M. A.

general recent conversations

Room for M. A. and Glorfindel

M. A.

Tue 09:50

@Glorfindel Thank you. I sent an e-mail replying to the network staff message directly from the e-mail. Is this correct, or should I have sent it somewhere else?

M. A.

Mon 15:48

@Glorfindel I received a reply from a moderator saying that she would only be able to remove my block if my questions were positively received, and all questions with a negative score would be removed from my ownership and assigned to an anonymous. What do questions have to do with suggested edits? By the way, none of my questions on StackExchange have a negative score.

M. A.

Mon 15:41

"Since every attempt to communicate with the TeX.SE moderators was unsuccessful, I decided to contact the Staff for further clarification and explanation.
I was suspended for a month from suggesting edits on TeX.SE, but received no prior warning or communication. I only realised this when I approached to suggest an edit.
I would like to know the objective requirements for a one-month suspension without warning, as there has been no second offence (the standard suspension period is usually seven days) and the approved edits amount to an approximately 96.65% realisation rate (317 suggested ed…

M. A.

Mon 15:41

I sent this to StackExchange staff:

M. A.

Mon 15:39

@Glorfindel Thank you. I just received a reply from StackExchange staff, but it has nothing to do with my complaint. I am very confused.

M. A.

Sun 17:21

@Glorfindel Thank you. About eight hours ago I sent a complaint to the staff via the Contact link at the bottom of the page as per the post you sent me, but I have still not received a reply. Is the waiting time very long?

M. A.

Sat 11:49

@Glorfindel According to the objective requirements, are 317 out of 328 approved edits sufficient for a one-month suspension from suggesting edits?

M. A.

Sat 11:37

@Glorfindel Thank you.

M. A.

Sat 11:34

@Glorfindel I was suspended for a month from suggesting edits on TeX StackExchange, but received no prior warning or communication. I only realised this when I approached to suggest an edit earlier today. I would like to know what the objective requirements are for a one-month suspension without warning, since there has been no second offence and the approved edits amount to a 96.53% realisation rate.

M. A.

Sat 11:28

@Glorfindel Since I received a moderation action against me on TeX.SE, I contacted a moderator of that site. He told me: "If you have concerns, you should contact the network staff. They can see all mod actions and have more data than we do." I am also writing to you because I have not received any official communication regarding the suspension.

M. A.

Sat 11:23

Hello @Glorfindel. Since you are a Meta SE moderator, can I talk to you?

Room for M. A. and Joseph Wright

M. A.

Sat 16:32

@JosephWright I did not understand. Could you write with clearer and more understandable formatting? Thank you.

M. A.

Sat 11:51

@JosephWright It seems that a moderator can implement a manual ban even though the standard suspension period is seven days. Right?

M. A.

Sat 11:10

@JosephWright And what is the difference between someone who is suspended for a week and someone who is suspended for a month?

M. A.

Sat 11:06

@JosephWright Then I try to keep everything as general as possible. What are the objective requirements for suspending a user for a month from suggesting edits?

M. A.

Sat 11:04

@JosephWright How can I contact you to find out more about a moderation action taken against me?

M. A.

Sat 11:01

Hello @JosephWright. Since you are a TeX.SE moderator, can I talk to you?

$Mathematics$ Mathematics

Associated with Math.SE; for both general discussion & math qu...

M. A.

Jul 2 19:16

Q: Prove $\int_0^{\sqrt2/4}\frac1{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)(x-1+x\sqrt{9-16x})}{1-2x}}dx=\frac{\pi^2}{8}$ (from a probability question)

Let $$I=\int_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)(x-1+x\sqrt{9-16x})}{1-2x}}dx$$ Prove that $I=\dfrac{\pi^2}{8}$. Wolfram suggests that it's true but does not find the antiderivative. Here is the graph of the function being integrated. Equivalent forms Using $\space \arcs...

M. A.

Jul 2 19:16

@XanderHenderson Here:

M. A.

Jul 2 19:11

@XanderHenderson Yes, sorry, I didn't see this. It is a version of an integral I have seen here on MSE. (a) does not hold, but (b) does. The answer I want is $\pi^2/16$, which is a nice result.

M. A.

Jul 2 19:01

@XanderHenderson Why? It is an integral that I must evaluate :-)

M. A.

Jul 2 18:56

@leslietownes OK, I don't beat around the bush. The original integral is $$\int_{3}^{2 \sqrt 2 - 1} \left(p(y)\log\left|\sqrt{1+q(y)^2}+q(y)\right|\right) \, \mathrm dy,$$ with $$p(y)=\sqrt{\dfrac{y^2}{(y^2-9)(y^2+7)}}$$ and $$q(y)=\sqrt{-\dfrac{(y^2+7)(y^2+2y-7)}{32(y+1)} }.$$ I integrated by parts and obtained a mess consisting of a constant, an elementary integral and that integral I posted earlier.

M. A.

Jul 2 18:39

@XanderHenderson No, $y$ is a function of $z$. How can I differentiate implicitly?

M. A.

Jul 2 18:38

@XanderHenderson It is a mess.

M. A.

Jul 2 17:55

OK. I have to evaluate this integral: $$\int_{y=\sqrt{9-2\sqrt{8}}}^{y=3}\left(\sqrt{z^2+1}\dfrac{\mathrm d}{\mathrm dz}\dfrac{p(y)}{q'(y)}\, \mathrm dz\right),$$ where $z = q(y) = \sqrt{-\dfrac{(y^2+7)(y^2+2y-7)}{32(y+1)} }$ and $p(y) = \sqrt{\dfrac{y^2}{(y^2-9)(y^2+7)}}$. This is the context from which my question arises.

M. A.

Jul 2 17:20

@XanderHenderson Firstly, is it possible?

M. A.

Jul 2 17:12

Hi everyone, I have a question. If we have $z = \sqrt{-\dfrac{(y^2+7)(y^2+2y-7)}{32(y+1)} }$, how do we express $\sqrt{\dfrac{y^2}{(y^2-9)(y^2+7)}}$ as a function of $z$?

M. A.

Jun 29 07:10

Please, can someone help me with this?

M. A.

Jun 28 23:51

Could anyone help me with this?

M. A.

Jun 28 21:58

Hi everyone. Could someone explain to me how, in this answer, they go from a double integral to a single integral with a logarithm? Where does the logarithm come from?

M. A.

Jun 10 17:22

@XanderHenderson You're right, sorry.

M. A.

Jun 10 14:13

Any hint on how to solve this limit? $$\lim_{x \to + \infty} x \left(\arctan x - \frac{\pi \, x + 2}{2x+1} \right)$$

M. A.

Jun 8 21:57

@copper.hat I would have $\alpha_n + \lambda \, \beta_n = 4 \, \alpha_{n+1} + 4 \, \lambda \, \beta_{n+1}$. Yes?

M. A.

Jun 7 20:27

@copper.hat How?

M. A.

Jun 7 17:46

@copper.hat $a_{k+1} = 8 \, a_k - a_{k-1}$ is the recurrence relation. $a_1 = 1$ and $a_n = 4\, a_{n+1}$ are the boundary conditions.

M. A.

Jun 7 17:40

@copper.hat You said "how can $a_k$ satisfy the two conflicting update equations?". What does this mean?

M. A.

Jun 7 17:39

@copper.hat What do you mean?

M. A.

Jun 7 16:55

@leslietownes Are you in?

M. A.

Jun 7 16:01

@leslietownes Let us assume that this linear solution works. I have $a_{k+1} = 8 \, a_k - a_{k-1}$, $a_1 = 1$, $a_n = 4\, a_{n+1}$. Obviously, $a_k = A \, r_1^k + B \, r_2^k$. How do I determine $A$ and $B$?

M. A.

Jun 7 15:52

@leslietownes Here I meant that I would use the binomial expansion and only consider terms below the second degree (thus, a linear solution).

M. A.

Jun 4 17:38

@leslietownes Perhaps we can work with a linearized recurrence? What do you think?

M. A.

Jun 4 17:34

@leslietownes I obtain something like $16 \, a_k^4 + 32 \, a_k ^3 + 24 \, a_k^2 + 8 \, a_k +1 \ldots$

M. A.

Jun 4 17:31

@leslietownes We want to find all solutions to a recurrence like this.

M. A.

Jun 4 17:26

@leslietownes I end up with a messy expression with $a_{k+1}, \, a_k, \, a_{k-1}$.

M. A.

Jun 4 17:16

@leslietownes I made a mistake. The problem is to find a linear recurrence of the form $$f_{k+1} = \frac{(2 \, f_{k -1} -1)^4 \, f_{k-1}}{2 \, f_{k-1} -1},$$ where $f_{k} = 1 + a_k$.

M. A.

Jun 4 17:06

Hi everyone. How can I solve the following recurrence relation? $$2 \, a_{k+1} = 16 \, a_k ^4 + 32 \, a_k^3 + 24 \, a_k^2 + 8 \, a_k$$

Room for Bml and John Rennie

M. A.

Jun 8 08:23

@JohnRennie I have a question: what is the significance of the derivative of temperature with respect to volume?

M. A.

Jun 8 08:20

@JohnRennie Hi :-)

M. A.

Jun 5 09:10

@JohnRennie Thank you :-)

M. A.

Jun 4 16:16

@JohnRennie Hi :-)