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Zee
6:00 AM
It’s kinda mysterious and vague
 
Well, my impression on it is it makes a lot of pretty pictures, is very important in complex systems and also very hard to solve even numerically
 
Zee
One thing I noticed is the use of computers in the subject , do you have the impression that’s an essential tool ?
 
I guess computers are essential (especially more powerful ones). The ODE systems in dynamical systems are generally have no closed form solutions, with complicated boundary conditions (e.g. weather forcasts and climate models), so you pretty much have to churn them with computers
It might be possible for the theory part to be done by people, but the actual number crunching is way beyond of doing it manually
 
Zee
Mmmm I don’t like technology, so I suppose I wouldn’t like that subject
 
yeah, its quite applied for most of its branches
 
Zee
6:05 AM
I don’t mind applied , I just don’t like to use any tools except my brain , nothing wrong with using other things , but I always have this paranoia that I may potentially end up in prison or an isolated island
And I would like to continoue my work there
 
yeah, dynamical systems is definitely not something that can be done easily without computers because so many things are interacting at the same time except for the simplest systems
 
Zee
How old are you ?
 
26
 
Zee
For some reason I always thought you were much older...
 
well.. many think the opposite
 
Zee
6:14 AM
That’s alright , people of science tend to resist creativity
Science always fights new ideas
 
well its more like: Science creates a lot of ideas, then send them into the gladier pit and only few survives
(cf theoretical physics)
 
Zee
Yes but that process is continoues
So when new ideas come along they go into the arena , but they won’t have a chance since the other theories are more familiar to people and usually have more backing
 
well, that depends, some new ideas are so awesome sounding that a lot of people support them, others just fade easily if they cannot show they are competative
 
 
3 hours later…
8:59 AM
I wonder how we can model that...
 
9:47 AM
@Secret what's the difference between a weird and a non weird according to you :p ?
 
The core difference is whether they can withstood my super weird personality
since I am Patient Zero of The Weirdness
 
@Secret what is that?
 
It's a metaphor, you now how I am somewhat insane right, and people can only understood my ideas if they are somewhat in tune with my thinking and patient enough
 
4 hours ago, by Secret
You are lucky that you managed to find one of the achilles heels of our operations, so you win for now and we cannot do anything about it without destabilising this timeline, but let me warn you, your reign will not last and you will be erased just like most of the social problems by the end of the Promised Year
 
O, I like to wrap my frustrations into narrative format so that they look like folklore stuff
 
9:50 AM
Are you angry? And it seems like you are warning him that you'll be back?
 
That message is not really directed to Kasmir, for he is just one of the many nonweirds
I rarely have interest in individuals, and only the events and phenomenon they caused
 
@Secret tell me if I am a weird or a non weird :P ...
 
Well, you are part of the weird, otherwise you won't be so interested in what I think anyway
and also:
You don't annoy chat rooms with volley of boring questions like Kasimir does recently like a help vampire
Silent also felt like a help vampire at times, but at least he does not ask trivial questions that drags out the chat room that frequently
I do not have enough data to determine whether Silent is weird or not, however
Meanwhile, Tobias is definitely not weird because:
Feb 20 at 11:09, by Tobias Kildetoft
@Secret Or maybe I just ignore random questions that happen to be addressed to me
Weirdness generally decreases with age, thus I expect by the end of my PhD, all the people I met will be predominantly non weird
(o, and this age thing is a lot more complicated, it does not just involve the biological age, but also the underlying culture of the workplace or whatever the person is in)
In particular, the 40-50s are often the most boring group, where reality have crushed all their ability to dream and thus all that left is a hollow shell acting as a cog of the society
I have been taking measures throughout my life in order to not become like those groups
 
10:35 AM
@Secret I am in my late 20s and already feel so. But reading your comment gives me some encouragement. I would be very much interested to know what kind of measures. Speaking broadly, I believe such measures would, of course, be on a individual to individual basis.
 
10:56 AM
Let $f(x)=0$ if $x$ irrational and $f(x)=1/n$ if $x$ rational, $f(0)=1$. I have a proof that $\lim\limits{t\to x}f(t)=0$ for all $x$. But I can't 'feel' that, i mean, after all $f$ touches arbitrary large numbers, for integers right? so how come limit is $0$?
 
@zeal Try to keep an active social life of people from diverse backgrounds (outside your field of study), met more undergrads, go to conferences, art galleries, block off a section to go to gym every one a while, always think critically, do a lot of cross checking and above all, be curious and up to date with the world
 
@zeal Keep doing what you want to do. Read books, watch movies. Don't let academic pressure destroy creativity or imagination.
 
that also, Try your best to do things you like. I also like to go hands on in helping out instead of donating in order to take money completely out of the equation
 
I think you're making "imaginativeness" more grandiose a quality than it is
Certainly I think it's crucial to my lifestyle. But you're making a bigger deal out of it than it really is
Really the right suggestion here is to interact and expose yourself with artistic creations and not just scientific expositions. The purpose of art has always been to maximize imaginative freedom
 
Art also has a calming and relaxing effect for the mentally insane like me for example
btw, I do really hate spending money, thus e.g. my uni don't have good food thus I can choose to not eat lunch for a week in order to save the money for something more meaningful
 
11:09 AM
You need a fair amount of creativity in science too, but it's the scientific community that's really crushing for individuals involved. It straddles the boundary between "work" and "leisurely endeavors", where by work I mean something that puts you in a dominance hierarchy and you automatically get obsessed with climbing through that hierarchy (that's why publishing research in science is more of a big deal than the research itself)
all the arguments about publishing being a way to communalize scientific development notwithstanding: deep down one really thinks about abstract concepts like success and recognition when you publish stuff ...
 
yeah, and that is terribly unhealthy, the community itself is aware of that but progress on a reform is sluggish and funding agency are still looking for one dimensional metrics instead of the actual capabilities and passion of the research groups
 
Right.
 
open source and citizen science does help a bit in encouraging scientists to be more collaborative instead of competitive, and there are also industry programs that encourage cross collaborations between academia and industry, but I think until they find a more human way to do the metrics, I don't think that "publish or perish" will be gone very soon
 
Given that that's how human society has worked in pretty much any context in the last couple centuries, it's unlikely it has a very easy solution, yes
Pretty much everything's a big fucking race
Some will blame the wage system for it (eg, academia works in this way because funding mumble mumble). I think it's just a symptom of how we inherently think, not the reason.
 
11:24 AM
I sometimes wonder, am I always too naive to say that most of our social problems is due to power being spontaneously centralised to a few individuals, it does seemed this is even more complicated than I thought
 
It's a good observation but surely it's completely biological.
I "expect" we'd come up with a solution but I also "expect" a few couple hundred thousand years of evolution is needed for us to grow out of this in-built biological artifact :P
Unless we all die sooner
In which case fucking fantastic
 
(Huge block of Antihuman comments censored) Well, I felt like the coming summit between north korea and the US would be important, but like the critics, we have no idea what will happen on that day
 
I also expect that we'll live for a few more millennia :P
The worst scenario is not that the worst will happen sooner, but it's that it will happen much later, but soon enough that it happens before the good thing happens :P
 
12:27 PM
Let $f(x)=0$ if $x$ irrational and $f(x)=1/n$ if $x$ rational, $f(0)=1$. I have a proof that $\lim\limits{t\to x}f(t)=0$ for all $x$. But I can't 'feel' that, i mean, after all $f$ touches arbitrary large numbers, for integers right? so how come limit is $0$?
 
WTF?!
Silent: I think the image of $f$ is only values in [0,1]
 
@Silent What is $n$?
 
so nowhere is $f$ gone unbounded
 
@AlessandroCodenotti Yeah! sorry i missed to make it clear. each rational is expressed in lowest terms, ie $q=m/n$ and $\gcd(m,n)=1$
 
ok, so it can get unbounded
What does $t \to x$ mean, is $t$ tending to some specific $x$ or all $x$?
 
12:46 PM
What's your proof? I think that's not true for $t\to 0$
 
it's not clear if $t\to x$, $x$ is rational or irrational
 
@AlessandroCodenotti this is proof i have from Wisconsin Rudin solution.
 
1:00 PM
Wisconsin Rudin solution?
@BalarkaSen did you get my messages
 
i did
 
It might get a Not Good
We’ll see
 
@0celo7 these
 
1:18 PM
@Secret f*** that’s horrifying
I didn’t quite realize how bad it was until that first set got flung
 
@Semiclassical I saw that and then went skiing 5 mins later
 
1:33 PM
Hi
 
hi
Can anyone solve $$5x+9=y/20x$$
nvm
 
then $$100x^2+180x=y$$ it's a parabolic curve
@PintOfMilk
@PintOfMilk ok ?
 
2:26 PM
@AkivaWeinberger, will you please take a look at this?
 
Yep, sounds like a mistake in the source. — JonathanZ yesterday
Right, $f:\Bbb Q\to\Bbb Q$ given by the identity can't be extended to a map $F:\Bbb R\to\Bbb Q$
 
@AkivaWeinberger Why is that?
 
What would $F(\pi)$ be? If $x_n$ is a sequence of numbers approaching $\pi$, then $F(\pi)=\lim F(x_n)=\lim f(x_n)=\lim x_n=\pi$, but $\pi$ isn't in the codomain $\Bbb Q$
 
Thank you so much.
 
You need completeness of the range
 
2:40 PM
Hello everyone, could anyone explain why $0\le r< a$ is not needed in this answer about $x^a - 1$ divides $x^b - 1$ if and only if $a$ divides $b$
If you could help please ping me, and thanks!
 
Leaky Nun, my favourite Nun.
 
@Niing it's just a mess in the comments
 
@Niing It is needed
Well you could get the answer without using it, but to conclude quickly and painlessly, you can say $x^a-1 | x^r-1$ however $ r\lt a$ thus $\deg (x^r-1) \lt \deg (x^a-1)$ thus $x^r -1 =0$ thus $r=0$ thus $a|b$
 
3:01 PM
@Niing : you can explain that with the root
$r$ (a root of $x^a-1$) is a root of $x^b-1$ iff $a|b$
 
@BalarkaSen I ate at Shoney's yesterday
 
@Niing ok ?
 
What did you have, @0celo7
 
buffet
it was not very good
had a salad, chicken wings, steak, fried shrimp, catfish
literally heart attack on a plate
I don't see why Rick likes it
 
The only buffets I think that usually are good are eastern buffets.
 
3:15 PM
@LeakyNun : can I help you ?
 
"Boom roasted" actually sounds like a cool way to prepare a potato
 
@Dattier i'm fine
 
@LeakyNun : good
@AkivaWeinberger : can I help you ?
 
That precious Dattier.
 
I pay my debts
 
3:20 PM
Go and sin no more, precious Dattier.
 
thanks all, for your helps
 
3:32 PM
@BalarkaSen Can I tell you a cute theorem/puzzle (that you might have heard of before)?
 
Let $A$ be nonempty subset of group $G$, and $C_G(A)$ be centralizer of $A$. Please provide an example that shows that $x\in C_G(A)$ and $y\in C_G(A)$ does not imply that $xy=yx$.
 
@Silent hint: take $A = \{e\}$, the subset of $G$ containing only the identity of $G$.
 
@anakhronizein omg! thanks
 
Take $G=S_N$ for $N\ge5$ and $A=\{(1,2)\}$
Then $(3,4)$ and $(4,5)$ are both in the centralizer of $A$
Or: $G=A\times B$ for arbitrary groups $A$ and $B$
Centralizer of $A\times\{e\}$ contains $\{e\}\times B$, which might not be commutative
 
3:45 PM
@AkivaWeinberger to generalize it further, take a group where there is an example of that phenomenon, then there is an example of that phenomenon
 
Profound.
 
so in Set, the product is the left-adjoint of hom?
however in R-modules (commutative with unity), the tensor product is the left-adjoint of hom?
in R-algebras, the coproduct is the left-adjoint of hom?
 
@anakhronizein In fact that example solved 4 queries in last 5 minutes!
 
Presumably those words mean something @LeakyNun
 
3:49 PM
Often times the most trivial things ruin the most fun of math problems.
 
i see that now :)
 
@AkivaWeinberger ugh, why does nobody do category theory
 
@AkivaWeinberger Thanks for that!
 
On the contrary most people I know do category theory.
 
does that include you?
 
3:51 PM
TIL: While $v^\top w$ is called the inner product, $vw^\top$ is the "outer product"
Makes sense in retrospect but I never had reason to use it
 
@0celo7 In terms of the general definition of divergence on a manifold (i.e. that $(div X)\mu = L(X)\mu$ for a volume form $\mu$), can I get (assuming a metric $g$) a similar product rule to: $div(fX) = grad(f)\cdot X + f\cdot div(X)$ from multivariable calculus?
@LeakyNun I know a little, I guess.
 
@AkivaWeinberger and $v \wedge w$ is the "exterior" product"?
 
@LeakyNun also re your adjoint stuff: note that the tensor product coincides with the product in Set, and the tensor product coincides with the coproduct in R-alg.
 
@anakhronizein definition of "tensor product" being exactly the left-adjoint of hom?
 
Is it true that if $A\subset B$ and $A$ and $B$ are subgroups of $G$ then $A$ subgroup of $B$?
really?
 
3:57 PM
sorry, misread
 
It's true.
 
@Silent yes
 
Well in algebras it coincides with the usual definition (via universal property).
 
how do you define it then?
 
How do I define it?
It's probably something like "left-adjoint of Hom in closed categories" or some nonsense like that.
But personally, I always define the tensor product through the universal property. Analogous to modules and vector spaces.
Because I don't use a lot of category theory and that usually goes over my head.
 
4:04 PM
ok
 
But as far as I understand, in my limited understanding, those things are all just equivalent to the tensor product in the category you mentioned.
 
@anakhronizein Yes.
Do you want the proof?
 
Yes, I wouldn't mind seeing it, @0celo7
 
@anakhronizein Hmm, do you not know the coordinate expression for the divergence?
 
Not quite.
 
4:11 PM
 
@Dattier: I don't understand... a root?
 
Makes sense.
 
@anakhronizein from there it's a simple calculation
 
How is the identity formulated exactly (coordinate free)?
 
which identity
 
4:15 PM
the product rule thing for the div.
 
$div(fX)=X(f)+f\, div(X)$
 
Ah, of course.
 
there might be a coordinate-free proof for this, but why bother
 
Because coordinate free proofs make me smile.
Isn't it always said, a gentleman never uses bases or local coordinates? ;)
 
then figure out what $L_{fX}$ is
there might be a nice formula, I dunno off the top of my head
maybe use Cartan's formula
 
4:18 PM
$L(fX)\mu = f\,L(X)\mu + df\wedge\iota(X)\mu$
 
@Astyx: How did you get the answer without using, could you explain? thank you...
 
@anakhronizein yes, that follows from a coordinate computation
do you have an abstract one?
if so, then that proves what you want
 
I thought that was abstract?
 
well how did you prove it
 
I didn't, I just remembered it off the top of my head. :P
 
4:22 PM
well what you want should follow from it
 
So somehow $df\wedge\iota(X)\mu = X(f)\cdot\mu$.
 
multiply by $\mu$ on the right
 
Oh yeah.
 
idk how to prove that without coordinates or a basis. good luck.
 
Yeah, that seems tough.
Yeah it's obvious if you have a Levi-Civita connection.
 
4:36 PM
@Niing without using what ?
 
4:52 PM
@Astyx: Well you could get the answer without using it
 
Through recursion
 
Hello
 
if $q\ne 0$, $r\lt b$
And $x^a-1 | x^r-1$
It's basically the same thing
 
$\int_c^df^{-1}(y)\,dy+\int_a^bf(x)\,dx=bd-ac$
 
And the initialization is just for $r\le a$
 
4:55 PM
Am I allowed to write $x$ for $y$ or the other way around in that formula ?
 
It's basically combining recursion for proof of existence of euclidean division and what I said earlier @Niing (see above)
 
I feel like I shouldn't but I'm stuck at in a problem that really seems to suggest this..
 
You are allowed
 
To Astyx: Thank you... I think i need some time to understand your words, but thanks...
 
5:00 PM
Even if they are two different variables on different intervals ?
 
You're very welcome, don't hesitate to ask questions ! @Niing
 
this book is of Shanti Narayanan. Dass Publications.
 
@FuzzyPixelz You can't combine the two integrals
 
Do anyone know any better textbook than this with same content?
 
I see now
Thanks
 
5:01 PM
standard book with ISBN number.
I searched in famous textbooks of calculus, I couldn't find.
I could see only special cases. volume of revolution about x-axis and y-axis.
 
Problem: Let $f : E \to \Bbb{R}$ be a bounded measurable function, where $m(E) < \infty$. Assume $g : E \to \Bbb{R}$ is bounded and $f=g$ a.e. on $E$. Show that $\int_E f = \int_E g$. Proof: Since $f=g$ a.e. on $E$, there exists $A \subseteq E$ of measure zero such that $f(x) = g(x)$ for every $x \in E-A$. Then $$\int_E f = \int_{E-A} f + \int_A f = \int_{E-A} g + 0 = \int_{E-A} g + \int_A g = \int_E g$$.
 
@Astyx: So with additional conditional I've said, $0\le r < a$, it will be more slowly or quickly? Sorry I'm not very good at English... And I guess I know what you meant, you meant it's the same idea like Euclid's division that from (b,a) reduced to (a,r), and finally r will become zero, so I don't need to specifically add the condition?
 
How does that sound?
 
Yes, but it's better to directly state that $0 \le r\lt a$
By better, I mean faster and less tedious
 
To Astyx: Ohh, I misread it as my adding will make things complicated...
 
5:08 PM
Hello
 
@user193319 Works fine, I'd have done it by first proving that the integral of an almost everywhere zero function is 0, then simply stating $f-g$ is zero almost everywhere
@Niing So you understand now ?
 
@Astyx Ah. I see. Thank you!
 
I submitted an edit suggestion and it was rejected, so I ask here to check two things: 1. Whether my adding would make the proof wrong 2. Even it's correct, I didn't make things more complicated.
 
As long as you understand the answer it's fine. This answer wasn't a full answer anyway, just a guided hint, so it's not that necessary to state $0\le r\lt r$
 
I think I understand it now... I just don't want to memorize a wrong proof done by me, and if it were, I would consider delete my comments... Since somebody just said I made a mess on the comments.
 
5:15 PM
why the fuss about $0\le r < r$ when everyone knows what the proof is about
you've been on it for a bloody hour
no, I just said that the comment section is a mess
just get over it
 
To Astyx: Thanks for your helps... I'll remember you :) I'm going to sleep...
 
you're welcome
 
5:55 PM
If $C$ is a correlation matrix (i.e., a PSD matrix with $1$'s along the main diagonal) with $C = z_1 U + z_2 V$, where $z_i \in \Bbb{C}$ and $U$ and $V$ are unitaries, is it true that $z_1,z_2 \ge 0$ with $z_1 + z_2 = 1$? I can't seem to find a counterexample.
 
How to go about integrating while finding the centroid (read: centre of mass) of a sector?
Basically, I am trying to consider it made of elementary triangles.
Each triangle would have centroid at $r/3$ from base.
then, I'll have to integrate from 0 to angle of sector. - unable to do this part.
 
Is this a counterexample: $$\begin{bmatrix} 1 & i \\ -i & 1 \\ \end{bmatrix} = 1 \cdot \begin{bmatrix} 1 & 0 \\0 & 1 \\ \end{bmatrix} + i \cdot \begin{bmatrix} 0 & 1 \\ -1 & 0 \\ \end{bmatrix}?$$
 
Heya, I don't quite understand the definitions of hyperbolic trig functions in terms of trig functions. When I enter cos(sqrt(-1)) on my calculator, the answer isn't real, but when I enter cosh(1), the answer is real.
If cosh(x) = cos(ix) these should yield the same result.
 
@AkivaWeinberger Ping me with it and I'll check it out later
 
6:10 PM
@user10478 $\cos i = \dfrac{e^{i^2}+e^{-i^2}}{2} = \dfrac{1}{2}(e^{-1}+e^{1}) = \cosh 1 $
 
6:36 PM
Hi. Could someone please help me with my question? What I do not understand is why it is integrated. math.stackexchange.com/questions/2697043/…
 
7:21 PM
@Semiclassical yo
my advisor sent me some EM analogy
help
 
7:33 PM
Anyone here owned or had a hand at Dugundji?
 
the topology book?
 
Yes. It's no longer in print it seems
 
yeah
 
hmm there's a 3rd party vendor on amazon selling 1 new hardcover for over a grand
 
@Corellian that happens
it's the algorithms they use
the price will go higher if you keep clicking on it, too
 
7:46 PM
@0celo7 oh no :o
think I'll settle with Hocking and Young if I decide to buy a gen topology text, followed by Massey for algebraic
 
@Corellian Munkres!
 
@0celo7 yeah.... would you recommend it over others?
 
it's really good
 
alright, thanks!
 
юHello
is there someone who can help with testing integrals for convergence
?
 
8:03 PM
Why is it called mathematical analysis
What're we even analyzing?
 
functions
 
@Corellian I would suggest trying this: pdmi.ras.ru/~olegviro/topoman/eng-book-nopfs.pdf
It's a bit more oriented towards self-study than most books.
Also it is freely available online
 
That's a good question actually, analysis seems to be a rather generic term
 
I've got the classic $f(x) = e^{\frac{-1}{x^2}}$ for $x \not = 0$ and $f(x) = 0 $ for $x = 0$.
Actually, wait
 
8:20 PM
I think "analysis" is supposed to be for like... analysis of functions of topological spaces?
 
"analysis" was actually term they used a long time before we had topologies.
And even before we formalized what functions were (so before Bourbaki, Cantor, and friends.)
I think it has something to do with the progression of mathematics at the time.
Initially Newton and Leibniz had created differential/integral calculus and a lot of people set out to prove things within this calculus.
 
Plus you can also "analyze" homomorphisms on groups
 
But a lot of notions broke down because we were making too many geometric arguments.
And so analysis was born, where we rigorously set out the foundations for calculus and its offspring
So the "analysis" part I am guessing is that it was more of an "after thought" to bring rigour to a subject which was seemingly devoid of it.
 
actually yeah, how do you show that it is infinitely differentiable at x = 0?
 
By recursion?
Show that it is 1) continuous and 2) if it is $C^n$, it is also $C^{n+1}$ at $0$
Although... not sure the $n$th derivative has a pleasant form
 
8:31 PM
hm
 
8:44 PM
I think it would have a nice form.
 
I did this problem in analysis once
It's not good
 
8:58 PM
Last night dream:
On a blackboard some maths were written showing the left distributive law using nablas, I then started to wonder the geometric interpretation of it. Later my friend joined the discussion. I then asked him about why we have
absorbers. He suspects that they arises in order to balance the inverse elements in a structure. Later on, the nablas represent arbitrary operators and I suspect the geometric meaning of the distributive law might be that the transformation is homogeneous wrt its arguments
Reality check:
9
Q: Distributive Law and how it works

John SmithQuick question, which seems to irritate some people as it apparently strikes them as a nonsensical question: Why does the distributive law only work in one direction (in $(ℝ,+,*)$)? Why does it work this way: $$ a * (b + c) = (a * b) + (a * c) $$ But not this way: $$ a + (b * c) = (a + b) * (a...

Consider: $Aa+Ab=A(a+b)$ where A is some arbitrary operator
Then the distributive law is basically saying that the overall transformation can be decomposed into a sum where the transformation applies to each term of it
So whenever it fails to hold, it means the operator is nonlinear wrt its arguments and you can resolve the image in terms of the sum of images
For example, in polynomial maps P(x), you end up with cross terms
 
9:23 PM
@BalarkaSen Consider $(-1,1)^n$ in $\Bbb R^n$. It's convex, centrally symmetric, and it contains no lattice points other than the origin. It turns out that's the largest such set. In other words:
Any convex set in $\Bbb R^n$ which is symmetric with respect to the origin and with volume greater than $2^n$ contains a non-zero lattice point.
So try to prove it, assuming you haven't heard of the proof before @BalarkaSen
 
58 mins ago, by Slereah
By recursion?
what a beautiful use of curry-howard
 
Don't forget Lambek.
 
(curry-howard correspondence is the theorem that proofs correspond to programs, and under this correspondence, induction (proof) corresponds to recursion (programs))
 
It's one of the cardinal sins.
 
more accurately, functionally programming
 
9:29 PM
I use this to handle induction also. It makes it more hands on
 
proofs are geodesics in math space, and induction proofs are closed loop geodesics
 
Why closed. I thought they go to infinity?
 
they loop infinitely many times
 
And why must proofs be a stationary or of some functional?
 
@Daminark Like, I mention analysis and lots of people assume that's some kind of statistics course
 
9:30 PM
@Secret it's a mathoverflow meme
 
"Let $f$ be a continuous real-valued function on $0<x<\pi$ such that $f(0)=f(\pi) = 0$ and $f' \in L^2(0,\pi)$. Prove that $\int_0^{\pi} (f(x))^2 dx \leq \int_0^{\pi} (f'(x))^2 dx$ "
 
poincare ineq.
 
I see. I will instead think that a shortest proof of a given proposition will be worth calling a geodesic in proof space (assuming we can actually find a metric for such space)
 
@Secret geodesics don't have to be the shortest
 
@0celo7 is this related to HoTT?
 
9:32 PM
@LeakyNun no, it's a meme
 
ok
 
hasn't larky trained you better
 
Yeah but there really isn't a unique longest proof
 
@0celo7 I don't need anything that fancy?
 
@Lozansky there's a really easy proof of the poincare inequality known to the Italians
 
9:34 PM
I sometimes have this weird thought on what happens when Lozansky asked a question about lorentzian geometry
His username often give me an impression that there is a lorentzian map that is rotating
 
@0celo7 Oh marone
I think there is some neat trick
 
there is
I've only seen it in italian books tho
it's just the fund thm of calculus
 
Like doing odd expansion over $(-\pi, \pi)$ and then Fourier it
 
oh god no
 
:>
 
9:39 PM
though that might work, dunno
 
Maybe something something Parseval's formula
 
emulate this and tell me what you get i.gyazo.com/1aeea34dc71cac006ee998b3d4260d4f.png
 
I could write $\int_0^{\pi} (f(x))^2 dx = \dfrac{1}{2} \sum_{n=1}^{\infty} a_n^2$ where $a_n$ are the sine coefficients
@0celo7 what is $W_{0}^{1,p}$??
 
doesn't matter
the statement isn't what you want anyway
they have a diameter thing there, you don't
(I'm not quite convinced what you wrote is true without the diameter, but ok)
 
You mean the sum thing?
 
9:47 PM
I'm pretty sure you need a diameter^2 on the RHS of your inequality.
 
If $Y$ is defined by $\beta = \iota(Y)d\beta$ and $\beta_x = 0$, then does $Y$ have non-zero div at $x$?
 
@Lozansky Oh, wait, have you done Rayleigh quotients
 
@0celo7 Very little, but not in this course
 
And you're sure there's no $\pi^2$ on that thing
 
Mhm
 
9:52 PM
Ok I get it with a 1 if I use the Rayleigh method.
Maybe something with Fourier series works, idk.
@Lozansky Another method is argue that it's true with $\pi^2$ using the link I gave you, then show it holds for 1 by giving an example of equality.
And then use $1\le\pi^2$.
(the case of equality is attained by sin(x))
ignore that one.
 
10:42 PM
What does collinear mean in term of geometry?
 
@quallenjäger linear but the arrows are the other way around.
 
Does it refer to point or lines?
 
points are collinear if they lie on a line
 
Do you have an example?
 
the vertices of a triangle aint collinear
 
10:47 PM
Can it also refer to lines?
I am reading this paper. The definition 6.1 on Page 37 says something of collinearity between lines.
I don't really understand what does his mean by that. Are two lines collinear if their concatenation is again a line?
 

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