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15:00
@TedShifrin hello
I wish I had a single friend @Abeau
have seen my edit please @TedShifrin
hi Vrouvrou
no, Vrouvrou ... I haven't had time.
@Answer I can be your friend if you want.
Me too :-)
15:01
@Chris'ssis Are you talking about math?
I'm not sure virtual friends are as good as irl friends
I appreciate it guys, but I need a friend I can be around and meet people through
@ABeautifulMind I'm investigating some multiple integrals coming from my research.
@TedShifrin I have at least one online friend who is much better than any of my real life friends.
please if you have time see it, because i will be crazy if i don't find it it seems 3 days i don't find any thing @TedShifrin
15:02
@ABeau How did you make IRL friends?
I know what you mean, Jasper, but still ...
1
Q: Question about computing a Complicated integral

Vrouvrou where $\beta$ is defined like this: I'm trying to prove (2.18) but i don't know how to do, i calculated the integral but i don't find anything %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% EDIT1: $\beta(\phi_{\lambda,p,\rho}(y))=\displaystyle\frac{\int_{\mathbb{R}^N} x |\nabla(\phi_{\lambda,...

@Ted How should a uni student make friends?
@Answer Well, you just talk to people and try to share about your life.
@ABeautifulMind And I also believe that Ramanujan himself would be in trouble with them.
15:04
@Answer Just go and talk to people, classmates or whatever.
@Chris'ssis I told you you are a genius.
@Abeau They just think I am weird if I approach them
@Answer Try to overcome your fears of approaching people.
@TedShifrin Yeah, I know what you mean too...
@Abeau Maybe join a sport?
try talking to people in your classes about what you're working on in class ... my students get to know each other quite a bit in my office hours. Find a club to go to and meet some people with common interests, @Answer.
@Answer Yes, also good.
@Answer Also, if you want a partner, join a dating site if you can't meet anyone.
15:06
Dat tooltip:
Guten Tag, @DanielF
Bonjour @Ted. Comment va t'il?
Comment ça va? :) ou Comment vas-tu? :)
Thank you all, I will go and read my universities clubs and sports page and find something and maybe I can make a friend
Bien, merci, et toi?
good, @Answer :)
15:07
@Answer If you need someone to talk to, email me.
@Abeau Where?
0
Q: Proof of the proposition $V(S)=V(\langle S \rangle )$

user159870In my lecture notes we have the following: Proposition: $$V(S)=V(\langle S \rangle )$$ Proof: $$\langle S \rangle=\left \{\sum_{i=1}^m g_i f_i | f_i \in S, g_i \in R=K[x_1, x_2, \dots , x_n]\right \}$$ It stands that $S \subseteq \langle S \rangle$ $\Rightarrow V(S) \supseteq V(\langle ...

@Answer jasperloy at outlook dot com
@Abeau Got it, thank you. Talk soon
@robjohn My proof to that limit I showed you yesterday also covers the case $|\sin(b x)|$. I mean in my book at the same problem I might add $2$ variants, also this one.
15:09
@user159870: $S\subset\langle S\rangle$ because for any $f\in S$, $f = 1\cdot f$.
@TedShifrin Ok.

Why do we write $\langle S \rangle$ as $$\langle S \rangle=\left \{\sum_{i=1}^m g_i f_i | f_i \in S, g_i \in R=K[x_1, x_2, \dots , x_n]\right \}$$ ?
Do you know about ideals in rings, @user159870?
@TedShifrin $$I=\langle a_1, a_2, \dots , a_n \rangle$$

If $x \in I$ then $x=a_1 \cdot r_1 + a_2 \cdot r_2 +\dots +a_n \cdot r_n, r_i \in R$
Correct?
Right. The $m$ in your formula is arbitrary, unless they told you that $S = \{f_1,\dots,f_m\}$.
@Chris'ssis which one?
15:21
@TedShifrin So there are $m$ functions at the family of functions, where each has $n$ variables?
@robjohn I'll upload it as soon as I'm done with it.
@robjohn but I try to make it rigorous for my book.
@TedShifrin and why does it stand that if $S \subseteq \langle S \rangle$ then $ V(S) \supseteq V(\langle S \rangle )$ ?
@robjohn If I had to post it as an answer on MSE, i would be done in a few lines. I cannot do that in my book.
Write down the definitions, @user159870. Remember that $g_i$ can be chosen arbitrarily. I don't know if there are just $m$ functions. I think $m$ can vary, unless they said specifically that there were $m$ functions.
@robjohn I'm trying to present a nice flow of the whole proof thing. Working on that.
15:28
@Chris'ssis okay... I have to head to the park. BBL
@robjohn OK
hi/bye @robjohn
Ok. @TedShifrin why does it stand that if $S \subseteq \langle S \rangle$ then $ V(S) \supseteq V(\langle S \rangle )$ ?
What's the definition of $V(S)$?
Hi @robjohn@TedShifrin@Chris'ssis
15:31
salut, @Gato
@Gato Hi
Hi all
hi Doe John
Ted is very naughty.
@TedShifrin It's the algebraic set.

$$S=\{f_a \in K[x_1, x_2, \dots , x_n] | a \in A\}$$

$$V(S)=\{(a_1, a_2, \dots , a_n) \in K^n | f_a(a_1, a_2, \dots , a_n)=0, \forall a \in A\}$$
15:32
naughtier than usual, Jasper?
OK, @user159870. Now write down the definition of $V(\langle S\rangle)$.
@Chris'ssis how did you say hi in Romanian? (not the 'official' work more the one in the street)
@TedShifrin Are you worried about getting Alzheimer's? I just want to understand how a person like you feels about it.
@Gato "Salut" or "Ceau"
Not a pleasant subject, Jasper. My mother hasn't known who I am in 3 years.
@Chris'ssis Thanks.
15:33
@Gato ;)
@TedShifrin I see. I guess it is a problem for you too, just like my mental problems.
I'm not going to waste good years being obsessed with it, Jasper. Maybe the cancer will come back and get rid of me. :)
@TedShifrin sorry but please if $x\in B_{\rho}(y)$ then $|x-y|=\rho$ right ?
@TedShifrin Sometimes, I wish I had cancer that would take my life away, then I don't have to struggle anymore.
I doubt it, @Vrouvrou. It should be $|x-y|\le\rho$.
15:35
@TedShifrin My cousin and my aunt both got cancer of the womb and removed their wombs.
lovely thought, Jasper
Does anyone know if the following is valid, or some variation of it, for sequences $a_{n}$ and $b_{n}$, if $a_{n} \rightarrow a$ then $$\liminf\limits_{n \rightarrow \infty}(a_{n}+ b_{n}) = \lim\limits_{n \rightarrow \infty}a_{n} + \liminf\limits_{n \rightarrow \infty}b_{n}$$?
@TedShifrin Haha, that response was atypical, lol.
If you have two continuous functions $f$ and $g$ on an interval, is it true that $\min(f(x)+g(x)) = \min(f(x))+\min(g(x))$, @JohnDoe?
@Chris'ssis Some word in Romanian are very close to french
15:38
@Gato Yeah, latin languages.
Is English closer to French or German?
lots of connections with both, Jasper
@Chris'ssis yep:), i like it it's a beautiful language. I think I need to learn about it.
@TedShifrin Most people would say I should not think that way, but you said lovely thought. That really surprised me. I must say yours is a lovely thought, lol.
I was being somewhat (!) sarcastic, Jasper.
15:40
I think it's OK to joke a little about these things.
@Ted uhm yes
Draw some pictures, @JohnDoe, i.e., graphs of functions.
@Gato Wait to see our girls ... ;)
@Chris'ssis Does that include you?
@ABeautifulMind hmmm, I let me think for long ... ;)
15:41
@Chris'ssis I have one in my "class" this semester, this is why I want to lean it ;P :D
@Gato :D
@Ted Okay will start drawing, but are you saying it's not true?
Yes, @JohnDoe, that's what I'm saying.
@Ted the original question
If you understand my graphs question, you'll understand your original question.
15:45
We have that $S \subseteq \langle S \rangle$. $V(S)$ is the set of roots of the functions of $S$ and $V(\langle S \rangle )$ is the set of roots of the functions of $\langle S \rangle$. So that $ V(S) \supseteq V(\langle S \rangle )$ it must be that the roots of the functions of $\langle S \rangle$ are also the roots of the functions of $S$ , correct? How can we show that it is indeed so? @TedShifrin
The key thing is that it's roots of all functions in $\langle S\rangle$, @user159870. Pick the right $g_i$.
@JayeshBadwaik How is life?
I didn't get it. I'm mixed up. @TedShifrin :/
o.O @TedShifrin
@ABeautifulMind Life is good.
@JayeshBadwaik I have no more words to say about mine. I am lost for words.
16:05
yo yo @TedShifrin
right when i get online you leave the chat?
thats like a dagger to my heart
16:24
Hi guys, I have a question about probability, suppose a person can assign at the same time 3 different tasks to an employer: A with 80%, B with 60%, C with 40% and at most 1 task will be assigned. Task A cost X minutes, B y minutes and Z y minutes. I need to calculate the average time that the employer will spend doing the tasks assigned. How can I achieve it?:-
Morning
@MikeMiller Hello mister.
16:46
0
Q: Algebraic Set-Radical Ideal-Nullstellensatz

user159870In my lecture notes there is the following: $$I \rightarrow V(I) \rightarrow I(V(I))$$ It stands that in general $I \subsetneq I(V(I))$. The equality stands if and only if $I$ is a radical Ideal. Can you explain why this stands??? $$V \rightarrow I(V) \rightarrow V(I(V))$$ The equalit...

17:35
@PedroTamaroff I am looking at irreducible algebraic sets.

$V \subseteq K^n$ is an algebraic set $\Leftrightarrow$ it is of the form $V(I)$, where $I=$radical Ideal of $K[x_1, x_2, \dots , x_n]$.

At my lecture notes there is the following:

$V((x^2-y^2))=V_1 \cup V_2$, where $V_1=V(x-y)$ and $V_2=V(x+y)$ algebraic sets.

How do we know that these two are algebraic sets??
Do we have to show that the ideals $\langle x-y \rangle$ and $\langle x+y \rangle$ is radical? How can we do this?

@PedroTamaroff
Please like/follow
more comics coming very soon
Since when did Emmy Noether become a physicist?
Nevermind. I just googled. Never knew she also had remarkable contributions to physics.
Hi guys
Hi@BalarkaSen
17:55
@user159870 Geometrically, $x^2=y^2$ is the union of two lines, $x=y$ and $x=-y$.
You want to see that these lines are algebraic sets. In fact they are algebraic varieties.
You can achieve this by showing that $(x+y)$ and $(x-y)$ are not only radical but prime.
@PedroTamaroff which field of mathematics is this
@SayanChattopadhyay Algebraic geometry.
The basics of it, at least.
Well u r the moderator now right@PedroTamaroff
Oh I thought group theory
@SayanChattopadhyay I am one of the many moderators.
@user159870 To achieve the above, show that the two of $k[x,y]/(x\pm y)$ are isomorphic to $k[x]$. This is a domain, hence they are prime ideals.
To do so, you can carry division in $k[x][y]$ by the monic polynomial $x\pm y$, to conclude that the kernels of the morphisms $k[x,y]\to k[x]$ with $x\to x,y\to \pm x$ are the desired ideals.
@PedroTamaroff You're familiar with homology a bit?
18:05
@BalarkaSen Can you be more specific?
Well, I was rereading the proof of homotopy invariance of singular homology groups, and I think I don't really understand the prism operator. Can you provide some insights?
I don't know algebraic topology.
Oh, OK.
At least, I don't know about singular homology.
ADG
ADG
18:20
@PedroTamaroff hi!
ADG
ADG
Pedro do you know I'm banned at Chem.SE
@ADG Now I do.
Why r u banned @ADG
18:24
@SayanChattopadhyay hi
ADG
ADG
The're banning people of "plagiarism" but they shouldn't be so critical on not citing "Chemistry 101"
@SayanChattopadhyay Why don't you write "are" and "you"? It's not like they're 10 letter words. =)
ADG
ADG
Many users give THesis type ansswers there so people think this is mandatory
Well I won't be coming on math stack exchange from tomorrow
cus hes lzy @PedroTamaroff
ADG
ADG
18:25
@SayanChattopadhyay just because you were kinda offeneded by this?
My mom and dad are not allowing me
To do maths
ADG
ADG
@SayanChattopadhyay why?
How unlucky I am
ADG
ADG
why?
They want me to be a dov
Doctor
ADG
ADG
18:26
dov=dove?
ADG
ADG
lel=??
So I have to biology from tomorrow for my exams
laughed out enormously loud
ADG
ADG
this is bad but i can't do anything try convincing them
oh you meant lol?
18:28
I am so unlucky....well I want to do maths and they tell me to do bio
Well goodnight then @BalarkaSen
you don't need to come at MSE to do math @SayanChattopadhyay
Hey guys if the quotient criteria for series is not fullfilled, does it mean the series diverges?
Or does it not say anything about my series then?
ADG
ADG
@PedroTamaroff r u there?
ADG
ADG
so what do you think about the thing
bann?
18:31
But my "R U" meter is filling up.
ADG
ADG
Oh you don't like "ru"?
R U JELLY BRO
ADG
ADG
@BalarkaSen don't know what that means?
Ignores Balarka.
ADG
ADG
18:32
Pedro did you wrote that or server decided on it's own?
hi @user158139
@ADG Well, if you didn't "plagiarize" purposedly, I find the ban at least an overreaction. But since I am not in the mod team of chem.SE, you must follow their policy.
In particular, I won't tell them to something about it, if that's something that crossed your mind.
ADG
ADG
I like your first line
I'm new at here
ADG
ADG
"purposedly" = great word
Oh @user158139 how coming here, need help?
Once a user asked, "Where can I find the reactivity order of ligands?" *ligands=(chemistry moeity). and I answered "You're looking for Spectrochemical series. Google up there's many around" and they deleted my answer saying it was too brief. They couldn't understand properly the help center and what it means to be briedf and when it's ok to have a hint.
AFAIK there's only one or(three atmax )moderators and site is beta. I'll revoltutionize the siet when I can become a MOD
@user158139 wanna say something
@PedroTamaroff btw CHem.SE is beta site
ADG
ADG
18:40
@PedroTamaroff would you tell me atleast how can I become a MOD here? Is it interseting?
WHat I need to have? to do?
@ADG Why do you want to become a mod?
ADG
ADG
I think it's interesting
Interesting? In what sense is it interesting?
ADG
ADG
To be like a GOD!
That's not the point of being a moderator.
But I get your point.
18:42
Hello @PedroTamaroff will you say something nice to me?
ADG
ADG
BTW english is my second tounge and can you keep that in mind (treat it just like a note)
@ABeautifulMind depends on what you want to hear
@ADG OK. It is my second tongue, too.
ADG
ADG
Oh you're originally...?
@PedroTamaroff I have only one tongue in my mouth.
ADG
ADG
Man I earned only around 7K in a year GOD how can you be having 66K?? 66K? 66K? that'll take 66/7\approx9 years!!
18:44
@ABeautifulMind OK... the ring of holomorphic functions over a region in $\Bbb C$ is not noetherian, nor a UFD yet it is a Bézout domain.
@Pedro That is not particularly nice.
@ADG Participation. It's not something crazy.
@MikeMiller There's still hope Mike.
ADG
ADG
are you both PhD student?
@MikeMiller Will you say something nice to me?
ADG
ADG
you talk high sky high level maths
18:45
@ADG I'm Argentinean.
@ADG No, not really.
ADG
ADG
@ABeautifulMind are you some kind of collector?
@ABeautifulMind Mike speaks only hurtful words.
ADG
ADG
@ABeautifulMind "sayings" collector? book maker?
You don't want to talk to him.
I am feeling very anxious now, I just wanted to hear some nice things.
18:45
That's true.
ADG
ADG
I'll tell you $e^{ik}=\cos(k)+i\sin(k)$?? isn't it great
greater is $e^{i\pi}+1=0$
@ABeautifulMind saw that?
I guess I should not have asked.
ADG
ADG
@ABeautifulMind you're irritating me? don't you talk normal?
@ADG What do you mean? I am talking normally.
ADG
ADG
@PedroTamaroff don't argentineans talk english first
18:48
@ADG Come again?
ADG
ADG
Why would someone come a long way just to ask everyone "Say something nice", feels wierd? @ABeautifulMind
come again=??
@Pedro
@ADG I don't understand your question.
ADG
ADG
oh, so what's your first language there?
@PedroTamaroff do I need to tag you repeteadly, are you busy with multiple tabs?
@ADG Not really, no. Just do so if I take too long to answer.
Our first language is Spanish, like almost all countries in America.
Except Brazil, USA, Canada and some little countries up north of South America.
ADG
ADG
jawdrop Spanish? America? not English? this is my year's dose of surprise
what about great New York? Washington DC? everyone Spanish?
18:52
@ADG By America I mean the continent.
ADG
ADG
BTW I'm Asian @PedroTamaroff
so I don't knwo much about there far away
Where in Asia are you?
ADG
ADG
bonus, I today gave second stage of International Chemistry Olympiad
@PedroTamaroff Asia->India->my home
ADG
ADG
I was first going to give my address but you know then...
Would you mind telling about your educational qualificiations? I'm not yet in college and seeking advice from everyone possible.
18:55
@ADG Yeah, don't do that.
ADG
ADG
What are other people doing here? system problem? or they all still here? or the google-chrome-tab thing? or or or are you people sneeking?
@PedroTamaroff EduQuali
You can hover over their avatars to see when they were last seen. If they don't leave the room, the avatar sticks here.
@ADG I'm an undergraduate.
@PedroTamaroff Sometimes, we leave the room but it still stays.
ADG
ADG
@ABeautifulMind I found a sneeker!
@ADG What is a sneeker?
ADG
ADG
18:58
@PedroTamaroff does UG mean high school?
Well, we do have clingy users.
@PedroTamaroff What do you mean?
@ADG No. It means university.
@ABeautifulMind I mean what I say, Jasper. And say what I mean.
I smell troll in the air.
@PedroTamaroff I guess I need to check the dictionary then.
ADG
ADG
18:59
@ABeautifulMind like some theif who sneeks into houses or someone sneeking others info, maybe it's not the corrcet word, Eng.Prob.
@BalarkaSen Who is the troll?

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