TITLE: Some examples of virtually cyclic groups
QUESTION [8 upvotes]: The only virtually cyclic groups (ie. groups containing $\mathbb{Z}$ as subgroup of finite index) I really know are : the groups $F \times \mathbb{Z}$, where $F$ is a finite group, and the infinite dihedral group $D_{\infty}$ (isomorphic to $\mathbb{Z}_2 \ast \mathbb{Z}_2$).
But all these groups are finitely presented, just-infinite (ie. their proper quotients are finite) and residually finite (ie. for all element $g$, there exists a morphism $\varphi$ onto a finite group such that $\varphi(g) \neq 1$).
So I am looking for examples of virtually cyclic groups without one of these properties. I only know that there exists a virtually abelian group not just-infinite but without having an explicit example.
As other virtually abelian groups, there is also the generalized dihedral groups $\text{Dih}(G)$ where $G$ is an infinite finitely generated abelian group, but I don't know them really. Are they virtually cyclic ?
NB: The groups I consider are finitely generated.

REPLY [3 votes]: The existing answer is restricted to virtually-cyclic groups, but more general things can be said: Finite presentability and residual finiteness are both preserved when moving from finite index subgroups to the big group. That is, suppose $H$ is a finite index subgroup of $G$. Then,

If $H$ is finitely presentable then so is $G$. This can be proven using covering spaces.
If $H$ is residually finite then so is $G$. The way to prove this is to remember that a group $T$ is residually finite if for each $x\in T$ there exists a subgroup $K_x$ of finite index in $T$ such that $x\not\in K_x$. So, suppose $x\in G$ and we shall find such a finite index subgroup of $G$. if $x\not\in H$ then we are done, by taking $K_x=H$, while if $x\in H$ then there exists some $K_x\in H$ such that $x\not\in K_x$ and $K_x$ has finite index in $H$. As $K_x$ has finite index in $H$ it also has finite index in $G$, as required.

Therefore, every virtually-cyclic subgroup is both finitely presentable and residually finite. So the groups you are looking for do not exist!


TITLE: How to fit fixed data from two linear functions
QUESTION [2 upvotes]: I have a set of points $(x, y)$ where each one comes from either one of two linear functions:
\begin{align*}
    y &= m_1 x + b_1\\
    y &= m_2 x + b_2
\end{align*}
Is there a fitting method to find such functions, without knowing from which function each of the points come from?
PS. can somebody add fit (or fitting) to the existing tags

REPLY [1 votes]: Imagine that your data is presented as a binary two dimensional image (assuming a unit value whenever the point coordinate (x,y) is present in your data). Then the problem is equivalent to straight line detection in binary images. The equivalent problem can be solved using the Hough transform for straight line detection.
The basic idea of the hough transform is as follows:

It transforms an array from the geometrical coordinates (x,y) to the space of initial points and slopes.
Every point contributes a unit to all sets of slopes and initial points of straight lines passing through this point.
The Hough transform of the whole array is the linear superposition of the transform of it's individual points.

The detected lines appear as peaks in the hough transform image


TITLE: Why are some extra dimension theories known as strongly coupled and others as weakly coupled?
QUESTION [3 upvotes]: I was looking at pdf file of the presentation of a conference talk. The speaker discusses two types of "mechanisms" for stabilizing the weak scale and calls them "weakly coupled" and "strongly coupled". The examples are:
Weakly coupled: SM with a light Higgs, SUSY, Little Higgs, Twin Higgs, Large extra dimensions, Universal extra dimensions.
Strongly Coupled: Technicolor, Topcolor, Top See Saw, Composite Higgs, Randall-Sundurm warped extra dimension models.
I need to understand what makes one such beyond standard model theory to be weakly coupled and another to be strongly coupled, in general and especially why large/universal extra dimension and Randall-Sundrum models belong to two separate groups.
Perhaps I should emphasize where my doubt is: I understand why theories like technicolour are thought to be "strongly coupled" but I do not at all understand why theories involving say Randall-Sundrum extra dimensional models are also so. When we say Randall-Sundrum warped extra dimension model, we mean a particular kind of background on which different types of interactions take place. But why should all interactions taking place in such a background will have to be strongly coupled? How does background spacetime decide whether an (or all) interaction(s) will be strongly coupled?
Or does this have anything to do with the fact that in (classical) GR, spacetime is dynamically determined and is coupled with matter and energy content of the universe and may be due to quantum gravity effects this strong/weak nature of interactions are/is manifested. But what is the way to explain that?

REPLY [3 votes]: Strongly coupled theories are those whose coupling constant is strong i.e. greater than a particular number of order of one, $g\gt {\mathcal O}(1)$. It means that perturbative expansions (and the leading simplest Feynman diagrams) are not good approximations for the most elementary physical processes that may occur in these theories. One must find other methods because the influence of quantum mechanics is important.
This is obviously the case for technicolor whose key processes depend on the confinment of the technicolor and confinement always means that the coupling is getting strong (and even stronger, arbitrarily stronger). But the other strongly coupled models in your list are analogous. On the other hand, the weakly coupled theories in your list are analogous to the Standard Model at high energies, a few leading terms i.e. simple Feynman diagrams are enough to get a good enough approximation.


TITLE: Partial fractions for a fraction with n-terms in the denominator
QUESTION [0 upvotes]: I was working doing some integrals for my project and i get stuck in how they use the partial fraction to obtain the following
$$
\dfrac{1}{(x+1)(x+2)(x+3)\cdots (x+n)}=\sum_{k=1}^{n} \dfrac{c_k}{x+k}
$$
where $c_k=\dfrac{\pi^{-k}}{k}$.
Can anyone help me with this? Or give me some ideas in how to obtain it.
Thank you in advance for your help.

REPLY [1 votes]: The correct expression for $c_k$ is:
$$c_k=\frac{1}{\prod\limits_{1\leq i\leq n,\,i\ne k}(i-k)}$$
This can be obtained most easily from the so-called "cover up method" for partial fractions.
To see all the steps:
$$\frac{1}{(x+1)\cdots(x+n)} = \frac{c_1}{x+1}+\cdots+\frac{c_n}{x+n}$$
Multiply both sides by the denominator on the left:
$$1 = \sum_{k=1}^n \left(c_k\prod_{1\leq i\leq n,\,i\ne k}(x+i)\right)$$
Setting $x=-k$, every term on the right except for the $k$th term becomes $0$, so we have:
$$1=c_k\prod_{1\leq i\leq n,\,i\ne k}(-k+i)$$
This leads directly to our solution.


TITLE: Prove that if polynomial $f(x) = x^6 + ax^3 + bx^2 + cx + d$ null places are real, then$ a = b = c = d = 0$
QUESTION [0 upvotes]: if polynomial  null places are all real, then how can i show that $a = b = c = d = 0$?

$x(x(0x + 0 + x^4) + 0) + 0 => x^6 = f(x)$ ?
But that doesn't prove anything for me i think

REPLY [0 votes]: As already stated in a comment, you can use Rolle's theorem: If
$$ f(x) = x^6 + a x^3 + b x^2 + c x + d$$
has 6 real zeros (counted with multiplicity) then
$$ f'(x) = 6x^5 + 3a x^2 + 2b x + c$$
has 5 real zeros. Repeating this argument twice it follows that
$$ f'''(x) = 120x^3 + 6a $$
has 3 real zeros (with multiplicity). This is possible only if $a = 0$.
Now work backwards: If
$$ f''(x) = 30x^4 + 2b$$
has 4 real zeros then $b = 0$, and so on ...


TITLE: Elementary question on Dynkin system.
QUESTION [1 upvotes]: I have got a question while I am solving a problem.
Show that $B\setminus A$ is in a Dynkin system if $A$ and $B$ is in the Dynkin system.
I was trying to:
$B\setminus A=B\cap A^c=B\cap(A^c\cup B^c)=B\cap(A\cap B)^c\\
=\{B^c\cup(A\cap B)\}^c $
Thus, we can argue that it is in the system because $B^c$ and $A\cap B$ is disjoint. However, I was confused that how can I ensure $A\cap B$ is in the system since there is no finite intersection property. How can I draw this using the definition of Dynkin system? or Should I find another way?

REPLY [1 votes]: It is in general not true that a Dynkin system $\mathcal D$ is closed under set differences. Consider $X = \{1,2,3,4\}$ with the Dynkin system $\mathcal D = \{S \subseteq X \mid \#S \text{ is even}\}$. Then $\{1,2\},\{1,3\} \in \mathcal D$, but $\{2\} = \{1,2\} \setminus \{1,3\} \not \in \mathcal D$.
However, if $A \subseteq B$, your proof shows that $B \setminus A \in \mathcal D$, provided $A,B \in \mathcal D$ for any given Dynkin system $\mathcal D$.


TITLE: The vector space formed on C[0,1] and the norm $(\int_{0}^{1}|f(t)|^2 dt)^{1/2} $
QUESTION [4 upvotes]: How does one show that $(C[0,1], ||.||_{2})$ where $||.||_2=(\int_{0}^{1}|f(t)|^2dt)^{1/2}$ and $C[0,1]$ is the space of all continous function which are mapped from $[0,1]\rightarrow \mathbb{R}$, is not complete?
First to show that $(C[0,1], ||.||_{2})$ is a normed vector space:
$||.||_2$ fulfills $||f(t)||=0 \Rightarrow f(t)=0$
$||.||_2$ fulfills the  scalar multiplication property: $$||\lambda f(t)||_{2} = (\int_{0}^{1}|\lambda f(t)|^2dt)^{1/2}=|\lambda|(\int_{0}^{1}| f(t)|^2dt)^{1/2}=|\lambda|||f(t)||_2$$
Thirdly, it also fulfills the triangle inequality:$$||x(t)+y(t)||_2^2 = \int_{0}^{1}|x(t)+y(t)|^2dt \le \int_{0}^{1}|x(t)|^2dt + 2\int_{0}^{1}|x(t)y(t)|dt +\int_{0}^{1}|y(t)|^2dt \le ||x(t)||_2^2 + 2||x(t)||_2||y(t)||_2+||y(t)||_2^2 = (||x(t)||_2+||(y(t)||_2)^2$$
this also shows the closure under addition in $C[0,1]$
so , $(C[0,1],||.||_2)$ is a normed vector space , but how can we show that it is not complete?

REPLY [5 votes]: You can prove that a metric or normed space is not complete by finding a Cauchy sequence whose limit is not in the space.
For example consider the sequence $f_n$ where $f_n = 0$ on $[0, \frac12]$, $1$ on $[\frac12 + \frac1n, 1]$ and linear on $[ \frac12,  \frac12 +  \frac1n]$. Then $f_n$ is Cauchy with respect to $L^2$ (prove it) but its limit $f$ which is $0$ on $[0, \frac12]$ and $1$ on $[\frac12,1]$ is not in $C[0,1]$.


TITLE: What number is midway between 2/4 and 3/4?
QUESTION [0 upvotes]: Sorry for posting this question but appreciate if you could help me.
What number is midway between 2/4 and 3/4?

REPLY [0 votes]: If you look at some measuring tape and look and the sixteenth hash-marks you can see that 10/16ths is in between 2/4 and 3/4


TITLE: evaluate $\lim_{n\to\infty}\left(\frac{n^3-n^2+2n+3}{n^3+2n^2-1}\right)^{\left(\frac{8n^3-3n^2+2n-2}{10n^2+3n-1}\right)}$
QUESTION [1 upvotes]: $$
\lim_{n\to\infty}\left(\frac{n^3-n^2+2n+3}{n^3+2n^2-1}\right)^{\left(\frac{8n^3-3n^2+2n-2}{10n^2+3n-1}\right)}
$$
I have tried the following:
$$
\lim_{n\to\infty}\left(\frac{n^3-n^2+2n+3}{n^3+2n^2-1}\right)^{\left(\frac{8n^3-3n^2+2n-2}{10n^2+3n-1}\right)}=\\
\lim_{n\to\infty}\left(\frac{n^3}{n^3}\right)\cdot\left(\frac{1-\frac{1}{n}+\frac{2}{n^2}+\frac{3}{n^3}}{1+\frac{2}{n}-\frac{1}{n^3}}\right)^{\left(\frac{n^2}{n^2}\right)\left(\frac{8n-3+\frac{2}{n}-\frac{2}{n^2}}{10+\frac{3}{n}-\frac{1}{n^2}}\right)}$$
So I have a limit of $1^{\infty}$

REPLY [2 votes]: Just write

$\frac{n^3-n^2+2n+3}{n^3+2n^2-1} = 1 - \underbrace{\frac{3n^2-2n-4}{n^3+2n^2-1}}_{a_n :=}$ and note that
$\lim_{n\to \infty}(1-a_n)^{\frac 1{a_n}} = \frac 1e$

Now, you need only the limit

$\lim_{n\to\infty}a_n\frac{8n^3-3n^2+2n-2}{10n^2+3n-1} = \lim_{n\to\infty}\frac{3n^2-2n-4}{n^3+2n^2-1}\frac{8n^3-3n^2+2n-2}{10n^2+3n-1} = \frac{12}{5}$

Putting all this together you get
$$\lim_{n\to\infty}\left(\frac{n^3-n^2+2n+3}{n^3+2n^2-1}\right)^{\left(\frac{8n^3-3n^2+2n-2}{10n^2+3n-1}\right)} = \lim_{n\to\infty}\left(\left(1-a_n\right)^{\frac{1}{a_n}}\right)^{a_n\left(\frac{8n^3-3n^2+2n-2}{10n^2+3n-1}\right)}$$
$$ = \left( \frac{1}{e}\right)^{\frac{12}{5}}$$


TITLE: Elements in sigma algebra generated by sets (A,B)
QUESTION [11 upvotes]: I understand that the number of elements generated by sigma(A,B) is 16. But I'm not able to find all of them.
I have found: $A$, $B$, $A^c$, $B^c$, $A ∩ B$, $A^c ∩ B^c$, $A^c ∪ B^c$, the empty set, the sample space, $A^c ∩ B$, $A ∩ B^c$, $A^c ∪ B$, $A ∪B^c$, $A ∪ B$. What are the last two?

REPLY [1 votes]: The last two sets are $(A\cap B^c)\cup(A^c\cup B)$ and  $(A\cap B)\cup(A\cup B)^c$.


TITLE: What is the average distance of a combination set?
QUESTION [1 upvotes]: I'm working on a genetic algorithm and would like to map each function to a set of "codons".  So + -> 011. Given this, I would like to figure out how easy it would be for any given codon set to mutate into another codon set.
If you were to take two combinations, say 012 and 111, they would have a distance of 2 since because it would take 2 mutations to move from one to the other. Eg. 012 - > 011 - > 111.
So, here's my question : given an (n, k) combination set, what is the average distance between any two combinations?
I think you might be able to model this as a small world network, but if there is a better approach I'm all ears.

REPLY [1 votes]: You seem to be talking about
$k$-tuples with $n$ possible values.  If two of those are randomly chosen,  the probability that they differ in the $j$'th position is $1 - 1/n$, and the expected Hamming distance between them is $k(1-1/n)$.


TITLE: In Verlet integration, why is $a*dt*dt$ considered "force vector"?
QUESTION [0 upvotes]: In Verlet integration, why is $a*dt*dt$ considered "force vector"?
Intuitively, since Verlet integration is:
$$x_{i+1}=x_i+(x_i - x_{i-1}) + a * dt * dt$$
then this doesn't have other "force" terms than $a * dt *dt$. However I'm unsure as to what the $dt*dt$ does here.

REPLY [2 votes]: It takes the centered difference approximation to the second derivative of $x$, namely $\frac{x_{i+1}-2x_i+x_{i-1}}{\Delta t^2}$, and sets it equal to $a(x_i)$ (implicitly assuming a force that is dependent only on position), and then solves for $x_{i+1}$.


TITLE: Finding a basis and dimension of a subspace
QUESTION [0 upvotes]: I wish to find the basis and dimension of the subspace $U = \{p \in P_4(\mathbb{R}) \mid p(1) = p'(1) = 0\} \subseteq V = P_4(\mathbb{R})$
In my solution I find conditions on the coefficients of the fourth degree polynomials: $a_4 = -a_3 - a_2 - a_1 - a_0, \; a_3 = -2a_2 -3a_1 - 4a_0$. So any polynomials in $U$ must have the tuple of coefficients: $(a_1, a_2 , -2a_2 - 3a_1 - 4a_0, -a_3 - a_2 -a_1 - a_0)$.
But I am confused because the set of vectors I form from the above tuple has no monomials in it. So it can't span $U$ since $U$ is a subspace of polynomials.
Where have I gone wrong in my logic in my solution?
The below is my solution

REPLY [1 votes]: $2$ restrictions on a $5$-dimensional space should indeed lead to a $3$-dimensional subspace.
$\{(x-1)^2,(x-1)^3,(x-1)^4\}$ is a set of linearly independent elements of $P_4(\mathbb R)$ each of which satisfies the requirements...
It is easy to see that $\{1,x\}$ spans a $2$-dimensional subspace whose intersection with $U$ is $\{0\}$.  This shows the dimension is indeed $3$...


TITLE: prove the limit inferior of $(x_n)$ where $n \in\mathbb{N}$
QUESTION [1 upvotes]: The problem states let $(x_n)$ be a bounded sequence for each $n \in\mathbb{N}$. Let $t_n=inf\{x_k: k\geq n\}$. Prove that $(t_n)$ is monotone and convergent.
After a little research because I was confused what limit inferior really was this was the proof I tried coming up with. Proof: Since $(x_n)$ is bounded then it has a supremum and info um. Let $t_n=inf\{x_n: k\geq n\}=inf\{x_n,x_{n+1},...\}$. It follows that $t_{n+1}=inf\{x_{n+1}: k\geq n+1\}=inf\{x_{n+1},x_{n+2},...\}$ If we let set $A= \{x_{n+1},...\}$ and set $B=\{x_n,...\}$ (Note* I assume that set A and B have greatest lower bounds since these are subsets of $(x_n)$ and $(x_n)$ is bounded). Then it follows that $A \subset B$ and $inf A\geq inf B$ Otherwise if $inf A<inf B$ then $inf A \leq x_n$ a contradiction since $x_n \notin B$.  It follows that $x_{n+1} \geq x_n$ otherwise if $x_n> x_{n+1}$ then $inf A \leq x_n$ a contradiction from above. Hence $t_n$ is increasing and by the Monotone convergence theorem $t_n$ is monotone and convergent. Not sure If I have all this correct.

REPLY [1 votes]: It is well known that an increasing sequence bounded above converges.
Since $(x_n)$ is bounded above $(t_n)$, always being a lower bound of some subset of $\{x_n : n \in \mathbb{N}\}$, is also bounded above.
Let $A_n = \{x_k : k \ge n\}$.  We have $A_{n+1} \subseteq A_n$, and so for any $x \in A_{n+1}$, $t_n \le x$.  Therefore,  $t_n$ is a lower bound of $A_{n+1}$, so by definition,
$$t_n \le \inf A_{n+1} = t_{n+1}$$


TITLE: Question about the binomial distribution
QUESTION [5 upvotes]: $X$ is a random variable with binomial distribution parameters $n$ and $p_1$.
$Y$ is a random variable with binomial distribution parameters $n$ and $p_2$.
$p_1 < p_2$
How can I show that $P(X \leqslant k) \geqslant P(Y \leqslant k)$?
Please just give me a hint.
I tried comparing the terms ${n \choose k}p_1^k(1-p_1)^{n-k}$, but this doesn't work (after trying this, it's obvious in hindsight that it doesn't work.

REPLY [5 votes]: Hint: define the two variables on the same probability space. Specifically, let $X = \sum_{i=1}^n A_i$ and $Y = \sum_{i=1}^n B_i$, where $\{A_i\}$ and $\{B_i\}$ are each collections of independent Bernoulli variables. However, $A_i$ and $B_i$ should be dependent variables; specifically, they should each be defined on the common sample space $[0, 1]$ in the following way:
$$(A_i, B_i) \mapsto \begin{cases}
  (1, 1), & x < p_1\\
  (0, 1), & p_1 < x < p_2\\
  (0, 0), & x > p_2.
\end{cases}$$


TITLE: How to extract velocity and acceleration from $N$-body simulation data?
QUESTION [0 upvotes]: I have data from an $N$-body simulation and want to do some data analysis. I would like to extract the velocity and acceleration of each particle. The data consists only of the positions of each particle for different points in time. For a single particle, the data looks as follows
step x y
1 0.952 -0.336
2 0.918 -0.333
3 0.925 -0.325
4 0.912 -0.332

What is the best approach to extract these quantities from my data?

REPLY [2 votes]: You will have to use numerical differentiation to obtain velocities and accelerations if this is all that you have. For the numerical methods you also have to decide how to treat the edge points. Finally, to obtain the velocity and acceleration in meaningful units, you also need to know what value and unit $dt$ has and in what unit your position coordinates are.
Here is a small Python code snippet how to calculate the velocity and acceleration based on an array containing positions at equidistant time spacing. The positions are array x and would correspond to your x column. vx is then the array with the velocities, obtained by applying the numerical derivative as provided by numpy.gradient, to obtain the correct values you have to divide by the timestep. The acceleration ax is obtained in the same way using the velocities as input. In the end, its just the discretized version of the equations
$$
v_x = \frac{dx}{dt}\\
a_x  = \frac{d^2x}{dt^2} = \frac{dv_x}{dt}
$$
# -*- coding: utf-8 -*-

import numpy as np
import matplotlib.pyplot as plt

t, dt = np.linspace(0, 2*np.pi, 1000, retstep=True)
x = 2*t*np.sin(t)
vx = np.gradient(x)/dt
ax = np.gradient(vx)/dt

fig, (ax0,ax1,ax2) = plt.subplots(3,1)
ax0.plot(t,x)
x_abs_max = np.abs(x).max()
ax0.set_ylim(-x_abs_max, x_abs_max)
ax1.plot(t,vx)
vx_abs_max = np.abs(vx).max()
ax1.set_ylim(-vx_abs_max, vx_abs_max)
ax2.plot(t,ax)
ax_abs_max = np.abs(ax).max()
ax2.set_ylim(-ax_abs_max, ax_abs_max)
plt.show()

This only works for equidistant data.
Alternatively you could also do an interpolation of your position values and take the derivatives of the obtained interpolation functions. Most interpolation functions provide analytic derivatives and thus "analytic" velocities and accelerations. This approach is more convenient if you need callable functions that provide velocities and accelerations at any value within your boundaries.


TITLE: Constructing two independent random variables from two given random variables
QUESTION [0 upvotes]: Suppose I have two random variables $X_1$ and $X_2$ with a joint exponential pdf, say
$$
f(x_1,x_2)=abK\cdot \exp\left\{-(ax_1+bx_2+cx_1 x_2 )\right\},
$$
for $x_1>0$ and $x_2>0$, and for some $a,b,c>0$, and $K$ is a normalizing constant.
I want to construct a random vector $Y=(Y_1,Y_2)^T$ by some transformation
$$
Y=AX,\text{ where }X=(X_1,X_2)^T,
$$
such that $Y_1$ and $Y_2$ are now independent, (but not necessarily exponential).
What should  be the components of (the square matrix) $A$ here, if this is possible?
Can anyone also recommend an article or a book that might have discussed my problem or anything similar to this?

REPLY [0 votes]: You don't even need two variables.
First, transform $X_1$ to uniform distribution on $[0; 1]$ via some function $f$. Now let $\gamma: [0; 1] \to [0; 1]^2$ be Hilbert version of Peano curve, and $\gamma_1$ and $\gamma_2$ been it's $x$ and $y$ coordinates. Then $Y_1 = \gamma_1(f(X_1))$ and $Y_2 = \gamma_2(f(X_1))$ are independent (see "Stochastic Independence and Space-Filling Curves" by John A. R. Holbrook for proof; the main idea is that probability of getting into any square with side $2^{-k}$ and angle with coordinates $(n \cdot 2^{-k}, m\cdot 2^{-k})$ is $4^{-k}$ starting from the $k$-th iteration).


TITLE: Group of order 35, and normal subgroup of order 7
QUESTION [1 upvotes]: G a group of order 35, H a normal subgroup of order 7. Prove that if g in G has order 7, then g is in H.

What I'm doing is invoking Sylow's third theorem to show that there exist a single subgroup of order 5 and a single subgroup of order 7. And then the direct product of these has order 35, and then somehow... this means G has an element of order 35, which would mean G is cyclic, and if G is cyclic, each of its subgroups are cyclic. So H has an element of order 7 and if there's an element g in G of order 7, then it generates the subgroup H?
This is my train of logic? I could use some help on the somehow... part, or if that's even a valid way of proving that a group of order 35 is cyclic. Please help steer me back on track.

REPLY [3 votes]: Let $K=\langle g\rangle$ be a subgroup of $G$ of order $7$. Since $H$ is a normal so $HK\leq G$ and so $|HK|\bigg||G|$. This means that $$\frac{7\times 7}{s}\mid 7\times 5$$ in which $|H\cap K|=s$. But $s=1$ cannot be happen.


TITLE: What counts as a standard model of arithmetic?
QUESTION [4 upvotes]: In my research so far, I've found that the canonical standard model of arithmetic is $\mathbb{N}$ under the addition and multiplication operations. However, I've been unable to find much on any other standard model of arithmetic.
Is $\mathbb{N}$ the only standard model of arithmetic, or are there others? I intuitively imagine that $\mathbb{Q}$ under addition and $\mathbb{R}$ under addition and multiplication are standard models, but I've yet to find any support for this. Is a model either standard or nonstandard, or is there an in-between where stuff like $\mathbb{Q}$ and $\mathbb{R}$ would go? If there's more than one, then what precisely defines a standard model of arithmetic?

REPLY [9 votes]: There is a standard model of arithmetic: any model of the Peano axioms + the second order axiom of induction. Any two such models are isomorphic (a fact known as the categoricity of the theory). The standard models of arithmetic are precisely these models.
A similar story holds for the reals. Any model of the axioms of a complete ordered field is a standard model of the reals. Any such models are isomorphic, so again the theory is categorical.
In light of this, the obtain nonstandard models one must relax some of the axioms (for each theory). Nonstandard models of arithmetic are basically models of the Peano axoims + the induction scheme axioms (one axiom for every relevant sentence, so countably many axioms, but all are first order). Any such model that is not isomorphic to the standard model of arithmetic is called a nonstandard model. There are plenty such models just by virtue of the Lowenheim-Skolem theorem + the compactness theore: There is a nonstandard model of any infinite cardinality. So there are plenty of non-isomorphic models of nonstandard arithmetic.
A similar story holds for nonstandard models of the reals. It's also worth mentioning that models can also be constructed using ultrapowers and all constructions rely on some form of the axiom of choice. Also, historically, nonstandard models appeared in logic via the Lowenheim-Skolem theorem, and made their serious appearance in (more) mainstream (at least at that time) mathematics in the work of Robinson on nonstandard analysis.

REPLY [3 votes]: The ordinary natural numbers, under the usual addition and multiplication, are a standard model of arithmetic. Any model not isomorphic to these is called non-standard. The usual convention in logic is that by $\mathbb{N}$ we mean the numbers $0,1,2,\dots$.
We have left the term arithmetic undefined. One usually has a particular theory in mind, such as first order Peano arithmetic. Or perhaps the theory whose axioms are all sentences of the usual language of arithmetic that are true in $\mathbb{N}$. That is a theory incomparably stronger than first order Peano arithmetic.
The term arithmetic may be confusing here. The word has many meanings. In particular, it is an old-fashioned term for number theory. That is what arithmetic means when we refer to non-standard models of arithmetic.
Remark: The structures $\mathbb{Q}$ and $\mathbb{R}$, under the usual operations, are not models of arithmetic. To see this, let $\varphi$ be the sentence $\forall x(\lnot(x=0)\longrightarrow \exists y(xy=1))$. Then $\varphi$ is false in $\mathbb{N}$, but true in $\mathbb{Q}$ and $\mathbb{R}$. There are many other similar sentences.


TITLE: Helix equation around vector
QUESTION [2 upvotes]: I'm looking to get the equation $(x,y,z)$ of a 3d helix according to a vector.

I'd like create helix around one vector (for example around the $X,Y $or $Z$ vectors on the picture). I know how to draw it around one axis :

$x = t, y = \cos(t), z = \sin(t) $ for the helix around the $x$-axis $(1,0,0)$
$x = \cos(t), y = \sin(t), z = t$ for the helix around the $z$-axis $(0,0,1)$
$x = \cos(t), y = t, z = \sin(t)$ for the helix around the $y$-axis $(0,1,0)$

But I'd like to generalize it to every vector (for example $(1,1,0)$ or $(1,1,1)$).
Thanks for the help!

REPLY [0 votes]: Your elix has a fixed radius $r=1$, a fixed step $a=2\pi$, and its axis passes through the origin $O$.
I show a more general version.
Simply choose three orthogonal univ vectors $\mathbf{i},\mathbf{j},\mathbf{k}$, where $\mathbf{k}$ is the direction of the axis of the elix, and you have
$$
P=P_0+r\cos(t)\mathbf{i}+r\sin(t)\mathbf{j}+\frac{a}{2\pi} t\mathbf{k}
$$
or in coordinates:
\begin{align}
x &= x_0 + r\cos(t)i_1 + r\sin(t)j_1 + \frac{a}{2\pi} t k_1, \\
y &= y_0 + r\cos(t)i_2 + r\sin(t)j_2 + \frac{a}{2\pi} t k_2, \\
z &= z_0 + r\cos(t)i_3 + r\sin(t)j_3 + \frac{a}{2\pi} t k_3
\end{align}
You can obltain your elix with $P_0 = O = (0,0,0)$, $\mathbf{i} = (1,0,0)$, $\mathbf{j} = (0,1,0)$ and $\mathbf{k} = (0,0,1)$ and $r=1$, $a=2\pi$:
\begin{align}
x &= 0 + 1\,\cos(t)\,1 + 1\,\sin(t)\,0 + 1\, t\, 0 &&= \cos(t), \\
y &= 0 + 1\,\cos(t)\,0 + 1\,\sin(t)\,1 + 1\, t\, 0 &&= \sin(t), \\
z &= 0 + 1\,\cos(t)\,0 + 1\,\sin(t)\,0 + 1\, t\, 1 &&= t
\end{align}


TITLE: Two-dimensional random walk
QUESTION [7 upvotes]: Suppose we have a particle in the plane at the origin $(0,0)$. It moves randomly on the integer lattice $Z^2$ to any of the adjacent vertexs with equal probability $1/4$. What's the probability of reaching a fixed point $(x,y)$ before returning to the origin?
The analogous problem in one dimension is easy. The probability is:
$
\dfrac{1}{2|x|}
$
I have read some related articles working on finite graphs; but I am not be able to obtain the answer for my problem.
Thank you very much for your attention

REPLY [1 votes]: http://puhep1.princeton.edu/~mcdonald/examples/EM/atkinson_ajp_67_486_99.pdf
I haven't checked its validity, but it has Mathematica code for calculating what you want. As a side notice, the asymptotics of this probability is C/log r, (where r is the distance to (x,y)).


TITLE: Order of equation that is used to get the matrix for calculating eigenvalues?
QUESTION [0 upvotes]: My professor in class has taught this equation as being $det(\lambda I - A)$ mostly with regards to finding the eigenvalues and eigenvectors of liner transformations but I have seen it online as $det(A - \lambda I)$. Does the order matter in these cases? I tried to google this but I couldn't find anything

REPLY [1 votes]: It doesn't matter. Recall that multiplying a single row (or equivalently, column) of a matrix $M$ by a scalar $a$ multiplies the determinant by $a$. Note also that $\lambda I - A = (-1)(A - \lambda I)$ is the result of multiplying all rows (or columns) by $-1$. Therefore we have,
$$\det (A - \lambda I) = (-1)^n\det(\lambda I - A)$$
where $n$ is the number of rows/columns.
So, we therefore have
\begin{align*}
&\det (A - \lambda I) = 0 \\
\iff \, &(-1)^n\det(\lambda I - A) = 0 \\
\iff \, &\det(\lambda I - A) = 0.
\end{align*}


TITLE: Help on mathematical notation and formalisation for the following description
QUESTION [2 upvotes]: I would like to describe a hypothesis, formally. So I have defined $p = {t_1,t_2,t_3, \cdots, t_n}$
and I would like to say
"A pattern $p_i$ is more important than $p_j$, if $t \in p_i$ is more important than $t \in p_j$", I know that it is a bit vague to describe in this manner, so what will be a better option, the ideas I have in mind.

To define an importance function or weight function saying $f(p) \propto f(t) $
To describe purely in words.

I kind of want to avoid 2nd option as I want to find a way and learn how to formally define relationship between variables.
Thank you.
Regards,
Andy

REPLY [1 votes]: Let $P$ and $Q$ be unordered $n$-tuples of real numbers. More formally, $P\subset \mathbb{R}$ and $Q\subset \mathbb{R}$ with $|M|=|N|=n$.
Let $f$ be a function which assigns an importance weight to each element in $P$ and $Q$.
Formally, $f$ is a function $f:P\cup Q\rightarrow \mathbb{R}^+$.
Definition
$$P \text{ is more important than } Q :\Leftrightarrow \sum_{t\in P}f(t) \ge \sum_{u \in Q}f(u)$$


TITLE: Is it possible to simplify this expression $ \frac{(x^4 \cos x)+(4x^3 \sin x)}{(x^4 \sin x)} $?
QUESTION [0 upvotes]: I am having trouble figuring out if I can simplify this expression anymore. This is the most I was able to simplify it with. Let me know if anyone has ideas please! Thanks
$$
\frac{(x^4 \cos x)+(4x^3 \sin x)}{(x^4 \sin x)}
$$

REPLY [0 votes]: Hint:
let $a=x^4 \cos x, b=4x^3 \sin x, c=x^4 \sin x$
,For $a, b \in \mathbb{R}$ and $c\in \mathbb{R^{*}}$  we have: $\displaystyle\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$


text:
{\bf Problem.} Suppose that $x = 2-t$ and $y = 4t + 7$.  If $x = -3$, what is $y$?
{\bf Level.} Level 2
{\bf Type.} Prealgebra
{\bf Solution.} If $x=-3$, then $-3 = 2-t$, so $t = 5$.  Therefore, $y = 4(5) +7 =\boxed{27}$.

set_name: MATH


TITLE: If the derivative of a function is bijective, is the function itself bijective as well?
QUESTION [0 upvotes]: Or the inverse: Is it possible for a function to  exist whose derivative is bijective, but is not bijective itself?

REPLY [1 votes]: The answer to the title question is "no."
Consider $f(x)=x^2$.
Is it bijective?
What is its derivative, and is that bijective?


TITLE: Vector version of concentration of Lipschitz functions on sphere (Levy's Lemma)
QUESTION [1 upvotes]: Levy's Lemma asserts Lipschitz functions of vectors chosen uniformly from the unit hypersphere concentrate:
Lemma.
Suppose $f:\mathbb{S}^{d-1} \to \mathbb{R}$ is $L$-Lipschitz on the unit hypersphere. Then, if $\vec{x}$ is drawn uniformly from the $d$-dimensional unit hypersphere, for some constant $C>0$,
\begin{equation*}
    \mathbb{P}[ |f(\vec{x}) - \mathbb{E}[f(\vec{x})]| > \epsilon ]
    \leq 2\exp\left( \frac{-C(d+1)\epsilon^2}{L^2} \right).
\end{equation*}
I am interested in 2-norm bounds for the vector case ($f:\mathbb{S}^{d-1}\to\mathbb{R}^n$ and $f$ satisfies $\|f(\vec{x}) - f(\vec{y})\|_2 \leq L \| \vec{x}-\vec{y}\|_2$). In particular, I am wondering whether a $d$ dependence is necessary? I have tried searching for such a result as I imagine this problem is standard, but haven't seemed to find the right keyword.

REPLY [1 votes]: One can avoid the  $n$-dependence. Let $H$ be a Hilbert space, endowed with the norm
$\|\cdot\|$.
Given $f:\mathbb{S}^{d-1}\to H$ which is $L$-Lipschitz, the goal is to prove a concentration inequality for $f$. Let $Y=(Y_1,\ldots, Y_d)$ be a Gaussian vector in $\mathbb R^d$, with mean $0$ and covariance matrix  $I$. Recall that $Y/|Y|$
is uniformly distributed on $\mathbb{S}^{d-1}$, where $|Y|$ is the Euclidean norm of $Y$.
The $\chi^2$ variable $|Y|^2$ satisfies the inequality:
$$P(|Y|^2<d/2)=P\Bigl(d-|Y|^2 \ge 2\sqrt{d \cdot d/16} \Bigr)\le e^{-d/16} \tag{*}$$
by [1], eq. (4.4).
Define $g:\mathbb R^d \to H$ by
$$g(y):=f\circ \psi, \quad \text{where} \quad  \psi(y)=\frac{y}{|y| \vee  \sqrt{d/2} }  \,.$$
Since $\psi$ is Lipschitz with constant $c_1=\sqrt{2/d}$, we infer that $g$ is Lipschitz with constant $c_1 L$.
Next, consider the Doob Martingale
$M_k:=E[g(Y) |Y_1,\ldots,Y_k]$ for $k=0,1,\ldots,d$.
We can apply Pinelis' Theorem 3.3. from [2] to this Martingale to obtain the desired concentration for $g(Y)$. Finally note that $(*)$ implies that
$$P[g(Y) \ne f(Y/|Y|)] \le e^{-d/16} \,.$$
[1] Laurent, B.  ; Massart, P.
Adaptive estimation of a quadratic functional by model selection.
Ann. Statist. 28 (2000), no. 5, 1302–1338. https://projecteuclid.org/journalArticle/Download?urlid=10.1214%2Faos%2F1015957395
[2] Pinelis, Iosif. "Optimum bounds for the distributions of martingales in Banach spaces." The Annals of Probability (1994): 1679-1706. https://projecteuclid.org/journals/annals-of-probability/volume-22/issue-4/Optimum-Bounds-for-the-Distributions-of-Martingales-in-Banach-Spaces/10.1214/aop/1176988477.pdf


TITLE: How to proceed this definite integral?
QUESTION [1 upvotes]: $\int_{\frac{1}{2014}}^{2014}\frac{\arctan x}{x}\text{d}x$

REPLY [0 votes]: Here's a hint: $\arctan(x) + \arctan(1/x) = \pi/2$ for $x > 0$. Try adding the integral to itself (perhaps with some changes to it by substitution).


TITLE: Let $P(X),Q(X)\in K[X]$, with $P(X)$ irreducible. Suppose there is $\alpha$ in an extension of $K$ such that $P(\alpha)=Q(\alpha)=0$.
QUESTION [0 upvotes]: "Let $P(X),Q(X)$ $\in K[X]$, with $P(X)$ irreducible. Suppose there is $\alpha$ in an extension of $K$ such that $P(\alpha)=Q(\alpha)=0$. Show that Q(X) is multiple of P(X). If $Q(x)$ is also irreducible, then $P(X)$ and $Q(X)$ are associated."
right, if $\alpha$ is the root of $P(X)$ and $Q(X)$, then there are $t_{1}(X)$ and $t_{2}(X)$ in $K[X]$ so that
$$P(X)=(x-\alpha)t_{1}(X)$$ and $$Q(X)=(x-\alpha)t_{2}(X)$$ And from here, I can't show that $P(X)|Q(X)$. Someone can help me?

REPLY [1 votes]: In the extension $F$, $X-\alpha$ is a common divisor of $P$ and $Q$ and therefore a divisor of the GCD $P\wedge Q$. Since the GCD is the same computed in $K[X]$ and $F[X]$, we see that $P$ and $Q$ have a common factor in $K[X]$.
But of course, since $P$ is irreducible, this means that $P|Q$.


TITLE: Proving $ord_p(ζ_p-1)=1/(p-1)$
QUESTION [1 upvotes]: After proving this, I was able to deduce an even more general result that $ord_p(ζ_p-1)=ord_p(ζ_p^2-1)=...=ord_p(ζ_p^{p-1}-1)$. Now, according to Lubin, $ord_p(ζ_p-1)$ should be $1/(p-1)$, but this isn't obvious to me so I need to prove it.
My thought was to multiply out the product $(ζ_p-1)(ζ_p^2-1)...(ζ_p^{p-1}-1)$ and hope that I get something with p-order $1$ (since this would finish the proof), but I'm stuck.
I know that $ord_p(ζ_p^k)=0$ for any $k \geq 0$. So if we multiply out the product, we get a sum with all of its terms having p-order $0$. I see from this that $ord_p[(ζ_p-1)(ζ_p^2-1)...(ζ_p^{p-1}-1)] \geq 0$.

REPLY [1 votes]: $|a|_p = p^{-v_p(a)}$
Since $\zeta_p^p = 1$ then $|\zeta_p|_p=1$
$\sum_{m=0}^{p-1} x^m =\frac{x^p-1}{x-1}=\prod_{k=1}^{p-1} (x-\zeta_p^k)$
$\prod_{k=1}^{p-1} |1-\zeta_p^k| = |\sum_{m=0}^{p-1} 1^m|_p = p^{-1}$ so there is some $k$ such that $|1-\zeta_p^k|_p <1$.
$1-\zeta_p^{kn} =1- (1+ (1-\zeta_p^k))^n =1-(1+n(1-\zeta_p^k)+O((1-\zeta_p^k)^2)$
whence (for $p \nmid n$)
$|1-\zeta_p^{kn}|_p = |n|_p |1-\zeta_p^{k}|_p=|1-\zeta_p^{k}|_p$ and $|1-\zeta_p^{k}|_p^{p-1} = p^{-1}$
letting $kn \equiv 1\bmod p$ you get your result $|1-\zeta_p|_p =p^{-1/(p-1)}$
Note this is also a proof that $\sum_{m=0}^{p-1} x^m$ is irreducible since we need to multiply all the $1-\zeta_p^k, k \in 1 \ldots p-1$ to obtain something of integer valuation