TITLE: What is formal intersection?
QUESTION [0 upvotes]: In this answer, formal intersection appears in this context:

Let each intersection $U_{i_1} \cap \cdots \cap U_{i_n}$ be denoted as
  $U_{i_1 \cdots i_n}$. Treat these sets as symbols (i.e., formal
  intersection), and distinguish $U_{ij}$ from $U_{ji}$ even though they
  are the same set.

How can the notion of "formal intersection" be made precise? I found nowhere a definition of it. I mean a similar definition as that of formal sum.
Edit: I mean a definition using only well-defined mathematical objects, such as set, function, etc, and no reference to the notation. Just like in the case of the definition of formal sum in the referenced Wikipedia article.

REPLY [1 votes]: When something is said to be a "formal sum" or "formal intersection" or "formal" anything, it is the same thing. It just means, treat it as a symbols only instead of as having the normal meaning.
So you are familiar with the formal sum
$$
2 + 3,
$$
which means the sequence of symbols $2, +, 3$ (with "$+$" in particular having no defined meaning, and being just a symbol.)
Similarly, here we have
$$
U_{ij}
$$
with $U$ having no defined meaning, and being just a symbol with two subscripts. (In particular, the order of the subscripts is significant.)


TITLE: Why does the frequency of a Transformer not change?
QUESTION [0 upvotes]: I was reading about the transformer, when I came across a note by the author that stated that the Frequency is not changed as the flux linked to the secondary coil in a transformer changes. Is there any mathematical proof, why the frequency remains unchanged as the flux changes?
Thank You.

REPLY [2 votes]: "The frequency doesn't change" is only true when the core is perfectly linear. For a real transformer, there will be some nonlinear effects (saturation) meaning that the sinusoidal input waveform will create harmonics in the output - second harmonics and higher frequencies will appear.
But if you ignore those, then the flux change will vary sinusoidally at a particular frequency and the induced current in the secondary must follow these changes - it cannot either "run ahead" (follow a change that hasn't happened yet) or "fall behind" (not change when the input is changing).

REPLY [1 votes]: Let $\color{blue}{i=i_0\sin(\omega t)}$ be the input alternating current to the primary coil with a frequency $\color{red}{\omega}$ then the voltage induced in the secondary coil of transformer is given as $$V_{in}=-M\frac{di}{dt}$$
Where, $M$ is the mutual inductance
setting the value of $i$,  $$V_{in}=-M\frac{d}{dt}(i_0\sin(\omega t))$$
$$=-Mi_0\omega \cos(\omega t)$$$$=\color{red}{-Mi_0\omega\sin \left(\omega t+\pi/2\right)}$$
It is clear that the frequency of output/induced voltage $V_{in}$ is $\color{red}{\omega}$ which is same as that of the input current.
Hence, in a transformer,  the frequency remains unchanged (constant) as flux changes.


TITLE: Find the exponential value of a number.
QUESTION [3 upvotes]: I'm wondering if there is a way to add two numbers with the same base but different exponents without converting the base of the numbers. For example lets say your adding 2^2 and 2^4 the value of 2^2 is 4 and the value of 2^4 is 16 so the total of the two numbers is 20. The closest I could get my calculator to finding a base of 2 with a value of 20 was 2^4.3219281 which equaled 20.000000071. So I have a few questions.

Can 2 be raised to a power that's exactly equal to 20?
I know that with division and multiplication you can just add or subtract the exponents. Are there any similar tricks for adding or subtracting values with exponents?
Is there anyway to arrive at the result, or a similar result of 2^4.3219281 without changing the base in the process?

REPLY [3 votes]: There is a particular function just designed for solving problems of the form $b^x=a$.  The solution to said problem is given by$$x=\log_b(a)=\frac{\log(a)}{\log(b)}$$where we use the second form if your calculator doesn't specify the base of the logarithm.  For example,$$2^x=20\implies x=\log_2(20)$$
No, definitely no simple tricks.  $\ddot\frown$
As to calculating what $\log_2(20)$ is without a calculator, one can implement the following algorithm:$$2^x=20=5\times2^2=2^{\log_2(5)}2^2=2^{2+\log_2(5)}$$$$x=2+\log_2(5)$$So the problem now is to find $\log_2(5)$, which is much easier to approximate.  Clearly, it's larger than $2$, so maybe $\log_2(5)\approx2.3$, giving us $x=4.3$


TITLE: Is there any other method to show that $\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x =-\frac{\pi^{2}}{8} \ln 2+\frac{7}{16}\zeta(3)?$
QUESTION [7 upvotes]: Noting that the evaluation of the integral can be simplified by the Fourier series of $\ln(\sin x)$,
$$\ln (\sin x)+\ln 2=-\sum_{k=1}^{\infty} \frac{\cos (2 k x)}{k}$$
Multiplying the equation by $x$ followed by integration from $0$ to $\infty$, we have
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x+\int_{0}^{\frac{\pi}{2}} x\ln 2 d x&=-\sum_{k=1}^{\infty} \int_{0}^{\frac{\pi}{2}} \frac{x \cos (2 k x)}{k} d x\\
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x+\left[\frac{x^{2}}{2} \ln 2\right]_{0}^{\frac{\pi}{2}}&=-\sum_{k=1}^{\infty} \frac{1}{2 k^{2}} \int_{0}^{\frac{\pi}{2}} x d(\sin 2 k x)\\
\int_{0}^{\frac{\pi}{2}} x \ln (\sin x) d x&=-\frac{\pi^{2}}{8} \ln 2-\frac{1}{2} \sum_{k=1}^{\infty} \frac{1}{k^{2}}\left[\frac{\cos 2(x)}{2 k}\right]_{0}^{\frac{\pi}{2}} \\
&=-\frac{\pi^{2}}{8} \ln 2-\frac{1}{4} \sum_{k=1}^{\infty} \frac{(-1)^{k}-1}{k^{3}}\\
&=-\frac{\pi^{2}}{8} \ln 2+\frac{1}{4}\left(\sum_{k=1}^{\infty} \frac{2}{(2 k+1)^{3}}\right)\\
&=-\frac{\pi^{2}}{8} \ln 2+\frac{1}{2}\left[\sum_{k=1}^{\infty} \frac{1}{k^{3}}-\sum_{k=1}^{\infty} \frac{1}{(2 k)^{3}}\right]\\
&=-\frac{\pi^{2}}{8} \ln 2+\frac{7}{16}\zeta(3) \blacksquare
\end{aligned}
$$
Furthermore, $$
\begin{aligned}
\int_0^{\frac{\pi}{2}} x \ln (\cos x) d x&=\frac{\pi}{2} \int_0^{\frac{\pi}{2}} \ln (\sin x)-\int_0^{\frac{\pi}{2}} x \ln (\sin x) d x \\
&=-\frac{\pi^2}{4} \ln 2-\left(-\frac{\pi^2}{8} \ln 2+\frac{7}{16}\zeta(3)\right) \\
&=-\frac{\pi^2}{8} \ln 2-\frac{7}{16} \zeta(3)
\end{aligned}
$$
and $$
\int_0^{\frac{\pi}{2}} x \ln (\tan x) d x=\int_0^{\frac{\pi}{2}} x \ln (\sin x) d x-\int_0^{\frac{\pi}{2}} x \ln (\cos x) d x=\frac{7}{8}\zeta(3)
$$

REPLY [1 votes]: Here's a direct attack on the integral
$$i=\int_0^{\frac{\pi }{2}} x \log (\sin (x)) \, dx$$
using only Euler's formula $e^{i x} = \cos(x)+ i \sin(x)$, properties of the $\log$-function $\log(x y)=\log(x) + \log(y)$ including the series expansion $-\log(1-x)=\sum_{k=1}^{\infty}\frac{x^k}{k}$, and the definition of Riemann's zeta function $\zeta(s) = \sum_{k=1}^{\infty}\frac{1}{k^s}$, also the derivative of an integral with respect to a parameter is used. The solution is lengthy because it is elementary and detailed. No tricky formulas or external references are used. Notice that parts of this solution are already contained in other solutions here.
We start by writing
$$\begin{align}
\log (\sin (x))& =\log \left(\frac{e^{i x}-e^{-i x}}{2 i}\right)\\
& =\log \left(\frac{e^{i x} \left(1-e^{-2 i x}\right)}{2 i}\right)\\
& =i x+\log (\frac{1}{2 i})+\log \left(1-e^{-2 i x}\right)\end{align}\tag{1}$$
Hence the integral can be written as $i=i_1+i_2$ with
$$\begin{align}
i_1 & = \int_0^{\frac{\pi }{2}} x \left(i x+\log (\frac{1}{2 i})\right)\,dx\\
& =-\frac{1}{8} \pi ^2 \log (2)-\frac{1}{48} \left(i \pi ^3\right)\\
i_2 & = \int_0^{\frac{\pi }{2}} x \log \left(1-e^{-2 i x}\right) \, dx\end{align}\tag{2}$$
Now
$$\begin{align}
i_2 &=\int_0^{\frac{\pi }{2}} x \log \left(1-e^{-2 i x}\right) \, dx\\
& =-\int_0^{\frac{\pi }{2}} \sum _{k=1}^{\infty } \frac{x e^{-2 i k x}}{k} \, dx\\
& =-\sum _{k=1}^{\infty } \int_0^{\frac{\pi }{2}} \frac{x e^{-2 i k x}}{k} \, dx\\
& = -\sum _{k=1}^{\infty } \frac{1}{k} \left(\frac{1}{-2 i k}\right)\frac{\partial}{\partial{k}}\int_0^{\frac{\pi }{2}} e^{-2 i k x} \, dx\\
& =-\sum _{k=1}^{\infty } \frac{-1+(-1)^k (1+i \pi  k)}{4 k^3}\\
& =\frac{1}{4} \sum _{k=1}^{\infty } \frac{1-(-1)^k}{k^3}-\frac{1}{4} (i \pi ) \sum _{k=1}^{\infty } \frac{(-1)^k}{k^2}\\
& =\frac{1}{4} s_1-\frac{i \pi}{4}s_2 \end{align}\tag{3}$$
The first sum is
$$\begin{align}
s_1 & =\sum _{k=1}^{\infty } \frac{1-(-1)^k}{k^3}=2 \sum _{k=1}^{\infty } \frac{1}{(2 k-1)^3}\\
& =2 \left(\frac{1}{1^3}+\frac{1}{3^3}+\frac{1}{5^3}+\dots\right)\\
& = 2\left(\left( \frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+\dots\right)-\left(\frac{1}{2^3}+ \frac{1}{4^3}+\frac{1}{6^3} +\dots\right)\right)\\
& = 2\left(\left( \frac{1}{1^3}+\frac{1}{2^3}+\frac{1}{3^3}+\dots\right)-\frac{1}{2^{3}}\left(\frac{1}{1^3}+ \frac{1}{2^3}+\frac{1}{3^3} +\dots\right)\right)\\
& = 2 \left(\zeta(3) -\frac{1}{8}\zeta(3)\right)=\frac{7}{4} \zeta(3)
\end{align}\tag{4}$$
and the second sum, similarly,
$$\begin{align}
s_2 &=\sum _{k=1}^{\infty } \frac{(-1)^k}{k^2}=\sum _{k=1}^{\infty } \left(\frac{(-1)^k}{k^2}+\frac{1}{k^2}\right)-\sum _{k=1}^{\infty } \frac{1}{k^2}\\
& =2 \sum _{k=1}^{\infty } \frac{1}{(2 k)^2}-\sum _{k=1}^{\infty } \frac{1}{k^2}=\left(\frac{1}{2}-1\right) \zeta (2)\\
& =-\frac{\zeta (2)}{2}=-\frac{\pi ^2}{12}\end{align}\tag{5}$$
So that
$$\begin{align}
i_1 & =-\frac{1}{8} \pi ^2 \log (2)-\frac{i \pi^3 }{48}\\
i_2 & =\frac{1}{4}\left(\frac{7}{4} \zeta (3)\right)-\frac{i \pi }{4}\left(-\frac{\pi ^2}{12}\right) =\frac{7}{16} \zeta (3) +\frac{i \pi^3 }{48}
\end{align}\tag{6}$$
and finally
$$i=i_1+i_2=-\frac{1}{8} \pi ^2 \log (2)+\frac{7 }{16}\zeta (3)\tag{7}$$
QED.


TITLE: Center point of parallelogram of two 2D vectors
QUESTION [1 upvotes]: I have two 2D vectors starting at $(0,0)$: $a(x_1,y_1)$ and $b(x_2,y_2)$ and I need to calculate the coordinates of the center point of the parallelogram that they "create". Does anyone know the formula?

REPLY [0 votes]: In general(also in $\mathbb{R^3}$) given 2 vectors $\vec A$ and $\vec B$ their midpoint $\vec M$ is given by: $$\vec M=\frac{\vec A+\vec B}{2}$$
More in general a point $\vec P$ along the line through $\vec A$ and $\vec B$  is given by:
$$\vec P=(1-\lambda)\cdot\vec A+\lambda\cdot \vec B \quad with \quad \lambda \in\mathbb{R}$$
NOTE
$\lambda=0 \implies \vec P = \vec A$
$\lambda=\frac12 \implies \vec P = \vec M$
$\lambda=1 \implies \vec P = \vec B$


TITLE: Why is finding the roots of a polynomial equation so important? What is to gain?
QUESTION [11 upvotes]: I have just started a pre calculus class, and our first lessons have been reviews on polynomial equation, quadratics and finding roots or solutions to equations. The topic is fairly simple but I just have to know why is it so important to find an X value of a function that results in it y value to become zero? Why do mathematicians go to such great lengths like creating a quadratic formula? There must be some reason and my lessons have failed to tell why.

REPLY [5 votes]: In a word:  factoring.  In logical terms:  $ab=0\implies a=0$ or $b=0$.
To the extent that algebra concerns itself with setting up and solving polynomial equations, bringing everything to one side, leaving $0$ on the other side, is often an effective way to find all the solutions of an equation -- and, importantly, to know you haven't overlooked any.  For example, consider the equation
$$(x^2+11)x=6(x^2+1)$$
If you play around with it, you might notice that $x=1$ is a solution, since $12\cdot1=6\cdot2$.  If you play some more, you might find a couple of other solutions.  And if you know some theory, you might know when you've found all the solutions.  But it becomes systematic, and easier, if you rewrite the equation as
$$x^3-6x^2+11x-6=0$$
The observation that $x=1$ is a solution means that the cubic on the left hand side factors:
$$(x-1)(x^2-5x+6)=0$$
In fact, the cubic in this case factors completely, giving us
$$(x-1)(x-2)(x-3)=0$$
so this tells us that $\{1,2,3\}$ is the complete solution set to the equation $(x^2+11)x=6(x^2+1)$.
Not every equation works out so nicely, of course.  But at the very least, setting things up to solve for roots is a systematic way of chipping away at problems involving polynomials.


TITLE: Peaks of quasi-periodic functions
QUESTION [0 upvotes]: Let $\lambda_1,\cdots,\lambda_n$ be real numbers. Let us consider $f : t \mapsto \frac{1}{n}\sum^n_{j=1} e^{i\lambda_j t}$. We have that $f(0) = 1$; in general, $f$ is not periodic.
Soon after $0$, one might expect that the $e^{i\lambda_j t}$ spread around the unit circle enough so that the sum has a lot of compensations, so that $f(t)$ is small, in modulus. However, for exceptional $t$, the $e^{i\lambda_j t}$ may realign and the modulus of $f(t)$ would be close to $1$ again.
If, for example, $\vert f(t) \vert$ is thought as a magnitude of risk of something, at time $t$, one is interested in estimating the first time $t$ at which $\vert f(t) \vert$ is close to $1$ again.
So, one could try to study $g_{\epsilon} : t \mapsto \{\vert f(t) \vert \ \mbox{if} \ \vert f(t) \vert  > 1 - \epsilon, \ 0\ \mbox{if not}\}$, and maybe try to estimate $\int^T_0 g_\epsilon(t) dt$ and figure out how it grows, but this seems to be a hard problem.
My question is: how can one formulate this problem in a rigorous mathematical way, such that the answer is both reachable and relevant to the informal problem? Are there efficient, analytical tools to study such problems?

REPLY [1 votes]: Let WLOG $\;\lambda_1\le\lambda_2\le\dots\le \lambda_n.\;$
The given function is periodic, if
$\lambda_j=m^\,_j\lambda_0,\quad m_j\in\mathbb N,\quad\lambda_0\in\mathbb R.\tag1$
In this case, $\;f(t) = P_{m_n}(e^{\lambda_0\,t}),\;$ where $\;P\;$ is a polynomial.
Looks convenient to approximate the given function by the periodic one.
The possible algorithm is the next.

The starting values setting: $\;\lambda_0 = \lambda_1,\;$
$$ m_j=\left\lfloor \dfrac{\lambda_j}{\lambda_0}+\dfrac12\right\rfloor,\quad j=2\dots n.\tag2$$
Calculating of the worst pair index
$$w=\underset{j=1\dots n-1}{\operatorname{argmax}}\left|\dfrac{\lambda_{j+1}}{\lambda_j}-\dfrac{m_{j+1}}{m_j}\right|.\tag3$$
Approximation of the ratio
$$\dfrac{\lambda_{w+1}}{\lambda_w}\approx\dfrac pq$$
by the Euclidean algorithm.
Divide $\;\lambda_0\;$ to $\;p\;$ and apply $(2).$
If $\;\gcd(m_j)\ge1,\;$ then multiply $\;\lambda_0\;$ to $\;p\;$ and apply $(2).$
To increase the approximation accuracy, go to step 2.


TITLE: Estimating the number of connected components of a curve contained in a given set
QUESTION [1 upvotes]: Let $X$ be a metric space and $\Omega\subset X$ an open set. Take $x\in\Omega$ and choose $r>0$ such that the open ball $B(x,r)\subset B(x,2r)\subset \Omega$. Let $\gamma:[0,1]\to X$ be a Lipschitz curve.
Is it possible to divide $\gamma$ in a finite number of subcurves $\gamma_i :[a_i,b_i]\to X$, with $0\le a_i\le b_i\le 1$ ($\gamma_i=\gamma_{|[a_i,b_i]}$), such that each curve $\gamma_i$ lies entirely either in $\Omega$ or $X\setminus B(x,r)$?
If we can prove that for some $\lambda\in [1,2)$, the cardinality of the set $\partial B(x,\lambda r)\cap \gamma([0,1])$ is finite then, the result should be true, however I can't formalize it.
If I assume on the contrary that for every $\lambda \in [1,2)$ such inersection is enumarable, then it seem to imply that the length of $\gamma$ is infinite, which is an absurd. Does this makes any sense?

REPLY [2 votes]: We don't even need the Lipschitz continuity, the uniform continuity you get from the compactness of the parameter interval suffices for the partition.
Let $C = \gamma^{-1}(\overline{B(x,r)})$ and $D = \gamma^{-1}(X\setminus B(x,2r))$. Call $s,t \in C$ (without loss of generality $s \leqslant t$) equivalent if $[s,t]\cap D = \varnothing$, and similarly for $D$. By the uniform continuity, there is a $\delta > 0$ such that $s\in C, t\in D \Rightarrow \lvert s-t\rvert \geqslant \delta$, and hence we have $\operatorname{dist}(E_1,E_2) \geqslant 2\delta$ if $E_1,E_2$ are two distinct equivalence classes of either $C$ or $D$. Therefore, there are only finitely many equivalence classes in each, $C$ and $D$.
For such an equivalence class $E$, let $I(E) = [\inf E,\sup E]$. That gives you finitely many disjoint subintervals $[\alpha_i,\beta_i]$, $1 \leqslant i \leqslant m$ with $\alpha_i \leqslant \beta_i < \alpha_{i+1}$, and $\alpha_i \in C \iff \beta_i \in C \iff \alpha_{i+1} \in D$. Let $t_0 = 0$, $t_n = 1$, and for $1 \leqslant i < n$, let $t_i = \frac{1}{2}(\beta_i + \alpha_{i+1})$.
Then for $1 \leqslant i < n$ we have $\gamma(t_i) \in B(x,2r)\setminus \overline{B(x,r)}$, and for all $i < n$ either $\gamma([t_i,t_{i+1}]) \subset B(x,2r)$ or $\gamma([t_i,t_{i+1}]) \subset X\setminus \overline{B(x,r)}$.


TITLE: inequality between norms of vector
QUESTION [3 upvotes]: Let $a=(a_1, \ldots, a_n)$ be a vector in $R^n$.
It is known that $\|a\|_{\infty}\leq \|a\|_2\leq \sqrt n\|a\|_{\infty}$.
Let $k<<n$. For which kind of vectors the following would be true:
$$
\sqrt{k}\|a\|_{\infty}\leq \|a\|_2?
$$
Thank you.

REPLY [1 votes]: Equivalently, we can look for vectors with $\|a\|_\infty/\|a\|_2\leq 1/\sqrt k$. Start by considering vectors with $a_i\geq 0$ for all $i$ and $\|a_2\|=1$. The set of such vectors is $[0,1/\sqrt{k}]^n\cap S^{n-1}$, which is a simply connected semialgebraic set. Allowing the sign of each component to be positive or negative gives us a union $X$ of $2^n$ reflected copies of this set. Depending on $k$ and $n$ the resulting set is either (non-simply) connected or has $2^n$ disconnected components. Allowing scaling gives $X\times (0,\infty)$. This accounts for all solutions except the lone trivial solution $a=0$.


TITLE: writing $M : \Gamma_{n,0} \backslash \Gamma_n$
QUESTION [3 upvotes]: Let $\Gamma_n = \operatorname{Sp}_n(\Bbb Z)$ and $\Gamma_{n,0} \subset \Gamma_n$ be a subgroup. We write
$$ M = \begin{pmatrix}A & B\\ C & D \end{pmatrix} \in \operatorname{Sp}_n(\mathbb{Z})$$
where $M$ is a $(2n)\times(2n)$ matrix and $A,B,C,D$ are $n\times n$ matrices.
I am reading some function theory script and I encountered the writing (definition of the Eisensteinseries for Siegel modular forms)
$$ E_k(Z) := \sum_{M : \Gamma_{n,0} \backslash \Gamma_n} \det(CZ + D)^{-k} . $$
What does $M : \Gamma_{n,0} \backslash \Gamma_n$ mean? What is $M$?
I know "$\backslash$" as a set-substraction but that doesn't seem to make sense here.
It is probably meant as some form of Quotient space but I'm not sure what exactly. Also, I only know "$/$" for the notation of Quotient spaces.
Also, if $\Gamma_{n,0} \backslash \Gamma_n$ means something (a set), why does the author write "$M:$" and not "$M \in$"? (The same author uses "$\in$" everywhere else.)

REPLY [2 votes]: In the theory of automorphic forms, quotients or coset spaces on both sides arise, and $X\backslash Y$ is written for example when $X$ is a subgroup of $Y$ and we want cosets of the form $Xy$ for $y\in Y$.
The group $Sp_n(\mathbb Z)$ is the group of $2n$-by-$2n$ integer matrices $g$ such that $g^t J_n g=J_n$ where $J_n=\pmatrix{0_n&-1_n\cr 1_n&0_n}$, where $1_n$ and $0_n$ are $n$-by-$n$ identity and $0$ matrices. Here, $\Gamma_n$ is another notation for $Sp_n(\mathbb Z)$, and $\Gamma_{n,0}$ is elements $g=\pmatrix{*&*\cr 0&*}$ therein.
The use of "$M:$" is indeed a bit inconsistent, but not surprising as a descriptor of summation. Here, $M$ is meant to range over representatives for that quotient, with lower $n$-by-$n$ block entries $C,D$.
That quotient has representatives uniquely determined by the lower halves $\pmatrix{C&D}$, and the pair $(C,D)$ is left-modulo $GL_n(\mathbb Z)$.


TITLE: Cardinality of Primes in UFDs over Infinite Sets
QUESTION [1 upvotes]: As we know, there are infinitely-many Integer primes per, e.g
, Euclid's proof.
Is it also true that there are infinitely-many primes in every UFD defined over an infinite set? I tried to see if Euclid's proof extended to UFDs, but I was not able to make it work. Thanks in Adance.

REPLY [2 votes]: As GreginGre points out, the answer is no, since every field is a UFD that does not contain any prime elements. However, fields are indeed the only UFDs in which this can occur. To see this, let $R$ be any UFD that is not a field; then $R$ is automatically infinite, and $R$ contains at least one prime element $p$. We will show that $R$ has infinitely many prime elements. If the group of units $R^\times$ is infinite, then we are done, since the family $\{\lambda p:\lambda\in R^\times\}$ is a collection of distinct prime elements of $R$. Thus assume that $R^\times$ is finite; since $R$ is infinite, we are now in a position to replicate the proof of Euclid's theorem. Indeed, suppose $p_1,\dots,p_n\in R$ is any finite collection of prime elements of $R$. Since $R$ is infinite, the set $\{1+p_1\dots p_n s:s\in R\}$ is infinite as well, and since $R^\times$ is finite there thus exists some $s\in R$ such that $1+p_1\dots p_ns$ is not a unit. Since $R$ is a UFD, there is then some prime element $q$ that divides $1+p_1\dots p_ns$, and we cannot have $q=p_i$ for any $i\leqslant n$, so $q$ is a prime element of $R$ not among $p_1,\dots,p_n$. In particular, $R$ has infinitely many prime elements, as desired. So, to answer your question, any UFD that is not a field has infinitely many prime elements.

On the other hand, again as GreginGre comments, if you are willing to consider examples that have finitely many prime elements modulo units, then there can be counterexamples that are not fields. Here is a recipe for constructing such examples. Let $R$ be any UFD that is not a field and let $p_1,\dots,p_n\in R$ be any prime elements of $R$. Then the subset $S=R\setminus\bigcup_{i=1}^n\langle p_i\rangle$ is a complement of prime ideals, hence multiplicatively closed, and by Kaplansky's criterion the localization $S^{-1}R$ is a UFD. But every prime element of $S^{-1}R$ is associate to one of the $p_i/1$ (why?), so $S^{-1}R$ has only finitely many primes up to associativity, as desired.
For a very tractable example of this construction, let $R=\mathbb{Z}$, so that $p_1,\dots,p_n\in\mathbb{Z}$ are just integer primes. Then the ring $S^{-1}R$ indicated above is just the subring $$\left\{\frac{a}{b}:p_i\nmid b\text{ for each }i\leqslant n\right\}$$ of $\mathbb{Q}$; as an exercise, try to check directly that this ring is a UFD and that all of its prime elements are associate to some $p_i$.


TITLE: Eigenvalues and Diagonalization
QUESTION [3 upvotes]: The typical definition given for a diagonalizable matrix is:
Given $A\in M^F_{n\times n}$
$A$ is diagonalizable $\iff$ A has $n$ linearly independent eigenvectors.
Is it also true that
$A$ is diagonalizable $\iff$ A has $n$ unique eigenvalues.

REPLY [10 votes]: No. It is not true that $A$ is diagonalizable $\iff$ A has $n$ unique eigenvalues.
For instance, the identity matrix is diagonalizable but all eigenvalues are the same, namely $1$.
However one way in your claim is true i.e.
A has $n$ unique eigenvalues $\implies$ $A$ is diagonalizable.


text:
\begin{document}

\title{On Representer Theorems and Convex Regularization}

\author{Claire Boyer$^1$, Antonin Chambolle$^2$, Yohann De Castro$^{3,4}$, Vincent Duval$^{4,5}$,  Fr\'ed\'eric de Gournay$^6$ and Pierre Weiss$^7$.\\
\footnotesize $^1$ LPSM, Sorbonne Universit\'e, ENS Paris
\footnotesize $^2$ CMAP, CNRS, Polytechnique, Palaiseau \\
\footnotesize $^3$ LMO, Universit\'e Paris-Sud, Orsay
\footnotesize $^4$ MOKAPLAN, INRIA, Paris \\
\footnotesize $^5$ Ceremade, Universit\'e Paris Dauphine
\footnotesize $^6$ INSA, Universit\'e de Toulouse \\
\footnotesize $^7$ ITAV, CNRS, Universit\'e de Toulouse
}

\maketitle

\begin{abstract}
We establish a general principle which states that regularizing an inverse problem with a convex function yields solutions which are convex combinations of a small number of \textit{atoms}. These atoms are identified with the extreme points and elements of the extreme rays of the regularizer level sets. An extension to a broader class of quasi-convex regularizers is also discussed. As a side result, we cha\-rac\-terize the minimizers of the total gradient variation, which was still an unresolved problem.
\end{abstract}

\noindent\textbf{Keywords:}
Inverse problems, Convex regularization,  Representer theorem,  Vector space, Total variation

\input{Introduction}
\input{Preliminaries}
\input{MainResults}
\input{Applications}
\input{Conclusion}
\input{Proofs}

\section*{Acknowledgments}
This work was initially started by two different groups composed of A. Chambolle, F. de Gournay and P. Weiss
 on one side and C. Boyer, Y. De Castro and V. Duval on the other side. The authors realized that they were working on a similar topic when a few of them met at the Cambridge semester in mathematical imaging during the Isaac Newton Institute (Cambridge) semester ``Variational methods and effective algorithms for imaging and vision'', supported by EPSRC Grant N.~EP/K032208/1. They therefore decided to write a joint paper and wish to thank the organizers for giving them this opportunity.
 This work initially started with the help of T. Pock through a few numerical experiments and with discussions with J. Fadili and C. Poon.

 The work of A.C. was partially supported by a grant of the Simons Foundation.

\bibliographystyle{abbrv}
\bibliography{Biblio}

\end{document}

config: arxiv


TITLE: Why Should I Care about Solvability by Radicals?
QUESTION [1 upvotes]: The Egyptians liked studying numbers in terms of Egyptian fractions. The Greeks liked straightedge and compass constructions. Both really liked rational numbers, especially the Pythagoreans before the proof of the existence of irrational numbers. Rational numbers are obviously very important, but knowledge of the existence of irrationals means they're held in slightly lower esteem today. Egyptian fractions became a mere novelty, and later straightedge and compass constructions met the same fate, although the focus in Europe and the Middle-East on Greek classics kept them alive much longer. They're just not particularly natural concepts. To me, solvability by radicals seems to belong with them. Before the Abel-Ruffini Theorem, it could have been a much more important concept, just like before the proof that $\sqrt{2}$ is irrational, rational numbers could have been somewhat more important. But without a statement like the negation of the Abel-Ruffini Theorem, solvability by radicals seems completely unmotivated. Studying it has certainly given us many important, natural constructs, but that doesn't make it important by itself; part of being natural is showing up in many places, so if we hadn't studied solvability by radicals, we'd have come across Galois Theory some other way, and indeed we did, with ruler and straight-edge constructions. So I'd like to know reasons that people would be interested in solvability by radicals, or indeed solvability of groups, for their own sake.

REPLY [1 votes]: The "simplest" operations are addition, multiplication, taking powers of numbers and their "inverses" (e.g. subtraction, division, and taking roots).  It's always been nice to express roots in terms of these intuitive processes, and they are moreover all readily computable e.g. roots by power series.


TITLE: Find the area of the parallelogram formed by the vectors (1, 1), (1, 3).
QUESTION [0 upvotes]: I believe the answer is 2.
the second part is the part I can't figure out.
Suppose we transform the plane (with the parallelogram in it) by multiplying by a matrix
A where det(A) = 5. Compute the area of the transformed parallelogram.
Is it just 10?

REPLY [1 votes]: $$A=|{\bf a}\times {\bf b}|=
\begin{vmatrix}
i&j&k\\
1&1&0\\
1&3&0
\end{vmatrix}=|2 k|=2.
$$
The second answer is also coorect as $\det(AB)=\det(A)\det(B)$.


TITLE: Does the series $\sum_{k=1}^{\infty}\arctan{\frac{1}{k}}$ converge or diverge?
QUESTION [1 upvotes]: You guys have helped me a lot with series and series expansions, thank you all. Let's see if something stuck. I'm going to do this myself and hope on good feedback from you guys.

We know that $$\arctan\frac{1}{k}=\frac{1}{k}-\frac{1}{3k^3}+O\left(\frac{1}{k^5}\right)=\frac{3k^2-1}{3k^3}+O\left(\frac{1}{k^5}\right)=\left(\frac{3k^2-1}{3k^2}+O\left(\frac{1}{k^4}\right)\right)\frac{1}{k}.$$
Denote $a_k=\arctan(1/2)$ and $b_k=1/k$. We know that $\sum b_k$ is divergent and by the limit comparison test we have
$$\lim_{k\rightarrow \infty}\frac{a_k}{b_k}=\lim_{k\rightarrow \infty}\left(\frac{3k^2-1}{3k^2}+O\left(\frac{1}{k^4}\right)\right)=\frac{3}{3}=1>0,$$
which ultimately implies that $\sum a_k$ has to be divergent too.
Will this do?

REPLY [0 votes]: $\arctan(x)$ is concave on $\mathbb{R}^+$, hence for any $x\in(0,1)$ we have $\arctan x\geq \frac{\pi x}{4}$ and the given series is trivially divergent by comparison with the harmonic series.


TITLE: Negative power of inequality
QUESTION [0 upvotes]: We know the following inequality,
\begin{equation*}
     (a+b)^{\alpha}\leq a^{\alpha} + b^{\alpha},\quad a,b \geq 0, \quad \alpha \in [0,1].
 \end{equation*}
Is it possible to write the following elementary inequality. If so, how we know.
\begin{equation*}
     (a+b)^{-\alpha}\leq a^{-\alpha} + b^{-\alpha},\quad \text{if}\quad a,b > 0,\quad \text{and}  \quad \alpha >1,
 \end{equation*}
Thanks in advance.

REPLY [0 votes]: $\dfrac{1}{(a+b)^{\alpha}}\leq\dfrac{1}{a^{\alpha}}+\dfrac{1}{b^{\alpha}}$ if and only if $1\leq\dfrac{(a+b)^{\alpha}}{a^{\alpha}}+\dfrac{(a+b)^{\alpha}}{b^{\alpha}}$ if and only if $1\leq\left(\dfrac{a+b}{a}\right)^{\alpha}+\left(\dfrac{a+b}{b}\right)^{\alpha}$ if and only if $1\leq\left(1+\dfrac{b}{a}\right)^{\alpha}+\left(1+\dfrac{a}{b}\right)^{\alpha}$.
But we see that $1+b/a>1$ and $1+a/b>1$ so $\left(1+\dfrac{b}{a}\right)^{\alpha}+\left(1+\dfrac{a}{b}\right)^{\alpha}>2$ actually.