input.) A function that is mapped into a vector is vector-valued. It is possible for the function to be both, as the following examples show:
(i)Multivariate: $f(x,y) = x^2 + y^2$
(ii)Vector-Valued: $g(x) = x^3\hat{i} + 4x\hat{j}$
(iii)Multivariate and Vector-Valued: $h(x,y) =  (x^2 + y^2)\hat{i} + (x^2 - y^2)\hat{j}$
So as far as the function you defined it would be like (iii) (so closest to $\vec{v}$ in your example)


TITLE: Problem related to the ring $\mathbb Z[\sqrt{d}]$
QUESTION [5 upvotes]: Let $d \in \mathbb Z$ and let $\sqrt{d} \in \mathbb C$ be a square root of $d$. Consider the subring of $\mathbb C$, $$\mathbb Z[{d}]=\{a+b\sqrt{d}: a,b \in \mathbb Z \}$$ and we define the norm of an element in $\mathbb Z[\sqrt{d}]$ by $N(a+b\sqrt{d})=a^2-db^2$.
Prove the following:
1) $z \in \mathbb Z[\sqrt{d}]$ is a unit if and only if $N(z)=\pm1$.
2) The ring $\mathbb Z[\sqrt{d}]$ is prefactorial, i.e., every non zero element which is not a unit can be written as product of irreducible elements.
I'll write what I could do:
1) Suppose $N(z)=\pm 1$, where $z=a+b\sqrt{d}$. Then $(a+b\sqrt{d})\dfrac{a-b\sqrt{d}}{a^2-b^2d}=1$, so the candidate for $z^{-1}$ is $\dfrac{a-b\sqrt{d}}{a^2-b^2d}$. From here it is clear that if $N(z)=\pm 1 \implies z^{-1} \in Z[\sqrt{d}]$. I couldn't prove the other implication.
I don't know what to do in 2), I would appreciate suggestions.

REPLY [3 votes]: The proof for 2) is very similar to the analogous theorem in $\mathbb{Z}$. Consider the set $S= \{x \mid x\text{ cannot be written as a product of irreducibles}\}$. Suppose the set is non-empty and notice that $\{|N(x)| \mid x\in S\}$ is a subset of $\mathbb{N}$ so you can use the Well-Ordering princple to choose a least element. Note that the least element that you pick can factor into two numbers that are strictly less than it (otherwise it would not be in the set).
Can you find the contradiction from there?


TITLE: blow-up of multiplicity 2 point on a curve is smooth
QUESTION [1 upvotes]: Suppose we have a smooth hypersurface $S \subset \mathbb{P}^3$ and a curve $C \subset S$ that is singular at $P \in C$ with multiplicity $\mu_P(C) = 2$ and not singular anywhere else. Then if let $\widetilde{C}$ be the blow-up of the curve $C$ at $P$, is $\widetilde{C}$ smooth?
From Corollary V.3.7 of Hartshorne we get that $p_a(C) - p_a(\tilde{C})= 1$, so the arithmetic genus decreases by one, does this mean that the multiplicity of $P$ also decreases by one?

REPLY [4 votes]: No, it is not. For instance, take $S$ to be a plane and $C$ a hypercusp, written in local coordinates as
$$
y^2 = x^n.
$$
A simple computation shows that the blowup of $C$ at the origin is given by
$$
{y'}^2 = x^{n-2},
$$
hence it is singular if $n \ge 4$.


TITLE: Natural example of finitely generated group with $\mathbb{Q}$ as a subgroup.
QUESTION [2 upvotes]: I know from an answer to this that every countable group embeds into a finitely generated group, but I'm curious if there is a nice explicit example that demonstrates this is possible with the rationals. I was able to exhibit the dyadic rationals as a subgroup of a finitely generated semi-direct product of them and the integers, but my method for that fails to generalize to $\mathbb{Q}$ due to the existence of infinitely many primes.

REPLY [3 votes]: The oldest "natural" example is probably the one due to Philip Hall, 1959. It's very elementary, here is it:
Let $u=(u_n)_{n\in\mathbf{Z}}$ be a bi-infinite sequence of nonzero rationals. Assume (one can ask less but nevermind) that every finite sequence of nonzero rationals appears as subword of $u$.
Let $V=\mathbf{Q}^{(\mathbf{Z})}$ be the vector space over $\mathbf{Q}$ with basis $(e_n)_{n\in\mathbf{Z}}$. Define two operators of $V$: the shift $s(e_n)=e_{n+1}$, and the diagonal map $d_u(e_n)=u_ne_n$. Define $\Gamma=\langle s,d_u\rangle$. Note that $s^nd_us^{-n}=d_{u'}$ where $u'$ is the $n$-shift of $u$, in particular $s^nd_us^{-n}$ and $d_u$ commute, so $\Gamma$ is naturally a quotient of the wreath product $\mathbf{Z}\wr\mathbf{Z}$ (actually, it is isomorphic to it). Next, observe that $V$ is a simple $\mathbf{Z}\Gamma$-module, that is, $V\neq 0$ and the only $\Gamma$-invariant subgroups of $V$ are $\{0\}$ and $V$ (easy from the assumptions).
Hence, the semidirect product $V\rtimes\Gamma$ is generated by 3 elements (the 2 generators of $\Gamma$ and any nonzero element of $V$), and this is a finitely generated (solvable) group containing a copy of $\mathbf{Q}$ (and of even of $\mathbf{Q}^{(\mathbf{Z})}$).
Actually, here one just needs $V$ to be generated by $e_O$ as $\mathbf{Z}\Gamma$-module, so it's enough to assume that the $u_n$ and $u_n^{-1}$ generate $\mathbf{Q}$ as additive group. This holds, for instance, if $u_n=\max(1,|n|)$.


TITLE: Left and Right Cosets
QUESTION [1 upvotes]: Let $H$ be a subgroup of a group $G$ and let $a, b \in G$. I need to give a counterexample or proof of the following statement:
If $aH = bH$, then $Ha = Hb$
Proof:
For every $h \in H, ah = bh$
$
\begin{align}
ah &= bh \newline
ahh^{-1} &= bhh^{-1} \newline
a &= b \newline
ha &= hb
\end{align}
$
Could someone critic my proof?
Thanks in advance.

Edit
Look at the answer below.

REPLY [2 votes]: Counterexample: We want to use a non-Abelian group such as $S_{3}$
$\mu_{1} = \pmatrix{1&2&3\\1 &3 &2}$, $\mu_{2} = \pmatrix{1&2&3\\3 &2 &1}$,
$\mu_{3} = \pmatrix{1&2&3\\2 &1 &3}$
$\rho_{0} = \pmatrix{1&2&3\\1 &2 &3}$, $\rho_{1} = \pmatrix{1&2&3\\2 &3 &1}$,
$\rho_{2} = \pmatrix{1&2&3\\3 &1 &2}$
Let $H = \{\rho_{0}, \mu_{2}\}$, $a = \mu_{1}$ and $b = \rho_{1}$
$aH = \mu_{1}\{\rho_{0}, \mu_{2}\} = \{\mu_{1}, \rho_{1}\}$
$bH = \rho_{1} \{\rho_{0}, \mu_{2}\} = \{\rho_{1}, \mu_{1}\}$
But
$Ha = \{\rho_{0}, \mu_{2}\}\mu_{1} = \{\mu_{1}, \rho_{2}\}$
$Hb = \{\rho_{0}, \mu_{2}\}\rho_{1} = \{\rho_{1}, \mu_{3}\}$
So $Ha \neq Hb$


TITLE: Ionic crystals and conduction of electricity?
QUESTION [0 upvotes]: Why do solutions of sodium chloride conduct electricity but crystals of sodium chloride do not?

REPLY [1 votes]: In a solid, we need free electrons for conduction, as positive ion cores are fixed at their lattice sites. In a sodium chloride crystal, the $Na^+$ and $Cl^-$ ions are held together by the electrostatic attractive force between the two ions. It is this strong attractive force that keep the crystal hard. The electrons are tightly fixed at their lattice sites. So there is no charge in the $NaCl$ crystal qualified for conduction.
In the aqueous state, the $NaCl$ dissociates into $Na^+$ and $Cl^-$ ions and hence the solution is polar since the solvent dissolving it (water) is polar too. The presence of $H^+$ and $OH^-$ ions in the solution weakens the electric attractive force between the $Na^+$ and $Cl^-$ ions, thereby increasing their separation distance. By applying a suitable electric field, you an overcome this lesser attraction, thereby causing the ions to displace under the field, resulting in a conduction.


text:
\begin{document}

	\begin{abstract}
		In this paper, we give
		an optimal estimate of an average of Hurwitz class numbers.
		As an application, we give an equidistribution result of the family $\{\frac{t}{2q^{\nu/2}} \ | \ \nu \in \NN, t \in \ZZ, |t|<2q^{\nu/2}\}$
		with $q$ prime, weighted by Hurwitz class numbers.
		This equidistribution produces
		many asymptotic relations among Hurwitz class numbers.
		Our proof relies on
		the resolvent trace formula of Hecke operators
		on elliptic cusp forms of weight $k\ge 2$.
	\end{abstract}
	\setcounter{tocdepth}{1}

\maketitle

\section{Introduction} \label{Sec1}

For a positive integer $D>0$ with $D\equiv 0,3 \,(\text{mod } 4)$,
let us define $H(D)$ as
$$H(D)=\sum_{Q}\frac{1}{\#\Gamma_Q},$$
where $Q$ runs over $\PSL_2(\ZZ)$-equivalence classes of positive
definite integral binary quadratic forms with discriminant $-D$,
and $\Gamma_Q$ denotes the stabilizer of $Q$ in $\PSL_2(\ZZ)$.
The positive rational number $H(D)$ is commonly called the $D$-th {\it Hurwitz class number} or {\it Kronecker-Hurwitz class number}.
There are some relations among $\{H(D)\}_D$:
\begin{align}\label{rel of H prime}
\sum_{\substack{t\in \ZZ \\ t^2<4q}}H(4q-t^2)=2q
\end{align}
for a prime number $q$,
or more generally,
Hurwitz's formula
\begin{align}
\sum_{\substack{t\in \Z \\ t^2<4m}} H(4m-t^2)=2\sigma(m)-\sum_{0<d|m}\min(d,m/d) \label{HWformula}
\end{align}
for $m\in \NN$, where $\s(m)$ is the divisor function (\cite{Kronecker}, \cite{Gierster}, \cite{Hurwitz}).
In \cite{Zagier0} and \cite{HirzeburchZagier}, it was shown that the generating function of $\{H(D)\}_D$ is a mock modular form of weight $3/2$ on $\Gamma_0(4)$, and this discovery inspired
many mathematicians to produce new relations of the Hurwitz class numbers
via modular forms.
Relations \eqref{rel of H prime} with condition
``$t\equiv c (\text{mod } a)$'' were given in \cite{BCFJS}
for $a=2, 3, 4$, and conjectured for $a=5, 7$.
Later, Bringmann and Kane \cite{BringmannKane} proved the conjecture for $a=5, 7$
by using mixed mock modular forms.
In recent years, many other relations were revealed by Mertens \cite{Mertens}, \cite{Mertens2}.

For $m\in \N$, let us consider a discrete measure
$$\mu_{m} = \sum_{\substack{t\in\ZZ \\ t^2<4m}}H(4m-t^2)\delta_{\frac{t}{2\sqrt{m}}}$$
on the interval $[-1,1]$, where $\delta_a$ is the Dirac measure supported at $a$. Observing that the total mass $\langle\mu_{m},1\rangle$ is exactly given by the formula \eqref{HWformula}, one may naturally raise the following two questions concerning the measure $\mu_{m}$:
\begin{enumerate}
\item Can one give an exact formula of the $n$-th moment  $\langle \mu_m, x^n \rangle$ for any $n \in \ZZ_{\ge 0}$ ?
\item What can one say about the limiting behavior of $\mu_m$ as $m\rightarrow \infty$ ?
\end{enumerate}
In a sence, the first question (1) has been completely answered by the Eichler-Selberg trace formula of Hecke operators on the space of cusp forms on $\SL_2(\Z)$ due to the fact that every monomial $x^{n}$ is a linear combination of Chebyshev polynomials of the 2nd kind. Indeed, \eqref{HWformula} is identical to the trace formula of the $m$-th Hecke operator $T(m)$ on the space of weight $2$ modular forms $S_2(\SL_2(\Z))=\{0\}$. For a prime number $q$, the Eichler-Selberg trace formulas of $T(q^{\nu})$ $(\nu=0,1,\dots)$ on $S_k(\SL_2(\Z))$ are collectively written by using a generating series:
\begin{thm}\label{SL2RTF} Let $k\geq 4$ be an even integer and $S_k(\SL_2(\Z))$ the space of cusp forms of weight $k$ on $\SL_2(\Z)$. Let $\{f_i\}$ be the normalized Hecke eigen basis of $S_k(\SL_2(\Z))$ and $f_i(\tau)=\sum_{n=1}^{\infty}a_i(n)e^{2\pi i n\tau}$ its Fourier expansion at $i\infty$. Let $q$ be a prime number. Then we have the following identity in the formal power series ring of indeterminate $X$: \begin{align}
\sum_{i} \frac{1}{q^{(1-k)/2}a_i(q)-(X+X^{-1})}
&=-\frac{k-1}{12}\frac{X}{1-q^{-1}X^2}
-\frac{1}{2} \frac{X}{1-X^2}
+\frac{X}{(1-X^2)(1-q^{\frac{1-k}{2}}X)}
 \label{RTF1}
\\
& \quad +\frac{X}{2}\sum_{\nu=0}^{\infty}\langle \mu_{q^{\nu}},U_{k-2} \rangle \,(q^{-1/2}X)^{\nu},
 \notag
\end{align}
where $U_{l}(x)$ is the Chebyshev polynomial of the 2nd kind of degree $l$.
\end{thm}
The formula \eqref{RTF1} is a simplest case of the resolvent trace formula \eqref{RTF} of Hecke operators explained in \S~\ref{Rtf}.
In this paper, we prove the following theorem, which answers the question (2) at least when $m$ is a power of a fixed prime number $q$.
\begin{thm}\label{equidist}
	Let $q$ be a prime number. For any $\a,\b \in [-1,1]$ with $\a<\b$, we have
	\begin{align*}
	\lim_{\nu\rightarrow \infty}\frac{{\mu}_{q^\nu}([\a, \b])}{\mu_{q^\nu}([-1,1])} =\lim_{\nu\rightarrow \infty}
	\frac{1-q^{-1}}{2q^\nu}\sum_{\substack {t\in \Z \\\a\leq \frac{t}{2q^{\nu/2}}\leq \b}} H(4q^{\nu}-t^2) = \frac{2}{\pi}\int_{\a}^{\b} \sqrt{1-x^2}\,\d x.
	\end{align*}
\end{thm}
Note that this theorem exhibits a weighted equidistribution of points $\{\frac{t}{2q^{\nu/2}}\}_{,\nu\in\Z_{\ge 0},\ |t|<2q^{\nu/2}}$ in $[-1,1]$ with weight factor $\{H(4q^\nu-t^2)\}_{\nu, t}$. The proof of this theorem relies on a good estimation of the $n$-th moments $\langle \mu_{q^\nu},x^{n}\rangle$ as $\nu\rightarrow \infty$ for each $n$. For $n\in \ZZ_{\ge 0}$ and $\a>0$, let us consider the following statement asserting a bound of the $n$-th moments of $\mu_{q^{\nu}}$ by $(q^\nu)^{\a+\e}$:
$$({\rm E}_{n,\a}) : \,(\forall \e>0)\,(\exists C>0)\,(\forall \nu\in \N)\,
\left|\sum_{\substack{t \in \ZZ\\ t^2<4q^{\nu}}} H(4q^\nu-t^2)\left(\frac{t}{2 \sqrt{q^{\nu}}}\right)^n\right| <C q^{\nu(\a+\e)}.
$$
The trivial bound $({\rm E}_{n,1})$ is easily obtained by the class number formula and the upper bound $|L(1,\chi)|\ll \log |D|$ of the Dirichlet $L$-functions
associated with non-principal Dirichlet characters $\chi$ modulo $D$ (cf.\ \eqref{trivial bound}). We have the following optimal improvement of the trivial bound.
\begin{thm}\label{best possible}
Let $q$ be a prime number. The statement $({\rm E}_{n,1/2})$ holds for all $n\in \N$. If $0<\a<1/2$, there exists $n\in \N$ such that $({\rm E}_{n,\a})$ does not hold.
\end{thm}
The proof of Theorem~\ref{best possible} in turn relies on the resolvent trace formula \eqref{RTF1}.

\section{Resolvent trace formulas} \label{Rtf}

 We shall state the {\rm resolvent trace formula} in a greater generality for Hecke operators acting on the space $S_k(N, \chi)=S_k(\Gamma_0(N),\chi)$ of elliptic cusp forms of weight $k$, level $N$ and of nebentypus $\chi$, where $k$ and $N$ are positive integers and $\chi$ is a Dirichlet character modulo $N$. The conductor of $\chi$ is denoted by $f_\chi$. For $m\in \NN$ relatively prime to $N$, the $m$-th Hecke operator $T(m)$ on $S_k(N, \chi)$
is
defined as
$$
 T(m)f(z) = m^{k-1}\sum_{\substack{a,d \in \NN \\ ad=m}}\sum_{b=0}^{d-1}\frac{\chi(a)}{d^{k}}f\left(\frac{az+b}{d}\right).
$$
The trace formula for $T(m)$ is well-known and can be stated in a couple of different forms. Here we quote the formula from \cite[Theorem 2.2]{Schoof}, which is convenient for our purpose.

\begin{thm}[Eichler-Selberg trace formula]
	\label{ES trace}
Assume $k\ge 2$ and $\chi(-1)=(-1)^k$.
	Let $m$ be a positive integer relatively prime to $N$. We have
	$${\tr}(T(m)|S_k(N,\chi))=A_1(m)+A_2(m)+A_3(m)+A_4(m),$$
	where $A_1$ is defined by
	$$A_1(m)=m^{k/2-1}\tilde{\chi}(\sqrt{m})\frac{k-1}{12}\psi(N)$$
	with $\psi(N)=N\prod_{p|N}(1+p^{-1})$, and $\tilde{\chi}(\sqrt{m})=\chi(\sqrt{m})$ if $m$ is a square, and $0$ otherwise.
	The term $A_2(m)$ is defined by
	$$A_2(m)=-\frac{1}{2}\sum_{\substack{t \in \ZZ \\ t^2<4m}}m^{\frac{k-2}{2}}U_{k-2}\left(\frac{t}{2\sqrt{m}}\right)
	H_{N,\chi}(4m-t^2),$$
	where

	$U_n(x)$ is the $n$-th Chebyshev polynomial of 2nd kind. The value $H_{N,\chi}(4m-t^2)$ is defined as
	$$H_{N, \chi}(4m-t^2)=\sum_{\substack{f \in \NN \\ f^2 | t^2-4m \\ \frac{t^2-4m}{f^2}\equiv 0,1 \pmod 4 }} h_w\left(\frac{t^2-4m}{f^2}\right)\mu(t,f,m),$$
	where $h_w(D)=\frac{2h_D}{\#\go_D^\times}$ with $\go_D$ and $h_D$
	being the order in $\QQ(\sqrt{D})$ of conductor $D<0$ and
	its class number,
	$$\mu(t,f,m) =\frac{\psi(N)}{\psi(N/N_f)}\sum_{\substack{x \in (\ZZ/N \ZZ)^\times
			\\ x^2-tx+m \equiv 0 \pmod  {N N_f}}}\chi(x)$$
	with $N_f=\gcd(N,f)>0$.
	The term $A_3$ is defined by
	$$A_3(m)=-\sideset{}{'}\sum_{\substack{d|m \\ 0<d\le \sqrt{m}}}d^{k-1}\sum_{\substack{0<c|N
			\\ \gcd(c,N/c)|\gcd(N/f_\chi, m/d-d)}}\varphi(\gcd(c, N/c))\chi(y).$$
	Here we set $\sideset{}{'}\sum_{d|m, 0<d\le \sqrt{m}}F(d)=\sum_{d|m, 0<d< \sqrt{m}}F(d) + \frac{1}{2}F(\sqrt{m})$
	for a function $F : \NN \rightarrow \CC$ with $F(\sqrt{m})=0$ unless $\sqrt{m}\in\ZZ$,
	an element $y=y_{m,N, c,d} \in \ZZ/(N/\gcd(c,N/c))\ZZ$ is taken so that $y\equiv d \ ({\rm mod}\ c)$ and $y\equiv m/d \ ({\rm mod}\ N/c)$, and
	the symbol $\varphi$ means the Euler totient function.
	The term $A_4$ is defined as
	$$A_4(m) = \delta_{k,2}\delta_{\chi,\bf1}\sum_{0<t|m}t.$$
\end{thm}
Let us explain the resolvent trace formula of Hecke operators. For simplicity, we restrict ourselves to the simplest case involving only one prime $q$ relatively prime to $N$. We fix a square root $\chi(q)^{1/2}\in \CC^\times$ and set $\chi(q)^{-1/2}=(\chi(q)^{1/2})^{-1}$ and $\chi(q^n)^{1/2}=\chi(q)^{n/2}$; although there are two choices of square roots, it does not influence the argument throughout this article. Set $T'(q)=\chi(q)^{-1/2}q^{(1-k)/2}T(q)$ and consider
$$\tr((T'(q)-\lambda)^{-1}|S_k(N,\chi)),
$$ where $\l \in \CC$ is not equal to any eigenvalues of $T'(q)$ on $S_k(N,\chi)$. Let $\{ f_i\}$ be the orthogonal basis consisting of normalized Hecke eigenforms of $S_{k}(N,\chi)$ with respect to the Petersson inner product.
Let $f_i(\tau)=\sum_{n=1}^\infty a_i(n)e^{2\pi i n\tau}$ be the Fourier expansion of $f_i$ at $i \infty$.
We remark that $a_{i}(q)=\chi(q)\overline{a_{i}(q)}$. The parameter $\{\a_i(q), \a_i(q)^{-1}\}$ is defined by $\chi(q)^{-1/2}q^{(1-k)/2}a_{i}(q) = \a_i(q)+\a_i(q)^{-1}$. Then, the Satake parameter of $f_i$ at $q$ is written as $\{\chi(q)^{1/2}\a_i(q), \chi(q)^{1/2}\a_i(q)^{-1}\}$. By the Ramanujan-Petersson conjecture (Deligne's theorem \cite[${\rm N}^\circ$ 5]{Deligne1}, \cite{Deligne2}), we have $|\a_i(q)|=1$, or equivalently, $\chi(q)^{-1/2}q^{(1-k)/2}a_{i}(q) \in [-2,2]$.
Furthermore, we have $|a_{i}(n)| \le d(n)n^{(k-1)/2}$ for any $n \in \NN$ relatively prime to $N$, where $d(n)$ is the number of the positive divisors of $n$. Since $S_{k}(N,\chi)$ is a finite dimensional $\C$-vector space, the operator $T'(q)-\lambda$ for $\lambda\in \C$ with large enough absolute value is invertible.

For $c\in \NN$ such that $c|N$,
set $l=\gcd(c,N/c)$.
Then, $c_1:=\gcd(c,f_\chi)$ and $c_2:=c/c_1$
satisfy
$c_1|f_\chi$, $c_1|c$, $\gcd(c,f_\chi/c_1)=1$ and
$\gcd(c_2, f_\chi)=1$.
By $\gcd(c_1, f_\chi/c_1)=1$,
we have the decomposition $\chi=\chi_c\times \chi_c'$
of Dirichlet characters so that
$f_{\chi_c}=c_1$ and $f_{\chi_c'}=f_\chi/c_1$.

For $q,l \in \NN$, let $m_{q,l}$ denote the order of $q$ in $(\ZZ/l\ZZ)^\times$.

The resolvent trace formula of $T(q)$ is described by $\chi_c'$ and $m_{q,l}$ as follows.
\begin{thm}[The resolvent trace formula of $T(q)$] \label{resolventTF}
Let $q$ be a prime number not dividing $N$.
If $|X|$ is sufficiently small, then $\tr (\{T'(q) - (X+X^{-1})\}^{-1}|S_k(N,\chi))$
is equal to
	\begin{align}\label{RTF}
	&  \sum_{i}\frac{1}{\a_i(q)+\a_i(q)^{-1}-(X+X^{-1})}\\
	= \,& -\frac{k-1}{12}\psi(N)\frac{X}{1-q^{-1}X^2}
	-\frac{1}{2}\frac{X}{1-X^2}\sum_{\substack{0<c|N \\ \gcd(c,N/c)|(N/f_\chi)}}
	\varphi(\gcd(c,N/c)) \notag\\
	&+\sum_{\substack{0<l|(N/f_\chi) }}\varphi(l)\sum_{\substack{0<c | N \\ l=\gcd(c,N/c)}}
	\frac{X}{(1-X^2)(1-\{\chi_c'(q)\chi(q)^{-1/2}q^{(1-k)/2}X\}^{m_{q,l}})} \notag \\

	\notag \\
	& -\delta_{k,2}\delta_{\chi,{\bf1}}\frac{X}{(1-q^{1/2}X)(1-q^{-1/2}X)}
	+\frac{X}{2}\sum_{\nu=0}^{\infty}A_{k,q}(\nu)(\chi(q)^{-1/2}q^{-1/2}X)^{\nu},\notag
	\end{align}
		where we set
	$$A_{k,q}(\nu)=\sum_{\substack{t \in \ZZ \\ t^2<4q^\nu}}H_{N,\chi}(4q^\nu-t^2)U_{k-2}\left(\frac{t}{2q^{\nu/2}}\right).$$
\end{thm}

\section{Proof of Theorem \ref{resolventTF}}

Theorem~\ref{resolventTF} is deduced from the trace formulas of $T(q^{\nu})$ with $\nu\in \N$ recalled in Theorem~\ref{ES trace} by considering the generating series
\begin{align}
\sum_{\nu=0}^{\infty}\tr(T(q^\nu)|S_k(N,\chi))\zeta^{\nu+1}
 \label{GENseries}
\end{align}
with $X=\chi(q)^{1/2}q^{(k-1)/2}\zeta$. Although the proof is completely elementary, we include it for convenience of the readers.

Invoking $U_n(\cos \theta)=\sin((n+1)\theta)/\sin\theta$,
a direct computation gives us the following.
\begin{lem} \label{L1}
	Let $\a\in \C$. Then, for any $X \in \CC$ such that $|X|<\min(|\a|,|\a|^{-1})$,
	\begin{align*}
	\frac{1}{\a+\a^{-1}-(X+X^{-1})}=-\sum_{n=0}^{\infty} U_{n}\left(\frac{\a+\a^{-1}}{2}\right)\,X^{n+1}.
	\end{align*}
\end{lem}
To prove Theorem \ref{resolventTF},
	it is sufficient to the case $X=\chi(q)^{1/2}q^{(k-1)/2}\zeta$ with $|\zeta|$ sufficiently small.
	By noting $U_\nu(2^{-1}(\a_i(q)+\a_i(q)^{-1})) = \chi(q^{\nu})^{-1/2}q^{(1-k)\nu/2}a_{f_i}(q^\nu)$ ($\nu \in \ZZ_{\ge0}$) deduced from the recurrence equation of Hecke operators, we have that \eqref{GENseries} equals
	$$-\chi(q)^{-1/2}q^{\frac{1-k}{2}}\sum_i\frac{1}{\a_i(q)+\a_i(q)^{-1}-(X+X^{-1})}.$$
	A direct computation also reveals $$\sum_{\nu=0}^{\infty}A_1(q^\nu)\zeta^{\nu+1}=\frac{k-1}{12}\psi(N)\frac{\chi(q)^{-1/2}q^{(1-k)/2}X}{1-q^{-1}X^2}.$$
	and
	$$\sum_{\nu=0}^{\infty}A_2(q^\nu)\zeta^{\nu+1}=-\frac{1}{2}\sum_{\nu=0}^{\infty}A_{k,q}(\nu) (\chi(q)^{-1/2}q^{-1/2}X)^{\nu}\times\chi(q)^{-1/2}q^{(1-k)/2}X.$$
	The series as above is absolutely and locally uniformly convergent on $|X|\ll q^{-1-\e}$ for a small $\e>0$.
	Indeed, by invoking the relation
	$$h_w(D)=\frac{h_{D_0}}{\#\go_{D_0}^\times}f\prod_{p|f}(1-p^{-1}\left(\tfrac{D_0}{p}\right)), \qquad (D=D_0 f^2),$$
where $D_0<0$ is the fundamental discriminant, the class number formula
and the estimate $L(1, (\frac{D_0}{p}))\ll \log |D_0|$,
we obtain
	the inequality
	\begin{align}
	\label{trivial bound}
|A_{k,q}(\nu)|\ll_{k,N} \sum_{\substack{t\in\ZZ \\ t^2<4q^{\nu}}}H_{N,\bf1}(4q^\nu-t^2)
\ll_\e \sum_{\substack{t \in \ZZ \\
	t^2<4q^\nu}}\sum_{\substack{f \in \NN \\f^2 | (t^2-4q^\nu)}}|\tfrac{t^2-4q^\nu}{f^2}|^{1/2+\e}\ll_{\e}
(4q^\nu)^{1+2\e}.
\end{align}

	The series involving $A_4$ is computed as
	\begin{align*}
	\sum_{\nu=0}^{\infty} A_4(q^{\nu})\zeta^{\nu+1} = &\delta_{k,2} \delta_{\chi,\bf1}\sum_{\nu=0}^{\infty}
	\sum_{\substack{0<t|q^{\nu}}}t\zeta^{\nu+1}
	= \delta_{k,2}\delta_{\chi,\bf1}\sum_{\nu=0}^{\infty}
	\sum_{j=0}^{\nu}q^j\zeta^{\nu+1}\\
	= & \delta_{k,2}\delta_{\chi,\bf1}\frac{1}{q-1}\left(\frac{q\zeta}{1-q\zeta}-\frac{\zeta}{1-\zeta}\right)
	= \delta_{k,2}\delta_{\chi,\bf1}\frac{\zeta}{(1-q\zeta)(1-\zeta)} \\
	= & \delta_{k,2}\delta_{\chi,\bf1}
	\frac{q^{-1/2}X}{(1-q^{1/2}X)(1-q^{-1/2}X)}.
	\end{align*}

To complete the proof of Theorem \ref{resolventTF},
we give an explicit formula of
$$F(\zeta)=\sum_{\nu=0}^{\infty}A_3(q^{\nu})\zeta^\nu.$$

\begin{lem} \label{explicit of generating A3} Set $X=\chi(q)^{1/2}q^{(k-1)/2}\zeta$. If $|X|<1$,
	then $F(\zeta)$ is absolutely and locally uniformly convergent, and we have
\begin{align*}
F(\zeta)
	=& -\sum_{\substack{0<l|(N/f_\chi) }}\varphi(l)\sum_{\substack{0<c | N \\ l=\gcd(c,N/c)}}
\frac{1}{(1-X^2)(1-\{\chi_c'(q)\chi(q)^{-1/2}q^{(1-k)/2}X\}^{m_{q,l}})} \\
&+\frac{1}{2}\frac{1}{1-X^2}\sum_{\substack{0<c|N \\ \gcd(c,N/c)|(N/f_\chi)}}
\varphi(\gcd(c,N/c)).
\end{align*}
\end{lem}
\begin{proof}
We start the proof of the following expression of $F(\zeta)$:
{\allowdisplaybreaks\begin{align}
	& -\sum_{\nu=0}^{\infty}\sum_{j=0}^{\nu}q^{(k-1)j}
	\sum_{\substack{c|N \\
			\gcd(c,N/c)|\gcd(N/f_\chi, (q^{2\nu+1-2j}-1)q^j)}}
	\varphi(\gcd(c,N/c))\chi(y)\zeta^{2\nu+1} \label{first} \\
	&
	-\sum_{\nu=0}^{\infty}\sum_{j=0}^{\nu}
	q^{(k-1)j}
	\sum_{\substack{c|N \\
			\gcd(c,N/c)|\gcd(N/f_\chi, (q^{2\nu-2j}-1)q^j)}}
	\varphi(\gcd(c,N/c))\chi(y)\zeta^{2\nu} \label{second} \\
	& +\sum_{\nu=0}^{\infty}\frac{1}{2}q^{(k-1)\nu}
	\sum_{\substack{c|N \\
			\gcd(c,N/c)|\gcd(N/f_\chi, 0)}}
	\varphi(\gcd(c,N/c))\chi(y) \zeta^{2\nu}. \label{third}
\end{align}
}The third term \eqref{third} equals
\begin{align}\frac{1}{2}\frac{1}{1-\chi(q)q^{k-1}\zeta^2}\sum_{\substack{c|N \\ \gcd(c,N/c)|(N/f_\chi)}}
\varphi(\gcd(c,N/c))
\label{third-formula}
\end{align}
for $|X|=|q^{(k-1)/2}\zeta|<1$ by $\chi(y)=\chi(q)^\nu$ and an easy calculation.

We consider the first term \eqref{first}.

Noting $y\equiv q^j (\text{mod } c_1)$ and
$y\equiv q^{2\nu+1-j} (\text{mod } f_\chi/c_1)$, we have
\begin{align*}
	\chi(y)=\chi_c(q^j)\chi_c'(q^{2\nu+1-j})=\chi_c(q)^j\chi_c'(q)^{2\nu+1-j}.
\end{align*}
From this, \eqref{first} is rewritten as
\begin{align*}
	-\sum_{\nu=0}^{\infty}
	\sum_{j=0}^{\nu}q^{(k-1)j}
	\sum_{\substack{c|N \\
			\gcd(c,N/c)|\gcd(N/f_\chi, q^{2\nu+1-2j}-1)}}
	\varphi(\gcd(c,N/c))\chi_c(q)^j\chi_c'(q)^{2\nu+1-j} \zeta^{2\nu+1},
\end{align*}
which is transformed into
\begin{align*}
	-\sum_{\nu=0}^{\infty}\sum_{\substack{l\in \NN \\ l|(N/f_\chi)}} \sum_{\substack{c | N \\
			l=\gcd(c,N/c)}}\sum_{\substack{0\le j \le \nu \\
			2\nu+1-2j\equiv 0 (\text{mod }m_{q,l})}} \varphi(l)
	\{q^{k-1}\chi(q)\chi_c'(q^{-2})\}^j \{\chi_c'(q)\zeta\}^{2\nu+1}.
\end{align*}
By changing the order of summations, this series equals
$$-\sum_{l|(N/f_\chi)}\varphi(l)\sum_{\substack{c | N \\ l=\gcd(c,N/c)}}
\sum_{\nu=0}^\infty \{\chi_c'(q)\zeta\}^{2\nu+1}
\sum_{\substack{0\le j \le \nu \\
		2\nu+1-2j\equiv 0 (\text{mod }m_{q,l})}}\{q^{k-1}\chi(q)\chi_c'(q^{-2})\}^j.
$$
If there exists $w \in \ZZ$ such that $2\nu+1-2j=m_{q,l}w$,
then $m_{q,l}$ and $w$ are odd and satisfy $m_{q,l}w\in\{1,3,5,\ldots,2\nu+1\}$.
Set $n_{q,l}=\frac{m_{q,l}-1}{2} \in \ZZ_{\ge0}$.
When $t\in\NN$ and $m_{q,l}t+n_{q,l}\le \nu <m_{q,l}(t+1)+n_{q,l}$,
$w$ ranges so that $w =2s+1$ with $0\le s \le t$.
Thus, the series as above equals
{\allowdisplaybreaks\begin{align}
	&-\sum_{\substack{l|(N/f_\chi) \\
			m_{q,l}: {\rm odd}}}\varphi(l)\sum_{\substack{c | N \\ l=\gcd(c,N/c)}}
	\sum_{t=0}^\infty
	\sum_{\nu=m_{q,l}t+n_{q,l}}^{m_{q,l}(t+1)+n_{q,l}-1}
	\{\chi_c'(q)\zeta\}^{2\nu+1}
	\sum_{\substack{s =0}}^{t}\{q^{k-1}\chi(q)\chi_c'(q^{-2})\}^{\nu-n_{q,l}-m_{q,l}s} \notag \\
	= &
	-\sum_{\substack{l|(N/f_\chi) \\
			m_{q,l} : {\rm odd}}}\varphi(l)\sum_{\substack{c | N \\ l=\gcd(c,N/c)}}
	\sum_{t=0}^\infty
	\frac{\{q^{k-1}\chi(q)\zeta^2\}^{m_{q,l}t+n_{q,l}}
		(1-\{q^{k-1}\chi(q)\zeta^2\}^{m_{q,l}}) }{1-q^{k-1}\chi(q)\zeta^2} \notag\\
	& \times \frac{1-\{q^{k-1}\chi(q)\chi_c'(q^{-2})\}^{-m_{q,l}(t+1)}}{1-\{q^{k-1}\chi(q)\chi_c'(q^{-2})\}^{-m_{q,l}}} \{q^{k-1}\chi(q)\chi_c'(q^{-2})\}^{-n_{q,l}}\chi_c'(q)\zeta \notag \\
	= &
	-\sum_{\substack{l|(N/f_\chi) \\
			m_{q,l} : {\rm odd}}}\varphi(l)\sum_{\substack{c | N \\ l=\gcd(c,N/c)}}
	\sum_{t=0}^\infty
	(\{q^{k-1}\chi(q)\zeta^2\}^{m_{q,l}t}- \{\chi_c'(q^2)\zeta^2\}^{m_{q,l}t}\{q^{k-1}\chi(q)\chi_c'(q^{-2})\}^{-m_{q,l}})\notag \\
	& \times	\frac{1-\{q^{k-1}\chi(q)\zeta^2\}^{m_{q,l}}}{1-q^{k-1}\chi(q)\zeta^2} \frac{\{\chi_c'(q)\zeta\}^{2n_{q,l}+1}}{1-\{q^{k-1}\chi(q)\chi_c'(q^{-2})\}^{-m_{q,l}}} \notag \\
	= &
	-\sum_{\substack{l|(N/f_\chi) \\
			m_{q,l} : {\rm odd}}}\varphi(l)\sum_{\substack{c | N \\ l=\gcd(c,N/c)}}
\left(\frac{1}{1-\{\chi(q)q^{k-1}\zeta^2\}^{m_{q,l}}}-\frac{\{q^{k-1}\chi(q)\chi_c'(q^{-2})\}^{-m_{q,l}}}{1-\{\chi_c'(q)^2\zeta^2\}^{m_{q,l}}}  \right)\notag \\
	& \times \frac{1-\{q^{k-1}\chi(q)\zeta^2\}^{m_{q,l}}}{1-q^{k-1}\chi(q)\zeta^2} \frac{\{\chi_c'(q)\zeta\}^{m_{q,l}} }{1-\{q^{k-1}\chi(q)\chi_c'(q^{-2})\}^{-m_{q,l}}} \notag \\
	=& -\sum_{\substack{l|(N/f_\chi) \\
			m_{q,l} : {\rm odd}}}\varphi(l)\sum_{\substack{c | N \\ l=\gcd(c,N/c)}}
	 \frac{\{\chi_c'(q)\zeta\}^{m_{q,l}}}{(1-\chi(q)q^{k-1}\zeta^2)(1-\{\chi_c'(q^2)\zeta^2\}^{m_{q,l}})}, \label{first-formula}
\end{align}
}where $X=\chi(q)^{1/2}q^{(k-1)/2}\zeta$. This computation is justified when $|\zeta|<q^{(1-k)/2}$, or equivalently
$|X|<1$.

In a similar fashion, the term
\eqref{second} equals
$$-\sum_{l|(N/f_\chi)}\varphi(l)\sum_{\substack{c| N \\ l=\gcd(c,N/c)}}
\sum_{\nu=0}^\infty \{\chi_c'(q)\zeta\}^{2\nu}
\sum_{\substack{0\le j \le \nu \\
		2\nu-2j\equiv 0 (\text{mod }m_{q,l})}}\{q^{k-1}\chi(q)\chi_c'(q^{-2})\}^j.
$$
When $2\nu-2j=m_{q,l}w$ for $w \in \ZZ$,
then $m_{q,l}w$ is even and $j=\nu-m_{q,l}w/2$.
Suppose that $m_{q,l}$ is even and that $2^{-1}m_{q,l}t\le \nu <2^{-1}m_{q,l}(t+1)$ for $t \in \ZZ_{\ge 0}$.
Then, $w$ varies so that $0\le w \le t$.
The term for even $m_{q,l}$ is given by
{\allowdisplaybreaks \begin{align}
&-\sum_{\substack{l|(N/f_\chi) \\
		m_{q,l} : {\rm even}}}\varphi(l)\sum_{\substack{c|N \\ l=\gcd(c,N/c)}}
\sum_{t=0}^{\infty}\sum_{\nu=2^{-1}m_{q,l}t}^{2^{-1}m_{q,l}(t+1)-1}\{\chi_c'(q)\zeta\}^{2\nu}
\sum_{w=0}^{t}\{q^{k-1}\chi(q)\chi_c'(q^{-2})\}^{\nu-m_{q,l}w/2}\notag \\
= & -\sum_{\substack{l|(N/f_\chi) \\
		m_{q,l} : {\rm even}}}\varphi(l)\sum_{\substack{c| N \\ l=\gcd(c,N/c)}}\frac{1}{(1-q^{k-1}\chi(q)\zeta^2)(1-\{\chi_c'(q)\zeta\}^{m_{q,l}})}. \label{second-m-even}
\end{align}
}When $m_{q,l}$ is odd, $w$ must be $w=2s$ with $0\le s \le t$.
Hence, the term for odd $m_{q,l}$ is given by
{\allowdisplaybreaks \begin{align}
&-\sum_{\substack{l|(N/f_\chi) \\
		m_{q,l} : {\rm odd}}}\varphi(l)\sum_{\substack{c| N \\ l=\gcd(c,N/c)}}
\sum_{t=0}^\infty
\sum_{\nu=m_{q,l}t}^{m_{q,l}(t+1)-1}
\{\chi_c'(q)\zeta\}^{2\nu}
\sum_{\substack{s =0}}^{t}\{q^{k-1}\chi(q)\chi_c'(q^{-2})\}^{\nu-m_{q,l}s}\notag \\
=	& -\sum_{\substack{l|(N/f_\chi) \\
		m_{q,l} : {\rm odd}}}\varphi(l)\sum_{\substack{c| N \\ l=\gcd(c,N/c)}}
\frac{1}{(1-q^{k-1}\chi(q)\zeta^2)(1-\{\chi_c'(q^2)\zeta^2\}^{m_{q,l}})}.\label{second-m-odd}
\end{align}
}By the consideration so far, we obtain the assertion from \eqref{third-formula}, \eqref{first-formula},
\eqref{second-m-even} and \eqref{second-m-odd}.
\end{proof}

\section{Proof of Theorem \ref{best possible} and Corollary \ref{equidist}}

By virtue of the resolvent trace formula (Theorem \ref{resolventTF}), we have the following estimate.
\begin{cor}\label{esti}
	For any $\e>0$ and $k\ge 4$, we have the bound
	\begin{align*}|A_{k,q}(\nu)|\ll_{q, k,\e} q^{\nu(1/2+\e)}, \quad \nu\in \N.
	\end{align*}
\end{cor}
\begin{proof}
	By the Ramanujan bound, the left-hand side of \eqref{RTF}
	in Theroem \ref{resolventTF} is holomorphic on the unit disc $|X|<1$. The first
	three summands of the right-hand side are evidently holomorphic on $|X|<1$.
	The fourth summand is vanishing by $k\ge 4$.
	Hence, the power series $\sum_{\nu=0}^{\infty}A_{k,q}(\nu)(\chi(q)^{-1/2}q^{-1/2}\,X)^{\nu}$ (which is convergent for $|X|\ll 1)$ becomes a holomorphic function on the unit disc $|X|<1$.
	The Cauchy estimate yields the desired bound.

\end{proof}

Let us prove Theorem \ref{best possible}.
The estimate $({\rm E}_{n, 1/2})$ for any $n\in\NN$ follows from Corollary \ref{esti} applied to the full modular group $\SL_2(\Z)$ with $\chi={\bf 1}$; we note that Chebyshev polynomials form a basis of the space of polynomials. By \cite[Th\'eor\`eme 1]{Serre}, there exist a large $k \in 2\ZZ_{\ge 2}$ and a normalized Hecke eigenform $f_i$ of weight $k$ such that $\a_{i}(q)\neq 1$.
Then,
$\sum_{\nu=0}^{\infty}A_{k,q}(\nu)q^{-\nu/2}X^\nu$ has a pole at $X=\a_{i}(q)$
by \eqref{RTF}, which is a special case of Theorem \ref{resolventTF}. Hence, its radius of convergence is $1$. Therefore for any $\a\in (0,1/2)$, the estimate $({\rm E}_{n,\a})$ does not hold for some $n \in \NN$.
\qed

\medskip
\noindent

Let us define a Radon measure $\mu_\nu^{N,\chi,q}$ on $[-1,1]$ by
$$\langle \mu_{\nu}^{N,\chi, q}, f\rangle = 2^{-1}q^{-\nu}\sum_{t \in \ZZ, \ t^2<4q^\nu}H_{N,\chi}(4q^\nu-t^2)\delta_{\frac{t}{2q^{\nu/2}}}.$$
Then, $\mu_{\nu}^{N,\chi,q}$ is positive if $\chi$ is principal.
\begin{lem}\label{formula for 1}We have
	$$\langle \mu_{\nu}^{N,\chi,q}, 1\rangle = \delta_{\chi,\bf1}\frac{1}{1-q^{-1}}+\Ocal_{N, \chi}(\nu q^{-\nu/2}).$$
\end{lem}
\begin{proof}By the Eichler-Selberg trace formula (Theorem \ref{ES trace}) for $k=2$, we have the equality
	$$\frac{1}{2}\sum_{t \in \ZZ, \ t^2<4q^\nu}H_{N,\chi}(4q^\nu-t^2) = A_1(q^\nu)+A_3(q^\nu)+A_4(q^\nu)-\tr(T(q^\nu)|S_2(N,\chi)).$$
	By easy calculation, we have two equalities $A_4(q^\nu)=\delta_{\chi,\bf1}\frac{q^{\nu+1}-1}{q-1}$ and
	$A_1(q^\nu) = \tilde{\chi}(\sqrt{q^\nu})\frac{\psi(N)}{12}$.
	The term $A_3(q^\nu)$ is estimated as
	$$|A_3(q^\nu)|\ll \sum_{j=0}^{\nu}\min(q^\nu, q^{\nu-j})\sum_{c|N}\varphi(\gcd(c,N/c))\ll \nu q^{\nu/2}d(N)\varphi(N).$$
	By the Ramanujan bound $|a_{f_i}(n)|\le d(n)n^{1/2}$, we have
	$$|\tr(T(q^\nu)|S_2(N,\chi))| \le \sum_{i=0}^{\dim S_2(N,\chi)} d(q^\nu)q^{\nu/2} = (\nu+1)q^{\nu/2} \dim S_{2}(N,\chi).$$
	By combining the evaluations as above, the value $\langle \mu_{\nu}^{N,\chi, q}, 1\rangle$, which is equal to $q^{-\nu}\{A_1(q^\nu)+A_3(q^\nu)+A_4(q^\nu)-\tr(T(q^\nu)|S_2(N,\chi))\}$ is estimated by $q^{-\nu}A_4(q^\nu)+q^{-\nu}\Ocal_{N,\chi}(\nu q^{\nu/2})$. Thus we are done.
\end{proof}

\begin{thm} \label{weak conv}
	As $\nu\rightarrow \infty$, the measure $\mu_{\nu}^{N,\chi , q}$ converges $*$-weakly to the measure
	$$
	\delta_{\chi,\bf1}\frac{1}{1-q^{-1}}\,\frac{2}{\pi}\sqrt{1-x^2}\d x.
	$$
\end{thm}
\begin{proof}Let us consider the case where $\chi$ is principal.
	By the positivity of $\mu_\nu^{N,{\bf1},q}$ and \cite[Proposition 2]{Serre},

	it suffices to show the following convergence:
	\begin{align}
	&\lim_{\nu\rightarrow \infty} \langle \mu_{\nu}^{N, {\bf1}, q},U_{n}\rangle=0, \quad (n>0) \label{mu-U}, \\
	&\lim_{\nu\rightarrow \infty} \langle \mu_{\nu}^{N,{\bf1}, q},1\rangle=\frac{1}{1-q^{-1}}.
	\label{mu-1}
	\end{align}
	Corollary~\ref{esti} gives us the bound $|\langle \mu_{\nu}^{N,{\bf1}, q},U_{k-2}\rangle|\ll_{\e}q^{(-1/2+\e)\nu}$ which is sufficient to have \eqref{mu-U}. Furthermore, Lemma~\ref{formula for 1} implies \eqref{mu-1}. This completes the proof of the assertion when $\chi$ is principal.

In the non-principal case of $\chi$,
although $\mu_\nu^{N, \chi, q}$ is not always positive,
we obtain the assertion in the same way as the principal case
from the inequality
$$\langle \mu_{\nu}^{N,\chi,q}, f\rangle \le \mu_{\nu}^{N,{\bf1},q}(1)\sup_{x\in[-1,1]}|f(x)|, \quad f \in C([-1,1])$$
and the vanishing of $\lim_{\nu\rightarrow \infty}\mu_\nu^{N,\chi,q}$ on the polynomial functions (Lemma \ref{formula for 1}).
\end{proof}

\medskip
\noindent
{\bf Remark} : To have the convergence of the measure, any improvement of the Hecke bound of the Fourier coefficients is sufficient (see the proof of Lemma \ref{formula for 1}).

\medskip
By the equalities $\mu_{q^\nu}=2q^{\nu}\,\mu_{\nu}^{1,{\bf1},q}$ and $$\mu_{q^\nu}([-1,1])=2\s(q^\nu)-\sum_{j=0}^{\nu}\min(q^j,q^{\nu-j})\sim 2q^\nu(1-q^{-1})^{-1}, \qquad (\nu \rightarrow \infty),$$
Theorem \ref{equidist} follows from Theorem \ref{weak conv} and \cite[Proposition 1]{Serre}.
\qed

With a bit more work, we have a generalization of Theorem \ref{equidist} as follows. Let $S=\{q_1,\dots,q_r\}$ be a finite set of prime numbers and $\N(S)$ the set of those positive integers whose prime divisors belong to $S$. Then the limit formula in Theorem~\ref{equidist} is true when $m$ inside $\N(S)$ grows to infinity, i.e., for any $[\a,\b]\subset [-1,1]$,
\begin{align*}
\lim_{\substack{\nu_q \rightarrow \infty (\forall q \in S) \\ m=\prod_{q \in S}q^{\nu_q}\in \N(S)}} \frac{\mu_{m}([\a,\b])}{\mu_{m}([-1,-1])}=\frac{2}{\pi}\int_{\a}^{\b}\sqrt{1-x^2}\d x.
\end{align*}
This is shown by the same argument as above by establishing the estimate
\begin{align}
|\langle \mu_{m},U_{l}\rangle|\ll_{l,\e} m^{1/2+\e}, \quad m\in \N(S)
 \label{Remark1}
\end{align}
for each $l\in 2\N$. This bound in turn is shown by the multivariable analogue of \eqref{RTF1}
for $k\ge 4$, which equates the rational function
\begin{align*}
\sum_{i=1}^{\dim S_k(\SL_2(\ZZ))}\prod_{j=1}^{r} \frac{-q_j^{(1-k)/2}}{q_j^{(1-k)/2}a_i(q_j)-(X_j+X_j^{-1})}
\end{align*}
with a sum of the following three power series:
\begin{align}
&\tfrac{k-1}{12}
\prod_{j=1}^{r}\frac{q_j^{\frac{1-k}{2}}X_j}{1-q_j^{-1}X_j^2},
 \notag
\\
& \sum_{(\nu_1,\dots,\nu_r) \in \Z_{\ge 0}^{S}} A_3(q_1^{\nu_1}\cdots q_r^{\nu_r})
\prod_{j=1}^{r}(q_j^{(1-k)/2}X_j)^{\nu_j+1},
 \label{Hypterm} \\
& -\tfrac{1}{2}\sum_{(\nu_1,\dots,\nu_r) \in \Z_{\ge 0}^{S}}
\langle \mu_{q_1^{\nu_1}\cdots q_r^{\nu_r}},U_{k-2}\rangle \prod_{j=1}^{r}\{(q_j^{-1/2}X_j)^{\nu_j}
\,q_j^{(1-k)/2}X_j\}.
\label{Ellterm}
\end{align}
It is not clear that the series \eqref{Hypterm} has a rational expression; but, it is easy to see its absolute convergence on the polydisc $\Delta=\{(X_1,\dots,X_r)\in \C^r|\,\max_{1\leq j \leq r}|X_j|<1\}$.
Hence invoking the Ramanujan bound, the holomorphic function defined in a neighborhood of the origin by the convergent multi-series \eqref{Ellterm} has a holomorphic continuation to the polydisc $\Delta$. Then the Cauchy estimate yields the bound \eqref{Remark1}. Details are left to the readers.

\section*{Aknowledgements}

The second author was supported by Grant-in-Aid for Scientific research (C) 15K04795.

config: arxiv


TITLE: The set of polytopes with given $f$-vector
QUESTION [10 upvotes]: Let $f=(f_0,\ldots f_n)$ be a vector in $\Bbb N^{n+1}$. Let $X$ be the set of all (ordered) $f_0$-tuples in $\Bbb R^n$ whose convex hull has $f$ as its $f$-vector. Assume that $X$ is non-empty. Is anything known about $X$? For instance, is it algebraic? Semi-algebraic?
Edit: By “is $X$ algebraic”, I mean: Is $X$ locally closed, or maybe just constructible, in the Zariski topology on $\Bbb R^{f_0 n}$?

REPLY [0 votes]: Hopefully this answer to my question is correct despite my minimal knowledge of semi-algebraic things.
$X$ is not in general algebraic: Consider the case of a square, i.e. $f=(4,4)$ (using the notation from my post, this should actually be $(4,4,1)$, but it seems that the number of  $n$-dimensional faces of an n-dimensional polytope is usually left out since it’s always one). Then $X$ is exactly the set of 4-tuples $(x_1,\ldots,x_4)$ of points in $\Bbb R^2$ such that $x_4$ is not in the convex hull of $\{x_1,\ldots,x_3\}$. Suppose $X$ is algebraic, and fix $x_1,\ldots,x_3$. Then so is the set of all $x_4$ not in the convex hull of $\{x_1,\ldots,x_3\}$. But this set is $\Bbb R^2$ minus a (solid) triangle, and such a set is not algebraic.
$X$ is semi-algebraic: By the definition of “face”, $X$ is exactly the set of tuples $(x_1,\ldots,x_{f_0})$ of points in $\Bbb R^n$ satisfying the following (lines are numbered and indented for clarity):

For all $k\in\{0,\ldots,n-1\}$,
$\quad$there exist exactly $f_k$ subsets $I\subseteq \{1,\ldots, f_0\}$ such that
$\quad\quad$there exists an affine linear functional $g\colon \Bbb R^n\to \Bbb R$ such that
$\quad\quad\quad g(x_i)\geq0$ for all $i\in \{1,\ldots, f_0\}$
$\quad\quad\quad$and $g(x_i)=0$ for all $i\in I$.

Lines 3-5 describe a semi-algebraic set when $I$ and $k$ are fixed. Since there are only finitely many possible $I$ and $k$, we get that $X$ is indeed semi-algebraic.


TITLE: Expressing the homomorphism condition with a commutative diagram.
QUESTION [3 upvotes]: I am asking myself how to express the homomorphic condition in commutative diagrams. For this let $(M, \cdot)$ and $(N, \cdot)$ be algebraic structures and let $\varphi : M \to N$ be a homomorphism, then for all $x,y \in M$
$$
 \varphi(x \cdot y) = \varphi(x) \circ \varphi(y).
$$
The only possiblity that comes to my mind is
$$\begin{matrix} M\times M & \overset{\varphi\times\varphi}{\longrightarrow} & N\times N \\  \downarrow {\scriptsize \cdot}& & \downarrow {\scriptsize \circ} \\ M \,\,\,& \overset{\varphi}{\longrightarrow} & N\end{matrix}$$
where $(\varphi \times \varphi)(x,x) := (\varphi(x), \varphi(x))$. But this seems quite complicated with all this products to express this simple idea, so are there simpler diagrams to express this (when mine is right, don't know if it is valid to assume that such product constructions are possible...)?

REPLY [2 votes]: That's correct, and it's as simple as it's going to get. This is a special case of a homomorphism of algebras for an endofunctor, which has exactly the same diagram but generalised. I'll explain below.
Let $F : \mathcal{C} \to \mathcal{C}$ be a functor from a category $\mathcal{C}$ to itself. An $F$-algebra is a pair $(M, \alpha)$, where $\alpha : FM \to M$ is a morphism, which may be required to satisfy some additional properties. In your case we have $FM = M \times M$ and $\alpha(x,y) = x \cdot y$.
A homomorphism of $F$-algebras $\phi : (M, \alpha) \to (N, \beta)$ is a morphism $\phi : M \to N$ making the following diagram commute:
$$\begin{matrix} FM & \overset{F\phi}{\rightarrow} & FN \\ {\scriptsize \alpha} \downarrow & & \downarrow {\scriptsize \beta} \\ M & \underset{\phi}{\rightarrow} & N\end{matrix}$$
That is, $f$ is a homomorphism so long as it kind-of-commutes with $\phi$, where by kind-of-commutes I mean you need to stick an $F$ in there.
In summary, here we have:

$\mathcal{C}$ is the category of whatever algebraic structures you're working with
$M$ and $N$ are objects of this category
$F$ is the diagonal product functor, defined by $FM = M \times M$ and $F\phi = \phi \times \phi$
$\alpha : M \times M \to M$ is the morphism $(x,y) \mapsto x \cdot y$
$\beta : N \times N \to N$ is the morphism $(x,y) \mapsto x \circ y$

[Added] I guess the point of me saying all this, which I forgot to make explicit, was to respond to the part where you said

"But this seems quite complicated with all [these] products to express this simple idea"

The fact of the matter is that 'all these products' all come from one thing, namely the diagonal functor ${({-})} \times {({-})}$, which plays the role of the endofunctor $F$ in the above construction.


TITLE: Why doesn't ordinary metal box shield magnetic field?
QUESTION [0 upvotes]: I have this metal box. I put a transformer inside.

Using a magnetic field meter. I can still measure 80 milligauss out of the original 100 milligauss.
Why didn't it complete block it? What would it take to complete block it?

REPLY [3 votes]: Assuming you're measuring the amplitude of magnetic field oscillations:
Radiation penetrates through a conductor to a depth on the order of the quantity known as the skin depth $\delta$, which, for good conductors and low frequencies, is given by
$$\delta = \sqrt{\frac{2\rho}{\omega \mu}}$$
for a conductor with resistivity $\rho$ and magnetic permeability $\mu$ and radiation at frequency $\omega$. For aluminum, $\rho=2.65\times 10^{-8}\;\Omega\text{m}$ and $\mu=1.26\times 10^{-6}\;\text{H/m}$, so $\delta\approx 3\;\text{cm}$. For steel, $\rho\approx 5\times10^{-7}\;\Omega\text{m}$ and $\mu$ ranges from $10^{-6}$ to $10^{-3}$ $\text{H/m}$, so $\delta$ ranges from $4\;\text{mm}$ to $12\;\text{cm}$, depending on the type of steel. In any case, it should be clear that it may take several centimeters of conductive shielding to insulate 60-Hz radiation, and it doesn't appear that the metal box is quite that thick. Low-frequency radiation is penetrating by nature, and therefore hard to insulate.
Assuming you're measuring the DC (steady) magnetic field:
First of all, the Earth's magnetic field is 250-650 milligauss, which is on the order of both of your measurements, so if you haven't zeroed your measurement properly or otherwise haven't replicated experimental conditions properly, you might just be measuring a local variation of the Earth's magnetic field, or the metal box's distortion of that field independent of the field of the transformer. In any case, though, you can't really "block" or "absorb" DC magnetic fields, you can only divert them, which is done through materials with high permeability. The permeability of metals varies widely (over three orders of magnitude for steel alone), and it's incorrect to assume that most metals are good at insulating DC magnetic fields. One of the more common high-permeability metals around is called Mu-Metal, with a permeability of around $10^{-2}\;\text{H/m}$, which is 10,000 times greater than that of aluminum and 10 times greater than high-permeability steel.


TITLE: Notation: function codomain is a set of functions
QUESTION [0 upvotes]: I know that the following notation works:
$f:A \rightarrow \mathcal{F}(B,C)$, where $\mathcal{F}(B,C)=\{g|g:B \rightarrow C\}$.
But does the following notation make sense? I mean, is it correct to write:
$f: A \rightarrow (B \rightarrow C)$?

REPLY [0 votes]: Short answer: no. Almost, if you replace the parentheses with curly braces. But don't. Stick with the original notation.

Thanks, Ethan Bolker!


TITLE: Source for roots of matrix polynomials?
QUESTION [8 upvotes]: A
matrix polynomial
is a polynomial whose variables are square $n \times n$ matrices,
let's say with entries in $\mathbb{C}$, and with coefficients in $\mathbb{C}$.
I am seeking a source of results on solving such equations.
For example, $X^2 =0$ has infinitely many solutions, because, e.g.,
$$ X =
\left( \begin{array}{ccc}
0 & 0 & x \\
0 & 0 & 0 \\
0 & 0 & 0 \end{array} \right)
$$
is a solution for all $x \in \mathbb{C}$.
(Second example removed, as it did not fit the definition.)
There is such a rich set of results on roots of polynomial equations over $\mathbb{C}$
and over $\mathbb{R}$, I am hoping there are analogs when the variables are matrices
rather than single elements of fields. But my (superficial) explorations did
not uncover a comprehensive source on this topic.
To pose a specific question:

Q. Are there theorems that yield the number of solutions/roots
  for such matrix polynomial equations?

REPLY [1 votes]: F.R. Gantmaher, The theory of matrices is a good reference of this subject, in my opinion.


TITLE: restriction a non compact operator to compact operator
QUESTION [2 upvotes]: If  $T\in\mathcal{B}(X,Y)$ is not compact can the restriction of $T$ to an infinite dimensional subspace of $X$ be compact?

REPLY [0 votes]: What the OP probably has is mind, is the question under which circumstances, for a given operator $T\colon X\to Y$ one can find a subspace of $X_0$ of $X$ which is infinite-dimensional and on which $T$ is compact.
This is true for operators which are not bounded below on finite-codimensional subspaces of $X$ (in other words, for operators $T$ which are not upper semi-Fredholm). In this case we have Kato's lemma which asserts that for every $\varepsilon > 0$ you will find a basic sequence $(x_n)_{n=1}^\infty$ in $X$ so that $T$ restricted to $X_0:=\overline{{\rm span}}\{x_n\colon n\in \mathbb{N}\}$ is compact and $\|T|_{X_0}\|\leqslant \varepsilon$.
This fact is necessary to prove that sum of two strictly singular operators is again strictly singular.


TITLE: Tensor Product Question
QUESTION [0 upvotes]: For a finite dimensional vector space $V$, is it true that $\bigwedge^{n - 1}V \otimes V = \bigwedge^{n}V \oplus \ker(\bigwedge^{n - 1}V \otimes V \overset{\psi}{\rightarrow}\bigwedge^{n}V)$ where $\psi$ is just the natural map?

REPLY [1 votes]: If you have a surjective map of vector spaces $f: W \rightarrow W'$, then you have an exact sequence $0 \rightarrow \text{ker}(f) \rightarrow W \rightarrow W' \rightarrow 0$.  Since every short exact sequence of vector spaces splits, you get an isomorphism $W \cong W' \oplus \text{ker}(f)$.


TITLE: Taylor expansion of arctan x
QUESTION [2 upvotes]: How can I prove that the tailor expansion of arctan x converges only at [-1,1]? I was able to prove that it converges at the given interval but I don't know how to prove that it does not converge anywhere else . Can someone help?

REPLY [2 votes]: The derivative of $\arctan x$ is
$$\frac d{dx} \arctan x =\frac 1{1+x^2}$$
From the formula for geometric series, we have that
$1+y+y^2+y^3+...=\frac 1{1−y}$ if and only if $|y|<1$.
Plugging in $-x^2$ for $y$, we get that
$$\begin{align*}
\frac{1}{1+x^2} &= \frac{1}{1-(-x^2)} \\
&= 1 + (-x^2) + (-x^2)^2 + (-x^2)^3 + \cdots + (-x^2)^n + \cdots\\
&= 1 - x^2 + x^4 - x^6 + x^8 - x^{10} + \cdots
\end{align*}$$
provided that $|-x^2| \lt 1$; i.e., provided that $|x|\lt 1$.
And if $|x|\lt 1$, then $x \in [-1,1]$. If $x$ isn't in this interval, then we do not get convergence.