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00:00 - 19:0019:00 - 00:00

12:44 AM
@Hayaku I am back for a short while before I have to go again. Did you get a chance to think about things?
 
1:03 AM
@robjohn so your final step was $\sum_{n=2}^\infty \int_0^1\frac{x^{n-2}}{(n-2)!}x\left(1-\frac{x}{2}\right)\,\mathrm{d}x$?
 
@Hayaku that is the final answer, yes
@Hayaku and unfortunately, I am leaving for a while again
 
my question is quick
let's say instead of 1 being the barrier i use 5
does the equation become
@robjohn $\sum_{n=2}^\infty \int_0^5\frac{x^{n-2}}{(n-2)!}x\left(5-\frac{x}{2}\right)\,\mathrm{d}x$
 
1:24 AM
I liked this!:)
@anon
 
1:46 AM
Good night everyone.
 
2:11 AM
@robjohn or actually say instead of 0 to 1 we start at .7 to 1
does't seem like i can just change the integral bound
or rather add the offset value to x
 
2:29 AM
hi, quick question, how do I comment on a question without posting an answer? Should I mark the Community wiki box or how?
 
2:54 AM
@ÁngelDavidRíosOrtiz There is a comment section below every question.
 
@skullpatrol Yes, but i dont know how to comment there
 
I think you need more rep :-)
 
@skullpatrol ohh i see now, thanks
 
3:13 AM
@Hayaku No, the convolution becomes more complicated the further in you go from the edge of the bell curve
 
i figured
 
@Hayaku up to $1$, things should be as before.
 
3:31 AM
yeah it probably has fewer "terms" in the second part
probably two integrals under one summattion?
 
@Hayaku what is the question you want to ask?
 
extending it past 1
i figure it is interesting since in this scenario the bound of the variable happens to equal the cutoff
in the other it's greater
i'm looking at mathworld.wolfram.com/UniformSumDistribution.html where they extend things to s=1,2,3..5
extending same logic finding xn for s>1
 
@Hayaku you can compute the convolution, but for $n$ rolls, there are $n$ piecewise defined parts
for $n=2$, you get the triangle...
 
then the pyramid etc
 
for $n=3$, you get three parabolic pieces
 
3:39 AM
but what if i wanted to just load the initial scenario
instead of starting the sum at 0 i start it at S
where S<=1
i tried changing it so the integral goes from S to 1 and then change all x's to (x+S)
@robjohn $\sum_{n=1}^\infty \int_S^1\frac{(S+x)^{n-2}}{(n-2)!}x\left(1-\frac{S+x}{2}\right)\,\mathrm{d}x$
 
4:35 AM
Wait, is there an election going on?
I don't see any.
 
5:24 AM
HELP
what's a good book on group theory?
 
@Hayaku The convolution at the edge, $x\in[0,1]$, is $\frac{x^n}{n!}$
@Hayaku it is more complicated for $x\in[1,2]$
@Hayaku and each interval on
 
5:50 AM
 
6:21 AM
@Hayaku I'm not sure what you are thinking this represents.
 
hey @robjohn
and good evening all
 
Ben
Good evening mixedmath.
 
6:39 AM
@mixedmath hi there... there are fires all over the LA area tonight
 
@robjohn oh, I didn't know
what happened?
 
@mixedmath One, at least, was started by someone's cigarette
 
Oh dear. :-o. I am sorry to hear that. Has anybody injured @robjohn?????
 
@mixedmath the winds have been high around here, so they might all be just subsequent fires
@BabakS. I don't think anyone has been killed
@BabakS. mostly, the fires are in untrimmed brush
 
I didn't think about death!!! God help them all, plz :-(((
I see
The world without Cigarettes feels peace than having the nuclear power. Idiot peoples.
 
6:52 AM
@anon hahahaha
 
@BabakS. there are people and horses in the affected areas
 
To show that the sorgenfrey line is a topology, we have to show it satisfies the basis properties?
 
7:08 AM
if you say the sorgenfrey line is the set of reals R equipped with the topology generated by the half-open intervals [a,b), then it is automatically a topological space
showing that the collection of half-open intervals [a,b) is actually a basis for the topology it generates (rather than just a subbase) is a stronger condition on this collection of sets, and to pove this one need only show it is closed under intersection
 
@anon Awesome, thanks for this reference!
 
7:25 AM
@lo
@Lord_Farin good morning
 
@Vrouvrou Good morning.
I had a look at your question. I'm afraid that I can't help you at this stage.
 
i don't understand
i must go bye
 
@Vrouvrou Bye.
 
Given a profinite group $G=\varprojlim G_i$, suppose we declare $$\left\{\lim_{l\longrightarrow}\,|{\rm Syl}_p(G_l)|: \{l\}\subseteq I ~\rm linearly~ordered\right\}\subseteq{\bf Z}_p=\varprojlim {\bf Z}/p^n{\bf Z}$$ to be the "Sylow spectrum" of $G$. I am curious if this provides any useful information about $G$, or conversely if there are nice characterizations of $G$ to correspond to nice characteristics of the spectrum
Or perhaps it would be more natural to speak of direct limits of finite groups.
 
@robjohn: What will be the mean square in Klein-Beltrami model look like?
 
user19161
7:39 AM
@BabakS. Who is the guy in the picture?
 
David Suchet @JasperLoy.
 
@BabakS. If Mathematica had a nice projection, we might find out.
@BabakS. Poirot
 
Aka Hercule Poiret
 
user19161
So now my target is to get 1k on MSE and 1k on ELU, and then I will retire, lol.
 
@JasperLoy you task me
 
user19161
7:43 AM
@robjohn I raised a question on ELU meta about the strange behaviour. meta.english.stackexchange.com/questions/3925/…
 
@JasperLoy your chat account was never deleted, I don't think
 
@JasperLoy Also, mail addresses once assigned are forever retired when deleted by services like gmail, yahoo.
(I have tried)
 
user19161
@robjohn Yes, it seems it can't be "deleted". But I think at least it should be hidden, like they do for gravatar accounts.
 
user19161
@JayeshBadwaik You have not tried "forever".
 
@JasperLoy Well, I have tried 6 years later, and the reply I got from the gmail guy was that "We cannot reassign you that email unless you are able to legally prove that you were the same guy."
 
7:48 AM
@JasperLoy The odd behavior, might be an effect of what the SE team did with your votes.
 
user19161
@JayeshBadwaik Right, but that statement doesn't mean they can't give it to another guy.
 
@JasperLoy It means that exactly.
 
user19161
@JayeshBadwaik Not purely logically.
 
@JasperLoy Not logically, but legally yes, because I would have had to show the "requirement" like some sort of court or law enforcement order, and even then, they would only give access to the mails, not the account itself.
 
user19161
@robjohn I think so too. I am thinking, what would happen if another JL a year later signs up for my deleted gmail address and signs up for SE as well! Then there would be chaos.
 
7:50 AM
@JasperLoy That would not happen.
@JasperLoy Here.
 
user19161
@JayeshBadwaik I am planning to try out a year later for myself.
 
Yes, you can do that.
 
user19161
@JayeshBadwaik Sure, but that is not meant to be totally precise. Anyway, yes I get the point already.
 
user19161
By the way guys, I tried out gmail, ymail and outlook.
 
user19161
7:55 AM
I have deleted my gmail and it cannot be recovered, but trying to delete ymail and outlook gives more allowance as it will only be deactivated and it still can be reactivated within a few months.
 
@JasperLoy: What is the difference between "wherein" "where" "in which" in Maths texts. I asked if from amWhy and she told that "where" is preferable. What do u think?
 
I don't know that there is any generic difference. Mostly they probably just feel different on the author's tongue.
 
I want to use them through paper, @anon.
 
user19161
@BabakS. I guess you can ask there, but you would need to include the exact context and sentence.
 
Ok. I 'll do that. I will be glad if I can see your suggestions. Thanks Jasper. Thanks anon. :-)
 
8:07 AM
@JasperLoy are you sure that there were no cookies still in your browser when you signed up again?
 
user19161
@robjohn Very sure. I store no history, no passowords, no cookies, no cache, etc... I am totally obsessed with minimalism.
 
user19161
Of course, how software and hardware actually work is complicated.
 
user19161
Strange things can happen, and maybe theoretical computer science studies some of them.
 
8:47 AM
Anybody got any ideas on how to find the max flow of a knights tour graph?
 
 
1 hour later…
9:56 AM
@anon hi, are you there?
 
yes
 
I linked my question, because seemed me related
 
yes, I noticed
 
if you think is not related tell me i'll delete the link
 
it's fine
 
9:58 AM
ok :P
 
 
1 hour later…
11:26 AM
Greetings
I just noticed that this integral can be computed without integration, but only geometrically
$$\int_e^{\pi}\sqrt{(x-e)(\pi-x)} \ dx $$
 
Greetings :D
 
@skullpatrol :D
 
@Chris'ssisterandpals How are you?
 
@skullpatrol fine, thank you. I'm writing some solutions to some problems. How about you?
 
@Chris'ssisterandpals Fine, thanks.
 
11:49 AM
@skullpatrol are you a math student?
 
@skullpatrol I didn't manage to read what you removed. I was afk.
 
12:17 PM
I'm pondering whether I post my question to mathoverflow.
 
12:27 PM
This guy here always posted nice questions like math.stackexchange.com/questions/347942/…. Unfortunately, many downvoted him and it seems he takes no interest in posting new questions.
If I had enough power, I'd remove permanently those that downvote people for fun and never let them come back. But this is the good part for them, I don't have that power.
Here people ought to come for the love to math not for other things.
2
The main reason for that I decided not to be an active user on MSE is that the math lovers aren't really protected. Of course, this is just my opinion. One day you see one guy that is not able to play with quadratic equations and tells you how to ask a question on calculus. Beside that, he modifies the question you post the way he wants.
For me it was too much and realized that my place is not there anymore.
I hope one day people that really love math will be protected.
3
 
12:51 PM
@Chris'ssisterandpals Have you been to artofproblemsolving.com ?
I think you might like it, it has lots of problems like that ones you are interested in and they never close down questions.
 
@RagibZaman Do you know combinatorics?
 
@PeterTamaroff Depends at what level you are asking about.
 
@RagibZaman I'm asked to count the number of equivalence relation on a $3n$-size set such that each equivalence class has size $n$. This is the same as partitioning a $3n$-size set into $3$ $n$-size subsets.
So I have $3n$ elements.
 
Do you know the right answer?
 
12:57 PM
let's figure it out together
 
It's well-known.
Stirling's numbers
 
@FrankScience Well, yes and no.
 
@PeterTamaroff ?
 
Stirling's nos. of the second kind count the number of ways of partitioning into nonempty $k$-size subsets an $n$ size subset.
 
I was thinking multinomial coefficients actually.
I think the answer is (3n; n,n,n)
 
12:58 PM
Okay.
 
@RagibZaman Yes, I am getting that.
But I'm not sure it is entirely correct.
Start with the $3n$ size set.
 
$\dfrac{(3n)!}{3!n!n!n!}$
 
We have $${3n \choose n}$$ ways of taking the first set.
Then we do $$2n\choose n$$
And we're done.
 
yea
 
1:00 PM
The order of sets.
 
So divide $3!$
 
aah yes
seems like that does it
 
So the total is $$\frac{(3n)!(2n)!}{n!(2n)!n!n!}\frac 1{3!}=\frac{(3n)!}{3!(n!)^3}$$
 
@RagibZaman Are you a mathematician?
 
1:03 PM
@FrankScience Depends on your criterion. I'm in the final year of my bachelors degree majoring in pure math, so I spend quite a lot of time studying mathematics so perhaps that makes me a mathematician. But im not paid for my study and I don't produce any research, so maybe that means Im not.
 
@RagibZaman Ugh, well, I'll ask for some questions. I wonder how much real analysis did you learn in your college.
For example, is the content of Big Rudin covered?
 
@FrankScience Big Rudin certainly not.
 
My classmate asked for a professor and she said that the real analysis course here was very superfluous.
 
I'm sorry, I don't quite understand you.
Is your classmate the "she" who said the course was superfluous?
 
Wait for a moment. I'll post a thread in meta.MO to ask whether a question is appropriate for MO.
No, "she" is referring to the professor.
 
1:08 PM
@FrankScience What do you want to know, exactly?
 
@PeterTamaroff In general, the same method proves that the number of ways to partition a set of $mn$ elements into $m$ $n$-element sets is $$\frac{(mn)!}{m!(n!)^m}$$
 
@user1 Yeah. I was thinking about generalizations, thanks.
=)
 
Generalization
 
@FrankScience "generalizations (of the problem)"
 
@FrankScience What is the question you want to ask?
 
1:10 PM
@user1 So, no username change so far?
 
@PeterTamaroff Why would I do that?
 
Suppose $N=d_1a_1+d_2a_2+\dotsb+d_ma_m$ where $a_1<a_2<\dotsb<a_m$. Partition $N$ into $d_k$ of $a_k$'s is $N!/d_1!\dotsb d_m!a_1!^{d_1}\dotsb a_m!^{d_m}$.
 
@user1 Don't know. Just asking.
@FrankScience No \frac?
 
@PeterTamaroff It would be another anonymous username, in case you thought I might reveal my identity.
 
@user1 Are you a renowned mathematician?
 
1:14 PM
@PeterTamaroff No.
 
@user1 Are you a decent mathematician?
 
@PeterTamaroff Hopefully, at some point, I will be.
 
@user1 What do you mean?
 
@PeterTamaroff I am still a student
 
@user1 Oh. But I guess you're at a higher stage than me.
 
1:17 PM
@PeterTamaroff Sorry
@RagibZaman I want to enrich my knowledge. Since the time is limited, I must decide how much I spend on analysis and algebra and so forth.
 
@FrankScience How much do you know about analysis, algebra, and so forth?
 
@user1 I'm an undergraduate.
 
What have you learned
 
@FrankScience That answer nothing =)!
 
and what do you find particularly interesting?
 
1:20 PM
I doubt I'll learn very little if I only follow the syllabus.
 
@FrankScience You need to actually know them to decide which ones you will want to study.
 
@FrankScience We can give you some advice, but you'll have to tell us what you already know and what you find interesting.
 
@user1 Before the choice, I wonder how much knowledge is sufficient for a good choice.
@RagibZaman I'm only standing at the entrance, a freshman. What I want is not the advice for choice, but the amount I need to learn before the choice. I'm not sure whether I have expressed clearly.
 
Ok, so you want to know what subjects to learn that will give you a solid grounding from which to specialize from. Yes?
 
Not only what subjects, but also the depth.
 
1:27 PM
Yes
What is your goal? To research in pure math?
 
It's the most probable aim.
 
Well, the number of mandatory courses is not small.
Let me enumerate part of them.
 
The main thing to remember is to be well rounded.
Never forget that.
 
rounded? Sorry, my english is not that good.
 
1:29 PM
In undergraduate, you should try to learn every offered subject and master the material
dont just get lazy on some subjects that you think you dont like when you are doing them
because later you will see how useful/important/interesting and linked they all are
 
Ah, I see.
Now I want to explain the situation.
 
Ok.
 
I have asked for some sophomores, ...(3rd, 4th, I don't know how to spell them out), and they said that they are busy with reciting proofs of theorems.
It's because the course don't give so much time for one to understand the whole stuff.
Now I found I'm not that busy as a freshman, so I decided to learn more.
 
@FrankScience If you understand the proofs there is no need to "recite" them.
4
 
@FrankScience If you're anything like me, you won't regret deciding to learn one thing over another at this point simply because of how little you will learn. I preferred analysis early on, but I would quickly say that I barely know it over fields I have spent much less time on.
 
1:36 PM
@user1 Sorry, my English is bad. As I've parsed your sentence, by learn one thing over another, you mean learn one thing and omit another?
 
@FrankScience Yes.
 
@user1 I didn't mean that. I meant, for example, now I have some spare time, then I'll spend them on learning new materials.
I'll split the spare time into many pieces, and each one for a material, for example.
 
@peter buenas
 
@FrankScience Sure, I always immersed myself in one topic, but I suspect others have the ability to study multiple fields.
 
There will be a bunch of time in summer holiday.
 
1:41 PM
@skull hi
 
@user1 I think you cannot do that here. We have a lot of mandatory courses.
At least, in order to do all the assignments, one cannot spend so little time on it.
 
@robjohn I am trying to just "shorten" the interval
 
@user1 Now I have a lot of materials to learn, so I should decide which one and how far I should learn. For example, I bought the books for real & complex analysis, functional analysis, topology, algebra and differential geometry. I should arrange them.
 
Hi @anorton
 
@FrankScience I am sorry, but I am not in a position to offer advice without saying which ones I would prefer.
 
1:48 PM
Oh, thanks.
I'll be off.
Goodbye.
 
2:18 PM
@Charlie hi/bye
 
2:30 PM
@Charlie hi! :)
 
@anorton wassup?
 
not much...
Finals week ended yesterday, so there's a lot less stress now... :)
 
Hmm
 
How are you doing?
 
Good :)
 
2:36 PM
That's always good to hear!
 
@anorton it's a beautiful day outside, the sky is very blue, no clouds, very sunny
A gentle breeze
 
That sounds like where I am! :D (except here, there's a couple of little white puffy clouds)
 
@anorton :D what time is it there?
 
@anorton Here there are no clouds, still the sky is not blue, more towards gray. :-(
Also, the day is beautiful but dull.
 
@Charlie 10:39 am. What about where you are?
 
2:40 PM
@anorton 11h41
 
@JayeshBadwaik :( The day isn't dull here, either... just not school related. :)
 
@anorton :-)
 
Hi @jayesh
 
@Charlie Hi!
 
Hello.
 
2:49 PM
Hello
 
3:40 PM
@RagibZaman yeah, that one is a nice place, but the feeling I have is that I do too less sport. I don't have the courage to go there yet because I suppose it will "eat" my whole time. :-)
@RagibZaman thanks for the suggestion!
 
hi people
 
how can I put a link here?
from the main without allocating big space
ok this is it
 
4:21 PM
Helloes
 
$\gcd(p,k)=a|\gcd(n,k)$
I didn't understand this line
here
 
gcd(p,k)=a and a divides n & k hence a divides gcd(n,k)
much like how we can concatenate equalities, a=b=c, or inequalities, a<b<c, we can also concatenate equalities and any other relations we want for abbreviation, namely we can say u=v|w to mean that u=v and v|w.
 
4:39 PM
@anon but a|gcd(n,k) is an expression not a number
 
I never said it was a number
 
@anon but how can this happen $\gcd(p,k)=a|\gcd(n,k)$ number = expression
 
you are not interpreting correctly. I already told you how to correctly interpret that. It does not mean 'number = expression.'
If I said 1=2-1=4-3, you wouldn't bother interpreting this as "the number one is equal to the equality 2-1=4-3."
Instead, you would interpret it as "one is equal to two minus one, and two minus one is equal to four minus three"
Similarly, u=v|w does not mean "u is equal to the relation v|w." Rather, it means "u=v and v|w."
Nor, for that matter, does it mean "the number w is a multiple of the equality u=v." This is just not how you go about interpreting the string of symbols.
 
@anon thanks I get he merged two ideas in one
thanks @anon
 
4:56 PM
Hiii @anon
 
hello charlie
 
@pourjour hi
 
@Charlie hi charlie how are you doing?
 
@anon how are you this saturday?
 
good. after mowing my grandparents lawn, I plan on going on a shopping spree and then watching the new iron man with some friends.
 
4:58 PM
@pourjour I am good, and you?
 
@anon good Iron man3
 
@anon oh, great
 
@Charlie feel better :D
 
@pourjour splendid
 
@Charlie any news?
 
5:00 PM
@pourjour no, Souf, none
 
@Charlie hmm the same here
 
@anon I watched it yesterday. Worth it.
 
we'll see
 
@anon I worked out the colouring problem.,
 
5:41 PM
@PeterTamaroff really?
 
How are you @Charlie?
 
@skullpatrol I'm good, and you?
 
@Charlie Fine thanks.
 
@skullpatrol great :)
 
@Charlie grrrrrrrreat :D
 
5:47 PM
@skullpatrol great³
 
hi ,i just need help for this question
1
Q: Deformation retract

VrouvrouHow to prove that $r_t$ is a deformation retract $M^a=\lbrace q\in M ; f(q)\leq a\rbrace$ We have the definition : $r_t$ is a difformation retract if: $r_t$ is a continius ,onto application and $r_t(q)=q , \text{for }q\in M^a$ , there exist an continuous application $H: M^b\times [0,1]\r...

 
@skullpatrol hahaha
 
@Charlie :D:D:D
 
@skullpatrol :D:D:D:D
@jasper hii jaspeer
 
user19161
5:57 PM
@Charlie Just came home, went out with a friend.
 
@JasperLoy good!
 
user19161
6:23 PM
@anon It's showing here too, but I am not a fan of Iron Man.
 
@JasperLoy O.o
 
:)
 
hi
 
how are you?
 
6:26 PM
fine and you?
Hi charlie how are you?
did you read my email
 
@Charlie What do you mean by that?
 
@somaye Fine thanks.
 
nice
 
@soma hi, yes I read
 
what is your idea?
 
6:28 PM
@PeterTamaroff is it good, i m 3
 
@Charlie Yeah.
 
Did you watch the first two?
 
@somaye what did you think of my suggestion?
 
it is bad idea?
no thing
i will send my gmail for you
i thought it is better
no problem
are you sad about me?
 
Oh
@somaye no, of course not
 
6:33 PM
:)
 
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