Mathematics - 2013-03-13 (page 1 of 4)

caveman

00:04

oh wait G/Z = C_{p^2} implies G is abelian so we can assume G/Z = C_p x C_p

Theta30

i cant solve it right away, I must think about it, but as I said do not assume anything except say, the inverse must be unique

caveman

im still working on it

Theta30

you have e,a and b; then ab cannot be a or b; 1)ab=e, then b is the inverse of a 2)ab not e, then ab must be a fourth element

then you ask about ba, case 2.1)ba=ab case 2.2)ba different than ab, so it must the fifth element (in which case there are three elements left!)

George V. Williams

Can anybody recommend a good combinatorics book?

Ethan

If $|k|_p$ is the p-adic norm of k then for all $n,p$, $$| \sum_{k=1}^n|k|_pe^{2\pi i \frac{k}{n}} |\leq 1$$
And if $n$ is a perfect power of $p$
$$\sum_{k=1}^n|k|_pe^{2\pi i \frac{k}{n}}=\frac{1-p}{n}$$

caveman

00:14

@Ethan, I don't understand how to deal with the $|k|_p$ part

wait a sec I got an idea

Ethan

$\frac{1}{n}=\prod_{p}|n|_p$

caveman

oh and mobius inversion?

on that product

Ethan

lol, thats just the fundamental theorem of arithmetic

caveman

you're just the fundamental theorem of arithmetic

Ethan

$$\sum_{p\leq x}(-1)^{[\frac{x}{p}]}\sim (1-2\ln(2))\frac{x}{\ln(x)}$$

the left sum is also never greater then zero

[.] is the floor function

Tobias Kildetoft

00:20

@caveman so what group theory are you working on now?

caveman

how do I get the formula with $|k|_p$ ?

Ethan

@caveman those exponential sums are useless lol dw

caveman

@Ethan, it looks nice I wanted to solve it :(

Ethan

i just thought it looked nice

caveman

@TobiasKildetoft, I've finished my classes so I'm just trying to answer my own questions and I'm going to go over all my theorems from my notes too

Ethan

00:21

$$\frac{-\sin(2\pi x)}{\pi}=\sum_{n=1}^\infty \frac{\mu(n)}{n}{ { \{n x}\} }$$

caveman

@TobiasKildetoft, if you meant in the chat we were trying to classify the p^3 groups

Tobias Kildetoft

ahh

it is interesting that there are two different ones regardless of the prime, but the description changes a lot when the prime is 2

caveman

that 2 thing comes up a lot

I've often not knownw hy 2 is different.. it's actually rare to have a reason for it I think

Ethan

$$\frac{-1}{\pi}=\sum_{n=1}^\infty \frac{\mu(n)}{n}{ { \{\frac{n}{4}}\} }$$

Tobias Kildetoft

@caveman well, it is easy to see in this case, since all elements having order 2 implies that the group is abelian

which us not the case for odd primes

caveman

00:27

oh it its just x = -x, for some reason I dismissed that as only valid in number only

Ethan

$$\frac{12}{d(n)}\sum_{d\mid n}\ln(d)^2-3\ln(n)^2=2\sum_{d\mid n}\Lambda(d)\ln(d)+\sum_{d\mid n}\Lambda(d)^2$$

$$(a+b+c)^5+(a-b-c)^5-(a-b+c)^5-(a+b-c)^5=80abc(a^2+b^2+c^2)$$

caveman

@Ethan, have you seen the proof that zeta has no zeros on the line?

Ethan

@caveman I don't know any complex analysis lol

$$[\text{# of perfect powers}\leq x]=\sum_{n=2}^\infty\frac{\ln(x)^n}{\zeta(n)n!} + O(\sqrt{\ln(x)})$$

caveman

the only complex analysis you need to continuing the zeta function to the strip

but the proof is so funny you have to see it

Ethan

I don't know any analytic continuation

caveman

00:32

@Ethan, teach me the $|k|_p$ please

Ethan

Also I am guessing on that O term lol

caveman

lol

Ethan

I really can only get it to $O(ln(x))$

:(

caveman

I wish I could understand all your sums

Ethan

what kind of math do you do

caveman

00:34

@Ethan im user58512 I changed name just incase you remember

Ethan

uhm...

caveman

im just saying we talked here before

Ethan

@caveman yes I remember

caveman

im mostly doing group theory rightnow

Ethan

whats that

caveman

00:36

it's about the symmetries of things

Ethan

$$\frac{\zeta(s-a)\zeta(s-b)\zeta(s-a-b)}{\zeta(2s-a-b)}=\sum_{n=1}^\infty\sum_{k=1}^n\frac{\gcd(n,k)^s\zeta(s-b,\frac{k}{n})}{n^{2s-a-b}}$$

caveman

this is my favorite example this is my favorite example clarku.edu/~djoyce/wallpaper/seventeen.html

just look at the pictures, there's 17 different ways to make patterns on walls

and thats all

Ethan

I don't understand

caveman

any pattern you make will be one of those 17 types

Ethan

whats your definition of pattern

$$\text{a(xy)=a(x)+a(y), for coprime x,y}$$
$$\lim_{s\to\ +1}\frac{1}{\zeta(s)}\sum_{n=2}^\infty\sum_{k=2}^\infty\frac{|a(n)|}{n^{ks}}=0$$
$$\lim_{x\to\infty}\frac{1}{x}\sum_{n \leq x}a(n)\mu(n)=0$$
$$\text{ then, }\lim_{x\to\infty}\frac{1}{x}\sum_{p\leq x}a(p)[\log_p(x)]=\lim_{s\to +1}\frac{1}{\zeta(s)^2}\sum_{n=1}^\infty\frac{a(n)}{n^s}$$

caveman

00:40

hmmm

if you tile the plane by a painted tile

then that's a pattern

but it has to repeat, I dont count aperiodic tiliings...

there's infinitely many of those

Ethan

lol I don't get it

caveman

sorry I'm bad at explaining :(

Alex J Best

Wallpaper group

A wallpaper group (or plane symmetry group or plane crystallographic group) is a mathematical classification of a two-dimensional repetitive pattern, based on the symmetries in the pattern. Such patterns occur frequently in architecture and decorative art. There are 17 possible distinct groups. Wallpaper groups are two-dimensional symmetry groups, intermediate in complexity between the simpler frieze groups and the three-dimensional crystallographic groups (also called space groups). Introduction Wallpaper groups categorize patterns by their symmetries. Subtle differences may place si...

Ethan

Are there any good sieves for obtaining the integer factorization of all the integers less then or equal to a given number?

caveman

I think you can use quadratic field sieve

wait are you programming it

Ethan

00:47

the quadratic field sieve, im quite sure is used for factoring particular integers

caveman

sorry I misread you

to factor all integers you should use just Eratosthenes

you wont get better

Ethan

Ye, I already have an adaptation of erathosthenes I thought of lol

caveman

but you can speed it up a little with a wheel

Ethan

but its garbage

caveman

im pretty stuck on this p^3 group

Ethan

00:49

$$M(n)=\sum_{q\in f_n}e^{2\pi i q}$$

farey sequences are nice

The farey sequence of order n, is all the rational numbers between 0 and 1, with denominator less then n, if you sum them over a complex exponential you get a nice formula for the mertens function

caveman

I got that evey element is of the form a^i b^j z (where a,b generate C_p x C_p and z lies in the center C_p) but the multiplication is a^{i+i'} b^{j + j'} z'' z z' where z'' is some horrible combination of central elements based on the i,j,i',j''s

Tobias Kildetoft

@caveman I guess you are not familiar with the classification of p-groups with a cyclic subgroup of index p?

caveman

@TobiasKildetoft, I suppose that's what I'm trying to do now, at least for |G|=p^3

@Ethan, yes those are amazing

Ethan

completely useless as far as I know the formula is tho lol

caveman

its relatedto continued fractions

Tobias Kildetoft

00:51

@caveman first, note that if the group has a cyclic subgroup of order $p^2$ then it is generated by 2 elements

caveman

@TobiasKildetoft, wait if I assume G is non-abelian, then can't I assume G/Z(G) is not C_{p^2}? since otherwise G/Z is cyclic so G is abelian

oopd

Tobias Kildetoft

sure, the center will be of order p

caveman

sorry I wrote it wrong, corrected now

Tobias Kildetoft

and yes, the quotient by the center will be generated by two elements

caveman

I've shown the center has order p, and I also think G/Z must be C_p x C_p

ok

Tobias Kildetoft

00:53

and those two generators then pull back to give generators of the group itself

since the center is actually the Frattini subgroup in this case

caveman

I don't know about this

Tobias Kildetoft

ok, it is a nice tool for p-groups but not essential for this

it is enough to note that your group will be generated by two elements (so you want to find the relations for those elements)

caveman

that's surprising

oh that gives me an idea

I got (ab)^p = a^p b^p (since (ab)^n = a^n b^n z^n) where a,b generate each side of the C_p x C_p

that doesn't seem to help much though

oh I haven't used that Z is generated by some z

anon

01:13

these suggested edit logistics can get ridiculous

caveman

just realize I was wrong before needed z^{n-1}

no its T(n)

Peter Tamaroff

@anon Yo, dawg.

caveman

I cantsolve this T_T ill keep trying though

Peter Tamaroff

@caveman What are you trying to solve?

caveman

classify groups of order p^3

Peter Tamaroff

01:24

@caveman Hmm.

anon

KCd wrote about it; you'll have to look at p odd/even separately

caveman

p and p^2 are trivial .. now im completely stuck that's a bit sad..

Peter Tamaroff

@caveman I was about to show you this

"Trivial"

caveman

if we suppose $C_4 \le G$ then we get $D_8$ easily! that's a good step

Tobias Kildetoft

not quite

the quaternion group also has an element of order 4

caveman

01:28

im really confused then!

if $c$ generates the $C_4$ in $G$ then surely the group is just $1,c,c^2,c^3$ and $z,zc,zc^2,zc^3$

Tobias Kildetoft

any non-abelian group of order 8 must have an element of order 4, since if all elements had order 2, it would be abelian

caveman

oh, I get it

yeah, that's a good point too

in fact that's really useful

Tobias Kildetoft

if p is odd, however, the difference between the two cases is precisely whether the group has an element of order $p^2$ or not

caveman

now im getting that the only group is Q8 haha

Tobias Kildetoft

it might be nice to know that Q8 is the only of the two to have just one element of order 2

caveman

01:35

@TobiasKildetoft, I know what you're saying is right but I don't get this: if $c$ generates the $C_4$ and $z$ is the central element of order 2 then $z c^2$ seems to have order 2 as well as $z$

Peter Tamaroff

@anon @TobiasKildetoft maybe that above is of your itnerest.

Tobias Kildetoft

@caveman you will have $c^2 = z$ in Q8

Peter Tamaroff

@caveman maybe you too.

(I'm just sharing, I'm not doing that now)

caveman

@TobiasKildetoft, oh mygoodness I didn't even thin k that was possible but of course it is!

In fact if I do case analysis on whether or not that's true I think that completes the proof

for p=2

Peter Tamaroff

@anon I am curious about the following. Let $G$ be finite, of order $pq$ a product of two primes. Then $G$ is not simple. By Sylow, $n_q\mid p$ and $n_p\mid q$, and $n_q\equiv 1(q)$; $n_p\equiv 1(p)$.

Now if $n_q=1$ or $n_p=1$, we're done.

So assume that $n_q=p$ and $n_p=q$.

That is $p\equiv 1 \mod q$ and $q\equiv 1\mod p$.

Tobias Kildetoft

01:46

@PeterTamaroff well, that is clearly not possible

since one will be smaller than the other

Peter Tamaroff

@TobiasKildetoft OK, WLOG assume $q<p$.

Then $p\mid q-1$, so $p\leq q-1$; which is impossible.

@TobiasKildetoft That is what you mean?

Tobias Kildetoft

yeah

Peter Tamaroff

@TobiasKildetoft These Sylow Theorems are dope =)

@TobiasKildetoft Did you see what I pasted above?

amWhy

Slow night again :-(

Peter Tamaroff

@amWhy What does that mean?

amWhy

01:52

Just slowish on main...or maybe it's my brain that's slow!

Peter Tamaroff

@amWhy Oh. I see.

amWhy

It's also sort of fast-ish...when I click to answer, someone has beat me to it! And I'm trying to be so very careful of not coming anywhere close to posting a similar answer to one that's already been posted; I even delete when I post simultaneously with another post, if my post is at all similar to it. I just haven't had a great day, that's all! Nothing to cry over :^)

caveman

omg!!

I've got G = {1,c,c^2=z,$c^{-1}$,...} and to be non-abelian I need another element y with $yc \not = cy$

it must be Q8 but..

it must be yc = cy^{-1} QED

I think this could be done better

anon

02:45

presumably the DV on your deleted answer was because free groups are not finite

Peter Tamaroff

Anyone knows about Erdös Szekres?

amWhy

03:49

Anyone around? @Peter - what about him?

@PeterTamaroff I meant to refer to your question about Szekres...

Peter Tamaroff

@amWhy They are two guys. Erdös and Szekeres. I was referring to one of they're theorems.

It says that if $S$ is a sequence of numbers of length $mn+1$ then it contains a decreasing subsequence of length $>m$ or an increasing subsequence of length $>n$.

amWhy

@PeterTamaroff Now that makes sense...it didn't click, I just read it as a name! I'm slow tonight! (time for bed, soon...maybe)

@PeterTamaroff Oh cool, thank for the reminder.

Peter Tamaroff

@amWhy The proof is nice a short, but I wondered if it was known or not.

(The theorem, not the proof.)

amWhy

04:04

@PeterTamaroff I recalled encountering it (the theorem but not the proof), once you mentioned it, but it's not something that was foremost in my memory. :-)

Peter Tamaroff

@amWhy I'm not being crazy here right?

amWhy

@PeterTamaroff Sounds reasonable...

Peter Tamaroff

@rob Rooooooob.

amWhy

I need to get more of a life!

Peter Tamaroff

@amWhy Hahaha. It is 1 am here. I should go...

amWhy

04:10

That's right, we're not too far off, time wise...

Peter Tamaroff

@amWhy Yes, that is interesting.

Most people here are having breakfast when I am going to sleep!

Or having some noon coffee.

amWhy

You're a bit east of me, but mostly south! Yes, I know, there's a lot of people 8 hours away from me give or take a little.

If you know where Milwaukee, WI, is, that's pretty much where I'm at, in between UW-Madison (UW Wisconsin) and UW Milwaukee...But Chicago is pretty close, if that gives you bearing...

Peter Tamaroff

@amWhy I can just look in a map! =)

You guys have too many states! We have 23 only.

Well, we call them provinces.

amWhy

Yes, lots of states, and the most populated ones are typically the smallest ones (East Coast). Wisconsin covers like 5 or six states on the East Coast, area-wise!

Ethereal

@amWhy How does your username originate?

robjohn

04:20

@PeterTamaroff Yeeeees?

Peter Tamaroff

@robjohn I wanted a sanity check! =)

robjohn

@PeterTamaroff I will look

Peter Tamaroff

@amWhy As you see my sleep enterprise is mostly a failure.

robjohn

@PeterTamaroff Doesn't the continuity when restricted to the boundary depend on whether A is open or closed or neither?

Peter Tamaroff

@robjohn Well, in either case we'll have some bad behaviour, methinks.

robjohn

04:36

@PeterTamaroff It A is open or A is closed, then I think it is continuous on the boundary. If A is neither, then it is not obvious

Peter Tamaroff

@robjohn Right. If $A$ is open, then $\partial A\subseteq A$, so the function is constantly zero. If $A$ is closed $\partial A\subseteq A$ so the function is constantly one.

I have to go to sleep.

FORCEQUIT.

robjohn

@PeterTamaroff okay...

Clayton

Hello!

amWhy

@Ethereal Sounding out the first two letters, A ... M... then adding the word "why", read it phonetically

A sounds like A, M sounds like M, "why" sounds like...?

Ethereal

@amWhy Amy.

amWhy

04:44

@Clayton hello!

@Ethereal who, me? ;-)

Ethereal

I knew that was your name. Hmm, interesting.

@amWhy Yesh, you!

Clayton

@amWhy: I always had wondered if that was your name. Clever trick!

amWhy

@Clayton I like word-play...language...I see letters and words all around me (just like numbers and operations and shapes, and....) . It's not so much trying to hide my name, as it is an attempt not to immediately be recognized as female...plus, I'm infatuated with the "why" of everything!

@robjohn Hello!

robjohn

@amWhy hello

Clayton

@amWhy Are you a graduate student? At times, I've thought so, but I'm really unsure

amWhy

04:50

@Clayton ABD...for too long!

Clayton

@amWhy What does ABD stand for? I'm apparently very uninitiated in these things...

amWhy

@Clayton Sorry...kind of graduate lingo for "All But Dissertation"...

Clayton

@amWhy I thought it might be something like that, but I didn't think of dissertation; thesis kept coming to mind.

amWhy

Yeah...we call it that too...but you'll see grad students designate themselves as ABD, post-prelims, etc.

Alexander Gruber

05:10

well

i chose to do group theory tonight instead of studying for my complex midterm tomorrow

Clayton

Well, I'm going to get some sleep.

Alexander Gruber

this is why they shouldn't have exams on wednesdays.

κρανίοπεριπολία

Hi @GarbageCollector how are you?

Garbage Collector

@κρανίοπεριπολία Of course fine. And you?

κρανίοπεριπολία

@GarbageCollector Fine thanks. You've changed your gravatar?

Garbage Collector

05:23

@κρανίοπεριπολία Not gravatar, but I uploaded a new one.

κρανίοπεριπολία

@GarbageCollector Who is that?

Garbage Collector

@κρανίοπεριπολία I am :D

κρανίοπεριπολία

@GarbageCollector ORLY!

Garbage Collector

@κρανίοπεριπολία :-)

Ethan

tineye.com/search/f9532d83717deb9cb8b7fb8721a2c96d5277fbd9

@GarbageCollector atleast 100+ people are using the same image

κρανίοπεριπολία

05:27

(removed)

Garbage Collector

@Ethan You have proven it. QED.

@JasperLoy: hONEY!

user19161

@GarbageCollector I am not your honey...

Ethan

@JasperLoy How old are you?

user19161

@Ethan Much older than you, hehe.

Ethan

30+?

user19161

05:40

@Ethan Have you gone to see the prof?

Ethan

@JasperLoy I sent you an email

user19161

@Ethan Oh, when? Just now?

Ethan

yes

user19161

@Ethan I will read later, I am not in the mood to read now... Anyway, have a good day.

amWhy

@JasperLoy Hi, Jasper! I see you've changed your user ;-)

user19161

05:42

@amWhy Yes, I am leaving this chat now.

Ethan

@amWhy what kind of math do you do?

Matt

I have a quick question, i'm asked to prove some basic division facts about algebraic numbers...but i'm going about this the same way i would for integers...should i be doing something else?

Ethan

whats the question

Matt

well just basic things like a | b and b | c implies a | c and such

i was thinking about representing them as some polynomial that they are the root of, but it seems equally as trivial

Ethan

what is them?

Matt

05:54

algebraic numbers

Ethan

and your trying to show what?

Matt

nevermind

user19161

06:26

(removed)

user19161

@ethan I have replied to your email.

Ethan

07:13

@JasperLoy I replied back, also can we talk in a separate chat?

user19161

@Ethan Well, there is no private chat here. All transcripts are visible, so I see no difference. We can talk in this chat or via email if you want. Yes, I got your reply.

Ethan

@JasperLoy I started having trouble writing, and reading things about several months ago', and thats when I started taking meds, but it doesn't really help, it makes it alot harder to study, do you still study math?

user19161

@Ethan I check my email daily, that's the best way to reach me! =)

Ethan

can you still study math?

is it that bad?

user19161

@Ethan I am just taking a long break now, not really studying, which is why I am on this site in the first place.

user19161

07:18

I won't be here answering 1+1=2 if I were working on the Riemann Hypothesis, lol.

Ethan

@JasperLoy I moved out of my dads house almost a year ago, because my step mom, and him were responsible for it, It can sometimes take me an hour to read like 15, 12 pnt. pages

@JasperLoy can you read easily?

user19161

@Ethan Well, I can read, no problem.

Ethan

Do you repeat things outloud?

user19161

@Ethan Well, only when I am thinking, not really repeating. So it's a different thing. Just thinking to myself.

user19161

@Ethan Anyway, I sense we are treading into sensitive topics. Let's continue over email. =)

Ethan

07:22

alright

user19161

@Ethan Don't panic if I don't reply too soon. I check it once a day though.

Ethan

@JasperLoy can I ask how you make a living?

user19161

@Ethan Well, I have people taking care of me now.

Ethan

family? or freinds?

user19161

Yeah, my mum.

user19161

07:25

@Ethan OK, I am leaving this chat now, we will email again.

Ethan

@JasperLoy ok, bye jasper

Matt N.

07:37

Good morning everyone.

Bad time, huh. Everyone's afk. Guess I'll go and do some work instead then.

Jonas Teuwen

@MattN. Hi.

Bye.

Jayesh Badwaik

08:12

@JonasTeuwen Hello. Wassup?

user19161

@JonasTeuwen Finally, your linkedin stopped sending me reminders, lol.

user19161

@MattN. Yes, doing work is better than chatting.

Jayesh Badwaik

@JasperLoy It is very difficult to control the monster that is LinkedIn.

user19161

@JayeshBadwaik Unless you are Jonas!

Jayesh Badwaik

@JasperLoy Even for him I guess. :P

user19161

08:22

@JayeshBadwaik Jonas is the Fourier God, lol.

Jayesh Badwaik

@JasperLoy Yeah, controlling LinkedIn requires being a Lawyer. :P

Okay, I am off for lunch.

See you guys later.

user19161

@JayeshBadwaik Jonas just says, bow down or I will give you a convolution!

user19161

09:29

openSuse 12.3 will be released in 5 hours.

user52932

anyone here?

i have a quick question why is the interval $[\sqrt{2},\sqrt{2}+\frac{1}{k}]$ closed in the rationals

robjohn

10:43

@user52932 I'm here now, are you still here and interested?

guess not

user19161

11:03

@robjohn LOL

user19161

@user52932 I don't understand your question actually.

robjohn

@JasperLoy why is that interval closed, when it doesn't contain the endpoints. it is the same as $(\sqrt2,\sqrt2+\frac1k)$

user19161

@robjohn Actually, I don't understand the question, so I won't try to guess!

robjohn

@JasperLoy Okay.

user19161

@robjohn Oh, I think I get the question now, but then again, I am guessing the meaning. =)

robjohn

11:16

@JasperLoy guessing at the meaning of which part?

user19161

@robjohn Well, first of all, he did not specify what k is, so I guess k is some positive integer. And second, I am guessing that the interval refers to only rational numbers as the set is only the rationals and not the reals.

robjohn

@JasperLoy Yes, I assume $k\in\mathbb{Z}$ and the topology is that induced on $\mathbb{Q}$ by $\mathbb{R}$. In $\mathbb{Q}$, $x\gt\sqrt2$ is the same as $x\ge\sqrt2$

user19161

math.stackexchange.com/a/329321/4594 Not sure who downvoted this, lol. Anyway deleted by owner.

user19161

Wow, the downvotes are really quite senseless these days.

user19161

Maybe the person thought he should not have assumed that M is the midpoint.

user19161

11:26

But in that case, a comment saying the assumption was made would be in order.

user19161

He should have been more confident. I would have upvoted the answer if he had not deleted, lol.

κρανίοπεριπολία

>8(

awllower

11:56

Hm, I wonder where is the mistake?

Old John

Hi folks

κρανίοπεριπολία

Hi

long time no see.

Old John

Hi Skull

I assume that Greek is a translation of skull and patrol?

κρανίοπεριπολία

Yep.

Skullpatrol

Old John

Where did you get the Greek version?

(or maybe you speak Greek?)

κρανίοπεριπολία

12:09

Charlie came up with it.

Old John

Good

κρανίοπεριπολία

@OldJohn How are you?

Old John

@κρανίοπεριπολία Fine thanks - been a bit busy with non-maths stuff recently. How are things with you?

κρανίοπεριπολία

@OldJohn Fine thanks.

Old John

I down-voted some answers yesterday, and today, I got the points back - so I assume the answerer left (hope I didn't cause that!)

loldop

12:14

Hello! How approximate with usage continuous distribution Poisson distribution with small parameter lambda?

Old John

@loldop sorry - I only know of the binomial and normal approximations to the Poisson, and they don't cover small lambda, I think

loldop

@OldJohn yes, so, I try to find small lambda. Maybe Gamma distributions are good enough?

Old John

@loldop Maybe - but I don't know

caveman

hi

user19161

Hi @old!

Old John

12:27

Hi Jasper

user19161

@OldJohn Maybe the answer got deleted.

Old John

@JasperLoy Maybe so

Since he was an unregistered user, I don't think I have any way to see if he is still here

user19161

When a downvoted answer is deleted, the answerer gets back the rep as does the downvoter, within a period of time at least.

Old John

He was a new user - posted 3 answers, and now all 3 seem to have vanished

κρανίοπεριπολία

Just like Jasper will one day...

Old John

12:31

we will all vanish one day :)

κρανίοπεριπολία

...true

user19161

@κρανίοπεριπολία LOL

caveman

how do I define a shorthand for \mathcal{}

user19161

A shorthand is a hand that is short.

κρανίοπεριπολία

Thanks OED

caveman

12:33

$\newcommand{bf}[1]{\mathbf{#1}}$We say that $U : \bf W

it doesn't work

Garbage Collector

@JasperLoy: Hi, bee :D

user19161

@GarbageCollector What happened to the old picture? Who was it?

Garbage Collector

@JasperLoy She is an artist, maybe :D

user19161

@GarbageCollector And what about the older picture, was that you?

Garbage Collector

@JasperLoy That is a very confidential thing to disclose here:D

user19161

12:39

@GarbageCollector Aha! Must be you then...

Garbage Collector

@JasperLoy I have very ugly face.

user19161

@GarbageCollector So that was not you?

Garbage Collector

@JasperLoy I am cdn.motinetwork.net/demotivationalposters.net/image/…

κρανίοπεριπολία

Do you want to see me?

user19161

You're all nuts.

κρανίοπεριπολία

12:43

You are a banana.

user19161

No, I am nuts too.

Garbage Collector

I am bolts.

user19161

Anyone seen mick?

user19161

Is he returning to this chat?

κρανίοπεριπολία

nope not me

awllower

12:48

Is this answer ambiguous?

caveman

@awllower, shouldn't it be $(\alpha+\beta)^p = \alpha^p + \beta^p$?

no, my mistake

awllower

And repeated argument gives us the result?

Garbage Collector

What is the name of ring-like object between nut and bolt?

awllower

@caveman Just take $p$ powers repeatedly!

κρανίοπεριπολία

@GarbageCollector Washer.

Garbage Collector

12:51

@κρανίοπεριπολία That is correct.

Jasper Loy is a nut, I am a bolt, who is the washer?

Vrouvrou

13:11

hi]

κρανίοπεριπολία

@GarbageCollector Who?

Garbage Collector

@κρανίοπεριπολία I am just a nut.

κρανίοπεριπολία

@GarbageCollector not to me.

@GarbageCollector Did you like my picture?

Vrouvrou

please who know something about the use of differential geometry in the critical point theory ?

Garbage Collector

@κρανίοπεριπολία Need more strong colors.

κρανίοπεριπολία

13:18

@GarbageCollector Why?

Garbage Collector

@κρανίοπεριπολία Because it is hard to see the main object. Need more contrast between the foreground and the background.

κρανίοπεριπολία

@GarbageCollector I see you have changed your "gravatar"?

Garbage Collector

@κρανίοπεριπολία Not the gravatar but I just uploaded a new one to SX server.

Transcript for

$Mathematics$ Mathematics