Deleted Question on Semi-Simple and Projective but not Injective Module

A backup of a deleted PSQ : OP : irfanmat It has a detailed answer by Atticus Stonestrom. It’s pity that his post got deleted. As there’s no reason for undeletion, I’m posting it here so as to preserve the contents. Question body Is there a semi-simple and projective but not injective module? I will be glad if you help. Response(s) In the non-commutative case, the answer is yes. Consider $R$ the ring of upper triangular $2\times 2$ matrices over a field $F$, and denote by $e_{ij}$ the element of $R$ with the $ij$-th entry equal to $1$ and all other entries equal to $0$. [Read More]

Ways to Draw Diagrams Displayed on Math.SE

I wanted to start a meta question, but I don’t see a point of that after viewing some related posts listed at the end of the next subsection. Intended question You may vote on your preferred way. Ways Advantages Disadvantages AMScd supported on Math.SE for a long time no diagonal arrowssyntax less well-knownTwo-way arrows $\rightleftarrows$ look odd array supported on Math.SE for a long timeeasier syntaxwrite basic diagonal arrows like $\nearrow$ fine tuning spacing is harddiagonal arrows only work for neighboring nodes ASCIIFlow WYSIWYG interface lines are rendered as slashes in code TikZ well known syntaxcan draw pretty diagrams not supported on SE, need to import as picture others? [Read More]

My Intended Trigo Answer

The Math.SE question $2\cos(2x) - 2\sin(x) = 0$ has attracted several answers from high-rep users.

I am expanding @rhombic’s comment into an answer.

$$ \begin{aligned} 2\cos(2x)-2\sin(x)&=0 \\ 2 - 4\sin^2(x)-2\sin(x)&=0 \\ 2\sin^2(x)+\sin(x) - 1&=0 \\ (2 \sin(x) - 1)(\sin(x) +1) &= 0 \\ \sin(x) = \frac12 \text{ or } \sin(x) &= -1 \\ x = \frac{\pi}{6}, \frac{5\pi}{6} \text{ (rejected) or } & \frac{3\pi}{2} \text{ (rejected)} \end{aligned} $$

JavaScript Copy Button

Goal To create a copy button for my Math.SE comment template in order to save the trouble of copying and pasting. My first attempt I put the boilerplate inside a Markdown codeblock to prevent them from getting interpreted by Hugo’s Markdown parser. Under each codeblock, I placed the copy button. Comment boilerplate goes here ... 📝 Another comment boilerplate goes here ... 📝 … My page’s original layout $(document).ready(function() { $('. [Read More]

My Dual Answer

Update: The question has been reopened. I intended to answer 김종현’s problem on Math.SE. However, the programs in the question body aren’t typeset in MathJax. As a result, I downvoted and closed this question because found it unclear. From the proposed dual, it seems that I shouldn’t interpret the primal as a linear program. Anyways, without further clarifications from OP, I found no reason to look at this further. Here’s my intended answer: [Read More]

Borel Cantelli Exercise 2019

I intend to post this for a Borel-Cantelli lemma exercise on Math.SE.

The target event is ${\exists i_0 \in \Bbb{N} : \forall i \ge i_0, X_i = 1}$, whose complement is

$$ {\forall i_0 \in \Bbb{N} : \exists i \ge i_0, X_i > = 0} = \limsup_i {X_i = 0}. $$

To apply Borel-Cantelli, one has to determine whether $\sum_i P(X_i = 0)<+\infty$.

Weak LLN Practice

My intended answer to a weak LLN problem on Math.SE. Problem: Suppose $(X_n)$ is a sequence of r.v’s satisfying $P(X_n=\pm\ln (n))=\frac{1}{2}$ for each $n=1,2\dots$. I am trying to show that $(X_n)$ satisfies the weak LLN. The idea is to show that $P(\overline{X_n}>\varepsilon)$ tends to 0, but I am unsure how to do so. My solution: As in the accepted answer in OP’s previous question, I’ll assume the independence of $(X_n)$. [Read More]

Solution to a $p$-test Exercise

I intended to answer Maddle’s $p$-test question, but T. Bongers has beaten me by two minutes, so I posted my answer here to save my work. The problem statement This is the sum: $$\sum\limits_{n=3}^\infty\frac{1}{n\cdot\ln(n)\cdot\ln(\ln(n))^p}$$ How do I tell which values of $p$ allow this to converge? The ratio test isn’t working out for me at all. Unpublished solution The integral test will do. $$ \begin{aligned} & \int_3^{+\infty} \frac{1}{x\cdot\ln(x)\cdot\ln(\ln(x))^p} \,dx \\ &= \int_3^{+\infty} \frac{1}{\ln(x)\cdot\ln(\ln(x))^p} \,d(\ln x) \\ &= \int_3^{+\infty} \frac{1}{\ln(\ln(x))^p} \,d(\ln(\ln(x))) \\ &= \begin{cases} [\ln(\ln(\ln(x)))]_3^{+\infty} & \text{if } p = 1 \\ \left[\dfrac{[\ln(\ln(x))}{p+1}]^{p+1} \right]_3^{+\infty} & \text{if } p \ne 1 \end{cases} \end{aligned} $$ When $p \ge 1$, the improper integral diverges. [Read More]