« first day (2093 days earlier)      last day (2945 days later) » 

12:01 AM
Have to stay focused. Need some excuses.
 
See you all!
 
hi
A convex, closed figure lies inside a given circle. The figure is seen from every point of the circumference at a right angle (that is, the two rays drawn from the point and supporting the convex figure are perpendicular).
is the diameter the only possible such figure?
 
12:23 AM
@quid echo ;-)
 
bah, trivial
 
 
1 hour later…
1:46 AM
hihihi
 
vzn
2:29 AM
anyone around working on math/ research prjs?
 
@MikeMiller Do you know where I can find the proof of "a fibred genus 1 knot is either the trefoil or the figure 8"?
 
2:51 AM
@PedroTamaroff mr Pedro: It's about damn time :P
hi @PVAL
 
@Ted Hi
 
Hope you're doing well.
 
3:05 AM
Good evening. Silly question: If $R$ is a ring, what is $R/(0)$? I am thinking that it is just $R$.
 
@PVAL I don't. I think I can write out a proof using Gordon-Lueck but this should be much more elementary than that.
 
@MikeMiller Whats the idea to using Gordon-Luecke?
 
@TedShifrin hi
do you know a good link of understanding how bch decoding works ?
 
When I heard this statement it seemed like it should depend on Harer's conjecture (or at least easily follow from that).
 
@TedShifrin ?
 
3:16 AM
@PVAL By fibered genus 1 we get a knot presentation that's an extension of Z by Z * Z, the meridian is the generator of Z, the longitude is aba'b' in Z*Z. Now try to get this in a standard form.
 
ahh groups
 
I don't know Harer's conjecture
Idk as groups go these seem pretty easy.
 
@MikeMiller have you heard of John Oprea?
 
Harer's conjecture is that all fibered links come from plumbing hopf bands.
 
topologist
 
3:19 AM
It's essentially the Giroux correspondence for S^3.
 
Is that open?
 
Define open
 
I know he has a paper about fibered links, maybe the result you want is proved in there.
But for some reason I feel like this is a 60s theorem.
 
It's widely accepted as proven (Giroux Correspondence that is), but I don't think there's a publicly available full proof.
Apparently this was solved by Acuna-Short in 1970
but I can't find the paper.
Likely Gordon or one of his students knows how to do this.
There is apparently a full proof written up of GC which is waiting for Giroux's approval.
 
 
1 hour later…
4:45 AM
Hello
quite empty...
anyway, for the integral of the vector field $\langle yz,xz,xy\rangle$ over the surface $z=\cos^2x+\sin^2y$ in $[0,\pi]\times[0,\pi]$ I obtained the result $\frac{1}{4}\pi^4-\frac{1}{4}\pi^2-\frac{3}{16}\pi$
what do I call this? a cuartic in $\pi$? lol
 
5:20 AM
No, @Adeek, sorry.
 
5:36 AM
i proved something at least :(
better than nothing
 
5:56 AM
Brandolini's Law: "The amount of energy needed to refute bullshit is an order of magnitude bigger than to produce it."
3
That's what it's like to grade tests sometimes, having to decipher a bunch of nonsense.
 
6:50 AM
@AriNubarBoyacıoğlu Do you still need help?
 
@Danu DanielF translated later.
@Danu You should teach me how to phrase the standard model mathematically.
 
\o @Danu
 
7:28 AM
hey raider
 
7:43 AM
@MikeMiller I will know soon-ish
Hi @skill
For now, we're actually proving the closed subgroup theorem (which I guess is nice), though I wonder how much we actually need it in gauge theory
 
8:07 AM
Hi All, I wanted to get a copy of Freidberg et al "Linear Algebra" but it is so outrageously expensive
What other books of that level (final year undergrad or pre-grad) would you recommend?
 
@BenjaminR Paul Halmos's Finite dimensional vector spaces
 
@Mambo is the title also called "linear algebra" or something similar? (googles...)
Ah thanks
wow, that looks really ideal. Thanks!
(plus it's 1/5 of the price of Freidberg)
 
It's a very standard book. Notations can be difficult to follow in the first read.
 
That's probably true of all grad-level / reference math books
I'm about to graduate with a minor in mathematics, want some books I can keep for my life as a reference and build on what I already know.
 
Then definitely you should read this
 
8:16 AM
Brilliant, thanks Mambo.
 
@Agawa001 hi
Interesting history @ForeverMozart
 
9:11 AM
Can anyone tell me what this symbol is:
I saw it in some slides I was looking at: $A^{k-1} - \mu^{k} /$
The division looking symbol right at the end, what does it denote?
 
@Owatch Are you sure it is not just a capital I that has been italicized?
 
It could be yeah.
 
otherwise I have no clue what it could mean
 
It must be then.
So does scalar addition/subtraction take precedence over dot products?
$A\cdot B + c$
Wolfram interprets as: $A \cdot (B+c)$
 
@Owatch there is only one possible interpretation, depending on whether $c$ is a scalar or a vector
 
9:25 AM
c is a scalar.
..
 
then $B + c$ makes no sense, so you should slap WA for the interpretation
 
@Owatch what did you actually type into W|A?
 
{Sqrt[2], Sqrt[2]} . {1/Sqrt[2], 1/Sqrt[2]} - 1
 
it's interpreting -1 as -1 in each component, dunno why
 
9:45 AM
This may be silly, but I'm reading a slide which tells me that to compute $\mu$, I need the lowest-rightmost $2\times2$ sub-matrix of A, called B. It's defined as:
$B =
\begin{array}{cc}
a_{m-1} & b_{m-1} \\
b_{m-1} & a_{m} \\
\end{array}$
Where $\mu = a_{m} - \frac{sign(\delta)b_{m-1}^{2} }{ |\delta| + \sqrt{ \delta^{2} + b_{m-1}^{2}}}$
And: $\delta = \frac{a_{m-1} - a_{m}}{2}$
But this has to assume that both $b_{m-1}$ are the same in that sub-matrix.
Because I can't see how it specifies which one to place where. I guess that answered my own question.
 
10:31 AM
hi all
 
11:03 AM
@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ echo :D where have you been when we did the ping the future session?
 
@quid ping the future?
 
you write down an ID, and hope that a future message would be that ID and you'd make it ping
it didn't work out
 
@TobiasKildetoft its cool that whenever some1 would have the ability to ping his future, he assures he can still remain alive for next x years
 
11:22 AM
Can anyone assist me with this linear-algebra problem?
I have a matrix:
0 1
1 0
Called A. I wish to find it's eigenvalues. You can normally use QR decomposition to find it, but this is a known case where that does not work, and you end up with your original matrix after executing it.
To solve it, you are to use a modification of QR that involves shifts.
I am trying to use a Wilkinson-Shift as a change to my algorithm. This is described as follows:
$\mu^{k} = a_{m} - \frac{sign(\delta)b_{m-1}^{2}}{|\delta| + \sqrt{\delta^{2} + b_{m-1}^{2}}}$
Where $\delta = \frac{a_{m-1} - a_{m}}{2}$
 
@Owatch why not just find the eigenvalues directly?
 
@TobiasKildetoft see the message after this one It links "back" to the subsequent one.
 
What do you mean? Solve for this particular case?
 
@quid Ahh, I see
@Owatch Yes, it is just a $2\times 2$ matrix
 
Because I am trying to get a general-case solution.
And so this is just the simplest example I can find to work with.
Anyways, I have to tell you the algorithm still. It's quite simple:
for k = 1 ...
$\quad select\ \mu^{k}$
$\quad [Q^{k}, R^{k}] = qr( A^{k-1} - \mu^{k} \cdot I)$
$\quad A^{k} = R^{k}\cdot Q^{k} + \mu^{k} \cdot I$
On this matrix, you get $\mu$ to be +/- 1
I picked 1, and when I follow through I get:
-1 -2
-2 -1
The result for $\mu$ are the eigenvalues.
But I thought the matrix was supposed to yield them on the diagonal. Which it does not. So how would I know when to stop my algorithm. The second iteration is just another scalar sum with A.
If it helps, I have got the slides I found here.
 
11:40 AM
@Owatch How is that one matrix supposed to be a QR decomposition?
 
Well it's not, the QR decomposition is two matrices.
But the algorithm doesn't just require you get those. I should have specified algorithm not decomposition.
Slides 11/17 for algorithm, Wilkinson shift respectively.
The idea is that you apply this algorithm until you get an upper-right-triangular matrix in A, and then the eigenvalues will lie along the diagonal.
 
@Owatch Well, then the matrix you got there was not what you were supposed to end on
 
Well it claims that:
For the example that “broke” the Rayleigh shift is μW = ±1, and we converge in one step.
(That example is this matrix)
 
@Owatch the failure is not that the eigenvalues are not n the diagonal though. It is that the matrix is not upper triangular
 
11:56 AM
Do you mean that A: I did not convert it in the beginning to Hessenburg form?
Or B: I did not continue until I had upper-triangular?
 
Hello @robjohn
I am looking at the proof that $C_C^{\infty}(\mathbb{R}^n) \subset S(\mathbb{R}^n)$ and the mapping is continuous:

If $u \in C_C^{\infty}(\mathbb{R}^n)$ then $\sup{|x^a D^{\beta} u|}< +\infty$.

Remark: $f$ continuous then if $x \to x_0$ then $f(x)-f(x_0) \to 0$

$\sup |D^{a'} \phi_j| \to 0 , \forall a'$ then also $\sup |x^a D^{\beta} \phi_j| \to 0$.

Could you explain to me how we apply the remark?

Why from the last line do we deduce that the mapping is continuous?
 
@Owatch I have no idea why it failed, but given the description the result was clearly not as it should be, ragardless of the diagonal
 
Okay. Well. The slides tell you that $\mu$ is +/- 1. So I know that. Then all I need to do is find Q, R for ($A^{0} - mu^{1}\cdot I$). This is (According to Wolfram QR decomposition): Q = $(\frac{-1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$, R = $(\sqrt{2}, -\sqrt{2})$.
Then $A^{1} = Q^{1}\cdot{R}^{1} + \mu^{1} \cdot I$
So: $(-2) + (1) \cdot I$
Which gives me:
-1 -2
-2 -1
I guess I'll try again and look for mistakes...
 
@Owatch that makes no sense. $Q$ and $R$ are supposed to be $2\times 2$ matrices, not vectors.
 
I had wolfram give me the results.
 
12:06 PM
And the QR decomposition for a scalar multiple of the identity is just the matrix itself together with the identity
 
What is the origin of the word modulo n?
 
QR decomposition {{-1, 1}, {1, -1}}
Was my input
 
@Owatch first, that is not what the matrix was
second, the output still makes no sense
 
Can I post a Venn diagram?
 
Well I'm not Wolfram.
And that input is not just A
It's $A^{0} - \mu \cdot I$
 
12:08 PM
 
@Owatch No, it is not
 
What is the origin of the word modulo n?
 
QR of ( {{0, 1},{1,0}} - 1 * {{1,0},{0,1}}) ?
 
@Owatch you forgot the ^0
 
@Shahab Have you tried searching Gauss and the modulo word?
 
12:10 PM
Anyway, if you want to use WA for this you should probably learn what the output means since it clearly does not give you $Q$ and $R$
 
Everything is something.
Everyone is someone.
All ducks are birds.
 
Sigh.
 
But there is nothing outside everything. So how can everything be something?
 
@MatsGranvik: I tried but did not find anything
 
So ^k was actually a power.
Not supposed denoting mu for step k?
Okay.
 
12:12 PM
@Shahab "Johann Carl Friedrich Gauss is usually attributed with the invention/discovery of modular arithmetic"
quote
 
@Owatch I don't know. You are the one reading about the algorithm
 
yes, that I know. But from where has to word modulo come and what did it originally mean
*to->the
 
Really frustrating.
 
@quid lol
 
This is a list of Latin words with derivatives in English (and other modern languages). Ancient orthography did not distinguish between i and j or between u and v. Many modern works distinguish u from v but not i from j. In this article, both distinctions are shown as they are helpful when tracing the origin of English words. See also Latin spelling and pronunciation. == Nouns and adjectives == The citation form for nouns (the form normally shown in Latin dictionaries) is the Latin nominative singular, but that typically does not exhibit the root form from which English nouns are generally derived...
modulus
 
12:18 PM
Well thank you for helping Tobias. I will try again
 
Everything is something. But if everything is something then according to the Venn diagram there is something outside everything, which can't be true because everything is everything. So the only correct conclusion is that everything is nothing.
 
12:46 PM
@MatsGranvik could you give an example?
 
@skillpatrol "All ducks are birds."
Draw a Venn diagram of that.
 
ok
ducks are a subset of birds
 
@skillpatrol Yes
@skillpatrol What is everything a subset of?
 
everything is the universe
 
@skillpatrol Yes. But what is the universe a subset of?
 
12:49 PM
everything is the subset of the magic word 'money'
 
by definition you can not leave the universe
 
How do you know?
 
you got money you got evrything
2
 
the definition says so
 
12:50 PM
Another Poincare Duality occurrence img.ctrlv.in/img/16/04/27/5720b56ec8b65.png
 
@skillpatrol But then there is nothing outside everything?
Hence everything is nothing.
 
you're getting the word "nothing" confused with the idea of "no thing exists"
 
Nothing can not exist.
Yeah, well anyways I guess I agree.
 
sure it can, we label it 0 and call it "zero"
 
12:58 PM
"nothing" does not equal "no thing"
 
Ok, but there is nothing in between the words "no" and "thing".
 
they're just words :)
 
think of it this way @MatsGranvik

compare "zero slope" with "no slope"

"Zero slope" means the slope = 0.

"No slope" means the slope is undefined.

I think people naturally associate "nothing" = "no thing."

But in this context, "nothing" means the value of 0,

while "no thing" means no slope.
 
@TobiasKildetoft It works fine, I should not have used Wolfram. It was doing extra stuff I did not notice, like returning a value 2 for a matrix product when it was actually a 2x2 with zeroes save for one value 2. I should be more careful. :(
It was also returning vectors because it was discarding zeroed rows.
 
1:09 PM
I am in desperate need of help:
0
Q: Singular point of $f(z)$ also a singular point of $1/f(z)$ and $f^{2}(z)$

Jessy CatSuppose $z_{0} \in \mathbb{C}$ is an isolated singular point of the function $f$ of a given type (removable, pole of order $N$, essential). I need to show that $z_{0}$ is an isolated singular point of $g(z) = 1/f(z)$ (here, additionally, assume that $f(z)$ has no zeros in some deleted neighbor...

@robjohn, if you're around, and you'd like to take a stab at it, that would be most welcome ;) You always have amazing things to say.
2
 
he is amazing, no doubt
 
Very much so.
@TedShifrin is also pretty amazing.
13
 
yup
let's not forget @DanielFischer
2
 
And @DanielFischer
2
Ack! I was just typing that!
 
;)
let's see who gets the most stars?
morning @Semiclassical
 
1:23 PM
morning
 
@Semiclassical, heya
Yeah, I'm not terribly happy with the answer I got for that question.
 
DanielF, robjohn, Ted are all pretty great.
2
 
1:39 PM
I don't think the answer the guy posted is right.
Okay, it's right, but it's not what was asked. So that upvote was not from me. LOL
I hate it when people do that.
Really, everybody knows how smart you are. Don't show off by answering a "better" question, just answer what the OP asked!!
 
"if you'd prefer to answer a different question than was asked, ask it yourself."
 
@Semiclassical, something like that ;)
 
Hey, honestly I answered the question the way I read it.. No need for snarky comments
 
> OK, the mutual admiration can cease - Ted Shifrin, 10hrs ago
everybody be circlejerking on my starboard. (also lold at Ted's allusion)
 
1:49 PM
say what?
 
Sorry @s.harp, I'm having a bad day :( It was a pretty awesome soltuion just not what i aksed.
Sorry for the snark.
@anon, whoa. "circlejerking on my starboard"?? And they said this was a family site...
 
where did who say what now? :P
anyway, it was a popular ELU question
 
Maybe I ought to change my name from Jessy Cat to something more of the female canine variety...
@TedShifrin should get a Louis Vuitton hat and a big gold necklace that says "Swag" on it.
 
you a girl?
 
 
1:53 PM
@skillpatrol, yeah...Last time I checked.
 
twas a family friendly show
 
twas it though?
 
actually I don't remember.
 
you been here too long
 
I just remember there was that one girl with the dark hair who was always hanging around even though nobody liked her. That was me.
 
1:54 PM
@JessyCat i vote for Temmie, if only because of the following Youtube remix. (Warning: explicit and bizarre)
 
I think we can start discussing math now?
 
yes sir
 
Yeah...I"m gonna go back and try working on that problem again.
 
have fun
 
Seems like this conversation is more off-topic than a single "these guys are great" comment, to be honest, so...
 
1:55 PM
Again, I'm sorry I got all snarky :( @s.harp
 
@JessyCat don't worry about it
the individual cases are a bit of text so it will take some time to write
 
@JessyCat wait, cats are feline, not canine
 
mine is definitely off-topic, but i've wanted to share the absurd glory of that link ever since i saw it
 
@anon, I realize this ;) It was because I'm snarky.
Think about it.
 
when a girl gets snarky, guys call it "attitude"
 
2:00 PM
that assumes that said guys are relatively classy.
not a guarantee.
 
true dat
but even the bad boyz like attitude ;)
 
@skillpatrol, @JessyCat, your icons are way to similar
 
@AkivaWeinberger, absolutely nothing alike
Mine is the Cheshire Cat from Alice in Wonderland
 
I mean, when shrunk down to that size, all I can see is a black-and-white square
Maybe it's larger on the desktop site? But I'm on mobile right now
 
you have a point
 
2:13 PM
See, right there, I didn't even realize that someone else responded right then.
I thought it was Jessy again.
 
it's different distributions of black-and-white, but yeah
 
Now this is confusing: Skill Patrol and Skull Petrol.
 
How's that? @AkivaWeinberger
 
How about Ski Patrol?
 
2:16 PM
@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ How's what?
 
Skill patrol (not you) and Jessy's icons look similar when shrunk down to half a centimeter wide.
 
That's better but we'd prefer a username which isn't highly inflammable.
 
Inflammable isn't negative
 
welp
 
2:18 PM
It's when you can set it on infire
 
@Evinda can you show that if $\phi$ is supported on a compact set, $\left|x^\alpha D^\beta \phi\right|\le C_\alpha\left|D^\beta \phi\right|$?
 
@anon: didn't you not know?
 
welp
now akiva's comment makes it look like I deleted after said comment
 
@BalarkaSen I think he didn't not not know
(She? It? They?)
 
2:24 PM
Why do people say welp ?
 
If the word "irregardless" was first used in 1912…
…and the Titanic sunk in 1912…
does that mean that "irregardless" sunk the Titanic?
Or maybe I have the causation backwards.
The Titanic would have sunk irregardless.
 
you chat a lot differently when you're on mobile :P
 
someone kill me
 
@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ I'm almost always on mobile.
 
2:29 PM
Orly?
must be nice
 
@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ What about her?
 
2:44 PM
Haha, excellent. I wonder who discovered the planimeter.
Wonderful application of Green's theorem.
 
how about scale patron
 
Actually it's not quite clear to me how to prove that a planimeter works, but I can intuitively see it.
 
hm? what can't you prove?
 
One's trying to measure the "swirly" around the boundary of the region.
@MikeMiller I mean I haven't explicitly tried to prove that it works.
Trying it right now.
 
Oh, locally defined functions ←→ line bundles, months wasted because that simple analogy was not obvious, ffs
 

« first day (2093 days earlier)      last day (2945 days later) »