abstract algebra - Definition of the multiplication matrix of element of finite field.

I've been reading this article XOR-counts and lightweight multiplication with fixed elements in binary finite fields and I came across the term Multiplication matrix and I've wondered if that's some terminology from finite fields theory I perhaps don't know (English is not my native language and I don't study math in English). It seems really odd that the author mentions the term in an abstract and doesn't define it properly (but then maybe it's just a widely used term).
From what I've understood multiplication matrix is matrix such that: $\forall \alpha \in \mathbb{F}_{2^n} \exists A \in \mathbb{F}_2$ such that $\forall x \in \mathbb{F}_{2^n}: \alpha \cdot x = A \cdot \hat{x}$, where $\hat{x}$ is element $x$ written as an n-dimensional vector over $\mathbb{F}_2$.
Thank you for any help.

Answer

Multiplication by an element of $\mathbb F_{2^n}$ on elements of $\mathbb F_{2^n}$ produces a $\mathbb F_2$ linear transformation on $\mathbb F_{2^n}$. As such, you can pick a basis of $\mathbb F_{2^n}$ (necessarily having $n$ elements) and find a transformation matrix (an element of $M_n(\mathbb F_2)$) for each element.

Then the explanation in the paper is quite clear (p 286, the second page of the pdf):

Our goal in this work is to explore some connections and properties of the direct and sequential XOR-count metrics and then to apply these to get some theoretical results regarding optimal implementations of matrices that represent multiplication with a fixed field element $α ∈ \mathbb F_{2^k}$. Optimal choices of these matrices (called multiplication matrices) can then be used for local optimizations of matrices over $\mathbb F_{2^k}$.

As mentioned in the paper in several places, they are optimizing to minimize the number of XOR operations required to perform multiplication.

limits - Find the sum of the infinite series $1+ frac{1}{2!}+ frac{1}{4!}+dotsb$

I wanted to find the limit of the series $1+ \frac{1}{2!}+ \frac{1}{4!}+\dotsb$. My approach:
Let $S$ be the required sum.

Then $S= (1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dotsb)- (1+ \frac{1}{3!}+...)$

i.e., $S= e - (1+ \frac{1}{3!}+\dotsb)$

But I don't know how to proceed further. I want to work the problem on my own. So please give me hint rather than the whole answer.

Thanks in advance.

Answer

HINT: If we define
$$G(x):=\sum_{n=0}^\infty a_n x^n$$
Then what is the series representation of
$$G(x)+G(-x)=\sum_{n=0}^\infty \space ?$$

taylor expansion and limit of a series??

$f(x)=\int_0^xtan^{-1}tdt$

what is the taylor expansion about the origin of this function?
and how do i use this to get the limit of the series

$1-\frac{1}{2}-\frac {1}{3}+\frac {1}{4}+\frac {1}{5}-\frac{1}{6}-\frac {1}{7}.......$

i could get the limit by using concepts like rearranging the terms and got a different limit since it is a conditionally convergent series and can be made to converge to any real number.but how do i get the limit using this taylor expansion.please somebody help?
Answer to the second part is $\frac {\pi}{4}-\frac {log2}{2}$

analysis - Measure theory limit question

Let $(X, \cal{M},\mu)$ be a finite positive measure space and $f$ a $\mu$-a.e. strictly positive measurable function on $X$. If $E_n\in\mathcal{M}$, for $n=1,2,\ldots $ and $\displaystyle \lim_{n\rightarrow\infty} \int_{E_n}f d\mu=0$, prove that $\displaystyle\lim_{n\rightarrow\infty}\mu(E_n)=0$.

Answer

Since $f$ is almost everywhere strictly positive, the increasing sequence of sets $$A_n=\{x\in X:f(x)>1/n\}$$
has the property that $$\lim_{n\to\infty} \mu(A_n)=\mu(X).$$ Now $\int_E f ~d\mu0$.

So let $\epsilon>0$. Choose $n$ so that $\mu(X\backslash A_n)<\epsilon/2$. For $N$ large enough, $\int_{E_N}f~d\mu<\epsilon/(2n)$ and hence $$\mu(E_N\cap A_n)

sequences and series - Evaluating Sum involving binomial coefficients and powers

I would like to evaluate the double sum $\sum\limits_{n=1}^{\infty} \sum\limits_{m=1}^{\infty} \dfrac{(n+m)!}{n!m!n^2 m^2}\left(\dfrac{1}{2}\right)^{n+m}$. My starting point was to consider $\sum\limits_{n=1}^{\infty} \sum\limits_{m=1}^{\infty} \dfrac{(n+m)!}{n!m!}x^n y^m = \dfrac{1}{1 -x -y} - \dfrac{1}{1-x} - \dfrac{1}{1-y} + 1$ $\;\;\forall\;\; |x|+|y|<1\;\;$ All what is left is to divide by $xy$ then integrate with respect to $x$ and then with respect to $y$ (process should be repeated twice) finally set $x = y = \frac{1}{2}$. I am however stuck in evaluating the resulting integrals. I expect logarithmic and polylogarithmic functions to show up in the final result. I would appreciate if you can help me formulating the value of this sum.
Thanks for your help...

Uniform Continuity of the given function

I would like to check whether $\sqrt{x} \sin(x)$ is uniformly continuous in its domain.

My attempt at question :

Since the given function is continuous everywhere therefore if the domain would have been bounded then the function would have been surely uniformly continuous , however since we are checking the uniform continuity in its whole domain therefore how should I approach the question ? Any help would be appreciated .

abstract algebra - Question about $operatorname{Aut}(S_6)$ and $operatorname{Aut}(A_6)$

From (1), (2), (3), $[\operatorname{Aut}(S_6):\operatorname{Inn}(S_6)]=2$.

My question:

$1$. How to prove $\operatorname{Aut}(S_6)\cong S_6\rtimes_\varphi \mathbb Z_2$?

$2$. How to prove $\operatorname{Aut}(S_6)\not\cong S_6\times \mathbb Z_2$?

$3$. How to prove $\operatorname{Aut}(A_6)\cong \operatorname{Aut}(S_6)$?

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My effort:

$1$. For 1, it remains to show there exists $\sigma\in \operatorname{Aut}(S_6)\setminus \operatorname{Inn}(S_6)$ s.t. $\sigma^2=\text{id}$.

$2$. For 2, $Z(S_6\times\mathbb Z_2)=\mathbb Z_2$, it's sufficient to show $Z(\operatorname{Aut}(S_6))\neq\mathbb Z_2$.

$3$. For 3, I proved $\operatorname{Aut}(S_n)\leqslant\operatorname{Aut}(A_n)$ (Is this correct?) and $[\operatorname{Aut}(A_6):\operatorname{Inn}(S_6)]\leqslant 2$.

Update:

I wrote my answer below, but there still remain three questions:

$1$. I copied the result from a book to give an explicit element $\psi\in\operatorname{Aut}(S_6)\setminus \operatorname{Inn}(S_6)$ of order $2$, and I wonder if there's a way to avoid doing so, i.e. find an element of order $2$ in $\operatorname{Aut}(S_6)\setminus \operatorname{Inn}(S_6)$ without writing it out explicitly.

$2$. I used the specific element $\psi$ to show $\mathbb Z_2\cong \langle \psi\rangle$ is not normal in $\operatorname{Aut}(S_6)$, I wonder if we can analysis the center of $\operatorname{Aut}(S_6)$ instead. And what is center of $\operatorname{Aut}(S_6)$?

$3$. Is there a better way to prove $\operatorname{Aut}(A_6)\cong \operatorname{Aut}(S_6)$?

Thanks for your time and effort!

Answer

For 1, there exists $\psi\in \operatorname{Aut}(S_6)\setminus \operatorname{Inn}(S_6)$ s.t. $\psi^2=\text{id}$.

$\quad\psi:(12)\mapsto(15)(23)(46), (13)\mapsto(14)(26)(35), (14)\mapsto(13)(24)(56),\\\qquad (15)\mapsto(12)(36)(45), (16)\mapsto(16)(25)(34).$

Therefore $\operatorname{Aut}(S_6)\cong S_6\rtimes\mathbb Z_2$.

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For 2, we have short exact sequence for groups: $1\to S_6\overset{f}{\to}\operatorname{Aut}(S_6)\overset{\pi}{\to} \mathbb Z_2\to 1 $, $\mathbb Z_2=\{\pm1,\times\}$.

This sequence right splits, so there exists homomorphism $g:\mathbb Z_2 \to \operatorname{Aut}(S_6)$ s.t. $\pi\circ g=\text{id}.$

Let $g(-1)=\psi\not\in \operatorname{Inn}(S_6)$, then $g(1)=\psi^2=\text{id}$.
$f:S_6\to \operatorname{Inn}(S_6)$, $g:\mathbb Z_2 \to \langle\psi\rangle$.

Claim: $\langle\psi\rangle$ is not normal subgroup of $\operatorname{Aut}(S_6)$, so $\operatorname{Aut}(S_6)\not \cong S_6\times\mathbb Z_2$.

For $\sigma\in S_6$, define $\gamma_\sigma \in \operatorname{Inn}(S_6)$ to be action by conjugation of $\sigma$.

It's sufficient to prove $\gamma_\sigma\psi\gamma_\sigma^{-1}\neq\psi$, i.e.$\gamma_\sigma\psi\neq\psi\gamma_\sigma$ for some $\sigma\in S_6$.

Let $\sigma=(12)$, $\gamma_\sigma\psi((12))=\gamma_\sigma((15)(23)(46))=(12)(15)(23)(46)(12)=(13)(25)(46)$.

$\psi\gamma_\sigma(12)=\psi((12))=(15)(23)(46)$. $\gamma_\sigma\psi\neq\psi\gamma_\sigma$ for $\sigma=(12)$.

Thus $\operatorname{Aut}(S_6)\cong S_6\rtimes\mathbb Z_2$ and $\operatorname{Aut}(S_6)\not \cong S_6\times\mathbb Z_2$.

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For 3, fix $1\neq\alpha\in A_n$, $c_\alpha\in\text{Inn}(A_n)$ is action by conjugation of $\alpha$.

Define $\varphi:\text{Aut}(S_n)\to\text{Aut}(A_n)$, $\varphi(\beta)=\beta c_\alpha \beta^{-1}$ for $\beta\in \text{Aut}(S_n)$.

Easy to check $\varphi$ is monomorphism, so $\text{Aut}(S_n)\leqslant\text{Aut}(A_n)$

Together with $[\text{Aut}(A_6):\text{Inn}(S_n)]\leqslant2$ and $[\text{Aut}(S_6):\text{Inn}(S_n)]=2$, we have

$\text{Aut}(A_6)=\text{Aut}(S_6)$.