Mathematics - 2015-11-08 (page 2 of 3)

skull petrol

10:03

@Agawa001 in a way you are right about them, but he is crying out for attention.

Mental illness is sad.

Raider nation is about being there when you need a friend.

:-)

Chris's sis the artist

@Jasper!!!!!!!!!!

user174558

@Chris'ssistheartist !!!!!!

Chris's sis the artist

@Jasper How are you doinnnnnnnngggggg? :D

user174558

@Chris'ssistheartist OK. I don't have to do military service anymore, because of my mental illness, so that is good news.

Agawa001

yay!!!

Chris's sis the artist

10:10

@Jasper hehe, good for you! :D

Agawa001

this word brings me a goosebump

user174558

@Chris'ssistheartist So when is the official date of publication for your book?

9814072356

@felipa, I tried to answer, let me know if it is of any help

my memory of norms is a little hazy, so point out if I made a mistake somewhere, please

Chris's sis the artist

@Jasper I don't have any official date so far. There is still work to do, it's not that simple. I told you, my book is like a neverending story. :-)

user174558

@Chris'ssistheartist My life too is a neverending story, lol.

9814072356

10:12

actually, I think I misunderstood what you meant by expected, so nevermind

Chris's sis the artist

@Jasper :-))))) btw, did you start doing some math?

user174558

@Chris'ssistheartist I will try to start on the first day of 2016. I must start on the first day of the new year, for various reasons.

Chris's sis the artist

@Jasper Oh, the magical day of the year ...hehe, OK~ :-)

user174558

@Chris'ssistheartist Are you still living alone with your dogs?

Chris's sis the artist

@Jasper Yes. I love to be alone.

user174558

10:16

@Chris'ssistheartist OK. I am still living with my mum, of course. We just had cod fish for dinner. It's delicious!

Chris's sis the artist

@Jasper Don't say it now!!! :-)

I still have work to do for 1-2 hours.

user174558

@Chris'ssistheartist Do you cook yourself or buy takeaway food?

Chris's sis the artist

@Jasper Both, but today I don't cook anything. I already bought some takeaway food.

@Jasper I miss eating some in the mountains, and I'll probably go there after finishing my book. My father lives in the mountains, it's an amazing place.

user174558

@Chris'ssistheartist I have never climbed a mountain. It would be nice for you to bring Monica there and eat together!

Chris's sis the artist

@Jasper lolll, be sure of that! :D

felipa

10:25

@Agawa001 it's defined in the question

@9814072356 did you post an answer? I don't see it

Agawa001

@Chris'ssistheartist so you frequenting the church, are you part of any chores ?

@felipa i ll see, as long as i can understand more notations

Chris's sis the artist

@Agawa001 I would have been a good singer, that's true. When I was younger I was singing in the church as part of a chore.

Agawa001

yes i like that sound when they sing in harmony, dunno what they utter thu

9814072356

@felipa, I realised I was wrong and deleted it, ignore that, sorry

felipa

10:40

@9814072356 ah ok.. no problem

@Chris'ssistheartist do you know anything about matrix norms?

Chris's sis the artist

@felipa I liked very much playing with matrix, but I didn't do that for a long time. Not the proper person to answer such questions.

9814072356

@felipa, by expected, you mean 'expected' in the probabilistic sense, right?

felipa

@9814072356 yes!

@Chris'ssistheartist ah.. shame.. do you know who is?

Chris's sis the artist

@9814072356 Interesting your username

By circular permutation we can get a phone number from Romania

9814072356

that is

0723569814 (this is a Vodafone number)

@9814072356 Are you from Romania? (maybe it's too personal asking about that)

9814072356

@Chris'ssistheartist thanks, I did not know that. That's quite interesting

I chose the username with the note that it was the largest pandigital square with no repeating digits

No, unfortunately, I am not

Chris's sis the artist

10:52

I see. OK.

9814072356

I like numbers :p

I'm from Canada

skull petrol

11:21

Welcome :-)

Agawa001

i have a solution just in head but couldnt know how to write it down, something alike happened to you !

skull petrol

Sometimes; yes.

Putting some thoughts into words can be difficult

Agawa001

yes, wish we can communicate by telepathy

putting thoughts into words , putting words into key words

its twice harder for a programer :D

Tien-Cheng Huang

Hello! I am (strictly-speaking) a math self-learner on undergraduate level. (I am auditing but not yet a registered student.) Which places on internet are suited for asking exercises hints for self-learners?

Huy

11:39

@Tien-ChengHuang: Why not ask here or on main?

Tien-Cheng Huang

@Huy oh thank you :-) but is it generally appropriate asking hints by directly posting a question on this site?

Khallil

May I ask what $\chi_{\{1\}}$ is in Pedro's answer linked here?

Agawa001

@Tien-ChengHuang answering at-reach questions repeatedly can improve your self-knowledge

Khallil

Is $\chi_{\{1\}}$ simply a step function at $1$? I can't see why it'd be a step function if it's only defined at one point.

Huy

11:54

@Khallil: Why is it not a step function?

Khallil

I've just been getting really muddled with uniform continuity, convergence and all kinds of things.

I've not seen the notation $\chi_{\{1\}}$ before either, @Huy.

Huy

@Khallil: $\chi_A$ for some set $A$?

Balarka Sen

Characteristic function, is it not?

Huy

yea

Khallil

Oh it's a characteristic function?

Huy

11:57

ah that's what you mean by step function

Khallil

I think I've only referred to them as indicator functions, $\mathbb{1}_A$ for some set $A$.

Tien-Cheng Huang

In fact I am trying to prove: "Given a subset of $\mathbb{R}$, by repeatedly taking closures and interiors, one can obtain at most 6 different sets". And in my attempt, I have to claim: "int(cl(int(cl(X)))) = int(cl(X)) and cl(int(cl(int(X)))) = cl(int(X)) for any subset X of $\mathbb{R}$. Now I am stuck at proving this claim.

Huy

$$\chi_A(x) :=\begin{cases} 1 & x \in A\\ 0 & \text{otherwise} \end{cases}$$

Tien-Cheng Huang

Does it seems suitable to post a question just for a hint?

Khallil

By a step function I mean a function that is constant on the open intervals of a set, @Huy.

Paradox 101

11:58

Can anyone explain how I show that g is differentiable?

Huy

@Paradox101 fundamental theorem ?

2

Paradox 101

@Huy yes but I don't know how to use it to prove the differentiability using FTC

Huy

@Paradox101: the fundamental theorem literally says "let ... be continuous and define ... Then ... is unif. cont. and differentiable and ..."

Martin Sleziak

@Tien-ChengHuang I'd guess it should be possible to find a few question about this already posted on the main. Let me search a bit.

@Tien-ChengHuang Have a look here - I guess some of those posts might help you.

Tien-Cheng Huang

12:15

@MartinSleziak I think my question is not quite duplicate: "Prove or disprove "int(cl(int(cl(X)))) = int(cl(X)) and cl(int(cl(int(X)))) = cl(int(X)) for any subset X of $\mathbb{R}$"

Paradox 101

@Huy so we just have to state that as the function whose integral we've taken is continuous then by FTC g is differentiable?

Tien-Cheng Huang

If X = (0,1) $\cup$ (1,2) the equalities still hold.

Martin Sleziak

@Tien-ChengHuang Have you look at those questions? Some of them ask exactly that. Some of them give answer to your question as an immediate consequence.

The only difference that your question has two questions in one.

There is a question with the proof of the first equality. And there is also a question about proof of the second equality.

Tien-Cheng Huang

@MartinSleziak Oh... I see this math.stackexchange.com/q/385774/275935 and this math.stackexchange.com/q/441049/275935 thank you :-) Let me digest them...

Martin Sleziak

And I was wrong about two questions which I posted there - they do not help you.

But at least the two you linked to now are what you need.

I'd guess that there are more posts about the same problem.

Tien-Cheng Huang

12:26

@MartinSleziak How do you search them? What key words do you use??

Martin Sleziak

I will look into my browser history and I will list the searches there. So that we do not flood this room.

Agawa001

i have a serious suggestion to rename congruence into hellgruence

not that kind of obnoxious hell but bittersweet

Américo Tavares

0

A: What is the purpose of this site?

I see this site mainly as an opportunity to help other users clarifying their mathematical doubts, and being helped by others in my own doubts, in an exchanging process. Normally, as a non-mathematician and a non-native English speaker, I only answer low-level questions. By reading questions and...

skull petrol

Shots fired.

Khallil

Pew pew.

Tien-Cheng Huang

12:38

If I have some complicated math "terms" and don't know how to search, who can I ask for help?

skull petrol

::makes popcorn::

Khallil

skull petrol

:D

Martin Sleziak

@Tien-ChengHuang Maybe yu can ask people in chat...? (Like with other math-related question.) Of course you§re not guaranteed that somebody answers. (As is always the case, if you ask something in chat or on the main site.)

Or you can start by reading some post here which give some advice for searching math: chat.stackexchange.com/transcript/message/25249702#25249702

it's probably better to copy that here.

Some tips on searching can be found on meta - have a look at question tagged search.

You can choose to sort those question by votes or use the frequent tab, to get the more relevant posts to the top.

In those post on meta you will also find some links to related question on the main site.

I am not sure to which extent such practice should be encouraged, but occasionally people simply ask on meta when they are looking for a specific question. Like here: meta.math.stackexchange.com/questions/21776/…

Tien-Cheng Huang

@MartinSleziak thanks! Maybe its better to have an article of "tips for searching math terms for new users" on the sidebar of the main.

main site*

Martin Sleziak

12:51

That would be a feature-request. I am not sure to which extent is it good to add new stuff to the already crowded main page.

There already are advanced search tips which are displayed when you search for something and help center contains this: How do I search?

Tien-Cheng Huang

@MartinSleziak Nice to chat with you :-)

Khallil

13:14

May I ask how I'd test the convergence of an improper integral that doesn't have an elementary antiderivative?

I'm thinking that I'd need to bound it by another function that has an elementary antiderivative. I know that my function $f$ is $> 0$ for all x.

Slereah

Is there a way to write parametric conic sections that are also parametrized by the eccentricity?

Like I can vary the eccentricity to go from the circle's parametric equation to the parabola to the hyperbola

Slereah

13:34

Oh I guess it is $r = \frac{l}{1 - e \cos\theta}$

Huy

13:48

@Khallil: depends on the integral really

Khallil

If I had the integral from $0$ to $\infty$ of $e^{-x^3}$, @Huy?

Huy

@Khallil: Do you know the one of $e^{-x^2}$?

Khallil

Yep, it's the Gaussian integral.

However, it doesn't bound $e^{-x^3}$ from above over $[0,1)$ I don't think.

Huy

@Khallil: you can still argue very quickly why in that interval the integral is bounded

Khallil

Could I break the integral up and bound by something else?

Huy

13:53

$\int_0^\infty = \int_0^1 + \int_1^\infty$

Khallil

Oh, ok that makes sense. :-)

Remember me

I had this question: Lets say some space $X$ is covered by $A_n|n\in J$. Consider the set $B_n|n\in J$ such that every element of $B_n$ is in $A_n$. We know that $B_n$ not necessarily always covers $X$. So what conditions should I impose on any of the following : $X,A_n,B_n$ such that $B_n$ happens to cover $X$?

Khallil

Would it suffice to just say that the Gaussian integral is convergent over $\mathbb{R}$, so $e^{-x^3}$'s integral is also bounded, @Huy?

Huy

@Khallil: what about the part from 0 to 1?

Khallil

We can simply bound that above by $1$, right?

Huy

13:58

yes but you have to
a) say that
b) explain why

Remember me

@Huy Can you look at the above question I posted?

Khallil

So I should probably go about using the increasing nature of the logarithm in an inequality, @Huy?

Huy

@Khallil: what???

Khallil

Saying that $1 \geq e^{-x^3}$, taking logs on both sides to get $0 \geq -x^3$, @Huy.

Is that a wrong way to think about it? :-/

Huy

always specify for which $x$

Khallil

14:02

Sorry! For $x \in [0,1]$. It should work for all $x \geq 0$ though, right?

Huy

yes I think so

it's the other way around though if I'm not mistaken

you want to get to $1 \geq e^{-x^3}$

so you actually consider $0 \geq -x^3$ for $x \geq 0$ and take $\exp$

@Rememberme: I don't know, sorry

Remember me

Okay. Thanks for looking

Huy

@Khallil: you can also argue by finding the global maximum

@Khallil: the derivative is non-positive, with only root at $x = 0$, so that's a global maximum

@Khallil: There might also be way to find an inequality similar to $e^{-x} < 1 - \tfrac{1}{2} x$

Khallil

Is that through a Taylor series exp, @Huy?

Huy

urm I don't think you need it

just note where the roots of $1-\frac{1}{2}x-e^{-x}$ are

I think the same argument works for $1-\frac{1}{2}x^3 - e^{-x^3}$

so you get the inequality for $0 < x < 1.168\dots$

I don't know if this is the best estimate though

@Khallil: in the end you should do it in whatever way you like most. :P just some things that I remembered and had the need to write down because I haven't done it in ages

Khallil

14:20

That's a very interesting way of doing it, @Huy!

user174558

14:43

It is interesting that there is a book referred to by Birkhoff/MacLane and another by MacLane/Birkhoff, and how neither are referred widely today.

user174558

Very sad that old books are becoming out of use, and new books with lots of colour are becoming the norm.

morphic

Hi @Jasper

How are you

Balarka Sen

New books provide newer perspective. I'd think Hatcher has more material than, say, Eilenberg-Steenrod.

user174558

@morphic Hi Bart. I am trying to start studying next year.

morphic

Hi Balarka

Balarka Sen

14:46

hi.

user174558

Appears that Pearson is making its hardback books softback while selling them at ridiculous prices.

user174558

Stewart's Calculus sells at over 200 USD. Absurd!

user174558

Wiley books are also too expensive.

user174558

These publishers just want lots of money, it seems.

user174558

Apostol's Calculus also over 200 USD per volume, ROFLMAO.

user174558

14:51

And the Cram101 series is just a scam, containing nothing but a glossary of terms that doesn't even help you in math.

user174558

Also, seems that some famous people are boycotting Elsevier journals.

Remember me

What all are the topics involved in computational topology?

hey@Jasper

user174558

Hello.

Remember me

so what all will you be studying when you again start doing math next year @Jasper

user174558

@Rememberme I am still deciding.

Remember me

14:58

Do tell me about it when you do.

evinda

Hello!!! Could you explain me why given the space $C([0,1], \rho)$, we can define a physiologic metric but this doesn't come from a norm?

Remember me

@evinda what do you mean by physiologic metric? Never heard of it

evinda

I think that just a metric is meant. @Rememberme

Remember me

I did not get you

For me a metric on a space X is just the function $d:X\times X \to X$ such that for any two points $x,y\in X$
$1) d(x,y)=d(y,x)
2) d(x,y)\geq 0
3)d(x,z)\leq d(x,y)+d(y,z)$

Huy

@Rememberme: how can you verify 3 on some unordered set $X$?

Alexander Bollbach

15:09

can any one help me with this khan academy example?

More limits | Limits | Differential Calculus | Khan Academy

►

at ~mins he is saying that the abs value of a -x is a negative value

how can that be?

Agawa001

i think im goin to work solo for a while, nothing here discussed about suit my field of knowledge

Remember me

@Huy In any given set $X$ I can define the metric $d$ such that $d(x,y)=1$ if $x \neq y$ and $d(x,y)=0$ if $x=y$. Here even for unordered set d happens to be a metric. I have never seen any definition which says $X$ has to be ordered.

Huy

@Rememberme: Then you must also change your definition that "a metric is a function $d: X \times X \to X$"

Remember me

Okay . I see. A metric is a function from $d:X\times X \to \Bbb{R}$

Huy

that makes more sense

Remember me

15:18

Thanks for pointing that out

@Huy Do you know of any place I can find problem sheets of real and complex analysis ?

Huy

I can send you some if you want

oh wait they're probably in German

Remember me

Sure.

Huy

yeah, sorry, my first 3 semesters were almost completely in German

Remember me

Bad luck.

Huy

@Rememberme: Do you have access to springerlink?

Remember me

15:27

what is springerlink?

0celo7

link.springer.com

free books!

Mike Miller

morning

Balarka Sen

morning @MikeMiller

do you think you have some time to listen to my philosophy (philosophy in the sense that i haven't thought it out, rigorously) on cup product vs. intersection? if not, it's ok - i understand.

Remember me

chat.stackexchange.com/transcript/message/25250690#25250690 @BalarkaSen can you look at this question if you are free?

Hey@PaulPlummer

Balarka Sen

15:44

hi @PaulPlummer!

Paul Plummer

Hello @Rememberme

Hello @BalarkaSen

Balarka Sen

what're you upto?

Paul Plummer

Eating breakfast, then going to work on some school work, maybe analysis, maybe diff geo, not sure yet

Yourself

Balarka Sen

cool, what are you learning in differential geometry?

I am chatting :P

Mike Miller

@Balarka: Sure.

Agawa001

15:55

i dont know either it is my misconseption for this problem or the op didnt clarify

i answered following a more advanced understanding of the prob

Mike Miller

Well, provided you can say it in finite time.

Balarka Sen

@MikeMiller Thanks. Let $M$ be an $n$-manifold. Triangulate $M$. $\varphi$ be a $k$-cocycle and $\psi$ be an $l$-cocycle (with codomain $\Bbb R$). Assume $k + l = n$. $\varphi$ and $\psi$ can be thought as piecewise linear functions on the triangulation of $M$. Then consider the premage $\psi^{-1}(1)$ and $\varphi^{-1}(1)$. One can pick chains from the preimages of each. These chains are something like level sets of $\varphi$ and $\psi$. Assume these chains are cycles. Assume, furthermore that they are actually submanifolds of $M$ and are subcomplexes of $M$, hence call them $M_1$ and $M_2$…

That's finite time!

I have assumed a lot of things which I suspect to be true for smooth manifolds. But that's not a proof anyway. And I don't think this should really work as all I have done is fiddling in the PL category.

Mike Miller

I am not sure I buy the first half but the second half is correct.

Probably.

The way I would personally do it: pick two smooth transversely intersecting submanifolds. A little care shows that you can triangulate them so that the intersection of the triangulations is a triangulation of the intersection. Some more care and maybe a little Morse theory shows you can extend this to a triangulation of the whole manifold.

Now this is precisely the setup you need to just literally look at the cochains and see poincare duality works out.

Paul Plummer

To be honest I am sort of lost in that class,so I will probably be trying to get not quite as lost today. Recently we have been talking about ($k$ dimensional) distributions, which roughly choosing $k$-dimensional from each $T_pM$ in a smooth manner (there is a vector field "basis" around each point, that are smooth). In the class we are going over when there is a submanifold that "fits" the distribution, so the tangent spaces that the distributions chooses are tangent spaces of that submanifold

@BalarkaSen

Mike Miller

(Which is roughly what you were doing at the end.)

Balarka Sen

16:05

@MikeMiller Yeah, it's not clear to me if one can extend that triangulation of $M_1$ and $M_2$ to the whole manifold, so I assumed it there. What's the proof by Morse theory you have in mind?

I actually asked this to someone, he said one probably needs a Riemannian metric on the total manifold and a bunch of other things I did not understand. But we did not talk much.

Mike Miller

First extend it to a tubular neighborhood by hand, then get a description of the whole manifold by adding handles to the tubular neighborhood. This is easy to do while respecting the triangulation.

Balarka Sen

Unsure how to add handles on the tubular neighborhood to get the whole manifold. Interesting idea, though.

Mike Miller

Morse theory says you can.

Balarka Sen

oh.

Thanks, that's helpful. It makes sense.

Mike Miller

One has to do some combinatorics with the triangulations to get this all to work but it shouldn't cause any trouble.

Wow, I really know how to fire up a room.

Balarka Sen

16:21

well, I'm still thinking about what you said.

Paul Plummer

@Rememberme What are you trying to do with these covers and spaces? In the generality you pose the question the answer is probably just "$\{B_n\}$ forms a cover of $X$." Some topologists study something called $\omega$-covers, which are covers where for any finite subset of the space, there is an open set in the cover that contains it. If you partition an $\omega$-cover into finitely many pieces, one of the pieces will be an $\omega$-cover.

Balarka Sen

maybe, if you're looking at a 3-manifold, there's a better way. if you have two transversely (i'm using this word very informally) intersecting submanifolds of dual dimension, one can pick a handlebody very far away from them. then take complement to get those two submanifolds inside a single handlebody

so all one needs to do now is to do this on a handlebody. no idea if this really works.

Mike Miller

That doesn't really make any sense to me.

Balarka Sen

I am trying to Heegaard decompose the 3-fold so that the two intersecting submanifolds lie inside one single handlebody in the decomposition. Is that even possible?

Mike Miller

I'm sure the construction I outlined is not standard. Most topologists don't want to think about triangulations in the necessary level of detail.

No. RP2 inside RP3. There's no reason you should be able to do this and you should expect it's generally impossible.

Balarka Sen

16:28

RPn is not orientable, so there's no Heegaard decomposition.

But I think I agree I am raving nonsense.

Mike Miller

How many times have we had this discussion? Yes it is.

Balarka Sen

OK, that confirms I am raving nonsense.

Mike Miller

In odd dimensions, because the antipodal map is orientation preserving there.

Balarka Sen

Sorry! Of course RP^3 is orientable.

I should not discuss any math for an hour lest I state crap again.

Mike Miller

Luckily I'm leaving for two.

Remember me

16:39

@PaulPlummer I want to see that what all should I impose on the space for which if we have a cover the subsets of the elements of the cover also forms a cover

Hmm. I guess I will read upon a bit about omega covers

Paul Plummer

If you had a cover and one of the elements was not the whole space, say $A_n$, you could take $B_n=A_n \neq X$ and the rest of the $B_i$ to be empty, so it won't be a cover.

@Rememberme

(Or you could take all the $B_i$ empty)

evinda

17:16

Hi @DanielFischer !!!
Suppose that we have the space $(C(0,1), \rho)$.

Then we can define a metric but this won't come from a norm.

Could you explain me the above proposition?

user174558

@evinda Do you want to understand the meaning of the theorem as stated?

user174558

@evinda First, you must know what a metric and a norm is.

user174558

@evinda Then, you must know how a norm induces a metric.

evinda

@Jasper I know the properties of a norm and a metric. The norm induces a metric if we pick the distance of two points ||x-y||... Or am I wrong? @Jasper

user147690

@evinda There are metric spaces in which there is no norm that induces this metric, $\|x-y\|=d(x,y)$, just as there are normed spaces, which don't satisfy the parallelegram law, thus their norm cannot be induced by an inner product $\langle x-y,x-y\rangle=\|x-y\|=d(x,y)$

user174558

17:21

@evinda The result says that you can have a metric there, but it won't be induced by a norm. I am not sure of the result, but that is what the statement says.

user174558

@evinda You are right.

user174558

Dear all, I have decided on the first book to read next year. It is Michael O'Leary's A First Course in Mathematical Logic and Set Theory. I will start on the first day of 2016.

evinda

Ok @AlexClark And how can we deduce that $(C(0,1), \rho)$ is such a metric space?

user147690

@Jasper What is the second book?

user174558

@AlexClark I will decide when I am halfway through the first book.

user147690

17:24

@evinda What is $\rho$?

evinda

The metric @AlexClark

user147690

What is the metric lol

user174558

LOL

evinda

You mean that I should mention the properties? @AlexClark

user174558

Now rereading the above statement I don't know what it means.

user147690

17:26

No write down $\rho(x,y) = xy^2$ or whatever it evaluates as

user174558

Hey @evinda I think the statement is not quite right.

evinda

It is not given. @AlexClark

user174558

I think they just want to say that you can define a metric on the space but it won't come from a norm. The rho is redundant @evinda.

user147690

Oh okay, let me think then, perhaps $C(0,1)$ necessarily satisfies absolute homogeneity

evinda

Of what form should $\rho$ be so that it won't be induced by a norm? @AlexClark @Jasper @anon

user174558

17:32

@evinda @evinda Can you quote the result exactly as stated and where did you find it?

anon

"what form should the metric on a infinite-dimensional vector space take so that it is not induced by a norm?" that seems like a very broad question

user174558

@anon I am thinking if the result means for all metrics, hehe.

user147690

I can't think of any metrics of $C(0,1)$ that can't be induced sorry @evinda

evinda

@Jasper It is only stated that what I wrote

user174558

Hey @AlexClark why you not sleeping?

anon

17:34

well, you can take $\rho(x,y)=\|x-y\|$ for some norm $\|\cdot\|$ and then consider $f(\rho())$ for some $f$ which won't scale linearly

user147690

@Jasper I just woke up

user147690

@Jasper I tried to sleep early, took a long time to fall asleep, and then I woke up and couldn't sleep at 3am, so here I am

evinda

@anon You mean that we could pick for example $\rho(x,y)=|x-y|^2$ ?

user174558

@AlexClark I think you miss me so much that you can't sleep xD

user147690

@Jasper I thought only girls write xD

user147690

17:36

:P

user174558

@AlexClark And, I am NOT user.

MickLH

@Jasper we all do, go easy on us

user147690

@Jasper Being a user is always nasty

user174558

@AlexClark I don't play games like skull or anon. I have only one account at a time.

user147690

@Jasper Who is user then?

user174558

17:37

@AlexClark How do I know. Probably someone called Twink long ago.

user147690

You are correct

user147690

Look at the address bar: chat.stackexchange.com/users/90263/twink

user174558

I prefer the chocolate bar.

0celo7

@AlexClark Jesus...

user174558

This user is a troll who does not deserve our attention. He keeps telling lies.

user147690

17:41

@Jasper I put them on ignore

user174558

What he said to me is horrid. I put it in the category of emotional abuse.

user147690

Well either you are Sarah, or User is

user174558

@0celo7 Christmas is coming.

user147690

(or you coincide)

user174558

@AlexClark No, me, Sarah and user are 3 distinct elements.

user147690

17:42

You mention that Sarah wasn't on, for about a year, and then she appears simultaneous to user and then leaves at the exact same moment as user

user147690

so at minimum they are one person

user174558

@AlexClark Poor reasoning.

user147690

They left at the exact same moment

user147690

Not poor reasoning

user174558

I think functional analysis has made you nuts.

user147690

17:43

Like within a one second period

user147690

Now I really think you are both of them

Huy

+1

user174558

I can assure you that we are 3 different people.

user147690

At minimum they are one person, I don't have enough evidence on you

user174558

Look. I email Sarah now and then, and she is nice to me while user is horrid to me. QED.

Remember me

17:45

Hey@AlexClark

user147690

Hey @Rememberme, how's it going?

Remember me

Not good. I am sick. So cant think much useful math@AlexC

user147690

@Rememberme Oh damn, being sick is the worst. Surely you can think of geometric problems atleast?

user174558

@AlexClark Your reasoning sucks, LOL.

MickLH

I actually chuckled out loud at that one

user147690

17:47

@MickLH Which one?

user174558

Hello @MickLH, LOL.

MickLH

Hey jasper loy my boy

user174558

Now that sounds wrong, LOL.

MickLH

Sorry, I am quite informal and make heavy use of Ebonics.

I intended it as the term of endearment which is roughly equivalent to... something unacceptable on stack exchange.

Not in the racist way! I am not saying they are unacceptable here, just the word for them!

user174558

@AlexClark How can you say the 3 are 1? LOLLOL.

Remember me

17:49

Well ya. Doing real and a bit of complex analysis revision. So getting tired easily. And then thinking a few topology problems @AlexClark

MickLH

@AlexClark this one

user147690

@Jasper, they argued amonst themselves for something like an hour, 'unrelated' to us, and then we ignored them, and they logged out of chat within a one second time interval, just read from ending here: chat.stackexchange.com/transcript/36?m=25235551#25235551

user147690

@MickLH Haha we sound like major nerds :D

MickLH

It was a nice little stab at geometers

user174558

@AlexClark I know. But that does not mean anything.

user147690

17:50

@Jasper Huh, two people leave within a one second time interval, by coincidence?

user147690

@Rememberme Complex analysis, revision?

user174558

@AlexClark Yes. Happens all the time.

user147690

@MickLH Oh hahaha that's not what I intended

user147690

@Jasper Yes, after a certain block of time, this was immediately after we blocked them

MickLH

Ugh, all this smiling and laughing, it literally hurts. I have to take my medication, so thanks for the harm guys.

user147690

17:51

@MickLH I just meant you don't need paper to think about it

MickLH

I mean, how dare you make me laugh?!

Huy

why do you have to take meds?

user147690

My bad @MickLH, I am a comedic genius(and I needed spell check for that last word)

MickLH

Actually I was jokingly serious, smiling and laughing is an anxiety trigger for me.

user174558

I am now LMAO at how @AlexClark can think 3 people are 1.

Remember me

17:52

Just basics. Not even that. I am going to read through everything again. I skimmed once . So a revision

user143442

Hello, it's me

MickLH

Oh my.

user147690

@Jasper Stop please

Remember me

What an amazing timing :P

user143442

I have to admit I'm @Jasper

user143442

17:53

I'm using two browsers at the same time

user147690

Well I better get back to work and ignore all this

Remember me

What on earth !!! ?

user174558

I think I should leave too. I don't like user.

user143442

@Jasper can you like one of my answers just as I liked all of your answers?

Remember me

@AlexC hows your altop fairing?

MickLH

17:55

@Jasper Do you not like them in a Disassociative Identity Disorder sense, or in a "This guy is impersonating me." sense of the notion?

user143442

@0celo7 did you say I'm a wierdo because I'm gay?

0celo7

@user Sure

Remember me

For gods sake, can the chat room please get back to maths ?

user143442

why? are you homophobic?

anon

Pirates of the Carribean - Stop Blowing Holes in my Ship

►

2

user143442

17:57

hi @anon

0celo7

@user That must be it!

user143442

he said he's gay

0celo7

@user I didn't notice that.

Transcript for

$Mathematics$ Mathematics