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8:01 PM
Amazing
And yeah nevermind I've reconvinced myself of your argument
So wait Balarka you want to go through the proof of the five lemma?
 
Sure!
 
The worst lemma
 
???
It's so useful
 
@EricSilva :(
 
Yeah but the proof is so boring
 
8:10 PM
Aight so we've got this
 
ah. well i only did it once in my life and forgot about it
the proof of snake lemma is better
 
We had to work it out for an alg top pset and it's just so dull
 
(I can barely TeX a one-line exact sequence so this is gonna take 3 years if I don't just screenshot it)
 
@Daminark can I suggest we just assume l, m, p, q are all iso's, and try to prove n is an iso?
 
8:12 PM
@Eric perhaps but it's apparently really useful, and everyone's gotta work it through at least once
 
It's definitely useful
The proof is like the basic exercise in diagram chasing
 
@AkivaWeinberger It should be $1$, that is if we let $u = t + \pi$, then $\frac{du}{dt} = 1$, which implies that $du = dt$.
 
Oh yeah we can do that for what I need it for. And I'm willing to take it on faith that it extends if necessary
 
@EricSilva It can be.
 
Is that an Ericism or is "diagram chasing" a thing?
 
8:13 PM
its a thing
 
Diagram chasing is a thing
 
@Daminark It's a thing.
 
It's a thing
 
lol
 
fuck fuck fuck
 
8:13 PM
sniped :(
 
maximally sniped :(
 
Oh Narcissus got more sniped than me
get rekt
 
I was watching USA network's television show "damnation" and they used the phrase "it's a thing", which was definitely not around in the 30s
 
There was a filtration on grammar though.
 
8:13 PM
@Narcissus even then you got sniped hard
 
@Daminark I... I have a full stop... So...
 
:thonk:
 
Insecurity intensifies
 
I also had to move my mouse to click the reply button!
 
Diagrams kind of make my head spin
 
8:14 PM
In a sense... I had to chase an arrow...
 
I'm trying to prove a function is identicaly zero in a region and following the hint I proved that the product $f(z)f(iz)f(-z) f(-iz)$ is identically zero. whta next?
 
@Eric they do for me as well, but I can understand them marginally better than I can pictures, and I need to have at least one thing I'm competent at so... yeah, I'm focusing there for now
 
@Daminark But yeah, if you want to see diagram chasing. Prove the three lemma.
 
also useful
 
8:17 PM
(for the record I don't dislike diagrams I just think diagram chasing is really boring/not very insightful)
 
Ah yeah, the proof on wikipedia does look like it's just bashing things out. Maybe I'll take it as given for now and afterwards I'll return to the proof
 
You don't even need snake lemma to find the boundary map in the homology LES
for the various topological versions of homology
It can be defined completely topologically
@Daminark It's a good exercise. You can also look in Hatcher, chapter 2 section 2?
Actually section 1
 
observation: the usual topological proof that complex polynomials have roots uses that the degree of $f/|f|$ on some large radius circle is one (by explicit calculation). the real version of this is the intermediate value theorem; it says that if the degree of $f/|f|: S^0 \to S^0$ is one, $f$ has a zero
 
Cute
 
this also clearly extends to a proof of a quaternionic fta for S^3 which is nice
 
8:20 PM
Does it mean anything meaningful in the other two division algebras?
oh come on
 
hahaha
 
Niiice
 
I guess it probably does it for the octonions too but I am a little paranoid
 
Very cool though
 
8:21 PM
@MikeMiller $n$, you mean
 
Wait there's a quaternionic FTA?
 
this is what "polynomial" should be taken to mean here
 
Fuck me
 
@AkivaWeinberger ah yeah
I was thinking mod 2 because of the real case
 
@MikeMiller This I don't believe
 
8:22 PM
and also forgot all the even things
@AkivaWeinberger where polynomial takes the specific form in the ams article I linked
:D
 
So Eilenberg wasn't the bad guy after all
 
That's pretty cool
 
@MikeMiller I'm reminded of the book "The Phantom Toolbooth." It includes the Island of Conclusions; you get there by jumping.
2
 
So just to be completely sure, the other LES that we use in order to apply the 5-lemma just comes from the homotopy fibration sequence $F \to Fp \to B$, right?
 
@Daminark Yeah
 
8:25 PM
And one of the arrows is an iso because $F_p \cong E$, I'm happy then
 
Yeppers
 
@MikeMiller So only one term of maximum degree
I see
 
We have that $$\int_{\Sigma}2\sqrt{1-x^2-y^2}dA=\iint_D2dxdy$$ where $\Sigma (x,y)=(x,y,-\sqrt{1-x^2-y^2})$. What will $D$ be?

Will it be the set of all $(x,y)$ such that the square root is defined? It must be $1-x^2-y^2\geq 0 \Rightarrow y^2\leq 1-x^2 \Rightarrow -\sqrt{1-x^2}\leq y\leq \sqrt{1-x^2}$. So that the square root $\sqrt{1-x^2}$ is defined, it must be $1-x^2\geq 0 \Rightarrow x^2\leq 1\Rightarrow -1\leq x\leq 1$.
So, we get that $D=\{(x,y)\mid -1\leq x\leq 1, -\sqrt{1-x^2}\leq y\leq \sqrt{1-x^2}\}$.
 
Otherwise you get counterexamples
(Things like $xix+jxkx$ are not of that form for example)
 
Yeah, I used to know this subtlety but I forgot ($ix + xi = 0$)
Thanks for the pointer
 
8:27 PM
Quaternions are fuckin dope mannnnn
 
it seems like this also applies to octonions because they're a normed algebra, yes? the same argument applies; all the parentheses one is paranoid about disappear after norming
 
what are the polynomials there though
they're not associative right? so you need paranthesis
and now i understand your comment
 
lol
yes
thanks for catching my errors this morning! i would have missed all those
 
sigh
 
8:29 PM
octonions are weird, but sedonions (or w/e the spelling is) are just goofy
 
@Balarka yo check this out: tikzcd.yichuanshen.de
 
Oh
nice
 
@MikeMiller Not ${}=1$, sorry
 
looks like a good spectral sequence drawer
 
${}=j$ would work
'Cause otherwise $1/(2i)$ would do it
 
8:30 PM
@AkivaWeinberger I was just trying to write down a polynomial which had an unexpected number of solutions
 
I suspect this is one of those places where knowing K-theory would make me appreciate it all better
but eh, I can't be arsed
 
@Mike do you know a good reference for rep theory of compact groups
 
Once one has that, I think it's easy to believe that you can get one with none
Brocker tom Dieck
What do you want to know
 
@MikeMiller Oh. $x^2=-1$ has an unexpected number of solutions (continuum infinitely many)
 
@AkivaWeinberger :D
 
8:33 PM
looks like it has what i need, thanks Mike
 
@AkivaWeinberger Hey, dude! So, what did you want to tell me with that last message?
 
@AkivaWeinberger I have a dope problem for you
 
Everyone here is talking to this poor guy.
 
lol
 
cause he da man
 
8:35 PM
@nbro Right so what's $dv$ then
(My battery's about to die)
 
@BalarkaSen do you remember a simple proof that hypersurfaces in $\Bbb R^n$ are orientable? My students want to do it using Jordan separation, which is quite nice, but I feel like there should be something simpler
I guess maybe I'm secretly just going to reproduce a proof of Jordan-Brouwer
Oh duh, you check that the normal bundle is trivial otherwise you would have nontrivial intersection with some loop
 
@MikeMiller What I have in mind is you assume it's not and then get a path on the hypersurface with a nontrivial section above it
 
Yes
 
That doesn't hit the surface
and take the bit of the normal that hits it
those are htpic
 
Yup, I think that's my proof
 
8:37 PM
but different mod 2 intersection
Ya
 
cool.
 
@AkivaWeinberger You defined $v = \frac{N}{2\pi} u$, so $\frac{dv}{du} = \frac{N}{2\pi}$
 
There's also a concrete proof somewhere by cutting the hypersurface off by a function on R^n
I don't remember how to do it
 
Yes but how does one get a defining function
 
It's Georges Elencewajg's favorite proof. IIRC it's a cocycle idea; locally cut it, then patch those up. Hm
 
8:39 PM
Oh, I see
 
too much work
 
True
 
You want something which simplifies that $2\pi$
 
It comes out in the change of variables
@BalarkaSen What is it
 
I am getting a lot of headshots today, which I am sad about
@AkivaWeinberger Prove the Riemann hypothesis
 
8:41 PM
Can't, it's false
 
GRH would be better
 
@AkivaWeinberger Well, but note that we would have $2 \pi$ at the denominator, which is then multiplied by the already existing $2\pi$ at the denominator, using your change of variables...
 
Apparently GRH is a very convenient assumption to make for various algorithmic problems
 
I'm kidding, it's that given a partition of R^2 by a bunch of straightlines, 2-color it (color it by two colors so that no two adjacent regions have same color)
@MikeMiller I agree
 
(removed)
 
8:42 PM
@nbro You sure? Make sure you don't have $dv$ and $dt$ confused in the integrals
@BalarkaSen Isn't that obvious though
 
No sure, also because I've not slept enough since months...
 
Finitely many, lines, right?
 
@AkivaWeinberger It took some thonking on my end to get it, but I can believe it's obvious to you
yea
 
Choose a side for each line
(a half-plane)
For each point, count how many of those half-planes it's in
Odds get black, evens get white
 
Hah. Quite nice.
My approach was different
perturb the whole thing so that every node (where lines intersect) is a double point
now make each crossing "X" into ") ("
 
8:45 PM
there are finitely many lines?
 
then you get a bunch of disconnected balls on the plane
color those black
and bring the whole thing back
and unperturb to make the regions formed by making n-order points into double points vanish
@MikeMiller Yeah
 
tfw you try to use tex commands in your humanities essay
 
\begin{proof}
2
 
Anyone has experience in Test Statistics, involving testing claims about population mean?
 
@AkivaWeinberger I had an inductive proof which is similar to yours. Assume by induction true for n lines. For n+1 lines, color the thing formed by n of them. Then reverse the colors on one side of the line
 
8:51 PM
That seems like his proof written recursively
It constructs the same coloring, right?
 
Yup
I guess it's an interesting question to ask how many ways there are to color it?
 
hi, if A,B is 3x3 matrix such that A is invertible, then is A + nB invertible for some n?I assumed the answer to be no, then used multilinear property of determinant and analysed the value for n-1 , n n+1 and arrive at a contradiction. But this doesn't seem to me to be a elegant solution (also I haven't verifyed if my method is right), so can someone answer this
 
lol
 
Math Facts is the best Twitter
 
8:52 PM
no Vsauce
no ketchup, just Vsauce
But what is Ting? cue Vsauce theme music
 
the guy from vsauce must have great thighs
 
@AkivaWeinberger You're right (as I guess most of the times)!
 
@Daminark "Category theory is the study of translating statements that are easy to prove into statements that are impossible to understand."
 
Right? It's amazing
 
My brain is working at turtle speeds
Sleep!! :D
 
8:56 PM
"Number theory is a subject that seeks to come up with as many proofs as possible for the fact that there are infinitely many primes"
 
"A K3 surface is a certain surface named after mathematicians Kummer, Kähler and Kodaira, which was thankfully not called a KKK surface."
 
ahahaha
 
@ManishKumarSingh well, it's certainly invertible when n=0
 
for non trivial case
 
8:58 PM
not enough information I presume
if you compute det(A-nB), you should get a quadratic polynomial in n
then A-nB is invertible iff the polynomial does not evaluate to 0
well, the polynomial can turn out to be linear or zero or a constant depending on A and B
I suppose the fact that p(0) is not equal to 0 tells us that the polynomial cannot be the zero polynomial
so the polynomial can have at most two roots
so there are at most two values of n which makes A+nB not invertible
@ManishKumarSingh does this answer your question?
 
are you considering the fact that its 3x3 matrix
 
yes
this is what makes the polynomial quadratic
 
give me a minute
 
wait, it can be cubic
so my conclusion becomes there are at most three values of n which makes A+nB not invertible
 
acutally we should get cubic
ignore my comment
 
9:03 PM
for the right value of B we can get quadratic or linear or constant
 
@BalarkaSen Yeah my thing is that thing unspooled I guess
 
you still didn't provide enough information
 
Your other one is more visually appealing than them though
 
Danke
 
@LeakyNun I see your point,
 
9:04 PM
It's like letting the ink run once you put it on a region and it flows along the crossings
kinda reminded me of making siefert surfaces
 
@LeakyNun I think it matches up with my solution that is to consider n-1, n , n+1 as solution and arrive at contradiction, could you confirm it
see my solution as my first comment
 
@ManishKumarSingh could you clarify your question
 
So i posted the question and arrived at an answer in a non-elegant way (1st comment), so i wanted to cross check my answer and also find a simpler answer, so could you recheck if the logic behing your and mine answers are same?
I think it is if I understood your solution
 
no, not clarify what you want
 
properly
 
9:07 PM
clarify your question
15 mins ago, by ManishKumar Singh
hi, if A,B is 3x3 matrix such that A is invertible, then is A + nB invertible for some n?I assumed the answer to be no, then used multilinear property of determinant and analysed the value for n-1 , n n+1 and arrive at a contradiction. But this doesn't seem to me to be a elegant solution (also I haven't verifyed if my method is right), so can someone answer this
there is one variable that you did not define
this makes the question unclear
 
which variable you are talking of?
B?
 
B any 3x3 matrix, need not be invertibel
 
if you read through my answer, you would have understood that A + nB is indeed invertible for all but finitely many n, so the answer is "yes", which nullifies your entire solution
 
the assumption of answer being 'no' was only made to arrive at a contradiction (which was det(A) = 0).
I think the n-1 , n,n +1 gave contradiction because the equation was quadritic?
 
9:11 PM
as I said, it is cubic
 
assuming B in not invertibel (if invertible then its easy)
 
why is it easy if B is invertible?
 
i can choose basis such that Ax not(=) -nBx,
x is basis
make this happen by manipulating n
 
I don't think "basis" is the right word
could you clarify what you mean?
 
@Alessandro I keep listening to the opening track of Departure Songs because it's so good.
 
9:15 PM
@BalarkaSen I relistened to the whole album while studying this weekend, it's great
 
using the fact that matrix and linear tranf. are essentially same, I choose certain basis
and say kernel is non-zero
 
@AlessandroCodenotti I loved the third or fourth track where they played over a passage narrated by Burroughs
I agree the the album is great in it's entirety
 
@ManishKumarSingh neither the kernel nor the value of Ax depends on your choice of basis, so I still do not understand what you are trying to say
 
Is the hyperboloid arbitrary close to a cone at infinity?
 
do you agree A + nB is a linear tranf. under suitable basis for vsp F^3
 
9:18 PM
@ManishKumarSingh it is a linear transformation regardless of basis.
 
@quallenjäger In a meaningful topology on the "space of hypersurfaces in R^3", it should be, yeah
 
yes but to write it in a form of a matrix, we need to specify the basis,
 
@quallenjäger yes, according to wikipedia: "Both of these surfaces are asymptotic to the cone"
@ManishKumarSingh but you do not need the matrix to find its determinant
determinant is still invariant under change of basis
 
How is the curvature of the cone? Are they negative?
 
No, it's 0.
 
9:20 PM
Under minkowski metric
 
Oh, I don't know.
 
You mean under standard Euclidean metric the curvature should be 0?
 
Yes.
It's locally isometric to the plane.
 
@quallenjäger i.e. the Gaussian curvature
 
9:22 PM
@LeakyNun yes, but i if we assume basis, all we have to show that Ax (not)= -Bx for that basis, and this would as you said will be true for all basis
 
@ManishKumarSingh then you are not choosing the basis; you are choosing three linearly independent vectors x1, x2, x3, so that Axi != -nBxi for each i=1,2,3, in order for {x1,x2,x3} to be a basis
much confusion I see between "basis" and its constituent vectors
 
Zero curvature means if I parallel transport a vector along a curve, the direction won’t change?
 
yes i am doing that
 
how do you know such three vectors exists?
 
@quallenjäger Well, no, it's locally isometric to the plane, not globally.
 
9:24 PM
because if it doens't, I change n
atmost i need to do this 3 times
 
how do you know?
 
Indeed, if you parallel transport along a path going around the cone you'll end up with a different direction
 
3 times?
 
And along the ruling?
 
@ManishKumarSingh yes
 
9:25 PM
Nothing happens along the ruling.
 
I am just bit confused, around the cone I have some kind of curvature
How can the curvature be 0?
 
assume the worst, say Ax_1 = -nBx_1, so increase n to n+1, now they are different, and repeat this proceducre 3 times
 
@ManishKumarSingh fair enough
@quallenjäger the Guassian curvature is defined as the product of the two principal curvatures, each of which has a meaning
I'm not sure if the product itself has any meaning
other than the fact that if the curvature is 0 then it is locally isometric to a plane, positive then to a sphere, negative then to a hyperboloid
 
I see, thanks
 
@ManishKumarSingh I think you can use your argument even when B is not invertible
to make it more rigorous:
let e1=(1,0,0), e2=(0,1,0), e3=(0,0,1)
 
9:33 PM
i shall try
 
since A is invertible, Ae1, Ae2, and Ae3 are all non-zero
 
as a result, Aei = -nBei can only have at most one solution
wait
even if Ae1+nBe1, Ae2+nBe2, Ae3+nBe3 are all non-zero, how does it follow that A+nB is invertible?
 
injective map
so kernel goes to 0
 
how do you know it is injective?
 
9:36 PM
Ae1+nBe1, Ae2+nBe2, Ae3+nBe3 are non zero any linear combo will be non zero
and that is your typical element
 
it does not follow
 
wait a minute
yes, there is a problem, we need special basis vector, such that first few goes to kernel, rest all go to range - 0
aahh it would be complicated
no actually we can simple assume tha basis to satify such a property
 
how can you?
 
because such a basis exsit
 
how do you know that?
 
9:40 PM
from rank nullity thm
 
how do you know that the nullity of A+nB is 0?
 
i dont, so I assume say e1 goes to kernel, rest go to range - 0
 
what if e1+e2 goes to kernel instead?
 
not possible because of choice of basis
but there is still another problem
such special basis keep changing as we change n
 
exactly
 
9:42 PM
so it won't worl
ya maybe quadrtic was the best way to solve
anyways thanks
 
The bounty is not taken yet, there is still a great opportunity to get some nice points in your pocket.
38
Q: The closed form of $\int_0^{\pi/4}\frac{\log(1-x) \tan^2(x)}{1-x\tan^2(x)} \ dx$

WaitingWhat tools, ways would you propose for getting the closed form of this integral? $$\int_0^{\pi/4}\frac{\log(1-x) \tan^2(x)}{1-x\tan^2(x)} \ dx$$ EDIT: It took a while since I made this post. I'll give a little bounty for the solver of the problem, 500 points bounty. Supplementary question: Ca...

 
@LeakyNun what program are you doing at imperial?
 
@quallenjäger pure mathematics, 3 years BSc
 
You are finishing this year?
I am at Imperial too
 
World is so small.
 
9:56 PM
no, just started
 
Lol u know really a lot for a first year Bsc
 
ok..?
 
Hey @LeakyNun
Did you see my question?
 
@Evinda ?
 
0
Q: Abelian monoids and integer polynomial composition?

mickConsider distinct integer polynomials of distinct degree. Also they are all univariate polynomials of the variable $x$. Let $*$ denote composition and $*$ is the operation for the monoid. Consider when such polynomials $a,b,c,d$ forms an abelian monoid. Many questions arise naturally. For insta...

Any ideas
?
 
10:01 PM
@LeakyNun Let $P(n)$ denote the largest prime factor of $n$.
Can we deduce something from the relations:


$\frac{|\{ p \leq x \mid \text{ p is prime and } P(p-1)\geq x^{\frac{2}{3}}\}|}{\frac{x}{\ln{x}}}\geq c_0$


and

$\frac{|\{ p \leq x \mid \text{p is prime}\}|}{\frac{x}{\ln{x}}}> (1-\epsilon)$


about how many prime numbers $p$ up tp $x$ will have the property that $p-1$ has a prime factor $q$ that exceeds $x^{\frac{2}{3}}$ ?
 
I know nothing about this
 
Ok no problem :) @LeakyNun
 
Evinda I think your problem is to hard
 
Let $f : X \to \Bbb{R}$ be some function, where $X$ is a measurable space. I have that $f^{-1}([q,\infty))$ is measurable for every rational $q$. I need to conclude that $f^{-1}([c,\infty))$ is measurable for all $c \in \Bbb{R}$. Let $(q_n)$ be a decreasing rational sequence converging to $c$. Is $[c, \infty) = \bigcup_{n=1}^\infty [q_n , \infty)$ or $[c, \infty) = \bigcap_{n=1}^\infty [q_n , \infty)$?
This is giving me so much damn trouble for some reason...
 
10:34 PM
@user193319 neither. $\displaystyle \bigcap_{n=1}^\infty [q_n , \infty) = [q_1, \infty)$ and $\displaystyle \bigcup_{n=1}^\infty [q_n , \infty) = (c, \infty)$
 
$e^{x+y} = e^x e^y$ and $e^{-x+y} = e^{(-x)+y} = e^{-x}e^y = \frac{e^y}{e^{x}}$
kkk
 
I want to use the formula of Stokes at an example where the normal vector of surface direct to the z-axis.

When we calculate the curve integral at the boundary do we consider a clockwise or an anticlockwise parametrization?
Hello @LeakyNun !! Do you have an idea?
 
no idea, sorry
 
tell me why! :D
 
10:50 PM
@nbro why what?
 
eh, just joking a bit :D
btw, it is the title of a song
check that out
lol
btw, I need to finish a g.d. proof I started two days and apparently is so simple, but given that I've not slept decently for months, my brain is f***** up
It is like basic calculus stuff
lol
So, here's the exercise I had to solve.
And this is what I have so far
Any help on how to finish this proof?
I am so tired...
 
I still need to derive $d_n$.
But it looks like I am close.
@Antonios-AlexandrosRobotis Hi
Is Antonios-Alexandros your real name?
 
It looks like a Latin-derived name combined with a Greek name
 

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