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4:00 AM
i put on some elevator music
 
@anon: ODEs, not PDEs :)
@Joshua: Any angle less than $30^\circ$ can't be right.
 
"What is an ODE but a particularly simple PDE?"
I should go.
 
@Joshua: $40.28^\circ$ is correct.
 
@TedShifrin It's based on a picture... I think I forgot to make the bottom 30$^{\circ}$ negative, though.
 
Wait. What?
 
4:12 AM
Wait. Who?
 
Mr @Pedro!
 
Hello, Theodore.
 
[The plot thickens]
 
Did you pass your topology exam, I presume?
 
Yes, yes.
I am reading some differential geometry from Spivak's book.
 
4:15 AM
With $-30^\circ$, the right angle is $23.29^\circ$, @Joshua.
 
@TedShifrin Yeah, that was the mistake I made. Thank you.
 
Not really differential topology in there, @Pedro, just differentiable manifolds. He doesn't talk about transversality and degree, etc.
 
Why are we computing things in degrees here?
 
Hush, mr @Pedro.
 
@TedShifrin I mean what I thought, not what I said.
 
4:16 AM
@PedroTamaroff because I have online homework and it's webassign. But I agree.
 
webassign oh no
 
You should take a gander at Guillemin & Pollack sometime, mr @Pedro. Some really cool stuff proved in there.
 
@TedShifrin Will do.
 
Or, if you're super sophisticated, look at Mo Hirsch's Differential Topology.
 
@TedShifrin Well, I am learning lots and lots of definitions.
Eventually I'll start doing stuff.
 
4:18 AM
You mean the definition of a smooth manifold and tangent space?
 
Mariano introduced today three ways to define orientability of a manifold.
 
OK ... have you learned about vector bundles yet?
 
Yes, paralellizability, tensors.
Lots of definitions.
 
DogAteMy .. it's past your bedtime!
 
@AkivaWeinberger you rick rolled me
 
4:20 AM
Well, one is a nowhere-zero $n$-form. One is chart overlaps having positive determinant.
One is that the tangent bundle is an orientable bundle.
 
@TedShifrin Very
I had a report I needed to finish
 
Hi
 
Aha ...
 
@TedShifrin Yep, all those three. I have a doubt about the definition of orientability of the bundle, though.
 
4:21 AM
i drew a picture
 
Are you trying to do all your studying for the whole course in two days?
 
exam is thursday
 
Spivak says a plane bundle $\pi:E\to B$ is orientable if for each connected trivilialising open set $U\subseteq B$, $t:\pi^{-1}(U)\to U\times \Bbb R^n$ induces either an orientation reversing iso in every fibre, or an orientation preserving iso in every fibre.
In fact one can check this in a cover, because of reasons.
 
Now you're doing polar coordinates, @Pallas.
 
yah
here's my drawing
 
4:23 AM
Yup, mr @Pedro. So this says we can assign compatibly globally an orientation on each of the fibers.
So in polar coordinates, what's the set-up, @Pallas?
 
you do r coordinates andtheta
 
@TedShifrin Because in overlaps things behave nicely?
 
(looking at picture) there's a double integral and multivariable functions going on there. Yep, you guys are beyond my level :)
 
Good so far, @Pallas. Now do the same game. Theta goes from what to what? And for fixed theta, what does r do?
 
@TedShifrin Now consider the tangent bundle $TM\to M$, seen as derivations. Thus given a chart $(x,U)$ with connected domain, one has a trivilization $t:\pi^{-1}(U)\to U\times \Bbb R^n$ that sends $(p,\sum \lambda^i \left.\frac{\partial}{\partial x^i}\right |_p)$ to $(p,\lambda^1,\ldots,\lambda^n)$.
So the basis of partials is sent to the canonical basis of $\Bbb R^n$.
 
4:26 AM
It's equivalent to covering with open sets on which the bundle is trivial and assigning orientations that are compatible on overlaps, yes, @Pedro. This means positive determinant transition functions on overlaps.
 
so theta makes an angle with hypotenuse 2 and the adjacent 1, so arccos(1/2)
so pi/3?
 
Yes, @Pallas.
 
-pi/3 to pi/3
 
Yes.
Now draw a ray for theta constant. Where does it enter, and where does it exit?
 
draw a ray?
 
4:27 AM
(theta = constant), @Pallas, just like we drew lines before (x = constant, etc.).
 
ok, so a ray from the origin to the entrance point of the domain makes a triangle with adjacent 1 and angle theta
 
My question is that the "substance" of the definition Spivak gives seems to lie in the overlaps of trivialisations, yet he doesn't talk much about this.
 
so, 1 /cos (theta)?
 
Well, you can talk about it without talking about transition functions. I tend to emphasize transition functions and also do vector bundles of which Mariano is definitely fond (universal bundles on projective spaces and Grassmannians, for example).
 
and then it goes to r = 2
 
4:29 AM
Yes, @Pallas. Good. And exits where?
 
so r ranges from 1/ cos(theta) to 2
 
Perfect. You're a master.
 
:D
 
@TedShifrin Well, Mariano did translate the general definition in terms of charts.
 
Just keep the same technique in mind whenever you're doing multiple integrals, no matter what coordinates, no matter how many dimensions.
 
4:30 AM
I'm trying to replicate what he did.
 
That compliment wasn't to you, @Pedro :)
 
I know that.
 
1/r is ln(r) right?
 
My point is, @Ted, that given a chart $(x,U)$ one gets an orientation $\partial/\partial x^i \mid_p$ of each $T_p U\simeq T_pM$.
 
So the overlap condition for the tangent bundle gives you positive determinant for the Jacobians.
That didn't quite make sense, @Pedro. You have to tell me what an oriented basis is.
Yup, @Pallas.
 
4:32 AM
@TedShifrin In order, $1,2,\ldots,n$.
The natural order of the partials.
 
Your power of $r$ isn't right, @Pallas. Look carefully at the problem.
OK, @Pedro.
 
oh, r^4
so then it's 1/ r^3
 
Right. This will turn out to be a relatively straightforward integral.
 
halo!
 
@Stan !! :)
 
4:34 AM
@TedShifrin Hey Ted! Question: I'm studying linear regressions. I don't understand what qualifies as "linear." Does that just mean stuff of the $y = a_1 x_1 + a_2 x_2 + ....$ form? Or can linear incorporate polynomials too?
 
Linear means linear. Perhaps with a constant to translate. (Like least-squares line fit all over statistics.)
 
If a wind is blowing "from the direction" N45$^{o}$W which is 135 $^{o}$, the vector must be 315$^{o}$, right?
 
Otherwise you're doing quadratic regression or exponential regression or whatever.
Yes, @Joshua.
 
hmm, answer was 1/4 but i got sqrt(3) - (pi/12)
 
You're integrating $\cos^2\theta$ from $-\pi/3$ to $\pi/3$?
With a constant.
 
4:38 AM
\int_{-pi/3}^{pi/3} \int_{1/cos(X)}^2 (1/r^3) dr dX
 
ah ok
good to know
 
Alright. I think the question I'm stuck on is just being unnecessarily tricky.
 
@TedShifrin Have you been following the election stuff?
I remember months ago we discussed trump being an idiot, but the problem seems to have ballooned since then lol
 
Oops, I messed up, @Pallas.
Yup, it sure has, @Stan.
 
4:40 AM
@TedShifrin Given any basis of $T_pM$, I can find a chart $(x,U)$ such that the canonical iso $T_p U\to T_pM$ sends the partials of $x$ to this basis. I mean, I can assume that the orientations in the fibres are given by charts.
 
@Pallas: I think the book's answer is wrong. Yours isn't quite right, but it's close.
 
does it have to be *2 because negative part?
 
Well, you can't necessarily do that for all $p$ at once, @Pedro. But once you've done it for one $p$, the nearby ones are determined.
 
test: $\mathbf{T}$
 
Yeah, symmetry doubles the answer @Pallas. What you typed above isn't what you just linked from Wolfram. Are you sure you got the right problem number from the book?
 
4:44 AM
yeah probelm 15 section 16.4
 
@Pedro: You aren't asking this, but given vector fields $X_1,\dots,X_k$ you can ask when they are of the form $\partial/\partial x^i$ for some coordinate system. Necessary and sufficient is that their Lie brackets be $0$.
 
yeah book says 1/4
 
Book's wrong, @Pallas. You did great. Move on.
 
@TedShifrin OK, but it is generally false.
 
ok, so for this one, dd i have to *2?
for the negative?
 
4:45 AM
But given linearly independent vectors at a point, you can always choose local coordinates so that $\partial/\partial x^i$ at the point agree with the given vectors, @Pedro.
 
if so, how come in other problems, i don't multiply by 2?
 
Your Wolfram Alpha had it right, @Pallas.
 
@TedShifrin Yes, that's what I said.
 
another one: $\det$
 
Not the more general one.
Spivak deals with that later.
 
4:46 AM
OK, @Pedro, I just elaborated.
 
ok cool. thanks for walking me through that one!
 
Hi there.
 
No, if you set up that integral, @Pallas, as you showed me, there is no *2.
 
I am trying to get from Spivak's definitions to "choose a chart at each point that gives an orientation at $p$ by partials, and this charts must be compatible, i.e. transition with positive determinant"
So yes, I can assume that an orientation is given by a chart.
 
Because charts give diffeomorphisms, so everything works locally, @Pedro. If you've chosen two different charts that agree on orientation at $p$, then they agree on any connected nbhd of $p$.
Ugh, it keeps giving me timeouts.
 
4:47 AM
@TedShifrin Yes.
 
Ugh, it keeps giving me timeouts.
 
What is a monic polynomial? Leading coefficient equal to unity?
 
Yes, @Brody. Google is your friend.
 
Definitely. A certain sentence on a Wiki article seems off, but it's probably my own poor understanding.
 
What sentence?
 
4:51 AM
Context is elementary linear algebra, on "characteristic polynomial"
"however the current definition always gives a monic polynomial, whereas the alternative definition always has constant term det(A)"
 
Yeah, that is very misleading.
 
One does not preclude the other, no?
 
There is a sign issue. That's what they're debating.
If you do $\det (A-tI)$, you don't have a monic polynomial when $n$ is odd.
If you do $\det(tI-A)$, then the polynomial is monic but the constant term is $-\det A$ when $n$ is odd.
 
Neat.
 
Second definition, second definition!
 
4:53 AM
I always do first definition @Pedro.
 
Thanks for clearing that up @TedShifrin. The original phrasing confused me a bit.
 
When $n$ is even, both hold fine.
 
Ted. No.
 
Too many minus signs for students to mess up with $-A$.
I'm making this on pedagogical grounds. I don't really care otherwise.
 
People have to learn how to deal with minuses.
 
4:55 AM
I'm a first year in introductory ODEs. Haven't taken LA yet but we're doing some matrix stuff for linear systems. We've only been taught the first definition.
 
It isn't that important, @Pedro. Chill out.
 
That is, $\det(\mathbf{A}-\lambda I_n)$ as our text notates it
 
I know you're thinking of module theory stuff, @Pedro, but it really is not a big deal.
 
whoops
 
@TedShifrin I'm thinking about orientations.
 
4:56 AM
You're probably only doing $2\times 2$ also, @Brody.
 
yep, for the most part
 
How could the sign on the characteristic polynomial matter for orientations, @Pedro?!
 
@TedShifrin I saying I'm thinking about something else.
Not about polynomials.
 
LOL, oh. Good. Brat.
I'm tired of all the delays and timeouts. I need to ice my neck and shoulder. Bubye all.
 
Adios.
 
4:57 AM
I appreciate the responses. Cya
 
@TedShifrin You sound like an athlete
heading to the training room
 
I guess to sum it up, the two definitions are equivalent for $n$ even. When $n$ is odd, $\det(\mathbf{A}-\lambda I_n)$ is (necessarily?) non-monic but the constant term is positive, namely $\det\mathbf{A}$. The other is strictly monic when $n$ is odd but the constant term is $-\det\mathbf{A}$.
Seems like a cost-benefit comparison. Not sure what's advantageous but at my level it's probably not important.
 
5:17 AM
you know whats better than radiation?
knowledge
 
Darn, I was thinking cupcakes.
 
they are good too
chernobyl is freaky stuff
they said the radiation could kill you in 20 minutes
but I wonder how
 
spontaneous dusting sounds cool
on second thought, the leading coefficient of any $p_A$ might generally have magnitude $1$. preserving the monic property is just a matter of sign perhaps
but that's enough of my late night ramblings. too sleep deprived. gg&gn
 
Hey guys, not trying to spam, but if you could take a look, I'd appreciate it.
0
Q: Show that a sequence of uniformly bounded continuous functions with Lipschitz condition is pre-compact in the space of bounded continuous functions

Jessy CatI am attempting to solve the following problem: Let the sequence of continuous functions $\{x_{n}(t) \}_{n=1}^{\infty}$, $0 \leq t < \infty$ be uniformly bounded on $t \in [0, \infty)$ and on each bounded interval $[0,A]$ also satisfy the Lipschitz condition: $\forall t_{1}, t_{2} \in [0,A]$...

 
5:55 AM
cool
 
6:30 AM
@AlexClark Hey, I just noticed a really silly mistake in the code I asked you to run, though it should still be fine for checking one specific order of group (basically, it will always return fail unless it finds a counter example of the last order checked). So I am now redoing it for the other orders and hoping that I had not overlooked something.
 
user147690
Oh okay, so I should just cancel this one @TobiasKildetoft?
 
@AlexClark It should be fine as long as you ran it with 768,768
 
user147690
I did
 
I had forgotten to make it stop checking if it found a counter example, so it would just overwrite that counter example with a "fail" if there were none for the next order
But this still works for a single order
I just really hope there are no counter examples of the other orders as that will make me feel somewhat stupid
 
user147690
So you mathematically proved your conjecture, and this is just computational checking to make yourself satisfied?
 
6:34 AM
@AlexClark No, I have no other evidence for the conjecture, but I have stated in a preprint that it is true for all orders up to 1000 except possibly 768, so I would need to correct that
 
user147690
Ahhh sure
 
And also comment to the reviewer that I was mistaken (he had asked for an explanation of why order 768 was problematic which is why I decided to try doing the calculation again)
I found the error because I was trying to use essentially the same code to find a non-solvable group with some of these properties (as I had claimed that one existed and the reviewer asked for an example), and it was not finding any.
I was then a bit unsure how to best describe the example until I found it again and it turned out to just be $A_5\times S_3$
 
7:10 AM
@PVAL: According to the construction you told me but which I haven't thought about yet, punctured 1/k-surgeries always embed in R^4. I wonder if I should believe that this is true for all homology spheres.
Other than trivialities like "Y is surgery on a knot" do you know any statements for 3-manifolds or homology spheres or whatever that are true for surgeries on a knot but false in general?
 
Is |x| a continuous function?
 
@ItachíUchiha Yes
 
Lets say you've proven that any $nxn$ matrix $A$ is equal to the sum of projection + nilpotent matrices in the form $A = \sum_{i=1}^m (\lambda_i P_i + N_i)$, where $\lambda_i$ is an eigenvalue without any real linear algebra theory, how would you get from this to the JNF?
 
@TobiasKildetoft Thanks
 
7:36 AM
@ItachíUchiha can you prove it?
 
@MikeMiller No I don't think any exist. It's really hard to show an irreducible homology sphere isn't surgery on a knot. You should look at Auckly's paper if you don't believe me.
@MikeMiller The construction appears to be wrong
or needs some thinking about.
 
8:42 AM
holy crap. it's 115F in my hometown.
Hi @AlexClark
 
This is starting to look weird. I define a new ordering on the naturals by letting $n \leq_p m$ if inequality holds for all "digits" in the $p$-adic expansions. And now I end up having to consider numbers $n$ such that $n\leq_p 2n + m - (p^r-1)$ for some fixed $m$ and where all the numbers are at most $p^r$.
this is a very weird condition
the ordering does really not behave well with respect to sums
 
9:21 AM
Hi, the direct product of $\mathbb Z_{12}$ with itself isn't a cyclic group right?
 
@Paradox101 Right
 
Hey @EricStucky.
 
eya
Does Stokes' theorem work if the form's rank is not one less than the manifold dimension?
 
Dunno Stokes.
 
ah :/
We haven't officially learned it in topology but it comes up on all the past prelims so...
 
9:26 AM
@TobiasKildetoft how do we determine whether a factor group is cyclic or not? I know the factor group is cyclic if the group itself is cyclic but how will we know if it's cyclic or not if the group itself is non-cyclic?
 
At some point of time someone should tell me what's the deal with Barratt-Milnor. It's not obvious to me what a nontrivial $3$-chain would be, but I am not enough interested to read the paper myself.
 
@Paradox101 In general, this can be tricky
 
@TobiasKildetoft should I simply try it by listing all the elements of the factor group and then check whether it's cyclic? Because I can't figure out a simpler or more elegant way to do that
 
@Paradox101 depends on the concrete case.
 
in my case the group $G$ is a direct product of $\mathbb Z_{12}$ with itself and $H$ is a cyclic group generated by $(2,2)$ @TobiasKildetoft
 
9:30 AM
@ForeverMozart Hey. You.
 
You should tell me what Barratt and Milnor says.
 
@Paradox101 So the subgroup is of order $6$?
 
who is that?
 
@TobiasKildetoft yes
 
9:32 AM
@Paradox101 Ok, then you can use that the order of the quotient is too large for it to be cyclic, since the largest element order of the original group is $12$
 
Does anybody know any cool symbols with which to label equations?
(I don't want to keep using numbers as it's a very short essay.)
 
@TobiasKildetoft I don't quite understand it yet. The order of the factor group is $24$ so how large is too large? And how is the largest element order of $G$ $12$?
 
@Paradox101 In a direct product of cyclic groups, the largest order of an element is the lcm of the orders of the groups
so in this case, this is $12$.
And the largest order of an element in a quotient is at most that of one from the original group
 
Trying to derive the idea of the cross ratio naturally by thinking about the projective plane $\mathbb{R} \mathbb{P}^2$: given 4 collinear points $A,B,C,D$ we note $C = aA + bB$ and $D = cA + dB$ so that the cross ratio is $(b/a)/(d/c)$. Why is this obvious? My guess is that you take $C = aA + bB = a[A + (b/a)B] \sim A + (b/a)B$ and $D \sim A + (d/c)B$ so that, for some reason,
we want to set $(b/a) = \lambda (d/c)$ so that $\lambda = (b/a)/(d/c)$ is worth even defining as a concept? Any thoughts?
 
user147690
@BalarkaSen You use F instead of C in India?
 
9:40 AM
No, we use C but I converted into F because most people in here are used to F :)
 
user147690
I C :P
 
@TobiasKildetoft so if we want $G/H$ to be cyclic it must have an element of order $24$ however the element with the highest order in the original group is of order $12$ and therefore the factor group cannot be cyclic? Thank you!
 
I am not going to reply with the two letter statement starting with an 'F'.
 
@Paradox101 Exactly
 
So, what's up, @AlexClark?
 
user147690
9:42 AM
@BalarkaSen Hahahaha I was actually considering how I could slip it in without seeming rude
 
user147690
@BalarkaSen What's up is massive sleep deprivation that I really want to end...
 
@AlexClark Answer is simple then: Go to sleep
 
user147690
8 allnighters in the last 3 and a bit weeks, and 6 hours sleep last two nights. I have 10 pages due tomorrow at 10am
 
user147690
and then I am sleeping 10pm to 6am from then on hopefully without fail
 
user147690
@TobiasKildetoft Can't :(
 
user147690
9:43 AM
Getting pizza now, be back in 10 :P
 
user147690
Atleast I am skipping cooking time
 
user147690
Super keen on being a human after tomorrow though :D
 
@AlexClark Yikes.
@AlexClark There's an interesting observation a person made in where I am. He's a geometer through and through, and jokingly sometimes says that algebra is vulgar and rude. So I was telling him that it's a bit of a work to prove that local ring of a variety at a point is unique factorization domain, and he retorted back with "no, only algebraists think it is - it's a beautiful corollary of the Weierstrass preparation theorem" or something like so.
I said that for curves it's easy because "by uniformization, it's a DVR. DVR means PID, and PID implies UFD".
To which he replied "see, that's how algebraists are. they are so vulgar that the only vulgar words they cook up are 3 letters, not 4"
 
@BalarkaSen did you read my paper?
 
Nope, @Forever. Link me?
 
9:48 AM
do you understand the Lelek fan now?
 
Yeah
I know what it means, at least.
 
oh ok
the link is in my special room
I explain it in the first 2 pages
 
Nice. What did you prove there?
 
in the first 2 pages I just show how to construct the fan
cause there were no good sources
 
ah, ok.
 
9:53 AM
the rest is just me rambling, trying to prove more stuff
 
Weren't you trying to prove a conjecture by Erdos or something like that?
 
yep
its tough
but I have at least one piece of information that Erdos did not have
which may be useful
 
what's that?
 
There is a closed non-disconnecting subset of the buckethandle continuum which meets every composant
1991 result
4 years after Erdos died
 
mmkay.
 
9:56 AM
I can maybe use that
its so frustrating
so many examples just barely fail
which means maybe its not possible
 
maybe there are no counterexamples
so you need to prove it
 
:(
:((((((((((((((
:|
:'(
:D
 
I need to grab hold of someone who can tell me how to do resolution of singularities for plane curves.
 
mustache
 
10:12 AM
ugghhh. I can't find a single useful reference on the internet.
Everyone's just crazy about normalization.
 
normalize
 
HI
may I ask here 3 short physics questions ?
 
"The question isn't who is going to let me, it's who is going to stop me." —Ayn Rand
 
10:30 AM
that is very aggressive
 
user147690
10:57 AM
@Algebra2015 Usually you won't get an answer I would imagine. People aren't that big on physics questions here
 
@AlexClark Yay, the conjecture still holds for orders other than 768
 
user147690
@TobiasKildetoft Oh good, so we are just waiting on me haha
 
user147690
Good good :D
 
I wonder how much time could have been shaved off if I had been better at the whole computational group theory thing
 
11:04 AM
@AlexClark Did you learn about blowups yet?
 
user147690
@BalarkaSen Not yet, but on the weekend I'll be looking over it for sure
 
Aw. Oh well.
 
user147690
@TobiasKildetoft I have friends in IT and software engineering, if I were in your situation I would partially defer to them haha
 
user147690
It would probably not make too much of a difference though
 
@AlexClark thing is that probably most of the speedup would involve knowing which parts take time which involves group theory
 
user147690
11:06 AM
Ahhh that's true
 
user147690
So the program is probably really efficient for a class of problems and yours is a subclass, is that what you are saying?
 
user147690
(and attacking just the subclass would be more efficient)
 
user147690
Also, is there a way to get my latex to use \bibliographystyle{Alpha} without making a .bib file?
 
@AlexClark Hmm, not sure, apart from manually naming everything
And yeah, probably the built-in functions in GAP are pretty fast, so it is just my own parts that might be trouble
 
user147690
This works:

`\begin{thebibliography}{9}

\bibitem{Test}
Leslie Lamport,
\emph{\LaTeX: a document preparation system},
Addison Wesley, Massachusetts,
2nd edition,
1994.

\end{thebibliography}`
 
user147690
11:10 AM
But it's not in the alpha form
 
I don't even remember how to manually name the entries
Why would you not want to use a .bib file anyway?
 
user147690
Ahh just out of paranoia of losing my files, I like to dump the long tex documents as text online
 
user147690
And if I can't dump it all as text it isn't self contained, but I guess it's fine
 
Do the same with the bibtex file
 
 
2 hours later…
1:42 PM
Hi guys!
I have a doubt.
Does being continuous imply being differentiabe?
I went through this answer math.stackexchange.com/a/140485/200649 But I have still confused.
Since one of the pre-requisite of being continuous is that limit exists and is same from right to left. It should imply that the function is differentiable?
The guy in the answer mentions "Because of the definitions, continuity is a prerequisite for differentiability, but it is not enough. A function may be continuous at a, but not differentiable at a."
What are these "really spiky" functions he talks about? I can't find them anywhere.
*But I am still confused.
Please reply whenever possible. :)
 
Hi, so the standard counterexample is the Weierstrass function, see en.wikipedia.org/wiki/Weierstrass_function
It is everywhere continuous but nowhere differentiable, but might be a bit hard to wrap your head around
it might be easier to think of a function that is continuous in 0 but not differentiable in 0, e.g. define $f=x$ for $x \leq 0$ but $f=0$ for $x>0$
 
1:59 PM
@AbhishekBhatia By spiky functions he means that when you are approaching that point from one direction(say the right) the slope is something and from another direction(left) the slope is something else. . So at that spike you really cant define the derivative because the right side and left side slopes do not agree
 
Hey guys. Anybody able to help me with this?
0
Q: Show that a sequence of uniformly bounded continuous functions with Lipschitz condition is pre-compact in the space of bounded continuous functions

Jessy CatI am attempting to solve the following problem: Let the sequence of continuous functions $\{x_{n}(t) \}_{n=1}^{\infty}$, $0 \leq t < \infty$ be uniformly bounded on $t \in [0, \infty)$ and on each bounded interval $[0,A]$ also satisfy the Lipschitz condition: $\forall t_{1}, t_{2} \in [0,A]$...

 
2:26 PM
Hi @robjohn
How can we find the weak derivative of $u(x)=|x|^{-\gamma}, 0< \gamma< \frac{n-p}{p}$ ?
 
2:51 PM
@EricStucky Those specific dimensions are chosen not because it's wrong for the other ones, but because it doesn't make sense - we need to integrate it, and you can only integrate n-forms on n-manifolds.
@PVAL I've read the paper, it's charming.
 
Hello,
i am having equation of a line on sphere as $ds=[(dx)^2+(dy^2)]^{1/2}$ no find the lenght of the circumference if radius of sphere is $r$, but we have to do this by calculus
 
3:13 PM
sorry but the equation i have given is not the equation of st. line but it is straight line distance between the points and $dx, dy$ are very very very small
 
3:26 PM
If somebody wants to share their opinion on these recently cretaed tags, they are welcome to do so in the tagging chatroom:
in Tagging, 33 mins ago, by pjs36
For example: are any of , , , , , , , and so on, worth preserving? I generally have mixed feelings about making a fairly large number of changes to tags in areas I'm not terribly familiar with.
 
3:39 PM
I'm reading a paper which performs the following operation on vectors: $[ 0\quad 1\quad 1] - \frac{1}{\sqrt{2}} [ \quad \frac{1}{\sqrt{2}} \quad \frac{1}{\sqrt{2}} \quad 0 ] \quad - \frac{1}{\sqrt{6}} [ \quad \frac{1}{\sqrt{6}} \quad -\frac{1}{\sqrt{6}} \quad \frac{2}{\sqrt{6}} ]$
I just multiply the inner vectors with the scalar to get: $[0\quad 1\quad 1] - [\frac{1}{2}\quad \frac{1}{2}\quad 0] - [\frac{1}{6}\quad \frac{-1}{6}\quad \frac{-2}{6}]$
Then subtract to get: $[ (0 - \frac{1}{2} - \frac{1}{6})\quad (1 - \frac{1}{2} + \frac{1}{6})\quad (1 - 0 -3) ]$
Or: $[\frac{-2}{3}\quad \frac{2}{3}\quad -2 ]$
However, the author gets: $[\frac{-1}{\sqrt{3}}\quad \frac{1}{\sqrt{3}}\quad \frac{1}{\sqrt{3}} ]$
What did I do wrong here?
 
@PVAL I just got funding for the thing in Rice so I guess I'll see you there.
 
As far as I know, scalar multiplication with a vector is just (elements * scalar). And the subtraction is just (a1 - b1, a2- b2, ... etc)
Also, what's the best way to write out a vector in this chat without getting some lengthy latex code?
 
4:23 PM
@BalarkaSen Torus?
 
4:33 PM
@MikeMiller Good news from my gauge theory lecture; we'll cover a lot more than the prof. did in previous incarnations of this course, and there will perhaps be a second course, too!
 
I'm going to assume the paper I'm reading is wrong..
 
@Danu hooray!
 
Indeed! Now I just need to remember how to diffgeo...
 
I want to write the following in latex:
over a fraction
Which is the symbol in latex?
Do you have an idea?
 
vee ?
\vee
 
4:46 PM
Hi one and all
 
$\vee$
 
That is for the symbol. But I want to write it over a function.
 
I dunno. Embedded fraction maybe? $\frac{\vee}{\frac{1}{2}}$
Nope. That looks bad.
 
Nope. Write it as $\check{f}$
This will give you a v over f
But you need to use package called mathabx @Evinda
 
I want to make \check bigger. Is it possible?
 

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