TITLE: Non-conjugated roots in the eigenequation
QUESTION [0 upvotes]: I have an ODE:
$$y'' - iky = 0, k > 0$$
I tried to solve it via the eigenequation, and the result is
$r^2 = ik$, $r = \pm k(\sqrt{2}/2 + i\sqrt{2}/2)$
And the roots are non-conjugated...
What should I do?

REPLY [1 votes]: Hint:
Assume the solution is:
$$y(t) = e^{m t}$$
Find the second derivative, substitute into $(1)$ and solve.
Spoiler Hover over the following area.

 $ \displaystyle y(t) = y_1(t) + y_2(t) = c_1 e^{-(1+i) \sqrt{k} t/\sqrt{2}} + c_2 e^{(1+i) \sqrt{k} t/\sqrt{2}} $


TITLE: Using int and floating gives different value of LegendreP
QUESTION [0 upvotes]: I should post this in mathematica forum, sorry for the inconvenience.
When I use LegendreP at mathematica 10.0, I found weird a thing happening.
In[693]:= N[LegendreP[300, Cos[50/360 2 Pi]]]

Out[693]= -1.04451*10^61

If
In[692]:= N[LegendreP[300, Cos[50.0/360 2 Pi]]]

Out[692]= -0.0402969

Obviously, $-0.0403$ is the right answer. Why has such big difference?
I also tried like
In[694]:= N[LegendreP[300.0, Cos[50/360 2 Pi]]]

Out[694]= -0.0402969

And also
In[695]:= N[LegendreP[30, Cos[50/360 2 Pi]]]

Out[695]= 0.126799

Looks if $n$ is smaller than $50$, also give a right value.
I totally confused....

REPLY [1 votes]: Catastrophic cancellation is happening here. The largest coefficient of $P_{300}(x)$ has magnitude larger than $10^{112}$; that's enormous! Even after plugging in the value of $x$, the largest term still has magnitude greater than $10^{76}$. You either need lots of precision or clever symbolic manipulation in order to get anything vaguely resembling the correct answer.
In particular, this cannot be computed in the naive way using machine precision arithmetic.
If you ask mathematica to specifically give you some number of digits of precision, it knows how to achieve it:
In[1]:= N[LegendreP[300, Cos[50/360 2 Pi]], 3]

Out[1]= -0.0403


TITLE: Interpretation of chi-squared test for contingency table
QUESTION [1 upvotes]: I am trying to solve an exercise where I applied the chi-squared test.

I have already calculated the test statistic, the degrees of freedom and the critical value.
At a question we have to determine whether the following sentence is true.
"That is, the bivariate relationship between these two features in the population is only with a probability of about x% (x is the critival value.) is different than that in her sample."
This means that we have to determine if the null hypothesis is accepted or rejected?
We have that when the chi-squared statistic exceeds the critical value then we reject the null hypothesis.
But at the sentence it is not clarified if it exceeds, is it? It says that it is about x%. Does this mean that it does not exceed the critical value?

REPLY [1 votes]: This is a chi-squared test for the independence of two categorical variables,
given data in a $3 \times 3$ contingency table.
Under the assumption that 'Religious Preference' and 'Party Affiliation' categories are independent,
one estimates the expected cell counts by using the multiplication rule
for independence, and the row and column totals. Here is how to find $E_{11}$ in the cell for 'Catholic and CDU-CSU'.
$$\hat P(\text{Catholic}) = 549/1739,\;
\hat P(\text{CDU-CSU}) = 756/1739,\,\text{ thus }\\
\hat P(\text{Catholic}\cap\text{CDU-CSU}) = \frac{549}{1739}\times\frac{756}{1739}.$$
Then the expected count for this cell is the grand sample size $1739$ times the
estimated probability, $E_{11} = \hat E(\text{Cath.}\cap\text{CDU-CSU})= 1739\hat P(\text{Cath.}\cap\text{CDU-CSU})$ $=
549(756)/1739 =  238.6682.$
While observed counts are integers, expected counts need not be integers;
do not round expected counts.
The general rule for an expected cell count is "its row total times its
column total divided by the grand total."
Then the contribution to chi-squared for cell $(1,1)$ is
$$\frac{(X_{11} - E_{11})^2}{E_{11}} = \frac{(308 - 238.6682)^2}{238.6682} = 20.141.$$
The chi-squared statistic for this test is the sum of all $3 \times 3 = 9$
contributions to the chi-squared statistic:
$$Q = \sum_{i=1}^3 \sum_{j=1}^3 \frac{(X_{ij} - E_{ij})^2}{E_{ij}}.$$
Provided that all $E_{ij} > 5,$ the statistic $Q \stackrel{aprx}{\sim}
\mathsf{Chisq}(\nu = 4).$ (Some authors would say it is OK for a few of the $E_{ij}$ to be as small as 3.)
In general $\nu = (r-1)(c-1),$ where the
table has $r$ rows and $c$ columns. A rigorous argument for this
computation of degrees of freedom $\nu$ is difficult to give at an
elementary level.

It may help to notice, in the current example, that if only the four observed counts $X_{11}, X_{12}, X_{21}, X_{22}$ were given along
  with the row and column totals, then it would be possible to reconstruct
  all of 9 of the observed counts. That is to say, given the structure of
  row and column totals used to find the $E_{ij},$ only four
  of the $X_{ij}$ provide additional information.

The Minitab 17 printout for this problem is as follows:
Chi-Square Test for Association: Relig, Worksheet columns

Rows: Relig   Columns: Worksheet columns

            CDU    SPD  NotVoted  All

Cath        308    162       79   549
          238.7  206.8    103.5
         20.141  9.699    5.820

Prot        284    273       80   637
          276.9  239.9    120.1
          0.181  4.559   13.415

None        164    220      169   553
          240.4  208.3    104.3
         24.284  0.658   40.129

All         756    655      328  1739

Cell Contents:      Count
                    Expected count
                    Contribution to Chi-square

Pearson Chi-Square = 118.885, DF = 4, P-Value = 0.000

First, notice that the expected count and contribution to chi-squared shown
for cell $(1,1)$ in the Minitab table agree with our earlier computations.
[In printouts, Minitab rounds numbers to make displays that fit nicely
on a page; internally about 12-place accuracy is maintained throughout
the computation.] Also notice that the smallest of the $E_{ij} \approx 104,$
very safely larger than 5.
Second notice that $Q$ is quite large. If the categories were independent
we would expect this statistic to be about 4 (the mean of a chi-squared
distribution is its degrees of freedom). The value that cuts 5% from the
upper tail of $\mathsf{Chisq}(4)$ is 9.4877. This is the critical value
for a test at the 5% level.
Sometimes, it is helpful to look at some of the larger contributions
to chi-squared. For example, in cell $(1,1),$ 308 Catholics voted for CDU-CSU,
whereas, under the null hypothesis of independence, one would have expected
only about 239 such votes.
However, the P-value for $Q = 118.885$ is smaller than 0.0005 (printed as 0.000). So we could also reject at the 1% level
and at the 0.1% level. (Maybe for "x%" in your question, you should
say something such as: "Reject at any significance level down to 0.0005.") Whatever phrase you choose, the data are very strongly inconsistent with the hypothesis
that voting behavior and religious preference are independent.


TITLE: Cubic splines better than quadratic splines?
QUESTION [6 upvotes]: I have read in a number of places that cubic splines are of more practical use than quadratic splines in general (there are exceptions of course). Anyone know specifically why they are more applicable/better?

REPLY [9 votes]: A cubic curve can twist in space (i.e. it can be non-planar). A quadratic curve is just a parabola, so it's always planar.
In real applications like graphic arts, engineering, manufacturing, nobody cares much about derivatives, they only care about curvature. And it's possible to get continuity of curvature without continuity of second derivatives (so-called G2 splines, versus C2 ones). So the C2 argument for cubics is a bit fragile.
For some applications, like design of car bodies or cams, cubic splines are not good enough, because you need continuity of the derivative of curvature (G3 continuity). So, in these applications, higher degree curves are often used.
In many situations, splines are used as approximations of more complex functions. If you use cubics, then you'll need fewer polynomial segments for a given approximation tolerance. But each segment will be more complex, so more difficult to handle in subsequent calculations. That's basically the trade-off -- number of segments versus complexity of segments.


TITLE: Real Analysis question that affects how to think about the Dirac delta function.
QUESTION [19 upvotes]: Okay, here are the ingredients to this question.
Me: 60 years old.  39 years ago I took two semesters of Real Analysis using the Royden textbook.  Rusty is an understatement.  But I am still quite anal and OCD.  I am also an electrical engineer, works in signal processing.  DSP and Linear System Theory are important to me.  I have also had two semesters (as a grad student) of Functional Analysis (using the Kreyszig text) and multiple courses in probability, random variables, and random processes (a.k.a. "stochastic processes").
Electrical Engineers (and I suspect many physicists) essentially treat $\delta(x)$ as a "function".  But it isn't.  One thing I remember from R.A. is that if
$$ f(x) = g(x) $$
almost everywhere in $E$, then
$$ \int_E f(x) \, dx \ = \ \int_E g(x) \, dx $$
problem is, of course, that electrical engineers (and their professors) like to think of
$$\begin{align} f(x) & = \delta(x) \\
g(x) & = 0 \end{align} $$
and that
$$ \int\limits_{-1}^{+1} f(x) \, dx = 1 \quad \ne \quad \int\limits_{-1}^{+1} g(x) \, dx = 0 $$
yet $f(x) = g(x)$ everywhere except at one single value of $x$ .
Now I have heard (or read) that the "Dirac delta function is not really a function but is a 'distribution' or a 'functional'."  And I understand the meaning of the terms "distribution" in the context of random variables and "functional" in the context of metric spaces, normed spaces, etc.  Is that the given usage of these two terms regarding $\delta(x)$?
Question 1:  Is the usage of the notation
$$ \int\limits_{-\infty}^{+\infty} f(x) \, \delta(x) \, dx $$
a misnomer?  There is no integration going on.  It's just a linear functional that maps the function $f(x)$ to the number $f(0)$.  Now EEs and maybe physicists will comfortably look at that as an integral that is the same as
$$ \int\limits_{-\infty}^{+\infty} f(0) \, \delta(x) \, dx = f(0)\int\limits_{-\infty}^{+\infty} \delta(x) \, dx = f(0) $$
But since $\delta(x)$ is not a function at all, what do mathematicians mean with that notation?
Question 2:  How fatal is it for electrical engineers and physicists to consistently treat the Dirac delta function simply as a limit of "nascent deltas" such as
$$ \delta(x) \ \triangleq \ \lim_{\sigma \to 0^+} \frac{1}{ \sigma} \operatorname{rect} \left( \frac{x}{\sigma} \right) $$
where $ \operatorname{rect}(x) \triangleq
\begin{cases}
1 \quad |x|<\frac{1}{2} \\
\frac{1}{2} \quad |x|=\frac{1}{2} \\
0 \quad |x|>\frac{1}{2} \\
\end{cases} $
or
$$ \delta(x) \ \triangleq \ \lim_{\sigma \to 0^+} \frac{1}{\sigma} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}\left(\frac{x}{\sigma}\right)^2} $$
What is gonna kill us to simple-mindedly treat the Dirac delta as such a function?  A function that is zero almost everywhere, yet it integrates to be equal to 1 (where the function that is zero everywhere integrates to be 0).
If we do that, within our own disciplines, what mathematical problem might crop up that kills us?
This is not exactly the same but smells a lot like this concern from Richard Hamming:

“Does anyone believe that the difference between the Lebesgue and Riemann integrals can have physical significance, and that whether say, an airplane would or would not fly could depend on this difference? If such were claimed, I should not care to fly in that plane.”

I might ask the same question regarding the mathematician's and the engineer's understanding of the Dirac delta function.  How might a mathematician answer that question?

REPLY [13 votes]: First, a comment about the property you mention.  It's true that if $f$ and $g$ are measurable functions and $f = g$ almost everywhere, then
$$
\int_Ef(x)dx = \int_Eg(x)dx
$$  This won't hold if $f$ and $g$ are distributions, because as you say distributions aren't functions.  So the integral notation is just that - it's notation.  Now...
A1. The answer to your first question comes from wanting the theory of distributions to functionally "look like" the theory of nicer linear functionals, i.e. we were really good at manipulating integrals, so we wanted the new thing to operationally work the same.  For physicists, who were used the the Riesz representation theorem, this meant wanting to think of linear functionals as being just integrals against a fixed function. So, hence the inner product/integral notation
$$
F(\varphi) = \langle F,\varphi\rangle = \int F(x)\varphi(x)dx
$$
Note that in distribution theory, this is not a standard Lebesgue integral - it's just formal notation. The integral notation is ubiquitous though - it's not just linear functionals that are written this way, but also linear operators
$$
g(x) = \int k(x,y)f(y)dy
$$  There is even a sort of generalization of the Riesz representation theorem called the Schwartz Kernel Theorem that says that any (nice enough) linear operator $g = Ky$ can be written using an "integral" like that, but where the kernel function $k(x,y)$ is possibly a distribution.  The moral of the story is, you should extend your understanding of the integral notation to include other linear operations, not just integration of standard functions.  Once you've proved that all the usual operations that you're used to, like integration by parts, make sense with distributions, you'll see that using the integral notation is very natural and 100% rigorous - as long as you remember that it's just notation for "apply the linear operation specified".
A2. It isn't fatal at all to think of the delta function in this way - in fact this is a preferred method to define the delta function and many other distributions.  The rect function represents a sort of local averaging, and you can think of the delta function as being an "infinitely local" averaging (i.e. sampling).  The one thing I would recommend is looking into "approximations to the identity" - the rect function construction is just one possible construction, and in order to show that the delta function is uniquely defined, one should show that any similar sequence of approximate deltas also gives the same result (e.g. triangles, Gaussians, etc).  In other words, you could either define the delta function as "that linear functional such that $F(\varphi) = \varphi(0)$, in which case you need to show that this is a well-defined, bounded linear operation on some function space, or alternately you could define the delta function as "the limit of $\langle \delta_\epsilon,\varphi\rangle$ as $\epsilon\rightarrow 0$, in which case you still need to prove that this is a well-defined, linear operation on some function space.  Either way, the result is the same.


TITLE: Find all eigenvalues and eigenvectors of $T$
QUESTION [0 upvotes]: Define $T \in \mathcal{L(F^2)}$ by $$ T(w,z) = (z,w)$$Find all eigenvalues and eigenvectors of $T$.

An eigenvalue is defined as $\lambda \in \mathcal{F}$ such that $Tv = \lambda v$ where $v \neq 0$. Expanding the definition, I get the following system of equations
$$z = \lambda w \\ w = \lambda z \tag 1.$$
Now I can simplify this system by substituting $w$ for $z$ values to get $$ z = \lambda^2 z \\
0 = \lambda^2 z - z \\
0 = z(\lambda^2 - 1)$$
Solving for $\lambda$ gives $ \lambda = \pm 1$ so the two eigenvalues are $1, -1$.
To find the corresponding eigenvectors, from (1) I think the vector I get is $(z, w) = (\lambda ^2 z, \lambda z) = (z,z)$ when $\lambda = 1)$ and
$(z, w) = (\lambda ^2 z, \lambda z) = (z,-z)$ when $\lambda = -1)$
However I am not sure if I found the corresponding eigenvectors. What would be a way I could check for myself that these are correct?

REPLY [1 votes]: Just check. If $(x,y)$ is an eigenvector of $\textsf T$ with corresponding eigenvalue $\lambda=1$, then $(x,y)$ satisfies
$$(y,x)=\textsf{T}(x,y)=1(x,y)=(x,y)$$
that is, $x=y$. The vectors $(x,y)$ that satisfies $x=y$ are exactly the vectors living in the span of $\{(1,1)\}$. Similarly, the eigenvectors with corresponding eigenvalue $\lambda=-1$ are the vectors in $\operatorname{span}\big(\{(1,-1)\}\big)$.


text:
\begin{document}

\title{{\bf On the composition of two spherical twists}}

\author{Federico Barbacovi}
\address{Department of Mathematics, University College London}
\email{federico.barbacovi.18@ucl.ac.uk}

\begin{abstract}
  \begin{comment}
  Spherical functors provide a formal way to package autoequivalences of enhanced triangulated categories.
  Moreover, E. Segal proved that any autoequivalence of an enhanced triangulated category can be realized as a spherical twist.

  When exhibiting an autoequivalence as a spherical twist, one has various choices for the source category of the spherical functor.
  We prove that, in the DG setting, given two spherical twists there is a natural way to produce a new spherical functor whose twist is the composition of the spherical twists we started with, and whose source category semiorthogonally decomposes into the source categories for the two spherical functors we began with.
  \end{comment}

   E. Segal proved that any autoequivalence of an enhanced triangulated category can be realised as a spherical twist.
   However, when exhibiting an autoequivalence as a spherical twist one has various choices for the source category of the spherical functor.
   We describe a construction that realises the composition of two spherical twists as the twist around a single spherical functor whose source category semiorthogonally decomposes into the source categories for the spherical functors we started with.

   We give a description of the cotwist for this spherical functor and prove, in the special case when our starting twists are around spherical objects, that the cotwist is the Serre functor (up to a shift).
   We finish with an explicit treatment for the case of $\mb{P}$-objects.
\end{abstract}

\maketitle

\tableofcontents

\input{Introduction}

\input{Generalities}

\input{Twist}
\input{Cotwist}

\section{Examples}
\input{Spherical-objects}
\input{P-objects}

\appendix
\input{Passing-to-add-closure}

\bibliographystyle{alphaurl}

\input{SF-Main.bbl}

\end{document}

config: arxiv


text:
\section{Symmetric Transitive and Serial Relation is Reflexive}
Tags: Reflexive Relations, Symmetric Relations, Transitive Relations, Serial Relations, Equivalence Relations

\begin{theorem}
Let $\RR$ be a [[Definition:Relation|relation]] which is:
:[[Definition:Symmetric Relation|symmetric]]
:[[Definition:Transitive Relation|transitive]]
:[[Definition:Serial Relation|serial]].
Then $\RR$ is [[Definition:Reflexive Relation|reflexive]].
Thus such a [[Definition:Relation|relation]] is an [[Definition:Equivalence Relation|equivalence]].
\end{theorem}

\begin{proof}
Let $S$ be a [[Definition:Set|set]] on which $\RR$ is a [[Definition:Relation|relation]] which is [[Definition:Symmetric Relation|symmetric]], [[Definition:Transitive Relation|transitive]] and  [[Definition:Serial Relation|serial]].
As $\RR$ is [[Definition:Symmetric Relation|symmetric]]:
:$x \mathrel \RR y \implies y \mathrel \RR x$
As $\RR$ is [[Definition:Transitive Relation|transitive]]:
:$x \mathrel \RR y \land y \mathrel \RR x \implies x \mathrel \RR x$
As $\RR$ is [[Definition:Serial Relation|serial]]:
:$\forall x \in S: \exists y \in S: x \mathrel \RR y$
Let $x \in S$.
Then
{{begin-eqn}}
{{eqn | lo= \exists y \in S:
      | l = \tuple {x, y}
      | o = \in
      | r = \RR
      | c = as $\RR$ is [[Definition:Serial Relation|serial]]
}}
{{eqn | ll= \leadsto
      | l = \tuple {y, x}
      | o = \in
      | r = \RR
      | c = as $\RR$ is [[Definition:Symmetric Relation|symmetric]]
}}
{{eqn | ll= \leadsto
      | l = \tuple {x, x}
      | o = \in
      | r = \RR
      | c = as $\RR$ is [[Definition:Transitive Relation|transitive]]
}}
{{end-eqn}}
Thus:
:$\forall x: x \mathrel \RR x$
and by definition $\RR$ is [[Definition:Reflexive Relation|reflexive]].
It follows by definition that such a relation is an [[Definition:Equivalence Relation|equivalence relation]].
{{qed}}
\end{proof}

config: wiki


TITLE: Tightness + Integrability does not imply uniform integrability?
QUESTION [6 upvotes]: Consider these definitions of tightness and integrability:
A sequence of real-valued random variables $f_n$ on a probability space $(X,\Sigma,\mu)$ is $\textit{tight}$ if for all $\epsilon$ there is a compact $K_{\epsilon}$ with $\mu(\{ x \mid f_n(x) \in K_{\epsilon} \}) > 1-\epsilon$ for all $n$.
A sequence of real-valued random variables $f_n$ on a probability space
$(X,\Sigma,\mu)$ is $\textit{uniformly integrable}$ if two conditions are satisfied:

$\sup_{n}\int_{X}|f_n|d\mu < \infty$
For all $\epsilon>0$ we can find a $\delta>0$ so that $\sup_n\int_{A}|f_n|d\mu < \epsilon$ whenever $\mu(A) < \delta$

I know that uniform integrability $\implies$ tightness. My first question was whether we can conclude the converse. Clearly the answer is no, since tightness does not require integrability. So my refined question is as follows:
Let $f_n$ be a collection of real-valued random variables defined on the probability space $(X,\Sigma,\mu)$, such that the $f_n$ are tight and  $\sup_{n}\int_{X}|f_n|d\mu < \infty$. Can we conclude that the collection is also uniformly integrable?

REPLY [6 votes]: (Edit: When first posting, I did not notice Ian's comment, so added some content).
No.
Problem is that while by tightness you require all functions to be large only on small sets, you still don't have an effective control on how large they can get on these small sets  by merely requiring boundedness in $L^1$.
Example: Ian has given one.
Uniform integrability is a very strong requirement. By Vitali's theorem, it is a  necessary and sufficient condition for a sequence in $L^1$ converging to $0$ a.s. (or in probability) to also converge in $L^1$ (dominated convergence is a only a sufficient condition, but much easier to verify).
You can get UI if you require a little more. For example, if you assume that $\phi$ is an nonnegative  function on $[0,\infty)$ satisfying $\phi(x)/x \underset{x\to\infty}{\to} \infty$, and $\sup_n \int \phi(|f_n|) d \mu< \infty$, then it is not hard to show that the sequence is indeed UI.


TITLE: Basis for solution space?
QUESTION [0 upvotes]: For the matrix:
$$ \begin{bmatrix} 1 & 0 & 2 & | & 0 \\
                   0 & 1 & 3 & | & 0 \\
                   0 & 0 & 0 & | & 0 \end{bmatrix} $$
which means $x_1 + 0x_2 + 2x_3 = 0$ and $x_2 + 3x_3 = 0$.
It has $3$ eqns, and $3$ variables are to be determined. This means $x_2$ is free to vary. So I get the solution : $[-2t, t, -t/3]$ for $t$ any real number.
But how do I determine the basis? How could this be described graphically?

REPLY [0 votes]: The solution is actually of the form $(\frac{2}{3}t, t, -\frac{1}{3}t)$. So one possible basis can be $(\frac{2}{3}, 1, -\frac{1}{3})$ or $(2, 3, -1)$ if you like integers...


TITLE: upper bound of a double sequence
QUESTION [7 upvotes]: Let $\{ a_{k,m} \}$ be a doublely indexed sequence of positive numbers satisfying:
$a_{1,n}\leq \frac{1}{n+1}\quad $
and $\quad a_{k,m} \leq \frac{1}{m+1}(a_{k-1,m+1}+L a_{k-1,m+2})\quad \quad (1)$   ,
where $L$ is a positive constant and indices $k,m,n$ are non-negative integers.
My question is:
what is the upper bound for $a_{n,0}$ in terms of n ?
I try to find the bound by induction using relation in $(1)$
to reduce the index $n$ in $a_{n,0}$ to $1$,
but I'm not able to compute the final representation in terms of $a_{1,k} (k\geq 1)$.

REPLY [0 votes]: As fedja mentioned, if $a_{k,m+1}\leq a_{k,m}$, we can easily arrive at a bound  in fedja's comment. Since this may not be the case, we do need some preparations to formalize the argument.
We define $\hat{a}_{k,m}:=\text{maximal achieveble upper bound for $a_{k,m}$ from inequality (1)}$, namely, $\hat{a}_{1,n}=1/(n+1)$, and
$\hat{a}_{k,m}=\frac{1}{m+1}(\hat{a}_{k-1,m+1}+L \hat{a}_{k-1,m+2})$.
Then it is easy to see from induction that $\hat{a}_{k,m+1}\leq \hat{a}_{k,m}$. Thus we deduce that
$a_{n,0}\leq \hat{a}_{n,0}\leq \frac{1+L}{1}\hat{a}_{n-1,1}\leq \frac{(1+L)^{2}}{2!}\hat{a}_{n-2,2}\leq\dots
\leq \frac{(1+L)^{n-1}}{(n-1)!}\hat{a}_{1,n-1}=\frac{(1+L)^{n-1}}{n!}$.


TITLE: Functional-analysis, continuous function
QUESTION [0 upvotes]: Let $\ell^1:=\{f:\mathbb{N}\to\mathbb{C}\colon ||f||_1:=\sum_{n=1}^\infty |f(n)|<\infty\}$ provided with $||\cdot||_1$.
  Show, that every $f\in\ell^1$ defines a continuous linear functional $S_f: c_0\to\mathbb{C}$ with $S_f(g)=\sum_{n=1}^\infty g(n)f(n)$ and it is $||S_f||_\text{op}=||f||_1$ for every $f\in \ell^1(\mathbb{N})=\ell^1$

It is easy to see, that $S_f$ is linear:
$(\ell^1, ||\cdot||_1)$ is a normed $\mathbb{C}$-vectorspace.
Let $g, h\in c_0$ and $\mu, \lambda\in\mathbb{C}$, then:
$S_f(\mu g+\lambda h)=\sum_{n=1}^\infty f(n)(\mu g+\lambda h)(n)=\sum_{n=1}^\infty f(n)(\mu g)(n)+f(n)(\lambda h)(n)\\=\mu\sum_{n=1}^\infty f(n)g(n)+\lambda\sum_{n=1}^\infty f(n)h(n)$ since the sum converges.
We get the desired $\mu S_f(g)+\lambda S_f(h)$.
Now I want to show, that $S_f$ is continuous.
Is it enough to show that $S_f$ is continuous at 0?
Else I can show, that for every sequence $(g_n)_n$ in $c_0$ with $g_n\to 0$ it is $S_f(g_n)\to 0$ or that $S_f$ is bounded. Therefore $||S_f(g)||_1\leq c||g||$ for some $c\geq 0$ and every $g\in c_0$.
Unfortunatly nothing really seems to work for me...
I wanted to show, that $S_f$ is bounded. I started like this:
$||S_f(g)||_1=||\sum_{n=1}^\infty f(n)g(n)||_1$ and now I already start to struggle.
How do I proceed from here. Do I get a "double sum"?
$\sum_{n=1}^\infty\vert\sum_{n=1}^\infty f(n)g(n)\vert$ but this does not make sense, because it does not converge...
And how can I show, that $||S_f||_\text{op}=||f||_1$, where I experience the same issue as above.
Hints are appreciated.
Thanks in advance.

REPLY [1 votes]: The norm on $\mathbb C$ is just the modulus and I assume that you are using the supremum norm on $c_0$ so to show that $S_f$ is bounded, you need to bound  $\lvert S_f(g) \rvert$ in terms of $\|g\|_\infty.$ This much is trivial. We see $$\lvert S_f(g) \rvert = \left \lvert \sum^\infty_{n = 1} f(n) g(n) \right \rvert \le \sum^\infty_{n=1} \lvert f(n)\rvert \lvert g(n) \rvert \le \| g\|_\infty \sum^\infty_{n=1} \lvert f(n) \rvert = \|f\|_1 \|g\|_\infty.$$ This shows that $S_f$ is bounded with $\|S_f\|_{\text{op}} \le \|f \|_1$. In order to prove that $\|S_f\|_{\text{op}} = \|f\|_1$, you need to shows that the above inequality is sharp. To do so, take $g_k$ defined by $g_k(n) = \text{sign}(f(n)),$ for $n \le k$ and $g_k(n) = 0$ for $n > k$. Then we see $$\lvert S_f(g_k)\rvert = \sum^k_{n=1} \lvert f(n) \rvert $$ and this becomes arbitrarily close to $\|f\|_1$ as $k\to \infty$ so since each $g_k$ has $\|g_k \|_\infty = 1$, this shows that $\|S_f\|_{\text{op}} = \|f\|_1$.


TITLE: Why does the specific heat of a real gas depends on temperature but not for the ideal gas?
QUESTION [3 upvotes]: The specific heat of a real gas, unlike an ideal gas, depends on temperature. How can we physically understand this? Thanks.

REPLY [0 votes]: The tempurature dependence of heat capacitance was one of the historical failures of classical physics, which predicts constant heat capacitance (i.e. $\frac{\partial}{\partial T}
 \frac{n}{2}k_\text{B}T$). The temperature dependence of solids and gases was not explained until the advent of quantum mechancis, in which the vibrational states are quantized.
An ideal gas is a classical gas and will have constant heat capacitance.


text:
\section{Bicriteria Approximations to Bounded Degeneracy for Edge Deletions}\label{appendix-degen-lp}

\subsection{$(5,5)$-approximation for edge deletion}
\label{section:positive_degeneracy_LP_edge}
In what follows we formulate an LP-relaxation for the problem of minimizing the number of required edge edits (deletions) to the family of $\Tr$-degenerate graphs. For each edge $uv\in E$, $x$ variables denote the orientation of $uv$; $x_{\arc{uv}}=1$, $x_{\arc{vu}} =0$ if $uv$ is oriented from $u$ to $v$ and $x_{\arc{vu}}=1$, $x_{\arc{uv}} =0$ if $e$ is oriented from $v$ to $u$. Moreover, for each $uv$ we define $z_{uv}$ to denote whether the edge $uv$ is part of the edit set $\editset$ ($z_{uv}=1$ if the edge $uv \in \editset$ and zero otherwise).

\begin{center}
\begin{minipage}{0.75\linewidth}
\begin{probbox}{\lpMinEdge}
	\emph{Input:} \hskip1em $G=(V,E), w, \Tr$
\begin{align*}
	\text{Minimize} &  & \sum_{uv \in E} z_{uv} w_{uv} & &\\
	\text{s.t.}     &  & x_{\arc{vu}} + x_{\arc{uv}}   &\geq 1-z_{uv}  &\forall uv \in E\\
						      &  & \sum_{u \in N(v)}x_{\arc{vu}} 		 &\leq \Tr 			&\forall v \in V\\
						 	    &  & x_{\arc{uv}} 								 &\geq 0			&\forall uv\in V\times V
\end{align*}
\end{probbox}
\end{minipage}
\end{center}

The first set of constraints in the LP-relaxation \lpMinEdge guarantee that for each edge $uv\notin \editset$, it is oriented either from $v$ to $u$ or from $u$ to $v$. The second set of the constraints ensure that for all $v\in V$, $\outDeg(v)\leq \Tr$. Note that if an edge $uv\in \editset$ and thus $z_{uv}=1$, then \WLOG we can assume that both $x_{\arc{uv}}$ and $x_{\arc{vu}}$ are set to zero.
\begin{lemma}\label{lem:lp-relaxation-edge}
$\lpMinEdge(G, w, \Tr)$ is a valid LP-relaxation of $\bDEE(G,w)$.
\end{lemma}

Next, we propose a {\em two-phase} rounding scheme for the \lpMinEdge.

\mypar{First phase.} Let $(x, z)$ be an optimal solution of \lpMinEdge. Note that since the \lpMinEdge has polynomial size, we can find its optimal solution efficiently.
Consider the following {\em semi-integral} solution $(x, \hat{z})$ of \lpMinEdge:

\begin{align}\label{rule:round_first}
\hat{z}_{uv} =
\left\{
	\begin{array}{ll}
		1  & \mbox{if } z_{uv} \geq \eps, \\
		0  & \mbox{otherwise.}
	\end{array}
\right.
\end{align}

\begin{claim}\label{clm:first-round}
$({x \over 1- \eps}, \hat{z})$ as given by Equation~\eqref{rule:round_first} is a $({1\over \eps},{1\over 1-\eps})$-bicriteria approximate solution of $\lpMinEdge(G, w, \Tr)$.
\end{claim}
\begin{proof}
First, we show that $({1\over 1- \eps}x,\hat{z})$ satisfies the first set of constraints. For each edge $uv$,
\begin{align*}
{x_{\arc{uv}}\over 1- \eps} + {x_{\arc{vu}}\over 1- \eps} = {1\over 1- \eps}(x_{\arc{uv}} + x_{\arc{vu}}) \geq {1\over 1- \eps}(1-z_{uv}) \geq 1-\hat{z}_{uv},
\end{align*}
where the first inequality follows from the feasibility of $(x,z)$ and the second inequality follows from Equation~\eqref{rule:round_first}. Moreover, it is straightforward to check that as we multiply each $x_{vu}$ by a factor of $1/(1-\eps)$, the second set of constraints are off by the same factor; that is, $\forall v\in V, \sum_{u \in V}x_{\arc{vu}}/(1-\eps)\leq \Tr/(1-\eps)$. Finally, since for each edge ${uv}$, $\hat{z}_{uv} \leq z_{uv}/\eps$, the cost of the edit set increases by at most a factor of $1/\eps$; that is, $\sum_{uv\in E} \hat{z}_{uv} w_{uv} \leq \frac{1}{\eps}\sum_{uv\in E} z_{uv} w_{uv}$.
\end{proof}

\mypar{Second phase.} Next, we prune the fractional solution further to get an {\em integral} approximate {\em nearly feasible} solution of \lpMinEdge. Let $\hat{x}$ denote the orientation of the surviving edges (edges $uv$ such that $\hat{z}_{uv}=0$) given by:
\begin{align}\label{rule:round-second}
\hat{x}_{\arc{uv}} =
\left\{
	\begin{array}{ll}
		1  & \mbox{if } x_{\arc{uv}} \geq (1-\eps)/2, \\
		0  & \mbox{otherwise.}
	\end{array}
\right.
\end{align}

We say an orientation is \emph{valid} if each surviving edge $(u,v)$ is oriented from $u$ to $v$ or $v$ to $u$.

\begin{lemma}\label{lem:valid-orient}
$\hat{x}$ as given by Equation~\eqref{rule:round-second} is a valid orientation of the set of surviving edges.
\end{lemma}
\begin{proof}
We need to show that for each $uv\in E$ with $\hat{z}_{uv} =0$ at least one of $\hat{x}_{\arc{uv}}$ or $\hat{x}_{\arc{vu}}$ is one. Note that if both are one, we can arbitrarily set one of them to zero.

For an edge $uv$, by Equation~\eqref{rule:round_first}, $\hat{z}_{uv} = 0$ iff $z_{uv}\leq \eps$. Then, using the fact that $(x,z)$ is a feasible solution of \lpMinEdge, $x_{\arc{uv}} + x_{\arc{vu}} \geq 1-z_{uv} \geq 1-\eps$. Hence, $\max(x_{\arc{uv}}, x_{\arc{vu}})\geq (1-\eps)/2$ which implies that $\max(\hat{x}_{\arc{uv}}, \hat{x}_{\arc{vu}}) =1$. Hence, for any surviving edge $uv$, at least one of $\hat{x}_{\arc{uv}}$ or $\hat{x}_{\arc{vu}}$ will be set to one.
\end{proof}
\begin{lemma}\label{lem:edge-edit}
$(\hat{x}, \hat{z})$ as given by Equations \ref{rule:round_first} and \ref{rule:round-second} is an integral $({1\over \eps},{2\over 1-\eps})$-bicriteria approximate solution of $\lpMinEdge(G,w,r)$.
\end{lemma}
\begin{proof}
As we showed in Lemma~\ref{lem:valid-orient}, $\hat{x}$ is a valid orientation of the surviving edges with respect to $\hat{z}$. Moreover, by Equation~\eqref{rule:round-second}, for each $uv \in E$, $\hat{x}_{\arc{uv}} \leq 2 x_{\arc{uv}} / (1-\eps) $. Hence, for each vertex $v\in V$, $\outDeg(v) \leq 2\Tr/(1-\eps)$. Finally, as we proved in Claim~\ref{clm:first-round}, the total weight of the edit set defined by $\hat{z}$ is at most $\frac{1}{\eps}$ times the total weight of the optimal solution $(x,z)$.
\end{proof}

Hence, together with Lemma~\ref{lem:degen-bounded-outdeg-orientation}, we have the following result.
\begin{corollary}\label{cor:edge-edit}
There exists a $({1\over \eps},{4\over 1-\eps})$-bicriteria approximation algorithm for \bDEE.

In particular, by setting $\eps = 1/5$, there exists a $(5,5)$-bicriteria approximation algorithm for the \bDEEfull problem.
\end{corollary}

\iffalse

\subsection{Integrality gap of \lpMinEdge and \lpMinVertex}
A natural open question is if we can obtain ``purely multiplicative'' approximation guarantees for $\bDEE$ and $\bDEV$ via LP-based approaches. In this section, we show that the existing LP-relaxation of editing to bounded degeneracy cannot achieve $o(n)$-approximation. These results are particularly important because they show that the best we can hope for are bicriteria approximations.

\begin{theorem}\label{thm:integrality-gap-edge}
The integrality gap of \lpMinEdge is $\Omega(n)$.
\end{theorem}
\begin{proof}
Consider an instance of $\bDEE(G)$ where $G$ is an unweighted complete graph of size $2n$ and $\Tr = n-2$. First, we show that $\lpMinEdge(G, r)$ admits a fractional solution of cost/size $O(n)$ and then we show that the size of any feasible edit set of $\bDEE(G)$ is $\Omega(n^2)$.

Consider the following fractional solution of $\lpMinEdge(G, r)$: for all $uv\in V\times V$ and $u\neq v$, $x_{\arc{uv}} = 1/2 - 1/n$ and for all edges $uv\in E$, $z_{uv} = 2/n$. Note that $x$ and $z$ satisfy the first set of constraints in $\lpMinEdge(G, \Tr)$:
\begin{align*}
\forall uv\in E, \quad x_{\arc{uv}} + x_{\arc{vu}} = 1- 2/n = 1-z_{uv}.
\end{align*}
Moreover, $x$ satisfies the second set of the constraints in $\lpMinEdge(G, \Tr)$
\begin{align*}
\forall v \in V, \quad \sum_{u\in V} x_{\arc{vu}} = (2n-1)(1/2 -1/n) < n-2.
\end{align*}
Finally, $\cost(x, z) = \sum_{uv\in E} z_{uv} = n(2n-1) \cdot (2/n) = 4n-2$ which implies that the cost of an optimal solution of $\lpMinEdge(G, \Tr)$ is $O(n)$.

Next, we show that any integral solution of $\bDEE(G, \Tr)$ has size $\Omega(n^2)$. Let $\editset$ be a solution of $\bDEE(G, \Tr)$. Then, there exits an ordering of the vertices in $G$, $v_1, \cdots, v_{2n}$ such that $\deg(v_i)$ in $G[v_i,\dots,v_{2n}]$ is at most $r\leq n-2$. This implies that for $i\leq n-2$,
$|\delta(v_i) \cap \editset| \geq n+2-i$, where $\delta(v)$ denotes the set of edges incident to a vertex
$v$. Thus,
\begin{align*}
|\editset| \geq {1\over 2}\sum_{i\leq n-2} |\delta(v_i) \cap \editset| \geq {1\over 2}\sum_{i\leq n-2} n+2-i \geq {1\over 2} (n^2 -4 - {(n-2)(n-3) \over 2}) \geq {n^2 / 4}.
\end{align*}
Hence, the integrality gap of \lpMinEdge is $\Omega(n)$.
\end{proof}

\begin{theorem}\label{thm:integrality-gap-vertex}
The integrality gap of \lpMinVertex is $\Omega(n)$.
\end{theorem}
\begin{proof}
Consider an instance of $\bDEV(G)$ where $G$ is an unweighted complete graph of size $2n$ and $\Tr = n-2$. First, we show that $\lpMinVertex(G, r)$ admits a constant size fractional solution and then we show that the size of any feasible edit set of $\bDEE(G)$ is $\Omega(n)$.

Consider the following fractional solution of $\lpMinVertex(G, r)$: for each $uv\in V^2$ and $u\neq v$, $x_{\arc{uv}} = 1/2 - 1/n$ and for each vertex $v\in V$, $z_{v} = 1/n$. First, we show that $x$ and $z$ satisfy the first set of constraints in $\lpMinVertex(G, \Tr)$:
\begin{align*}
\forall uv\in E, \quad x_{\arc{uv}} + x_{\arc{vu}} = 1- 2/n = 1-z_{u} - z_{v}.
\end{align*}
Moreover, $x$ satisfies the second set of the constraints in $\lpMinVertex(G, \Tr)$
\begin{align*}
\forall v \in V, \quad \sum_{u\in V} x_{\arc{vu}} = (2n-1)(1/2 -1/n) < n-2.
\end{align*}
Finally, $\cost(x, z) = \sum_{uv\in E} z_{uv} = 2n \cdot (1/n) = 2$ which implies that the cost of an optimal solution of $\lpMinVertex(G, \Tr)$ is at most $2$.

Next, we show that any integral solution of $\bDEV(G, \Tr)$ has size $\Omega(n)$.
Let $\editset$ be a solution of $\bDEE(G, \Tr)$. Since $G\setminus \editset$ is a complete graph of size $2n-|\editset|$, in order to get degeneracy $n-2$, $|\editset| \geq n+2$.

Hence, the integrality gap of \lpMinVertex is $\Omega(n)$.
\end{proof}
\fi


TITLE: Non-normal Subgroup of Quaternion Product
QUESTION [0 upvotes]: I needed to show that the $Q_8\times Q_8$ contains a non-normal subgroup where $Q_8$ is the quaternion group.
My example was the subgroup $\{(1,1), (i,k), (-i,-k)\}$.
I got the question wrong because, according to Lagrange's theorem, my subgroup cannot actually be a group because its order does not divide the order of $Q_8 \times Q_8$.
This makes sense; however, I do not understand why this set fails to be a group explicitly. It seems to me that it is closed and associative. It obviously has the identity and the other 2 elements are inverses of each other.
If this set is not a group it should fail one of these criteria, unless there is something I am not understanding.
Would anyone be able to shed some light on why my set fails to be a group explicitly other than Lagrange's Theorem?

REPLY [1 votes]: Each subgroup $H$ of a group $G$ must be closed under every product in $H$, including powers of elements. Since
$$(i,k)^2=(i^2,k^2)=(-1,-1)$$
is not in your example, it cannot be a subgroup.


TITLE: Why do cathode rays bend towards a positively charged plate? Please be a bit elaborate in the answer
QUESTION [0 upvotes]: I need help in this quite basic question. Your answers should be  bit elaborate and make the concept clear to understand.

REPLY [2 votes]: The simplest analysis is to assume that you have two parallel plates one with a positive charge and the other with a negative charge arranged so the electric field between the plates is approximately uniform.
Imagine a beam of negatively charged cathode rays coming in horizontally from the left between the parallel and the positive plate at the bottom.
You now have a situation which is essentially the same as the analysis of a projectile thrown horizontally in the Earth's gravitational field with no air resistance present.
In that case the motion of the projectile can be analysed in term of its constant velocity horizontal motion as there is no horizontal force and it constant acceleration vertical motion the constant force being the weight of the projectile.
So for the cathode rays their motion horizontally is a constant velocity as the force due to the charged plates only acts downwards, there is no horizontal component of force.
The cathode rays accelerate vertically downwards under the influence of the constant downward force downwards on them due to the electric field produced by the charged plates.
So the horizontal displacement $x$ after a time $t$ is given by $x = v_{ho} t$ where $v_{ho}$ is the constant horizontal velocity.
The vertical displacement $y$ of the cathode rays is $y=\frac 1 2 a t^2$ where $a$ is the constant downward acceleration of the cathode rays.
From this you can say that $x \propto y^2$ which is the equation of a parabola and that is the path taken by the cathode rays, the same shape as that of a projectile in the Earth's gravitational field.
The cathode rays move at constant speed horizontally but faster and faster towards the positively charged plate.


TITLE: Homology between a loop and a constant loop
QUESTION [0 upvotes]: Two paths $f:I\to X$ and $g:I\to X$ are path homotopic. Does it follow that the loop $f*\bar{g}$ and the constant loop are path homotopic?

REPLY [0 votes]: I assume that $\overline{g}$ is the path defined by $\overline{g}(t)=g(1-t)$ and $*$ is a classical path composition. In that case yes. It follows from the fact the fundamental group $\pi_1(X)$ is well, a group, with following properties:
$$[f][g] = [f*g]$$
$$[f]^{-1} = [\overline{f}]$$
where $[.]$ denotes homotopy class. Now for a constant path $c$ you have
$$[c]=1=[f][f]^{-1}=[f][g]^{-1}=[f][\overline{g}]=[f*\overline{g}]$$

Right, the proof above is for loops only. But the similiar proof can be given in general case.
First of all if $f$ and $g$ are homotopic via $H$ then $\overline{f}$ and $\overline{g}$ are homotopic via $H'(x, t)=H(x, 1-t)$. Now since the homotopy behaves well on composition of paths then
$$[f*\overline{g}]=[f*\overline{f}]=[c]$$
So all you need to know is that if $f_1\sim f_2$ and $g_1\sim g_2$ then $[f_1*g_1]=[f_2*g_2]$ (assuming compatible ends) and $[f*\overline{f}]=[c]$ where $c$ is constant.


TITLE: Maximizing/Minimizing two Integrals Subject to constraints
QUESTION [3 upvotes]: This is actually exercise 4.14 from Rudin's Real and Complex Analysis (also, yes I am aware of the AoPS solution manual but it actually skips this problem).
Problem statement:
Compute
$$ \min_{a,b,c} \int_{-1}^1 |x^3 - a - bx - cx^2|^2 dx $$
and find
$$\max \int_{-1}^1 x^3 g(x) dx$$
Subject to $$\int_{-1}^1 g(x)dx = \int_{-1}^1 xg(x)dx = \int_{-1}^1 x^2g(x)dx = 0$$
And
$$\int_{-1}^1 |g(x)|^2 dx = 1$$
To give some context, this problem is from a chapter on elementary Hilbert space theory. Indeed the first integral is really just minimizing $||x^3 - a - bx - cx^2||_2$ ($L^2$ norm). However, the method in my head honestly seems like it would be more complicated than could be reasonably expected from just the material of the chapters (if you have a copy of the book, let me know if you agree).
Of course on the given interval, the Legendre polynomials are an orthonormal set (after normalizing by a suitable constant). Actually, a side note: after looking through the chapter Grahm-Schmidt is not actually introduced, so I guess the reader is assumed familiar with it. Anyway, we can of course rewrite any polynomial as a linear combination of the Legendre polynomials. After doing this, we merely apply Bessel's inequality and differentiate w.r.t $a$, $b$, and $c$ and find our minimum.
As for the second part, I assume that this must be related to the first part of the problem, but I cannot see how. In fact, I am pretty much lost as to how to solve this. While I am aware of (possible) techniques from the calculus of variations, I feel like that must be well above the scope of the text's intended solution.
In terms of $L^2$ formulation, we see that $g(x)$ is orthogonal to $1$, $x$, and $x^2$, and has norm $1$. So we are asked to maximize the inner product with $x^3$. I cannot find any obvious utility with these facts though. Any help is appreciated.
Thanks!

REPLY [5 votes]: Let $X$ be the space of real-valued polynomials with degree $\le 2$. Note that the Legendre polynomials $e_0(x)=1, \ e_1(x)=x, e_2(x) = \frac{1}{2}(3x^2-1)$ form an orthogonal basis for $X$. To determine $$\min_{a,b,c} \int_{-1}^1 |x^3 - a - bx - cx^2|^2 \; dx$$ we minimize $\|x^3-p(x)\|_{L^2}$ with $p(x) = a+bx+cx^2$ by projecting $x^3$ onto $\{e_0,e_1,e_2\}$ via $$\displaystyle\sum_{n=0}^2 \dfrac{ \langle e_i, x^3 \rangle}{\langle e_i,e_i \rangle} e_i.$$
You should find $p(x) = \frac{3}{5}x$,
and for your final answer you substitute into the integral and get $\frac{8}{175}$.
For the second part, as you said we have $g \perp X$, and $\|g\|_{L^2}=1$. We're looking to maximize $ \langle x^3, g \rangle_{L^2}$. An initial thought is Cauchy-Schwarz: $| \langle x^3, g \rangle_{L^2} | \le \|x^3\|_{L^2} \|g\|_{L^2}$, with equality when $g$ is a multiple of $x^3$. But if $g(x) = \lambda x^3,$ then $g$ is not orthogonal to $X$; in particular, $g$ is not orthogonal to $e_1 = x$.
Instead with $p = 3x/5$, we have $\langle x^3, g \rangle_{L^2} = \langle x^3 -p, g \rangle_{L^2} + \langle p,g \rangle_{L^2} = \langle x^3-p, g \rangle_{L^2} $, and we apply Cauchy-Schwarz again so that $g(x) = \lambda (x^3 - p)$. Since $g$ has norm 1, we get $$g(x) = \dfrac{ x^3-p}{\|x^3-p\|_{L^2}} = \sqrt{\dfrac{175}{8}} \left( x^3- \dfrac{3x}{5} \right).$$
By construction, $x^3 -\text{proj}_{X}(x^3)= x^3 - p$ is $ \perp X$ so $g \perp X$ and we are finished except for actually computing the maximum integral. The final computation yields $\frac{2\sqrt{14}}{35}$.


TITLE: Given $C$ a field, $f$ a polynomial in $C[x]$ and $I = (f)$, describe $C[x]/I$
QUESTION [1 upvotes]: In my course it has been proven that the set $I_f$ whose elements are of the form $r+I$ ,where $r\in C[x]$ such that $r=0$ or $\deg(r) < \deg(f)$, is in bijection with $C[x]/I$.
I wonder whether it is possible to state a stronger fact: $I_f = C[x]/I$.
It may follow from the fact that $C[x]/I$ is an Euclidean Domain; if $g\in C[x]/I$ then there exist $h,r\in C[x]/I$ such that $g=fh+r$ and $r=0$ or $\deg(r)<\deg(f)$, implying that $g+I=fh+r+I=r+I$.
Is this observation sufficient to state that any class in $C[x]/I$ has at least one representative and thus to prove that $C[x]/I = I_f$?
In addition, if the field is finite and $n=|C|$ is it true that $|I_f|=n^{\deg(f)}$ ?

REPLY [1 votes]: Let $f(x)\in \Bbb C[x]$ and $I=(f(x))$.

Consider $\psi: \Bbb C[x]\to \frac{\displaystyle \Bbb C[x]}{\displaystyle I}$, defined by $\psi(g(x))=g(x)+I$.

Here $\psi$ is a surjection.


TITLE: Why is homomorphism defined as such?
QUESTION [0 upvotes]: I am studying ring homomorphism and was curious why it is defined as such.The definition I have is

Let $(R,+,*)$ and $(S,\oplus,\cdot)$ be two rings. A map $\varphi : R \to S$ is called a homomorphism if for every $a,b \in R, \varphi(a+b)= \varphi(a) \oplus \varphi(b)$ and $\varphi(a*b)=\varphi(a)\cdot\varphi(β)$

I get that since $a,b\in R$, then $a * b\in R$ as well, and that $\varphi(a),\varphi(b),\varphi(a*b)\in S$, as well as that $\varphi(a)\cdot\varphi(b)\in R$. But why does $\varphi(a*b)=\varphi(a)\cdot\varphi(b)$ though?
Thanks in advance :)

REPLY [2 votes]: Here's a slightly different way of getting an intuition about this. Let's say we did have a bijection $\varphi:R\to S$. Then the most natural way of "mapping" the operations would be to say $s\oplus s' = \varphi(\varphi^{-1}(s)+\varphi^{-1}(s'))$ and similarly for the multiplicative structure. This can be equivalently written $r+r'=\varphi^{-1}(\varphi(r)\oplus\varphi(r'))$. Finally, this can also equivalently be written $\varphi(r+r')=\varphi(r)\oplus\varphi(r')$. This last form no longer requires assuming that $\varphi$ is a bijection.
As a historical note, quoting from the second footnote of Colin McLarty's The Uses and Abuses of the History of Topos Theory (PDF):

Group theorists into the 1950s generally counted as homomorphisms only
  isomorphisms and projections onto quotient groups. In hindsight they
  lacked the idea of a codomain as opposed to an image so they
  recognized only surjective homomorphisms, where the image coincides
  with the codomain.

I suspect that the way this manifested is they wouldn't talk about a homomorphism between groups, but rather the group structure induced by a bijection via my first two equations in the case of isomorphism. For projections onto quotient groups, this just looks like considering the "same" operation only "mod $N$" (for $N$ a normal subgroup).


TITLE: Will a hole cut into a metal disk expand or shrink when the disc is heated?
QUESTION [83 upvotes]: Take a metal disc and cut a small, circular hole in the center.
When you heat the whole thing, will the hole's diameter increase or decrease? and why?
What will happen to the  diameter of disc?

REPLY [0 votes]: I would like to add a microscopic perspective to this question that were mentioned in comments but which I think deserves more attention. A macroscopic body expands when heated because the average separation between molecules increases.
The picture below shows a disc with a hole. The external and internal radius are showed by solid lines. The red dots along the internal circle represent the molecules in that circle. When we heat the disc the average separation between molecules must increase and of course this happens to molecules in the internal circle as well.  Since the numbers of molecules in the internal circle cannot change for a solid, the internal circumference becomes bigger after thermal expansion (dotted lines).


text:
\section{Teaching Convex Polytopes via Halfspace Queries: A General Model}\label{sec:formulation}

\paragraph{Convex polytopes induced by hyperplanes}
Let $h = \condcurlybracket{\h{}}{\eta \cdot \h{}= b,\: \h{}\in \mathbb{R}^d}$ be a hyperplane in $\Rd$, where $\eta\in\reals^d$
and $b\in\reals$. We say a point $\h{}\in \Rd$ \tt{satisfies} or \tt{lies} in $h$ if $\h{}\in h$. We define a \tt{halfspace} induced by a hyperplane $h$ to be one of the two connected components of $\big(\Rd - h\big)$ \tt{i.e.} sets corresponding to $\mathrm{sgn}\parenb{\eta \cdot \h{}- b}$. We define  $\Hnd \triangleq \curlybracket{h^{(1)}, h^{(2)}, \dots, h^{(n)}}$ as a set of $n$ hyperplanes in $\mathbb{R}^d$.
The arrangement of the hyperplanes in $\Rd$, denoted as $\cA\parenb{\Hnd}$, induces \tt{intersections of halfspaces} which create connected components.
Any connected component of $\Rd-\cup_{h \in \Hnd} h$ is defined as a \tt{region} or \tt{convex polytope} in $\Rd$.
Equivalently, any region $r$ can be exactly specified by the intersections of halfspaces induced by hyperplanes in $\Hnd$. We call the smallest subset $B_r \subseteq \Hnd$ that exactly specifies $r$ the \tt{bounding set of hyperplanes} for $r$. We define connected components induced on hyperplanes (e.g. $h^{(i)}-\cup_{h \in \Hnd\setminus h^{(i)}} h$ for any $h^{(i)} \in \Hnd$) by $\cA\parenb{\Hnd}$ as \tt{faces}. Thus, bounding set $B_r$ forms the faces to the polytope $r$.
\begin{example}[Convex polytopes induced by hyperplanes]
\figref{fig:example.polytope} provides an example of the arrangement of 5 hyperplanes in $\reals^2$, where arrows on the hyperplanes specify halfspaces. The bounding set for the highlighted region $r$, namely $\{h^{(2)}, h^{(4)}, h^{(5)}\}$, forms 3 faces to $r$.
\end{example}
\indent
We use $\regionset$ to denote the
regions \tt{induced} by the arrangement $\cA\parenb{\Hnd}$
and the number of regions
$\rnd$ $\triangleq$  $|\regionset|$. We define a \tt{labeling function} $\ell_r:\Hnd \rightarrow \{-1,+1\}$ for an arbitrary region $r \in \regionset$. Note that
$r$ uniquely identifies its labeling function $\ell_r$.

\paragraph{The teaching framework

}\label{section: teaching framework}
We study the problem of teaching \tt{target regions} (convex polytopes) induced by hyperplane arrangment $\cA\parenb{\Hnd}$ in $\Rd$. Our teaching model is formally stated below.

Consider the set of instances $\Hnd$, with label set $ \mathcal{Y} = \{1,-1\}$ corresponding to two halfspaces induced by a hyperplane. Our hypothesis class, denoted as $\regionset$, is the set of regions induced by $\cA\parenb{\Hnd}$. Consider a target region $r^* \in \regionset$.
Let $\mathcal{Q} \subseteq \Hnd \times \{1,-1\}$ be the ground set of examples (i.e. labeled instances).
We define a labeled subset $Q \subseteq \mathcal{Q}$ as \tt{halfspace queries}.
We assume that for any halfspace queries $Q$ $\mathrm{wrt}$ $r^*$, the labels are consistent, \tt{i.e.}, $\forall (h, l) \in {Q}$, $ \ell_{r^*}(h) = l$.
The \tt{version space} induced by ${Q}$ is the subset of regions $\mathbf{\mathrm{VS}}({Q}) \subseteq \regionset$ that are consistent
with the labels of all the halfspace queries \tt{i.e.},
\begin{align}
\mathrm{VS}({Q}) = \condcurlybracket{r \in \regionset}{\forall (h,l) \in {Q},  \ell_r(h) = l}, \nonumber
\end{align}
\tt{or} equivalently, set of convex polytopes which satisfy the halfspace queries ${Q}$. We define our version space learner as one which upon seeing a set of halfspace queries, maintains a version space containing all the regions that are consistent with all the observed queries. Corresponding to a version space learner and a target region $r^*$, we define a \tt{teaching set} $\mathcal{TS}(\Hnd,r^*)$ as a minimal set of halfspace queries such that the resulting version space exactly contains $\{r^*\}$. Formally,
\begin{align}
    \mathcal{TS}(\Hnd,r^*) \in \argmin_{Q \subseteq \mathcal{Q}} |Q|,  \text{~s.t.~}  \mathrm{VS}(Q) = \{r^*\}.\nonumber
\end{align}
Consequently,
we want to teach a target hypothesis (regions), say $r^*$  via
specifying halfspace queries in the teaching set $\mathcal{TS}(\Hnd,r^*)$ to a learner. Given a target region $r^*$, the teaching \tt{complexity} \citep{goldman1995complexity} is defined as the sample size of the teaching set  \tt{i.e.} $|\mathcal{TS}(\Hnd,r^*)|$.\vspace{6pt}\\
\noindent
In section \secref{sec:avgtd}, we analyze the teaching complexity of convex polytopes both in the framework of \tt{average-case} and \tt{worst-case}.
We define \emph{average teaching complexity} of convex polytopes via halfspace queries as the expected size of the teaching set \tt{i.e.} $\expctover{r\sim\cU}{|\mathcal{TS}(\Hnd,r)|}$, when the target region $r$ is sampled uniformly at random.
We define \tt{worst-case teaching complexity} as the worst-case sample size of a teaching set corresponding to target regions from the set of hypotheses.

\paragraph{Hyperplanes in general position}
We adopt a common assumption in computational geometry \cite{feldman2013neural,miller2007geometric}
that the

hyperplane arrangement is in \tt{general position}, and further provide a relaxed notion of general position hyperplane arrangement, as defined below.

\begin{definition}[General position of hyperplanes \cite{miller2007geometric}]\label{defn general position}
For a set of $n$ hyperplanes $\Hnd$ in $\mathbb{R}^d$, the arrangement $\mathcal{A}(\Hnd)$ is in general position if any subset $\mathcal{S}$ $\subseteq$ $\Hnd$ of $k$ hyperplanes where $1 \le k \le d$, intersects in a $(d-k)$-dimensional plane, otherwise has null intersection.
\end{definition}

\begin{wrapfigure}{h!}{.48\textwidth}
\vspace{-4mm}
\centering

    \subfigure[general]{
		\includegraphics[width=.25\linewidth]{fig/3general.pdf}
		\label{fig:example.3general}
		}
	\subfigure[2-relaxed]{
		\includegraphics[width=.25\linewidth]{fig/2general.pdf}
		\label{fig:example.2general}
		}
	\subfigure[1-relaxed]{
		\includegraphics[width=.25\linewidth]{fig/1general.pdf}
		\label{fig:example.1general}
		}
		\vspace{-2mm}
	\caption{relaxed general position}\label{fig:illu-relaxed-general}

	\vspace{-6mm}
\end{wrapfigure}
\begin{definition}[Relaxed general position of hyperplanes]\label{defn relaxed general position}
For a set of $n$ hyperplanes $\Hnd$ in $\mathbb{R}^d$ and $d' \in [d]$, the arrangement $\mathcal{A}(\Hnd)$ is in $d'$-\textit{relaxed general position} if any subset $\mathcal{S}$ $\subseteq$ $\Hnd$ of $k$ hyperplanes where $1 \le k \le d'$, intersects in a $(d-k)$-dimensional plane, otherwise has null intersection.
\end{definition}

As illustrated in \figref{fig:illu-relaxed-general},
\defref{defn relaxed general position} accounts for arrangements beyond general position (\figref{fig:example.3general}) \tt{e.g} parallel hyperplanes in \figref{fig:example.1general}. \defref{defn general position} is a \tt{special case} of \defref{defn relaxed general position} which we discuss in details in \iftoggle{longversion}{\appref{appendix: active learning}}{the supplemental materials}. \vspace{-3mm}


TITLE: limit with $\arctan$ and $\ln$
QUESTION [2 upvotes]: I have to find the limit
and want ask about a hint:
$$\lim_{x \to \infty} \frac{\frac{1}{2}\pi-\text{tan}^{-1}x}{\ln\left(1+\frac{1}{x^2}\right)}$$
I dont have idea what to do. Maple show me that the answer is $\infty$.

REPLY [1 votes]: You can also just use the l'Hôpital's rule:
$$\lim_{x\to\infty}\frac{\frac{1}{2}\pi-\text{tan}^{-1}x}{\ln\left(1+\frac{1}{x^2}\right)}=\lim_{x\to\infty}\frac{-\frac1{1+x^2}}{\frac{-\frac2{x^3}}{1+\frac1{x^2}}}=\lim_{x\to\infty}\frac1{1+x^2}\cdot\frac{x^3+x}2=\infty$$


TITLE: Show $\sum_{1\leq n\leq x}\sum_{d\mid n}f(d)=\sum_{1\leq d\leq x}\sum_{1\leq m\leq x/d}f(d)$
QUESTION [1 upvotes]: I have been trying to get my head around this step in a proof, but havn't been able to,
Question: Show $$\sum_{1\leq n\leq x}\sum_{d\mid n}f(d)=\sum_{1\leq d\leq x}\sum_{1\leq m\leq x/d}f(d)$$
Attempt:
I know we have to remove $n$ and replace it with $m$, so first let $n=md$
Then substituting in $m$ to remove $n$ we get, $$\sum_{1\leq n\leq x}\sum_{d\mid n}f(d)=\sum_{1\leq md\leq x}\sum_{d\mid md}f(d)$$
Also I know that $d\mid md \Leftrightarrow \exists e\in\mathbb{N}$ s.t. $de=md$
I am not really sure how to continue from here, any help would be greatly appreciated

REPLY [1 votes]: We have, as you showed
\begin{align}
\sum_{1\leq n\leq x}\left(\sum_{d|n}f(d)\right)&=\sum_{1\leq n\leq x}\left(\sum_{n=md}f(d)\right)\tag{1}\\
&=\sum_{1\leq md\leq x}f(d)\\
&=\sum_{1\leq d\leq x}\left(\sum_{1\leq m\leq x/d}f(d)\right).
\end{align}
For $(1)$, as you note, $d|n$ is equivalent to $n=md$, but it is not convenient to write as you tried to do, also, you wrote a condition which is ever true: $d|md$. For the las line, we want to sum over the $d$'s, so, if we want also that $md\leq x$ (in order to be equivalent to the last sum) we need $m\leq x/d$.

REPLY [1 votes]: In the first summation, you are summing up the values of $f$ over all divisors of all numbers between $1$ and $x$.
In the second summation, you are looking at it from a different point of view. You consider all numbers $d$ between $1$ and $x$, and look at how many times $d$ occurs as a divisor of an integer between $1$ and $x$: clearly this number is the integer part of $x/d$.


TITLE: How to Visualize Diagonally Opposite Vertices
QUESTION [0 upvotes]: Consider a cube that exactly fills a certain cubical box. As in Examples 8.7 and 8.10, the ways in which the cube
can be placed into the box corresponds to a certain group of permutations of the vertices of the cube. This is the group
of group of rigid motions (or rotations) of the cube. (It should not be confused with th e group of symmetries
of the figure, which will be discussed in the exercises of Section 12.) How many elements does this group have? Argue
geometrically that this group has at least three different subgroups of order 4 and at least four different subgroups of
order 3.
Fraleigh Solution The group has $24$ elements.
The first subgroup of order $4.$ For any one of the $6$ faces can be on top, and for each such face on top, the cube can be rotated in $4$ different positions leaving that face on top. The $4$ such rotations, leaving the top face on top and the bottom face on the bottom, form a cyclic subgroup of order $4$.
The second rotation group of order $4$ is formed by the rotations leaving the front and
back faces in those positions.
The third rotation groups of order $4$ is  formed by the rotations leaving the side faces in those positions.
One exhibits a subgroup of order $3$ by $\color{red}{\text{taking hold of a pair of diagonally opposite vertices and
rotating through the three possible positions}}$, corresponding to the three edges emanating from each
vertex.
There are $4$ such diagonally opposite pairs of vertices, giving the desired $4$ groups of order $3$.
Question 1. I feel this is easier than my other post. But I can't see how 'taking hold a pair of diagonally opposite vertices' causes 'three possible positions'? I tried the animation at that other post but no luck.
Question 2. How do you decide on the classifications of the rotations of a shape? These two solutions just say what they are. They never revealed how they prefigured this group has at least three different subgroups of order 4 and at least four different subgroups of
order 3. I don't mean just playing around with the shape. I tried that and it didn't help me here.
Question 3. Does this solution break down the type of rotations differently than the other post? That solution talks about " a line joining the centers of opposite faces" and "a line joining diagonally opposite vertices ". This one doesn't.
This is from John B. Fraleigh page 86 exercise 8.45 A First Course in Abstract Algebra

REPLY [2 votes]: I just want to add to the excellent answer (and to the question) that there are some groups of order 2 that deserve mention as well: if you look at an edge of the cube so that your line-of-sight through the edge passes through the cube's center, then there's another parallel edge, which you might say was "diagonally across from" your edge, also lined up with the center. Let's hold the cube so that the chosen edge sits horizontally in front of your face, as does its diagonal opposite. Let's call the left end of your edge "A"  and the right end "B". You can now rotate the cube 180 degrees about the line of sight so that "A" and "B" are swapped. Clearly doing this twice gives the identity. But it's also clear that this element of order 2 isn't in any of the order-3 subgroups (2 doesn't divide 3), nor any of the order-4 subgroups, since in  those subgroups, each non-identity element moves every edge to a new position.
Anyhow, what I'm saying is that there are the vertex-fixing subgroups, the face-fixing subgroups, and, as I've just described, the edge-fixing subgroups.