text:
\begin{document}

\title{Bare Advanced Demo of IEEEtran.cls\\ for Computer Society Journals}

\author{Michael~Shell,~\IEEEmembership{Member,~IEEE,}
        John~Doe,~\IEEEmembership{Fellow,~OSA,}
        and~Jane~Doe,~\IEEEmembership{Life~Fellow,~IEEE}
\IEEEcompsocitemizethanks{\IEEEcompsocthanksitem M. Shell is with the Department
of Electrical and Computer Engineering, Georgia Institute of Technology, Atlanta,
GA, 30332.\protect\\

E-mail: see http://www.michaelshell.org/contact.html
\IEEEcompsocthanksitem J. Doe and J. Doe are with Anonymous University.}
\thanks{Manuscript received April 19, 2005; revised December 27, 2012.}}

\markboth{Journal of \LaTeX\ Class Files,~Vol.~11, No.~4, December~2012}
{Shell \MakeLowercase{\textit{et al.}}: Bare Advanced Demo of IEEEtran.cls for Journals}

\IEEEtitleabstractindextext{
\begin{abstract}
The abstract goes here.
\end{abstract}

\begin{IEEEkeywords}
Computer Society, IEEEtran, journal, \LaTeX, paper, template.
\end{IEEEkeywords}}

\maketitle

\IEEEdisplaynontitleabstractindextext

\IEEEpeerreviewmaketitle

\section{Introduction}

\IEEEPARstart{T}{his} demo file is intended to serve as a ``starter file''
for IEEE Computer Society journal papers produced under \LaTeX\ using
IEEEtran.cls version 1.8 and later.

I wish you the best of success.

\hfill mds

\hfill December 27, 2012

\subsection{Subsection Heading Here}
Subsection text here.

\subsubsection{Subsubsection Heading Here}
Subsubsection text here.

\section{Conclusion}
The conclusion goes here.

\appendices
\section{Proof of the First Zonklar Equation}
Appendix one text goes here.

\section{}
Appendix two text goes here.

\ifCLASSOPTIONcompsoc

  \section*{Acknowledgments}
\else

  \section*{Acknowledgment}
\fi

The authors would like to thank...

\ifCLASSOPTIONcaptionsoff
  \newpage
\fi


TITLE: Does the following function admit a maximum?
QUESTION [1 upvotes]: I should determine if a maximum value for the function exist and, if it is so, the maximum points.
$$f(x, y, z) = (1 + x^
2
)e^{-z}$$
over the region:
$$D = \{(x, y, z) ∈ R^
3
|x = y^
2 + z^
2
, x^2 − x + y^
2 + z^
2 ≤ 2\}$$
The set D is clearly closed but I am not sure how to prove that it is compact here. Thank you..

REPLY [2 votes]: Your region $D$ consists of the intersection of the paraboloid $x=y^2+z^2$ and the interior of the sphere $(x-\tfrac12)^2+y^2+z^2=\tfrac94$ (after completing the square). So $D$ is closed and bounded, and thus compact by Heine-Borel.
As for the maxima/minima, since the roles of $x$ and $z$ are independent (and there is no $y$ in the formula, you need to maximize/minimize on $x$ and $z$ independently.
For max: you want $z=0$ and maximize $1+x^2$. In the paraboloid we have $x=y^2+z^2$, so the formula for the sphere because $x^2-x+x\leq 2$, so $x^2\leq 2$. Thus $1+x^2$ maximizes as $1+2=3$ when $x=\pm\sqrt2$. So the maximum is $3$, achieved at points, $(\pm\sqrt2,\pm2^{1/4},0)$ (the values for $y$ coming from $\pm\sqrt2=y^2+0^2=y^2$.
For the minimum you want $x=0$ to minimize $1+x^2$, and $z$ as big as possible; but from $0=y^2+z^2$ you get $y=z=0$, so the minimum is $1$ at $(0,0,0)$.


TITLE: Estimate a value of $\int_2^\infty \frac{x^k e^{-x}}{x \left( \log(x) \right)^2}\, dx $
QUESTION [1 upvotes]: How to estimate a value of the following integral
\begin{align}
\int_2^\infty   \frac{x^k e^{-x}}{x \left( \log(x) \right)^2} \,dx,
\end{align}
for $k=0,1, \dots$.

REPLY [1 votes]: This solution is guided by concepts from the asymptotic estimation of integrals. For $k$ large, the integrand looks like a gaussian with its center offset much further than its width from the origin.  Find the stationary point of argument of the exponential, $\exp((k-1)\log{x} -x)),$ which is $x_0=k-1.$ Since the logarithm is slowly varying, evaluate it at $x_0$ and pull it through the integral.  You are then left with the integrand of the gamma function. There is no reason why the lower integral limit can't be extended to 0 because by assumption $k$ is large enough that you are on the far left side of the peak, which is nearly zero for $x<2.$  Thus you get,
$$ \int_2^\infty e^{-x}\,x^{k-1}\frac{dx}{\log^2{x}} \sim \frac{\Gamma(k)}{\log^2(k-1)}. $$
The approximation is within 1% for $k$ as small as 13.  Use the Stirling approximation for the gamma function if you need a more elementary form.


TITLE: Cycle decomposition of an element of prime order
QUESTION [2 upvotes]: I am reading the proof of the theorem that every alternating group $A_n$ is simple for $n \ge 5$ in Artin's Algebra.
In one step, Artin said

We are given that $N$ is a normal subgroup and that it contains a permutation $x$ different from the identity ... Our first step is to note that a suitable power of $x$ will have prime order, say order $l$. We may replace $x$ by this power, so we may assume that $x$ has order $l$. Then the cycle decomposition of $x$ will consist of $l$-cycles and $1$-cycles.

I think I understand the first statement (correct me if I am wrong) because $N$ is finite and hence $x^L=e$ for some $L$. We can always find a prime factor $l$ of $L$ and write it as $(x^{(L/l)})^l=e$.
But I don't understand how he can arrive at the last statement, that the cycle decomposition of $x$ will consist of $l$-cycles and $1$-cycles.
Moreover, later in his proof, he said

Case 1: $x$ has order $l \ge 5$.
How the indices are numbered is irrelevant, so we may supposed that $x$ contains the $l$-cycle $(12345\cdots l)$, say $x=(12345\cdots l)y$, where $y$ is a permutation of the remaining indices...

But didn't he just stated that the cycle decomposition of $x$ consists only of $l$-cycles and $1$-cycles?

REPLY [2 votes]: Let $\sigma=\sigma_1\sigma_2\sigma_3 [...] \sigma_n$, where $\sigma_i$ are all disjoint cycles. If the order of $\sigma$ is $l$ for prime $l$ and the order of $\sigma_i$ is $r_i$, then $l$ is the least common multiple of $r_1, r_2, r_3, [...], r_n$ since that's how the order of products of elements works. Thus, $r_i \mid l$, so $r_i$ is $1$ or $l$. Therefore, $\sigma$ is the product of disjoint $l$-cycles or $1$-cycles. This is a pretty common theorem, so Artin must've assumed the reader already knew about it.
Also, in the last statement, he's simply taking out an $l$-cycle from the decomposition of $x$. Since you know the decomposition of $x$ is just $l$-cycles and $1$-cycles, then $y$ is just a bunch of $l$-cycles and $1$-cycles composed together. There is no contradiction with the first statement.

REPLY [1 votes]: $\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$First of all, the last statement does not contradict the previous ones: $y$ will be a product of $l$-cycles and $1$-cycles.
Then, if the permutation $x$ has prime order $l$, you may use the orbit-stabilizer theorem, which tells you that the orbits of $G = \Span{x}$, a group of prime order $l$, have size dividing $l$, thus either $1$ or $l$. As the orbit of $\alpha$ is
$$
\alpha, \alpha^{x}, \dots, \alpha^{x^{l-1}},
$$
you get the required decomposition.


TITLE: Number of solutions for $n^5 + 2 n^4 + n^3 - 3n + 2 $ mod $ 23^2 = 0$, where $0 \leq n < 23^2$ and $n\in \mathbb{N}$
QUESTION [0 upvotes]: $0 \leq n < 23^2$  and  $n\in \mathbb{N}$
For how many $n$
$n^5 + 2 n^4  + n^3 - 3n + 2 $ mod $ 23^2 = 0$

REPLY [4 votes]: $$ n^5 + 2n^4 + n^3 - 3n + 2 = (n+2)(n^4 + n^2 - 2n + 1) = (n+2)(n^4 - (n-1)^2) = (n+2)(n^2 - n + 1)(n^2 + n - 1) $$
By the quadratic formula, the last two factors cannot split in fields where $ \sqrt{-3} $ and $ \sqrt{5} $ are not members of the field, respectively. However, the quadratic residues in $ \mathbb{Z}/23\mathbb{Z} $ are $ 1, 4, 9, 16, 2, 13, 3, 18, 12, 8, 6 $ (obtained by squaring the first 11 elements.) Neither $ -3 = 20 $ nor $ 5 $ is in this list, therefore the final two factors are irreducible in $ \mathbb{Z}/23\mathbb{Z} $, and do not have roots lying in the field. (This means that they are never divisible by $ 23 $, and by extension are not zero divisors in $\mathbb{Z}/23^2 \mathbb{Z} $.)
As such, the only possibility is for the first factor to be divisible by $ 23^2 $, therefore the polynomial has only one root in $ \mathbb{Z}/23^2\mathbb{Z} $.


TITLE: Convert $x/y \cdot \frac{1-\delta_1-\delta_3}{(1-\delta_2)}$ to $x/y \cdot (1-\delta)$ using calculus
QUESTION [1 upvotes]: Want expression in the form $x/y \cdot (1-\delta)$

$\delta_3 \delta_1$ and anything $\delta^2$ can be excluded since they are too small

$=x/y \cdot \frac{(1-\delta_1)}{(1-\delta_2)}(1-\delta_3)\\=x/y \cdot \frac{(\delta_3 \delta_1 - \delta_1 - \delta_3 + 1)}{(1-\delta_2)}\\\approx x/y \cdot \frac{1-\delta_1-\delta_3}{(1-\delta_2)}$

As you can see its not in the correct form I'm not sure how to simplify the denominator in this case.
Here is an example with multiplication/addition from my textbook

REPLY [0 votes]: Suppose $\max_i(|\delta_i|)\leqslant\epsilon$:
$$\frac{1-\delta_1-\delta_3}{1-\delta_2} \leqslant 1-\delta \iff 1-\delta_1-\delta_3 \leqslant (1-\delta_2)(1-\delta) = 1-\delta_2-\delta+\delta_2\delta \iff$$
$$\iff \delta \leqslant \delta_1-\delta_2+\delta_3 \iff |\delta| \leqslant |\delta_1-\delta_2+\delta_3| \leqslant |\delta_1|+|\delta_2|+|\delta_3| \leqslant 3\epsilon$$
So if you set $|\delta| \leqslant 3\epsilon$, you get:
$$\frac{x}{y} . \frac{1-\delta_1-\delta_3}{1-\delta_2} \leqslant x/y.(1-\delta)$$


TITLE: Mathematical theory of aesthetics
QUESTION [20 upvotes]: The notion of beauty has historically led many mathematicians to fruitful work. Yet, I have yet to find a mathematical text which has attempted to elucidate what exactly makes certain geometric figures aesthetically pleasing and others less so. Naturally, some would mention the properties of elegance, symmetry and surprise but I think these constitute basic ideas and not a well-developed thesis.
In this spirit, I would like to know whether there are any references to mathematicians who have developed a mathematical theory of aesthetics as well as algorithms(if possible) for discovering aesthetically pleasing mathematical structures.
To give precise examples of mathematical objects which are generally considered aesthetic, I would include:

Mandelbrot set
Golden ratio
Short proofs of seemingly-complex statements(ex. Proofs from the Book)

I think the last example is particularly useful as Jürgen Schmidhuber, a famous computer scientist and AI researcher, has attempted to derive a measure of beauty using Kolmogorov Complexity in his series of articles titled 'Low Complexity Art'. Meanwhile, I find the following research directions initiated by computer scientists particularly fruitful:

Bayesian Surprise attracts Human Attention
Curiosity and Fine Arts
Low Complexity Art
Novelty Search and the Problem with Objectives

Note: From a scientific perspective, researchers on linguistic and cultural evolution such as Pierre Oudeyer have identified phenomena which are both diverse and universal. Diversity is what makes our cultures different and universality enables geographically-isolated cultures to understand one another. In particular, many aesthetics have emerged independently in geographically isolated cultures especially in cultures which developed in similar environments. Basically, I believe that if we take into account what scientists have learned from the fields of cultural
and linguistic evolution, embodied cognition, and natural selection I think we could find an accurate mathematical basis for aesthetics which would also be scientifically relevant.

REPLY [2 votes]: What about this paper where the aesthetics of fractal dimension is measured. The peak seem to be (according to the paper) near Hausdorff dimension 1.5.


TITLE: Invert double integral
QUESTION [2 upvotes]: I'm trying to reverse the order of integration of the following double integral:
$$\int _0^2\int _x^{2x}x^2dydx$$
I am aware that it is not possible to invert all double integrals, but my teacher told me this one can be inverted. I believe he is wrong but I want to make sure about that. I would highly appreciate if anyone can help me with this.

REPLY [3 votes]: The region of integration is a triangle. Crudely drawn here:
         (2,4)
         (2,2)
(0,0)

To reverse the order, $y$ would run from $0$ to $4$.
Within that, $x$ would at first run from $y/2$ to $y$. But halfway up, $x$ would start running from $y/2$ to $2$. Most people would break up the integral in two: $$\int_{y=0}^2\int_{x=y/2}^y+\int_{y=2}^4\int_{x=y/2}^2$$
You could express it as a single integral like $$\int_{y=0}^4\int_{x=y/2}^{f(y)}$$ where $f$ is a piecewise function that changes behavior at $y=2$. $$f(y)=\begin{cases}y/2&y\leq2\\2&y>2\end{cases}$$ You can be "clever" and find this way to express the same function $$f(y)=\frac{y+2-|y-2|}{2}$$ So you have $$\int_{y=0}^4\int_{x=y/2}^{\frac{y+2-|y-2|}{2}}$$


TITLE: Cauchy integration of a circle
QUESTION [0 upvotes]: Evaluate: $$\int_{C(0,1)} |z-1||dz|$$
I have been trying to apply Cauchy's integration formula to the integral above but I cannot seem get it to work with the absolute values. How do those play a part in the formula?

REPLY [2 votes]: Not sure how Cauchy's formula applies here, as the integrand appears nonanalytic.  Here's a direct approach.
$z=e^{i \phi} \implies dz = i e^{i \phi} d\phi \implies |dz| = d\phi$
Thus,
$$\oint_{|z|=1} |z-1| \, |dz| = \int_0^{2 \pi} d\phi \, \left | e^{i \phi} -1\right | = 2 \int_0^{2 \pi} d\phi \, \left | \sin{\frac{\phi}{2}} \right | = 8$$


TITLE: Understanding the quantum physics behind light polarization
QUESTION [3 upvotes]: I'd like to help use the following exercise to reinforce my understanding of some basic concepts of quantum states. Here's a picture of the setup:

Now, I'll try and make my series of questions about the insights taken from this exercise as clear as possible.
First question - why is the intensity after going through the first slit $1/2 \ I_0$ regardless of slit orientation?
Firstly, I attempted the following, which was flat out wrong (also I don't know how to use Bra-ket notation here, I'm going to use unit vectors instead.. hopefully that's okay? Please excuse this hideous 'notation', just trying to illustrate my confusions):
$$V' = \cos{(7\pi/12)} \hat V + \sin{(7\pi/12)}\hat H$$
Where $V'$ is the new vertical polarization state, and $\hat V$ and $\hat H$ are the vertical and horizontal states respectively.
I posited this because $\hat V$ and $\hat H$ represent the basis vectors $(0,1)$, $(1,0)$ that represent the vertical and horizontal aperture arrangements of the filters. We are instructed, as far as I could understand, to construct a new basis vector when there is an arbitrary aperture angle as a linear combination of the typical horizontal and vertical arrangements.. which led to something like this. This led me to state that the probability for absorption of the vertical filter to be..
$$cos^2(7\pi/12) \approx 0.06$$
It's clear that this stuff is fuzzy to me, but hopefully my confusion is laid relatively bare enough. $7\pi/12$ is merely the angle I took from $\theta_0 = 0$, so $\alpha = \pi/12 + \pi/2$.
This was an attempt to answer the first part of the question, which I'm aware was an answer of probability and not intensity, but I wasn't sure how to do this any other way. Anyway, that garbage attempt for physics aside, I'm told that the answer to this is that the intensity is halved from the first aperture, regardless of orientation of the slit. I don't really know why - my only guess is that it has something to do with the light being unpolarized. But, I mean, thinking off the cuff, if the light is a beam with a diameter roughly equal to the filter, then the filter would have to be a gap half the size of the circle in order for only half the light to make it through! I don't have any good justification as to why this doesn't make sense.
Second question - what is the sense behind the approach to find the probability for a photon to be detected by a PMT placed after the final filter?
My lecturer's answer states that after the light goes through the first slit, it is polarized along the $\alpha$ axis, which I think means it basically forms a line parallel and on top of the slightly diagonal, dotted line in the first filter. He contends that, since it makes an angle $\alpha + \beta$ with the transmission axis of the second filter (which is not apparent to me visually), the beam is reduced by a factor
$$cos^2(\alpha + \beta)$$
and by a similar argument for the third filter:
$$cos^2(\beta + \gamma)=cos^2(55^0) \approx 0.33$$
Which leads to a final intensity of $I \approx 0.068 I_0$.
So my specific confusions I need addressed are:

Why the intensity after going through the first slit is $1/2 \ I_0$ regardless of slit orientation?
Why we use a cosine squared term. I thought I knew why in my screw up above but I don't think I do anymore.
Why, in order to find the proper new orientation of polarization, we add the previous angle with the new one as our cosine argument

REPLY [4 votes]: Your question is interesting as the answer involves more than basic quantum mechanics.
Unpolarized light is actually best described as a mixed state, rather than a pure state.  Basically, unpolarized light is a statistical superposition of two perpendicular polarization in the sense that half of the photons have one polarization, and half of the photon the other.  Mixed states are described by a density matrix, which here would take the form
\begin{align}
\rho &=\frac{1}{2}\vert H\rangle\langle H\vert +\frac{1}{2}\vert V\rangle\langle V\vert\,  ,\\
&=\frac{1}{2}\vert \updownarrow \rangle\langle  \updownarrow\vert +\frac{1}{2}\vert \leftrightarrow \rangle\langle \leftrightarrow\vert \tag{1}
\end{align}
In mixed states, there is no possibility for the vertically polarized and horizontally polarized states to interfere, and the factors of $\frac{1}{2}$ are the statistical weights of each polarization.   Mixed states are represented by operators rather than kets.
There is a subtle difference between the interpretation of statistical weights, which are also probabilities, and the probabilities obtained by overlaps like $\vert\langle\psi\vert\phi\rangle\vert^2$: both types probabilities have different origins.   One is tied to incoherent averages, as referred to in the linked wiki page.
Moreover, by symmetry,  "which one" of the two polarization is completely undefined if the light is completely unpolarized.  In other word, the completely unpolarized light is equally well described by the mixture
$$
\rho =\frac{1}{2}\vert \nearrow \rangle\langle \nearrow\vert +\frac{1}{2}\vert \nwarrow\rangle\langle \nwarrow\vert\, .
$$
(unfortunately,  slanted double arrows are not so easily accessible so $\nearrow$ is a standin for $\updownarrow$ rotated by $45^\circ$ to the right, and $\nwarrow$ is a standin for $\leftrightarrow$ rotated by $45^\circ$ to the right) or for that matter, using any  slanting of $\updownarrow$ and $\leftrightarrow$.
If you set your polarizer along the vertical axis, you are making (in the parlance) a projective measurement and the outcome is always the pure state $\vert \updownarrow\rangle\langle \updownarrow\vert $.  Since the statistical weight of this state is $1/2$, you eliminate half of the intensity.  Since there is nothing special about $\vert \updownarrow\rangle\langle \updownarrow\vert $, this will be true of any orientation.  This is, in the formalism of mixed states, the content of the statement immediately below your figure.
After the first polarizer, the state is pure, and pure states are those that you are familiar with.  We can do away with all the density matrix stuff and think of the state as the ket $\vert \updownarrow\rangle$, or any slanted thereof where the slanting would be parallel to the axis of the polarizer.  Your $V'$ state is described by a ket
$$
\vert V'\rangle =\cos\theta \vert \updownarrow\rangle +\sin\theta \vert \leftrightarrow\rangle\, . \tag{2}
$$
What is the difference between (1) and (2)?  In (1) is it not possible to find an orientation of the polarizers that will result in no transmission of the light.  If there was such an orientation, light would be polarized at $90^\circ$ w/r to this direction.  In (2) on the other hand, the ket
$$
\vert H'\rangle = -\sin\theta \vert \updownarrow\rangle +\cos\theta \vert \leftrightarrow\rangle
$$
is orthogonal to $\vert V'\rangle$ so the probability of obtaining the outcome $H'$ as a result of having the system initially in $\vert V'\rangle$ is $
\vert \langle H'\vert V'\rangle\vert^2=0$, i.e. no light would pass through a filter aligned parallel to the $V'$ direction if light was initially described by the polarization ket $\vert H'\rangle$.
Note that, from (2), you can recover Malus' law.  The probability of light initially polarized as $\vert \updownarrow\rangle$ to exit in the state $\vert V'\rangle$ is
$$
\vert \langle \updownarrow\vert V'\rangle \vert^2 =\cos^2\theta
$$
if $V'$ makes and angle $\theta$ with the vertical.  The attenuation in intensity from Malus' law follows immediately from the (discrete) probability of a photon going though two polarizers sets at a relative angle $\theta$.
Finally,  it is possible to have partially polarized states, in the sense that (1) generalizes to
$$
\rho= a\vert H\rangle\langle H\vert +b\vert V\rangle\langle V\vert
$$
where $a+b=1$,  where $a$ is the statistical probability of having an "H" photon and $b$ the prob of finding a $V$ photon in the beam.  In this case, and contrary to (1), there is an orientation where the intensity of light will be minimal but non-zero.
The best reference on density matrix in 2-level systems, with discussion that is done in terms of Stern-Gerlach magnet but otherwise immediately applicable to polarization, is the first chapter of Blum, Karl. Density matrix theory and applications. Springer Science & Business Media, 2013, available as a google books.  Indeed, section 1.1.2 is on polarization (in terms of Pauli matrices).


TITLE: Conventions for definitions of the cap product
QUESTION [12 upvotes]: In singular (co)homology, if $\alpha\in C^*(X)$ and $x\in C_*(X)$, then the cap product $\alpha \cap x$ is generally defined by the following process:

Apply to $x$ the diagonal map $C_*(X)\to C_*(X\times X)$ followed by some choice of Alexander-Whitney chain equivalence $ C_*(X\times X)\to C_*(X)\otimes C_*(X)$ to obtain an element $\sum y_i\otimes z_i$.
Apply $\alpha$ to $\sum y_i\otimes z_i$ by the slant product, or, in other words and roughly speaking, apply $\alpha$ to ``half of the factors''.
Depending on your favorite conventions (and what you're trying to accomplish), there may also be a sign (which might also be part of your definition of the slant product - but let's ignore this).

The reason I'm being so cagey with wording in part 2 is directly related to my question: In almost all major textbook sources I have consulted, step 2 is performed by forming $\sum y_i \alpha(z_i)$, which strikes me as somewhat unnatural, forcing the $\alpha$, which starts off on the left to jump all the way over the $y_i$ terms to get to the $z_i$ terms on the right. Is there a good mathematical reason for this convention? Why not define the cap product to be $\sum \alpha(y_i) z_i$?
The one major exception to this convention seems to be Hatcher. He does form $\sum \alpha(y_i) z_i$, but he also writes cap products as $x\cap \alpha$, so his cochain also has to jump, but it jumps over the $z$s instead!
(For the record, I'm not asking this question out of idle pickiness. Jim McClure and I have been doing a lot of work with cap products recently, and we're trying to be consistent amongst various conventions for various issues, but preferably with good reasons thrown in!)

REPLY [8 votes]: This is just an expanded version of Tyler's comment, I think.
Let's use a, b, c for cochains, x, y, z for chains, [a,x] for the value of a cochain on a chain. I'll be lazy and write $ab$ for $a\cup b$ and $ax$ for $a\cap x$. Let $[a\otimes b,y\otimes z]=(-1)^{|b||y|}[a,y][b,z]$.
I like to define $bx$ in such a way that $[a,bx]$ is $[ab,x]$. Then associativity and unit of cup makes cap into a module structure: $(bc)x=b(cx)$ because $[a,(bc)x]=[a(bc),x]=[(ab)c,x]=[ab,cx]=[a,b(cx)]$, and $1x=x$ follows from $[a,1x]=[a1,x]=[a,x]$.
If you have a product $a\otimes b\mapsto ab$ of cochains and a coproduct of chains defined in such a way that $[ab,x]=[a\otimes b,y_i\otimes z_i]$ where the coproduct of $x$ is $\Sigma y_i\otimes z_i$, then that means I have to define $bx$ to be $\Sigma (-1)^{|y_i||b|}[b,z_i]y_i$, so as to get
$[a,bx]=[ab,x]=[a\otimes b,\Sigma y_i\otimes z_i]=\Sigma (-1)^{|y_i||b|}[a,y_i][b,z_i]=[a,\Sigma (-1)^{|y_i||b|}[b,z_i]y_i]$.
The sign is a bit unexpected, as is the jump you mention, but it's worth it for the neat formulas in the third paragraph above.
It's not a bad idea to define $xa$ as well. I'd do it by first declaring $[x,a]$ to be $(-1)^p[a,x]$ when $a$ is a $p$-cochain and $x$ is a $p$-chain, then defining $xa$ in such a way that $[xa,b]=[x,ab]$. This insures $x(ab)=(xa)b$ and $x1=x$. The chain-level formula is no better and no worse than the formula for $ax$.
Of course, $ax$ and $xa$ end up differing only by a sign when you get to homology, but the sign is hard to remember; in working it out you have to use the commutativity law for the cup product.


TITLE: Show that every contractible space is connected.
QUESTION [3 upvotes]: So I'm trying to show every contractible topological space is connected. At the moment I've established that $X$ is contractible iff there is a homotopy equivalence $f:\{x_0\}\to X$ for some $x_0\in X$, but I'm having trouble showing that if $X$ and $Y$ are homotopy equivalent and $X$ is connected, then so is $Y$ (from which the result would follow, as $\{x_0\}$ is clearly connected).

REPLY [5 votes]: Let $H: X \times I$ be a contracting homotopy to $p \in X$, so that

$H$ is continuous.
$H(x,0)=x$ for all $x \in X$.
$H(x,1)=p$ for all $x \in X$.
$H(p,t)=p$ for all $t$.

Then for each $x \in X$ we have a continuous path $p_x: [0,1] \to X$ such that $p_x(0)=x, p_x(1)= p$, namely $p_x(t)=H(x,t)$ where we restrict $H$ to $\{x\} \times I$, reallly, so $p_x$ is continuous as $H$ is.
Then clearly $X$ is path-connected: If $x,y \in X$ define $p: I \to X$ by
$$p(t)= \begin{cases} p_x(2t)& 0 \le t \le \frac12\\
                      p_y(2-2t) & \frac12 \le t \le 1\\
        \end{cases}$$
which is continuous by the pasting lemma for two closed sets (note that $t=\frac12$ is consistent as $p_x(1)= p = p_y(1)$, so they connect); we go from $x$ to $p$ via $p_x$ and then beckwards from $p$ to $y$ via $p_y$.
And a path-connected space is connected. (Follows from $I$ being connected and continuous paths preserving connectedness, so $p[I]$ also is connected.)


TITLE: Convergence of $a_n= (7/10+7/10i)^n$
QUESTION [3 upvotes]: I am new to real analysis and want to prove that $$a_n=\Bigg(\frac{7}{10}(1+i)\Bigg)^n$$ is convergent for $n\rightarrow \infty$.
This is what I have done:
$a_n= (7/10+7/10i)^n \\
= (49/100+2\cdot49/100i-49/100)^n\\
= (98/100i)^n\\
= \Bigg(\frac{-9604}{10000}\Bigg)^n$
By multiplying some of the terms it seems that the sequence converges to $0$, because $\frac{-9604}{10000}$ is less than $1$. Is this a correct approach? I'm not sure if I am allowed to treat sequences this way. My book only convers sequences very briefly.
Thank you

REPLY [1 votes]: You seem to have a somewhat right idea, but the computation is not correct. Verify that $$c:=\left|\frac7{10}(1+i)\right|=\frac7{10}\sqrt{1^2+1^2}=\frac{7\sqrt 2}{10}<1$$
because $\left(\frac{7\sqrt 2}{10}\right)^2=\frac{98}{100}<1$.
Then $|a_n|=|a_1^n|=|a_1|^n=c^n\to 0$.


text:
{\bf Problem.} Compute $\tan (-3645^\circ)$.
{\bf Level.} Level 2
{\bf Type.} Geometry
{\bf Solution.} Rotating $360^\circ$ is the same as doing nothing, so rotating $3645^\circ$ is the same as rotating $3645^\circ - 10\cdot 360^\circ = 45^\circ$.  Therefore, $\tan(-3645^\circ) = \tan (-45^\circ)$.

Rotating $45^\circ$ clockwise is the same as rotating $360^\circ - 45^\circ = 315^\circ$ counterclockwise, so $\tan(-45^\circ) = \tan (360^\circ - 45^\circ) = \tan 315^\circ$.

Let $P$ be the point on the unit circle that is $315^\circ$ counterclockwise from $(1,0)$, and let $D$ be the foot of the altitude from $P$ to the $x$-axis, as shown below.

[asy]

pair A,C,P,O,D;

draw((0,-1.2)--(0,1.2),p=black+1.2bp,Arrows(0.15cm));

draw((-1.2,0)--(1.2,0),p=black+1.2bp,Arrows(0.15cm));

A = (1,0);

O= (0,0);

label("$x$",(1.2,0),SE);

label("$y$",(0,1.2),NE);

P = rotate(315)*A;

D = foot(P,A,-A);

draw(O--P--D);

draw(rightanglemark(O,D,P,2));

draw(Circle(O,1));

label("$O$",O,NW);

label("$P$",P,SE);

//label("$A$",A,SE);

label("$D$",D,N);

[/asy]

Triangle $POD$ is a 45-45-90 triangle, so $DO = OP = \frac{\sqrt{2}}{2}$.  Therefore, the coordinates of $P$ are $\left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)$, so $\tan(-3645^\circ) = \tan (-45^\circ) = \tan 315^\circ = \frac{\sin 315^\circ}{\cos 315^\circ} = \frac{-\sqrt{2}/2}{\sqrt{2}/2} = \boxed{-1}$.

set_name: MATH


TITLE: Is the Euler–Mascheroni constant an EL-number?
QUESTION [7 upvotes]: This question is based on Chow - What is a closed-form number?.
The author of the linked paper had proposed a plausible definition of "elementary numbers" (which he calls "EL-numbers") concept, building it analogously to the concept of elementary function (author admits, "elementary number" would be a better name for his proposal, but the term is already occupied).
So, the author defines a set of $\mathbb{E}$ of "EL numbers" which stands for "elementary" and well as "exponentially-logarithmic".
He defines the set as any numbers that can be produced by applying finite number of field operations, exponential and logarithmic functions to the number $0$.
For instance, in his system
\begin{gather*}
1=\exp(0) \\
e=\exp(\exp(0)) \\
i=\exp\left(\frac{\log(-1)}2\right)=\exp\left(\frac{\log(0-\exp(0))}{\exp(0)+\exp(0)}\right) \\
\pi=-i\log(-1)=-\exp\left(\frac{\log(0-\exp(0))}{\exp(0)+\exp(0)}\right)\log(0-\exp(0)).
\end{gather*}
It turns out that any root of a polynomial with rational coefficients, expressible in the radicals, is also in $\mathbb{E}$.
So, my question is, does the Euler–Mascheroni constant $\gamma$ belong to the EL-numbers? I think no, but that it is "nearly-elementary" in the same way as the digamma function is a nearly-elementary function.
My thoughts on this revolve around these points:

There is symmetry between $\pi/4$ and $e^{-\gamma}$

$\gamma=\psi(1)$, but $\psi(x)$ is antidifference of $1/x$ while logarithm is antiderivative. $\psi(x)$ relates to $\log x$ the same way as Bernoulli polynomials relate to monomials (both Bernoulli polynomials and $\psi(x)$ are slices of Hurwitz Zeta function).

Many divergent integrals and their logarithms regularize to $\gamma$, particularly, $\operatorname{reg} \int_0^1 \frac1x dx=\gamma$ (which makes it in some sense the regularized value of logarithm at zero).

REPLY [10 votes]: Despite the simplicity and elegance of the definition of EL numbers, it is very hard to prove that an explicit number is not EL, as Chow points out throughout the paper.
In particular, I am certain that as of right now, nobody knows if the Euler-Mascheroni constant is or isn’t EL. It most likely isn’t, but we can’t even prove that it is irrational.