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12:04 AM
This is somewhat of a coding problem, but it relates to complex exponentiation, so I figured I'd ask here to see if anyone has any idea - with arbitrary real numbers $a,b,c,d$, how can I get $(a+bi)^{c+di}$? I'm working in JS, which unfortunately doesn't have built-in complex numbers like Python does.
I was looking at de Moivre's formula but that only seems to help for integer exponents, or maybe rational exponents, but I'm storing my numbers as just arbitrary floating-point numbers
 
mm, in general this will not have just one conceivable output. but, implement complex exp (think euler's formula, which gives real and imaginary parts), and some choice of complex log (subtler and where the choices come in), then do exp((c+di) log(a+bi))
and complex multiplication too, i guess, for (c+di) * log(a+bi)
 
oh, that's a good idea, thanks.
i have complex + - * / implemented cuz those are comparatively simpler i was just having trouble thinking of how to approach exponentation
 
the recipe for complex log basically boils down to real log and a choice of a value of the complex 'argument' function. en.wikipedia.org/wiki/Complex_logarithm has some of it although not quite in a usable format.
"calculating the principal value" section. atan2 makes an appearance!
 
hm.. alright, i'll have to dig around a bit in that article to see if i can figure smth out :p
 
@Yorch The corporate overloards are concerned about the difficulties of maintaining and improving the quality of info on the network. The devs have started working on things to help us clean up the sites. The 1st target is SO, which has a big problem of old, high scoring outdated / obsolete answers.
Obviously, obsolescence isn't much of an issue for Math.SE, or many other sites on the network. But it's expected that some of the new tools (whatever they turn out to be) will be applicable to other sites. Please see here for details: meta.stackoverflow.com/q/405302/4014959
 
12:20 AM
Speaking of logarithms, a cute thing came up a few days ago here: math.stackexchange.com/q/4172078/207316 $\log_{2}x \approx \ln x + \log_{10}x$, which is true because $1/\ln(2) \approx 1+1/\ln(10)$. That's easy to show by simple calculation, but it would be nice if there were an analytical way to do it. We can rearrange it to $\ln(5)=\ln(10)-\ln(2)\approx \ln(10)\ln(2)$. But how do you tackle the RHS?
It's easy to see that there can't be any other integer solutions to $1/\ln(x)-1/\ln(y)\approx 1$, but there are solutions for $1/\ln(x)-1/\ln(y)\approx 1/z$, with $(x,y,z)$ positive integers.
 
the corporate overlords have started working on things to help us clean up the sites?
sounds good
 
Eg, (8, 76, 4), (9, 131, 4), (29, 2152, 6), (108, 904, 15), (96, 4434, 10), (3670, 3796, 2004)
@Yorch I'm optimistic, but it's still very early in the process. At least they're consulting with the community, and getting (some of) us to help test stuff. We have no idea how long it will take before the process leads to useful tools, or how applicable the new stuff will be for other sites.
But even though the sites outside SO don't suffer from the same problem of obsolete answers, they do have the issue of what I call dinosaur answers: ancient answers that have a huge score but which have some flaws. The dinosaurs make it difficult for newer, more elegant answers to become visible.
 
12:36 AM
can't you just add another answer on the same post?
but I know what sort of thing you're talking about
 
Sure. But people may not notice it if there are multiple answers with huge scores on the same page. Of course, it's easy to see new stuff if you hit the Active tab, but casual & unregistered readers may not even notice that tab.
 
I'm a moderately experienced and competent user and I don't know most of the tricks.
 
i never noticed that tab either :)
 
There are tons of little details about how the sites work that aren't easy to learn about. Something came up a few months ago, I think it was on Meta SE, (sorry I don't remember the details) and one person commented that it was news to them, even though they've been a member for several years, on several sites, and they're even a moderator on one of those sites.
 
i have an active tab in a few bars, does that count?
 
12:47 AM
The SO corporation do want to improve that. They're working on ways to improve the "on-boarding" experience for newbies, so that the relevant info is easier to find. And so that the newbies have a better idea of what the goal of the site is, how to use it more effectively, and to avoid doing stuff that will attract downvotes, close votes, and negative comments.
 
it would be interesting to see how usage patterns correlate with certain exams, etc.
 
@copper.hat :D Here's a song for you Whiskey Legs, by the Tedeschi Trucks Band.
 
If I have to prove that A only if B, then do I have to prove A implies B or B implies A or A iff B?
 
@PM2Ring Nice :-). More my generation: youtube.com/watch?v=6WDSY8Kaf6o
 
@copper.hat It would. The best (worst) PSQ I ever saw was a few years ago on Math.SE. It was a bad quality phone photo of an exam paper, including the OP's name. IIRC, it had no other text, and "Urgent!!!", in the title. ;)
@copper.hat Know it well. And of course, the song itself is far older.
 
12:58 AM
@copper.hat take mine, too!
 
thin lizzy was a shot in the arm to a lot of older trad stuff
 
i saw one once where the question was a screenshot of a desktop or phone displaying an exam question, where they hadn't even bothered to crop out the OS stuff around the window that the exam was in
at least it was high quality because it wasn't a photo. it was very easy to read the question
and was the name of the class, school, and [that day's] date on the screen? yes, it was.
 
That's Dunning-Kruger grade incompetence. ;)
 
i naturally upvoted and left a comment "+1 for a screenshot and not a photo"
 
:)
@copper.hat It's great when musicians try to maintain a connection with the traditional body of work. But it can be hard to do it successfully so that it appeals to a young audience.
I'm also a fan of song writers who create new work that is faithful to the old tradition.
Eg, Bella Hardy: The Herring Girl
 
1:32 AM
There's was a covers band from Georgia who'd been plodding along for years, without much success. They were on the verge of calling it quits when the 2 guitarists asked their teenage daughters to do some vocal harmonies with them. They renamed the band "Foxes and Fossils", and they became a minor YouTube hit. They took a break for a few years, while the girls went to college. But they returned a year or so ago, with an extra fox, and a few more fossils.
Here's a clip from their 1st gig, Suite Judy Blue Eyes. A slightly later one, doing XTC's Senses Working Overtime. And a more recent one: Dedicated to the One I Love
 
senses working overtime isn't covered enough, in my opinion. without googling, i'm only aware of mandy moore's, and (now) that one.
and the fact that mandy moore covered an XTC song at that point in her career is still kind of amazing to me.
but that's another story.
 
The foxes were a bit dubious about doing it, until they heard Mandy Moore's version.
It's an outstanding song. I also like "Dear God"... but I guess it wouldn't do too well in the Bible Belt. ;)
 
2:02 AM
Interesting! Gabriella has just done a cover of Wonderwall, in an open minor tuning. I love her guitar playing, but I wish she'd collaborate with a singer occasionally.
 
2:59 AM
@leslietownes thanks for the help; I managed to implement it successfully :D
 
3:16 AM
awesome!
 
 
1 hour later…
4:30 AM
How do they make such diagrams? Using what software/website? The diagram as the answerer has posted in answer?
0
Q: Height and Distance problem from Intermediate trigonometry page 58 Q.no 16 by B.C Das

Sagar RawalI solve the all the height and distance problem form Book of intermediate trigonometry by B.C Das except one I am stuck with. I couldn't make a figure of it as I didn't quite understand it. Here is the question From station, two light-houses A and B are seen in directions North and $30^\circ$ Eas...

 
i sort of hate that diagram (the visual look, not the content). the lowercase italics look like Times New Roman, the upper case I don't recognize.
there is or was something within the office suite you could use to do this. Times New Roman makes me think of that.
i don't remember the name. i used to have it on my office computer.
there are packages within tex/latex/etc. that do it. tikz i think is a popular one.
whenever i had to do graphics i did find it easier to do them in something non-latex and then import in. but some people swear by it
 
Importing seems the quickest way :-)
 
Hello I have a question. If we have to prove "A is B iff it is C", is that the same as proving "C and A imply B, and A and B imply C"?
 
i have no idea what the first means
 
Hi copper :)
 
4:43 AM
@copper.hat Like "a closed set is metrizable if and only if it is countable" . (This is nonsense, just an example)
Then would I have to prove "A closed countable set is metrizable, and also a closed metrizable set is countable".
 
I would interpret it as: "A closed set S is metrizable" if and only if "S is countable".
 
yes, if you have an implicit universal quantifier over closed sets in there
i.e. fix an arbitrary closed set A, and prove both A countable => A metrizable and A metrizable => A countable
 
Thank you!
 
which certainly is nonsense but has the right structure :)
 
Hi @Koro!
 
4:52 AM
:)
 
Is there something in between fully algebraic logic and dependent type theory? I want dependent type theory but with something like prenex normal form
Because variables are really hard to formalize so I would like to group them all in one place.
 
 
3 hours later…
7:40 AM
Hey, guys! I have a somewhat simple calculus question: let f(x) be a differentiable function that satisfies the relation f(x+y) = f(x)f(y). Then, f(x) is either a constant of value 0 or the exponential function
However, doesn't f(x) = 1, as in f(x) is a constant of value 1, also work? In this case, f(x) is differentiable and also satisfies the relation f(x+y) = f(x)f(y)
Why isn't f(x) = 1 included in the solution?
 
8:23 AM
this might be stupid, but how do one mathematically argue that this is the same:
(my question conserns the change of the inequality by replacing x with -x, i dont know this trick sadly)
for example $1+x < e^x$ then $ 1-x > e^{-x}$
i know that if we multiply an inequality by -1 the sign switches, but this does not seem to be the case here
 
 
1 hour later…
9:45 AM
Can someone explain me this paragraph ;
On each iteration we will compute a diagonal preconditioning matrix $\mathbf{D}$ (we omit the subscript $n$ ). $\mathbf{D}$ is expected to be a rough approximation to the Hessian. In our experiments, following Martens (2010), we set D to the diagonal of the Fisher matrix computed over $\mathcal{A}_{n} .$ To precondition, we define a normalized parameter vector $\tilde{\boldsymbol{\theta}}=\mathbf{D}^{1 / 2} \boldsymbol{\theta}$, compute the Krylov subspace in terms of $\tilde{\boldsymbol{\theta}}$, and convert back to the "canonical" coordinates. The result is the subspace spanned by the ve
This paragraph is taken from paper danielpovey.com/files/vinyals_aistats12.pdf
 
 
1 hour later…
11:15 AM
@MadSpaces Look at the graphs. $y=1+x$ has a slope of 1, and passes through (0, 1). $y=e^x$ also passes through (0, 1) and has slope of 1 at that point. $1+x\le e^x$ for all $x$, with equality only at $x=0$
 
12:03 PM
Why is this question off-topic ?
https://math.stackexchange.com/questions/2290050/how-to-find-an-orthonormal-basis-for-p-2-mathbb-r-equipped-with-the-inner/
 
Probably "no attempt shown" or smth
 
12:18 PM
Hi y'all, not really a regular here, but does this site respond well to bountied questions?
or if I bounty something am I likely to have wasted the rep?
 
Hello,
6
A: Number of functions $f:\{1, 2, \ldots, n\} \to \{0, 1\}$ that assign $1$ to exactly one positive integer less than $n$

Jordan GlenThe functions to be counted are those that assign to $1$ exactly one (one and only one) of the integers less than $n$ (one of the elements in $\{1, 2, \ldots, n-1\}$. Once that assignment is determined $(n-1$ choice $)$, the remaining elements in $\{1, 2, \ldots, n-1\}$ must then be assigned to $...

I am a little bit confused about case (c) in the question above in case you have any hint for my problem. I don't know why we multiply in this case by 2 in (c) as $(n-1)\times 2$?
 
@AncientSwordRage Bounties can work here. Eg, this question math.stackexchange.com/q/4100315/207316 got 2 answers soon after a bounty was posted. It was sitting there unanswered for ~2 months. But it did get closed & reopened. Twice...
 
@PM2Ring thanks!
bounty started!
 
math.stackexchange.com/questions/4178702/… - I posted an answer to this question that was quite low quality even though the OP made some effort in the end and I was downvoted, then the question closed, so should I delete my answer?
 
@EdwardEvans I agree the question is not of the best quality. But it is certainly about maths, so "off-topic" seems weird. I would have rather put "Needs details or clarity"
Not that I care much, I was just surprised
 
12:29 PM
nah you're right, off-topic is the wrong label
 
@Axel I think people downvote answers to low effort question to dissuade people from helping low quality posters
 
@Astyx Ok, thank you for the explanation.
It is indeed dissuading...
 
Hello! I've been puzzling about the function of $$\cos\left(\sqrt{\left|x^{2}-1\right|}\sin\left(\frac{1}{2}\arg\left(x^{2}-1\right)\right)\right)\cosh\left(\sqrt{\left|x^{2}-1\right|}\cos\left(\frac{1}{2}\arg\left(x^{2}-1\right)\right)\right)$$ and getting a single continuous function out of it using only hyperbolic cosine. If that's an implausible desire, is there a way I can use the result of $\cosh(\sqrt{1 - x^2})$ to compute cosine from the reals without the Gudermannian function?
 
@Axel It's a PSQ homework dump with zero effort shown from the OP. You should try to encourage the OP to improve such questions before posting an answer. Otherwise, the anti-PSQ brigade may downvote you.
 
@PM2Ring That's what I did, the OP at first tried to show what he did, but the formatting was not there, so I told him to use MathJax to edit his question and he tried, but afterwards he realized that his attempt was wrong so deleted it. Therefore the question looks very low quality even though the OP tried a bit to improve his post.
 
12:41 PM
FWIW, I don't like PSQs either. But I don't believe in downvoting technically correct answers because that may mislead readers into thinking the answer is incorrect.
 
PSQ = Poor Standard of Quality?
 
Puma Soliciting Quadrants
 
@Astyx Problem Statement Question. It just states the (homework) problem without any further info or attempt. See math.meta.stackexchange.com/q/33508/207316
 
Ah ok
On that topic, I posted this answer math.stackexchange.com/questions/4178876/… and it received a downvote. I know the question has been asked before, but thought I could give hints instead of directing the OP to a complete answer where all the work is done for him. What should I do in this situation?
 
@Axel A wrong attempt is much better than no attempt. It gives us some insights into the OP's thought processes, and skill level. If the OP realises it's wrong, they should say so. And explain what's blocking them from fixing it.
 
12:47 PM
@PM2Ring I do agree with you on this point.
 
@Astyx Tricky. Many of us agree that it's better to give hints. And that dupe target is a complete worked solution. I agree that it would've been better for the OP to follow your hint, before looking at that dupe.
But some people will downvote you for answering a question that has a dupe target which is easy to find. It's a bit frustrating...
 
What makes me sad is that among low-quality posts is that some OPs are just completely rude, stubborn and expect people to do their homework for them so it's perfectly normal their post get closed or deleted, but others are not like that and are willing to do some efforts in order to receive help even though the quality of their post is not perfect, I think they just need to be sent in the right direction. In the end they often have the same treatment, their questions are downvoted and closed.
 
@Axel I just looked at the edit history of that post. I can't see much of an attempt there, but I guess there are a couple of lines of something... If he'd explained what he was trying to do, and then posted a few lines of algebra (even without MathJax), we could roll it back and convert it into a more acceptable question. But frankly, there isn't really much there that's worth salvaging. IMHO.
 
1:03 PM
@PM2Ring You are right, there is not much to salvage... and I wondered what he was trying to do to be honest.
 
Yep. Some OPs are willing to make the effort. But they just don't know what's expected of them, or how to do it.
However, one reason that OPs are clueless about site standards & procedures is that they haven't put in an effort to study the site. They haven't read the Help pages. And more importantly, they haven't browsed the site, and studied what good posts have in common, and what badly scoring questions are doing wrong.
 
Yes, conscientious OPs should not encounter those kind of problems.
Some of them are a little bit too careless I suppose.
 
We can encourage the new OPs to learn more about the site, and what is expected of them. But we can't force them to do it. ;) Allegedly, the Stack Exchange / Stack Overflow company is working on ways to improve the "on-boarding" process for new members. Hopefully, that will improve things a bit... but knowing human nature, I don't expect a huge improvement.
The bad OPs are like bad tourists, who go to a new country without learning anything about it beforehand, and get upset / confused / angry when the country has different rules to their country, and the locals don't behave like the folks back home. ;)
 
1:21 PM
That's great for Stack Exchange, I was thinking precisely about something like that! I guess that it won't change all people, but if it can help some of them, it would be great ;) Good analogy :)
 
Segregating the site between advanced and beginner and having a proving grounds for new users to mature would be a good solution in my opinion. "(homweork) deRiVatIve?" questions would be separated from questions like "Why does squaring x produce this result?"
 
There is an advanced site: Math Overflow. We're happy to help people with their homework, but we don't want to be (or create) a site that just does people's homework for them. I enjoy teaching people stuff, and helping them to understand how to attack problems.
 
:)
 
Hence the issue. Homework problems are the same problems reiterated but not usually seeking anything beyond surface-level understanding. It's something to be done with and never looked at again or really pondered to get something more meaningful from. Not all homework problems are like this, but a number of them in my opinion are.
 
If we create a beginner's site, then hardly any experts will visit it, and the few that do will most likely suffer burn-out very quickly. So you'd end up with a site of the blind leading the blind. This has been discussed to death on the metas of virtually every technical site on the network.
 
1:31 PM
Then perhaps the site(s) need to make it more apparent to new users what the purpose of stack exchange et al. is, and whether it is "not just for solving homework" or "not for solving homework at all".
In general, I tend to look for practical answers in practical questions. Homework is usually solving problems that have already been solved for the purposes of learning.
Sadly for me, I seem to be asking questions that no one wants to answer or at least has the time to answer because a good enough solution already exists. Well, I'm not asking questions to find what is good enough.
 
I have found mathoverflow to be kinder somehow
which I find confusing tbh
 
lol
 
especially when it comes to things like "soft questions"
or "opinion based"
But on stuff that looks like homework they'll close it quicker
which I completely agree with btw
unless it looks like graduate level homework in which case I think there won't be a problem
 
What does graduate mean please? When are you considered undergraduate exactly?
 
If it were up to me, I'd just close every homework-specific question, and anything that might be homework-related that offers a legitimate question which hasn't been asked before, I'd allow (or promote).
 
1:40 PM
how much rep was needed to close stuff?
 
3k no?
 
oh yeah
you're almost there
 
Homework is supposed to be for practicing what you've learned so that it sinks in. And to show the teacher that you actually understand what you're doing. However, in some places, there's this culture of "teaching to the exam".
So instead of people learning how to think like a mathematician, they learn, parrot-fashion, how to pass mathematics exams. Physicist Richard Feynman encountered this problem (in connection to physics), when he spent a year teaching in Brazil, not long after World War 2.
 
10k to delete questions
20k to delete answers
 
1:43 PM
Maybe I'll actively close people's stuff when I reach 3m rep
which likely won't happen ever
 
@Yorch Yes. And at 20k the delay before when you can delete vote questions is shorter. Plus you get more delete votes per day. (I have delete privileges on SO & Astronomy).
 
@PM2Ring Bruh, don't even get me started. The so-called public "schools" that I went to basically just spoon feed you information and you just spit it back out and you get a grade for it, mostly optimized for the grade, not for learning.
 
I've never had to use my deletion privilege
I think the things that I actually want to delete
can get deleted by normie flagging
but I don't remember exactly how it worked
 
As an inactive member that feeds off of the work of others like a parasite, I am proud to state that I have contributed little since I first joined SE. (I'm joking of course.)
 
goodbye
 
1:48 PM
goodbye
 
o_o
 
The system (i.e. the Community user) will automatically delete any post red flagged enough times as rude or abusive or spam. It takes six red flags from normal users, or a single one from a moderator, to delete a post for this reason.
that's what I was looking for
so 6 low reppers can delete bad stuff
 
Nah, but for real, I'd like to contribute more, but I can't really do much since I don't have much formal knowledge in subjects, just mostly informal knowledge, and also being productive in my own enterprises.
Any ideas about my question I posted earlier?
 
Sorry I don't have experience with that
why is the variable $x^2-1$?
 
What do you mean?
 
1:58 PM
why are you using $x^2-1$ as the variable instead of just $x$
 
The input to cosh and cos here is $\sqrt{|1 - x^2|}$ which produces half of a circle and a parabola.
Best I can put it, the circle and hyperbola are complex counterparts of some kind that when you take that counterpart and you put it into the respective functions of $x(t)$ and $y(t)$, you get some form of the complex counterpart's $x(t)$ and $y(t)$, hence these functions.
I obviously can't just use the inverse of $\sqrt{1 - x^2}$ in the case of $\cosh(\sqrt{1-x^2})$ to get cosine because it will give hyperbolic cosine, so the challenge here posited is to get circular cosine by modifying the result of this function somehow using only hyperbolic functions and the fundamental operations of mathematics but from within the reals.
 
$\cos\left(\sqrt{\left|a\right|}\sin\left(\frac{1}{2}\arg\left(a\right)\right)\right)\cosh\left(\sqrt{\left|a\right|}\cos\left(\frac{1}{2}\arg\left(a\right)\right)\right)$
 
Sorry, I haven't touched that stuff for a while... And it's after midnight here, and I was up before dawn. But yes, there are some neat connections between the circular & hyperbolic functions. The Gudermannian function is a way to connect them without explicitly using complex numbers. So it's kinda handy for people who are suspicious of complex numbers. ;) On a more serious note, the inverse Gudermannian is used in the Mercator map projection.
 
Indeed, but the issue is how to reproduce that exactly using only hyperbolic functions and real-valued exp.
 
yeah I understand that's the problem, I can't help :(
 
2:09 PM
@PM2Ring Thanks :) . Yeah I've known about that stuff for quite a while due to my research. The Gudermannian is great, but it still requires inverse circular functions to define it in the identities listed (or the complex domain which is computationally expensive in the absence of sine and cosine functions).
Since $\cosh(\sqrt{1 - x^2})$ yields a form of circular cosine, it must have in it some form of the (inverse) Gudermannian. Extracting that from this is the whole issue, apart from the simpler alternative of just finding cos(x) from this.
This here is my workspace where I've been playing around with this stuff: desmos.com/calculator/hbh0sbvgmb
Well, one of the workspaces. ;)
The worst case is that I just simply implement the matrix exponential and that's how I can compute sine and cosine at the same time, but I'm hoping for something even better than that.
 
2:29 PM
Oh, by the way, the use of cosine, sine, and arg here are mostly just acting as filters for the real and imaginary parts such that $$\sqrt{|1 - x^2|} \sin(\frac{1}{2} \arg(1 - x^2)) = \begin{Bmatrix}
\sqrt{x^2 - 1} \text{ for } -1 \le x \le 1
\\
0 \text{ otherwise}
\end{Bmatrix}$$
And vice versa
 
have you considered asking the question on main?
along with those observations?
 
I mean that's a possibility I suppose if only a scary one...
 
why scary?
 
The last time I asked a related question, it was closed because several people found it unclear. Turns out knowing something without knowing the language to communicate your thoughts to others makes it very difficult to communicate. Who would've thought...?
 
@PM2Ring Where did you hear about the Feynman thing?
 
2:39 PM
Is any one of you have any idea what the sum of the series $\sum_{n=0}^\infty (-1)^n$ could be?
 
Try putting it in wolfram alpha
 
do you think that series converges?
 
@AMDG Thankyou. That worked! No. It diverges.@Thorgott
 
:)
 
3:01 PM
why would you expect that sum to have any reasonable value?
 
There is theorem for test of convergence in Advanced calculus book by Taylor & Mann Theorem IV where, it states that, any positive, continious, non-decreasing series $sum_{n=N}^\infty u_n$, where N is any positive integer, converges or diverges according to the integral $\intfdx$(limit from N to $\infty$) is Convergent or divergent where $f=u_x$. So here i am assuming that "integral" to be a sequence made from putting values from N to $\infty$, am i wrong?
 
I do not understand what you have written. What does it mean for a series to be continuous? You need to provide more context.
 
@copper.hat because after rearranging terms it resembles a pattern i.e. beautiful to signify that there must be a value.
 
Sounds like rubbish to me.
 
Pardon me. I am typing on mobile, i forgot, so i forgot there is no option for edit.
It is not the series but the function $f$. Any way please ignore it, i will again type this in correct form through my laptop. And its bit hard because the function provided by admin to run mathjax for android chrome seems not working.
 
3:48 PM
@kunalCh. The Cesaro means converge to $\frac12$, but the series as given diverges (oscillates between $0$ and $1$).
@kunalCh. Another regularization is to look at $\frac1{1+x}=\sum\limits_{k=0}^\infty (-1)^kx^k$. Let $x\to1$.
 
I can feel a $-{1 \over 12}$ in the air...
 
@copper.hat I updated an answer about that recently. I added an approach that I was surprised no one had used.
 
Nice :-)
 
@copper.hat Thanks!
It seems safer to add to existing answers than to answer new questions sometimes.
I've updated two old posts since I've answered a new question.
 
4:17 PM
@robjohn Since my suspension in May I only answer when someone (one person in particular) requests my input or if there is a simpler answer to a simple question crying out to be added :-). I have not spent any time on questions that require thought since.
 
Given any path is there a way to construct an object that corresponds to the path s.t. the object traces out the path as it falls
 
what does that mean?
 
@copper.hat a large contingent of users did not react well to the complaints about answers to low quality questions before the EoQS. I did not expect that to change just because the policy was made official.
 
Why does suspension happen? And why do some users' accounts get banned for low quality contributions? What are low quality contributions? I saw accounts of many users that I admire a lot, suspended due to low quality contributions, it makes me really sad. @robjohn
The users I mentioned above have been using this site probably since it was invented
 
@robjohn It was the way it was communicated that deflated my interest in contributing. Almost entirely interaction style issues with one particular user.
 
4:23 PM
@copper.hat for example, a falling leaf traces out a path as it falls
 
Are you asking if there is a differential equation that a given path satisfies?
 
@geocalc33 sounds interesting...
Copper, he's asking to design an object that follows a certain path, I think
 
I don't know what 'design an object' means
 
@Koro The reasons are often not completely coherent. The motivation for the site is not always clear. Some people want an encyclopedic collection of answers to a variety of topics. Some want to help people who are having trouble with understanding ideas. The explanations needed when helping people muddy up the encyclopedia. The encyclopedia answers are not explanatory enough for helping those struggling with basic concepts.
2
 
it's not mathematics related probably @copper.
 
4:28 PM
@geocalc33 that path is affected by external influences as well, so not only do we need to specify the object, but also its surroundings.
 
My interpretation is that give a path they are looking for a vector field such that if you pick a point on the path and the it 'follow's the vector field it will trace out the path.
but unfortunately mathematics is not poetry, so some precision is required.
(unfortunately i had many ideas which seemed promising at the time but after careful analysis turned out to be known or not as exciting as they seemed when i woke up with the idea :-).)
 
hey chat
 
@copper.hat poetry benefits in that it can be beautiful, without being encumbered by details, whereas math needs those details. The beauty of math can therefore be harder to see.
 
my teacher asked us to prove that $Q_8 := \langle a, b | a^4 = 1, b^2 = a^2, bab^{-1} = a^{-1}\rangle$ has 8 elements
 
By designing an object, I meant to determine its dimensions, mass, shape etc..It may have more general meaning though
 
4:32 PM
@robjohn i suppose i spoke a mistruth, there is poetry in mathematics, but one needs to push through the bushes of precision first
 
it's not hard to prove that every element of this group can be written as $a^m b^n$ where $m \in \{0, 1, 2, 3\}$ and $n \in \{0, 1\}$, thus $|Q_8| \le 8$
 
Is the "trig power" notation considered correct/unambiguous/well-understood/etc., when applied to logarithms, i.e., $\ln^2 x$, or is $(\ln x)^2$ better/more clear?
 
however I can't see a good way to show that all of those are different
 
@copper.hat once one has abstracted the details enough, the beauty can be seen.
 
@LucasHenrique It is a good question to ask. One way it to find a concrete group that satisfies all the constraints and has order 8.
 
4:34 PM
Lucas, do you know quaternion group?
 
yeah, I know it is the quaternion group.
 
@robjohn for me the beauty is the reminder of how little we know and how shallow the depth of our analysis
 
@LucasHenrique something more needs to be said, since the group $\langle 1\rangle$ satisfies those constraints. Perhaps $a\ne b$ is enough?
 
so... by using the quaternion group, there is an injective morphism $Q \to Q_8$ and then $Q_8 \cong Q$, yeah?
 
Presuably $a,b,1$ are distinct elements
 
4:37 PM
@copper.hat yes
 
@copper.hat I think that $a\ne b$ implies that (given the constraints)
 
that is the presentation, so formally it's the free group generated by a,b quotiented by the rhs
 
@robjohn mental latency here
 
4:58 PM
@LucasHenrique If you know the quaternions, then show there is a homomorphism from the above group onto the quaterions.
 
Questions like this make me realize I don't remember how everything is usually definted in school
woops
sometimes I feel like I should have taken the logic courses in college
but it's only for a brief moment
 
@kunalCh. The mobile chat browser does allow you to edit, if you do so within 2 minutes. Click on the comment you wish to edit, then click the pencil icon.
 
How would you prove $|x| \geq x$ for $x\in \mathbb R$
 
@Yorch if $x\ge0$, $|x|=x$. If $x\lt0$, $|x|\gt0\gt x$
 
Or more generally, if you want to show all elements of $X= A\cup B$ satisfy a property. You can show the elements of $A$ satisfy it and the elements in $B$ also satisfy it
 
5:05 PM
yes
 
is there a name for that rule or something?
 
great
 
A dichotomy is a partition of a whole (or a set) into two parts (subsets). In other words, this couple of parts must be jointly exhaustive: everything must belong to one part or the other, and mutually exclusive: nothing can belong simultaneously to both parts.Such a partition is also frequently called a bipartition. The two parts thus formed are complements. In logic, the partitions are opposites if there exists a proposition such that it holds over one and not the other. Treating continuous variables or multicategorical variables as binary variables is called dichotomization. The discretization...
 
I find a lot of those trivial questions hard to answer
because I've seen different people in different classrooms do different stuff
 
5:09 PM
Would you expect different people in different classrooms to be identical?
@Yorch It's a case of case-checking.
 
I mean, for higher maths I feel more comfortable doing whatever I want
as long as it works
 
Well, I don't know that I trust you enough to say that's OK.
Dichotomy is more like: "If it's not A, then it has to be B." That's different.
 
You don't have to.
In probability theory and logic, a set of events is jointly or collectively exhaustive if at least one of the events must occur. For example, when rolling a six-sided die, the events 1, 2, 3, 4, 5, and 6 (each consisting of a single outcome) are collectively exhaustive, because they encompass the entire range of possible outcomes. Another way to describe collectively exhaustive events is that their union must cover all the events within the entire sample space. For example, events A and B are said to be collectively exhaustive if A ∪ B =...
that seems to be more suitable, it appears in the dichotomy page
 
I think it really is called case-checking. You could, for example, check that some statement holds for all even integers and holds for all odd integers and conclude it holds for all. (In general, $A\cap B$ needn't be empty.)
No, I don't think that is more suited.
Clubs and hearts are more suited.
 
you are showing ( p or not p ) implies q ?
or how does it work
if (p or not p) implies q ... then q?
 
5:15 PM
Where does this come from? The part in blue?
 
$A\cup B=A\cup(B\setminus A)$
 
(4) is direct but (5) and (6) are not
 
@TedShifrin $A\cup B=A\cup(B\setminus A)$
Stupid mobile browser, that was supposed to be on the same comment
 
@robjohn Sure: So you can reduce to the disjoint case. But I'm questioning the "outcomes" versus hypothesis part of the issue.
@Yorch: Sure, I'll accept that. You want to prove that Q holds universally. You write the universe as the union of two (possibly disjoint) sets and prove that each of those implies Q.
 
I know. Just being crotchety this morning ;-)
 
5:19 PM
Well, I was just trying not to dismiss Yorch's query.
 
is that constructivist friendly?
 
What, not dismissing you?
 
oh, the constructivists don't have real numbers
 
I thought constructivism didn't allow proof by contradiction.
 
@TedShifrin That should keep them running in circles.
 
5:25 PM
Fine by me.
@robjohn I didn't realize continents answered questions here.
 
Again, the mobile browser messed with me. I was trying to reply to Yorch, but things scrolled and I hit your comment by accident.
 
only europe
 
@TedShifrin we were dismissed as not being able to answer, but Europe could.
 
Oh.
Yes, I went and looked at your comment, @robjohn, and added my star of approval.
 
5:53 PM
Errands beckon. BBL
 
Is it okay to learn about cardinal & ordinal numbers as a beginner from "Theory of Sets" by E.Kamke ? If not , why ? (I am asking this before I dive more deeper into the book)
 
6:31 PM
Perhaps someone else has an opinion on this, but I certainly do not, @Prithu.
 
6:52 PM
I've never heard about that book unfortunately so I cannot say anything about it
 
Greetings, demonic @Alessandro. Long time!
 
Hi @Ted
Yeah I've been quite busy with my PhD lately
 
7:08 PM
Good for you!
Are you back in Germany?
 
yeah, I'monly travelling to Italy every 5 weeks or so
 
life is tough :-)
 
7:33 PM
i find it strange that you can completely change a comment and there is no indication, but if you remove it there is a big reminder left behind!
 
i don't
isn't there an indication? i get a little pencil graphic to the left of the comment
maybe that's just only visible to me for my own comments
 
oh yeah
 
and our moderator overlords
 
no I can see it
did u edit "I don't" ?
 
that's much too subtle for joe...
now all i can see is the silly little pencil
i still find it a bit odd that removal had a bigger visual impact than editing.
 
7:43 PM
@copper.hat I read about the Feynman thing in his book, Surely You’re Joking, Mr. Feynman!
 
I see that @copper.hat edited two comments immediately above, and, @leslie, I see you edited "i don't".
And Ted Shifren edited
2 hours ago, by Ted Shifrin
I thought constructivism didn't allow proof by contradiction.
 
I just didn't notice the little pencil icon.
 
@Yorch Edited as well!
 
@PM2Ring Thanks, loved that book.
 
@copper.hat One gets so used to it that it doesn't cause any dissonance, like the obvious "(removed)" does.
 
7:48 PM
I suspect there are all sorts of 'easter egg' sort of thingies in the code
 
There's even a comment history, but it's hiding. You you can access it from the transcript, via the permalink.
 
I just saw that. When it is a little quieter I will explore more to avoid annoying more than my usual.
 
I want to discuss a group theory question
I have a small doubt
2
Q: $|G|=p^n m$ , $p$ is prime and $\gcd(p,m)=1$. Let $H$ normal in $G$ of order $p^n$ and $K$ is a subgroup of $G$ of order $p^k$ then $K\subseteq H$

Akash Patalwanshi$|G|=p^n\ m$ where $p$ is prime and $\gcd(p,m)=1$ suppose that $H$ normal in $G$ of order $p^n$ and $K$ is subgroup of $G$ of order $p^k$ then show that $K\subseteq H$ this question is already asked here( Let $|G|=p^nm$ where $p$ is a prime and $\gcd(p,m)=1$.Suppose that $H$ is a normal subgroup...

Above linked is the question
There are answers to it which use things which I believe are not required
at all
So I propose the following solution
 
@copper.hat A fairly well known one is the "This is fine" Easter Egg. I think it works on all 3 chat servers.
Yep. :)
 
:-)
Group theory is close to incomprehensible to me.
 
7:54 PM
all 3 chat servers?
 
@Koro: A quick comment. You need to learn proper LaTeX notation. It's $\langle a\rangle$, not $<a>$, for the subgroup generated by $a$.
 
in a written document you would probably include aliases for this in the preamble. i used \< and \>.
 
@Yorch chat.stackexchange.com, chat.meta.stackexchange.com, and chat.stackoverflow.com.
 
Suppose on the contrary that $K$ is not a subset of $H$. So there exists $x\in K$ such that $x\notin H$. $xH\in G/H$ and $|G/H|=m$ and by Lagrange's theorem: $o(xH)|m$. Also since $o(x)|p^k$, we must have $(xH)^{p^k}=H\implies o(xH)|p^k$ and so $o(xH)|gcd (p^k, m)=1$ and therefore $xH=H\implies x\in H$, which is a contradiction. Hence our assumption was wrong and by contradiction we are done!
Hey Ted, Leslie, copper
Ted, I rarely use <a>
 
Oh, I didn't know that existed
 
7:58 PM
which usage of $\langle a \rangle $ by me are you referring to?
So my question is: what is wrong in my method?
 
On a group of order 60 post on main.
 
@Koro isn't your problem a corolary of sylow's theorem?
 
You've done the same thing in here, too.
 
@Yorch For details, see meta.stackexchange.com/q/297299/334566 & the posts linked there.
 
I've seen you use <> frequently.
 
7:59 PM
all maximal $p$ subg-groups (under inclusion) have size $p^n$ and are conjugates of each other.
 
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