« first day (1333 days earlier)      last day (3703 days later) » 

6:01 PM
@FernandoMartin
$$\eqalign{
& {\alpha _{1x}}f + {\alpha _{2x}}g + {\alpha _1}{f_x} + {\alpha _2}{g_x} = 2x \cr
& {\alpha _{1y}}f + {\alpha _{2y}}g + {\alpha _1}{f_y} + {\alpha _2}{g_y} = 0 \cr
& {\beta _{1y}}f + {\beta _{2y}}g + {\beta _1}{f_y} + {\beta _2}{g_y} = 2y \cr
& {\beta _{1x}}f + {\beta _{2x}}g + {\beta _1}{f_x} + {\beta _2}{g_x} = 0 \cr
& {\gamma _{1x}}f + {\gamma _{2x}}g + {\gamma _1}{f_x} + {\gamma _2}{g_x} = y \cr
& {\gamma _{1y}}f + {\gamma _{2y}}g + {\gamma _1}{f_y} + {\gamma _2}{g_y} = x \cr} $$
OK....
 
god
what have we done to deserve that
 
I confess.
I never rewinded any VHS.
Wonder if @BalarkaSen knows what a VHS is.
 
@PedroTamaroff Betamax for the win!
 
@DanielFischer Damn you. =D
 
6:06 PM
Any ideas on our prahwblem?
 
@PedroTamaroff I don't know stupid compactization of words, no sire.
And I forget things quick.
 
Hyallo poepal .
 
What is a VHS, by the way?
 
DINGDINGDING-
Good night.
@FernandoMartin
 
6:10 PM
so early @PedroTamaroff >D
 
Does V stand for vector?
 
@FernandoMartin I am doing something.
If $\Sigma =x^2,\Gamma=y^2,\Phi=xy$, then $\Phi^2=\Sigma\Gamma$.
I'll see if I get something with that.
Where the greek letters are inear combinations of the two generators.
 
I never heard of "inear" combination.
 
@BalarkaSen The $l$ is stuck up your... ear.
 
Then correct it, first mate.
Land Ahoy!
 
6:16 PM
@BalarkaSen Video Home System
 
@DanielFischer What the... ?
 
@BalarkaSen VHS - Video Home System.
 
@DanielFischer WAT TEH?
Lame abbreviation.
 
No, LAME is something else.
But you probably only know about MP3.
Or MP4.
 
@BalarkaSen We weren't so hot on abbreviations in the seventies.
 
6:20 PM
LAME ain't an mp3 encoder
 
@FernandoMartin wut?
 
@JayeshBadwaik GNU is Not Unix
 
@FernandoMartin I didn't say not.
 
@Pedro: That's what LAME stands for
 
@DanielFischer Ahh....
 
6:21 PM
@FernandoMartin Oh, LOL:
@FernandoMartin Do you happen to know the informal reason why $(x,y)^2$ cannot be generated by two elts?
 
If $N$ people take off their coat and jumble them , what is the probability that $r$ people get their own coat while choosing randomly ?
 
I have started a room explanation, this is because I want to reduce the chat under answer given in mathematica stackexchange. I am posting a question in regards a answer given in the stackexchange, would someone please join Explanation and help me out?
 
@ALEXANDER This is mathematics, not mathematica. =P
@FernandoMartin
I got $${\gamma _1}^2{f^2} + {g^2}{\gamma _2}^2 = {x^2}{y^2} - 2{\gamma _1}{\gamma _2}fg$$
This means $$\frac{{{x^2}{y^2}}}{2} \geqslant {\gamma _1}{\gamma _2}fg$$
Equality everywhere is impossible since it would yields $f=g=0$.
 
What's $\gamma_i$?
 
6:31 PM
$$\eqalign{ & {\alpha _1}f + {\alpha _2}g = {x^2} \cr & {\beta _1}f + {\beta _2}g = {y^2} \cr & {\gamma _1}f + {\gamma _2}g = xy \cr}$$
 
What does $\geq$ mean in this context?
 
Just usual inequality of numbers.
 
But we're working over an arbitrary field
 
Oh, fuck.
Well let's work on $\Bbb R$.
=D
 
I've given up on that for now
I believe there's a simple high-brow explanation
 
6:36 PM
We can ask Ted, maybe.
 
and maybe there's an horrific elementary explanation
Well, I know there's a simple high-brow explanation since Alicia said that
 
Ram
Hi All, I have a simple complex analysis question.
Say $f'(x)$ is continuous in an open set containing unit disk. Then is the derivative of the function $F(r) = \frac{1}{2 \pi i}\int_{|z|= r} \frac{f(z)}{z} dz $ zero for $0 < r \le 1$

This seems trivial for me since, $F(r) = a_0$ which comes from power series expansion of $f(z)$ and is independent of $r$ so, for $0< r< 1$ $F'(r) = 0$
Am I missing some thing?
 
Apostol's Analysis has a proof of that.
Your argument is kinda circular.
BADUM TS.
 
Ram
@PedroTamaroff you mean Apostol's Complex Analysis book?
 
@Ram No, the Analysis one.
The one that is purple.
 
Ram
6:41 PM
ok ok, let me check
 
@FernandoMartin "Let (A,I) be a commutative excellent normal local domain."
 
Hahah
What's 'excellent'?
 
No idea.
@FernandoMartin I think I have an idea on this.
@FernandoMartin The ideal has the particular property that if $f\in I$; then $xf_x,yf_x,xf_y,yf_y$ will be in $I$.
 
Ram
@PedroTamaroff, yes he uses invariance of domain, but he gave a similar problem under Taylor series exercises too. :-)
 
@FernandoMartin More generally I think it has the following property.
If $f\in I$, then $M f_{D}\in I$ if $D$ is differentiation some amount of times and $M$ is a monomial of total degree that amount of time you $D$'d.
 
6:52 PM
Yep, that follows since the ideal is exactly the set $\{f: f(0,0)=f_x(0,0)=f_y(0,0)=0\}$
 
Also, if $f_{xy}=f_{yx}$ might help.
 
Heya
What is it called when you rearrange the parts in a divergent sum ?
 
Rearranging?
 
I remember seeing a place to make the alternating harmonic series approach different values by grouping the terms in various ways.
 
Aha.
Riemann's rearrangement theorem.
 
6:57 PM
Ah! Just found it
A few friends of mine found the whole $1 + 2 + \cdots = -1/12$ thing, sigh
 
@N3buchadnezzar Will they still be friends thereafter, is the question.
 
no
@DanielFischer I am gathering gasoline, chloroform and ducttape atm.
 
@N3buchadnezzar Always Prepared. Good.
 
@DanielFischer To be honest the ideas are very interesting in their own right.
I am taking a introductory course in analytic number theory, and we have talked a lot about the Rieman Zeta function and it's analytic expantion.
 
@N3buchadnezzar You mean summation methods for divergent series? Yes, of course. Just the Numberphile shitstorm is horrible.
 
7:07 PM
@DanielFischer Exactly
 
Such language @Daniel :)
 
$$ \int_{-\infty}^{\infty} \frac{\mathrm{d}x}{x} = 0 $$
 
@TedShifrin Teeeeeeeeeeed.
 
@TedShifrin There are things, you know, that can only be aptly described by "WHAT THE FUCK".
 
@Peeeeeedro :)
 
7:09 PM
Is there a windowing technique to apply to data before fft?
 
I need some help with something.
 
I'm just amused that I get banned for "bitch" while all these words proliferate.
 
Prove the ideal $(xy,y^2,x^2)$ is not generated by two elements in $k[x,y]$. Now, I don't want a solution.
 
Probably not by me, @Pedro
 
I have tried some stuff, but it gets ugly quickly.
 
7:10 PM
Oh ... Let's think
 
That is, when I try to write things with coefficients and all. =)
 
@TedShifrin You got banned for "female dog"? What the ...
 
Intuitively, @Pedro, it has to do with $\dim_k k[x,y]/I$.
 
@TedShifrin Yes, dimension stuff.
 
@N3buchadnezzar I wonder if this is a perfect square 1223334444...999999999
 
7:13 PM
@TedShifrin Oh, quotient?
Let me see.
 
You will learn to think about the dimension of varieties and varieties defined by ideals. This problem is about an interesting scheme (not just a simple reduced variety).
 
No, it is not a perfect square. @Chris'ssis
 
@Chris'ssis no
 
Nothing ending 99 is a perfect square...
 
@TedShifrin I am trying to see if I can determine what the quotient is.
 
7:15 PM
@ThomasAndrews that's correct
 
Good @Pedro. You should get dim 3.
 
@TedShifrin I can consider the map to $k^3$ given by $(f(0,0),f_x(0,0),f_y(0,0))$.
So $I$ is the kernel of that.
 
That'll do ...
 
@PedroTamaroff $k[x,y]/I= d+ax+by : a, b,d \in K$ ?
 
7:17 PM
Well, @Complex, that's sloppy but intuitively correct.
 
So $\dim _k k[x,y]/(x^2,y^2,xy)=3$.
@Complexanalysis Not entirely correct though.
 
Can you argue each generator reduces that dimension by at most $1$?
 
@PedroTamaroff just intuitively , don't know how to do it formally :(
 
@TedShifrin What do you mean by that?
 
You need equivalence classes of $x$ and $y$ mod the ideal, @Complex.
 
7:21 PM
@TedShifrin I didn't understand what you mean by "generator reduces the dimension":
 
Play around with $\dim k[x,y]/(f,g)$, @Pedro.
 
@TedShifrin For various $f,g$?
 
Right ...
I have to write a uniform final for Calc I along with 4 of my own exams. Disappearing for a bit ...
 
@N3buchadnezzar This one is in the spirit of your previous question (does it look like a pie?). One needs to arrange numbers from 1 to 19 in such a way we get 30 on each line (that, surely, is consisted of 3 circles)
I wonder if we can explain that mathematically in a nice elementary way ...
(how about some generalization?)
 
@TedShifrin you mean $\langle [x], [y]\rangle= k[x,y]/I$ ?
 
7:33 PM
@PedroTamaroff tsk tsk
I-told-you
=P
 
@BalarkaSen ?
 
OK, nevermind.
 
@Complexanalysis $k[x,y]/(x^2,y^2,xy)$ can be seen as the set of classes $a+bx+cy+I$.
 
@PedroTamaroff yup
 
@robjohn I'm going to write up the solution to $$\int_0^{\infty} \sin(a x^2) \cos(2 b x) \ dx=\sqrt{\frac{\pi}{8a}}\left(\cos\left(\frac{b^2}{a}\right)-\sin\left(\frac{b^2‌​}{a}\right)\right)$$
(by real methods only)
brb
 
7:40 PM
Say
We have a number of white and black balls in a cup
Lisa and bob each draws a ball from the cup, without putting it back. If the balls have the same color Lisa wins, and if the have opposite colors bob wins.
The problem is now deciding the ratio of black and white balls in the cup, to make it a fair draw.
Seems like all the solutions involve triangle numbers, but I do not see intuititvely why this is true.
 
@N3buchadnezzar Not intuition, but $$2(W(W-1) + B(B-1)) = (W+B)(W+B-1).$$
 
That relation must hold for the game to be fair.
 
8:00 PM
@DanielFischer Daniel.
 
What is a neighborhood of a set ? I know what this means for a point of a set. But now they say let $f$ be holomorphic on a neighborhood of the closure of a domain.
 
@Kasper An open set containing the set.
Or a set containing an open set containing your set.
Some people call the first an "open" neighborhood.
And the second simply a neighborhood.
 
haha okay thx !
 
beiberhood
 
@Mike You said you were away.
 
8:03 PM
@Mike Pedro disagrees with you.
 
Damn it that's why I write nbhd.
 
I'm going back away.
 
@Mike You said Balarka's thing was iso to $\Bbb R^\times$.
 
@PedroTamaroff Oui?
 
I don't believe you, @Mike.
 
8:04 PM
@PedroTamaroff Tell the reason.
It's pretty cool.
But I think it has less to do with algebraic isomorphism.
 
I am not sure my argument is legit.
It is just one.
@DanielFischer Suppose $f=a_0+a_1X+\cdots+a_nX^n,g=b_0+b_1X+\cdots+b_mX^m$ are inverses of each other in $A[X]$. I have proven that $a_n^{r+1}b_{m-r}=0$ for each $r$.
Thus $a_n$ is nilpotent.
 
Good.
 
I have also proven the (rather easy) claim that if $a$ is a unit and $x$ is nilpotent, then $a+x$ is a unit.
Using this, I should show $a_1,\ldots,a_{n-1}$ are nil too, it seems.
 
@PedroTamaroff It seems your argument much points out that those two are not topologically isomorphic rather than algebraically.
 
Yes, sorry.
 
8:07 PM
@PedroTamaroff Sorry to whom?
 
@BalarkaSen Daniel. I typoed.
 
@PedroTamaroff What can you say about $f - a_nX^n$?
 
@DanielFischer This gives that $W = 2a^2 + a$, $B = 2a^2 - a \vee 1+2a^2+3a \quad a\in\mathbb{N}$, kinda cool.
 
@DanielFischer Oh. =P
Stoopid.
Ah.
So $a_nX^n$ is nil in $A[X]$; and $f-a_nX^n$ is a unit.
Hence $a_1,\ldots,a_{n-1}$ are nil by induction on $\deg f$.
 
Boo!
 
8:23 PM
-_-
 
@PedroTamaroff Now that I have 3k, may I be your friend?
 
@JasperLoy Damn it not again.
 
8:39 PM
@PedroTamaroff how much is it needed to be your friend ?
<>D
 
Does @Pedro have friends? :D
 
@PedroTamaroff I don't have that much.
But I am not Peter's friend either =P
 
Hello Professor @TedShifrin
 
hi mr @skull
 
Hello, @TedShifrin
 
8:42 PM
hi @BalarkaSen
 
@BalarkaSen But Peter's Friends are cool.
 
haha @DanielFischer
 
Well, anything with Emma Thompson is good.
 
That is soo last century ;-)
 
8:45 PM
@TedShifrin You want Emma Watson now.
Hehe.
Or Emma Stone.
 
Generally I don't yearn for Emmas ...
So, @Pedro, did you develop some intuition for your ideal quandry?
 
Watson is a good actress.
But I don't really see movies.
So I will keep outta this, on second thought.
 
@TedShifrin Not even Emma Peel?
 
Oh, Emma Peel was fantabulous.
 
Abso-freaking-lutely!
 
8:51 PM
Sounds like a "peeler" :D
 
Emma Watson is the best i would say =P
 
@TedShifrin Not yet. =/
I think?
I did finish some other exercises though.
 
Well, I assume that question is purely rhetorical.
I'm regretting volunteering to do all this work for our uniform exam. Ugh.
 
@TedShifrin Our professor mentioned she knows a high brow way to solve it.
And that she never tried to do it "manually."
 
Well, yes, there are things called Krull dimension and other notions ...
It's a nice question.
 
8:57 PM
Krull sucks
I like transcendence degree definition
Much more natural to me.
 
Not remotely relevant in such situations.
 
@TedShifrin Aha.
She mentioned that. Next class we'll learn about the prime spectrum of a ring and the Zariski topology.
 
@BalarkaSen you seem to be encyclopedia of Mathematics =P
 
@BalarkaSen Don't be a kid.
Oh, wait...
=D
 
@PedroTamaroff whats prime spectrum ?
 
8:59 PM
@PedroTamaroff Why the smug smile?
@Complexanalysis Yes, not a mathematician though.
Saying big words is not mathematics.
 
Smug? Meh.
 

« first day (1333 days earlier)      last day (3703 days later) »