Mathematics - 2016-04-19 (page 2 of 3)

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.

Ted Shifrin

Now you're doing polar coordinates, @Pallas.

Pallas

imgur.com/G99qmm3

yah

here's my drawing

Ted Shifrin

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?

Pallas

you do r coordinates andtheta

Pedro Tamaroff

@TedShifrin Because in overlaps things behave nicely?

Joshua Lamusga

(looking at picture) there's a double integral and multivariable functions going on there. Yep, you guys are beyond my level :)

Ted Shifrin

Good so far, @Pallas. Now do the same game. Theta goes from what to what? And for fixed theta, what does r do?

Pedro Tamaroff

@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 $\Bbb R^n$.

Ted Shifrin

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.

Pallas

so theta makes an angle with hypotenuse 2 and the adjacent 1, so arccos(1/2)

so pi/3?

Ted Shifrin

Yes, @Pallas.

Pallas

-pi/3 to pi/3

Ted Shifrin

Yes.

Now draw a ray for theta constant. Where does it enter, and where does it exit?

Pallas

draw a ray?

Ted Shifrin

4:27 AM

(theta = constant), @Pallas, just like we drew lines before (x = constant, etc.).

Pallas

ok, so a ray from the origin to the entrance point of the domain makes a triangle with adjacent 1 and angle theta

Pedro Tamaroff

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.

Pallas

so, 1 /cos (theta)?

Ted Shifrin

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).

Pallas

and then it goes to r = 2

Ted Shifrin

4:29 AM

Yes, @Pallas. Good. And exits where?

Pallas

so r ranges from 1/ cos(theta) to 2

Ted Shifrin

Perfect. You're a master.

Pallas

:D

Pedro Tamaroff

@TedShifrin Well, Mariano did translate the general definition in terms of charts.

Ted Shifrin

Just keep the same technique in mind whenever you're doing multiple integrals, no matter what coordinates, no matter how many dimensions.

Pedro Tamaroff

4:30 AM

I'm trying to replicate what he did.

Ted Shifrin

That compliment wasn't to you, @Pedro :)

Pedro Tamaroff

I know that.

Pallas

1/r is ln(r) right?

Pedro Tamaroff

My point is, @Ted, that given a chart $(x,U)$ one gets an orientation $\partial/\partial x^i \mid_p$ of each $T_p U\simeq T_pM$.

Ted Shifrin

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.

Pedro Tamaroff

4:32 AM

@TedShifrin In order, $1,2,\ldots,n$.

The natural order of the partials.

Ted Shifrin

Your power of $r$ isn't right, @Pallas. Look carefully at the problem.

OK, @Pedro.

Pallas

oh, r^4

so then it's 1/ r^3

Ted Shifrin

Right. This will turn out to be a relatively straightforward integral.

Stan Shunpike

halo!

Ted Shifrin

@Stan !! :)

Stan Shunpike

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?

Ted Shifrin

Linear means linear. Perhaps with a constant to translate. (Like least-squares line fit all over statistics.)

Joshua Lamusga

If a wind is blowing "from the direction" N45$^{o}$W which is 135 $^{o}$, the vector must be 315$^{o}$, right?

Ted Shifrin

Otherwise you're doing quadratic regression or exponential regression or whatever.

Yes, @Joshua.

Pallas

hmm, answer was 1/4 but i got sqrt(3) - (pi/12)

Ted Shifrin

You're integrating $\cos^2\theta$ from $-\pi/3$ to $\pi/3$?

With a constant.

Pallas

4:38 AM

\int_{-pi/3}^{pi/3} \int_{1/cos(X)}^2 (1/r^3) dr dX

Stan Shunpike

ah ok

good to know

Joshua Lamusga

Alright. I think the question I'm stuck on is just being unnecessarily tricky.

Pallas

wolframalpha.com/input/…

Stan Shunpike

@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

Ted Shifrin

Oops, I messed up, @Pallas.

Yup, it sure has, @Stan.

Pedro Tamaroff

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.

Ted Shifrin

@Pallas: I think the book's answer is wrong. Yours isn't quite right, but it's close.

Pallas

does it have to be *2 because negative part?

Ted Shifrin

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.

Brody

test: $\mathbf{T}$

Ted Shifrin

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?

Pallas

4:44 AM

yeah probelm 15 section 16.4

Ted Shifrin

@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 $0$.

Pallas

yeah book says 1/4

Ted Shifrin

Book's wrong, @Pallas. You did great. Move on.

Pedro Tamaroff

@TedShifrin OK, but it is generally false.

Pallas

ok, so for this one, dd i have to *2?

for the negative?

Ted Shifrin

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.

Pallas

if so, how come in other problems, i don't multiply by 2?

Ted Shifrin

Your Wolfram Alpha had it right, @Pallas.

Pedro Tamaroff

@TedShifrin Yes, that's what I said.

Brody

another one: $\det$

Pedro Tamaroff

Not the more general one.

Spivak deals with that later.

Ted Shifrin

4:46 AM

OK, @Pedro, I just elaborated.

Pallas

ok cool. thanks for walking me through that one!

Brody

Hi there.

Ted Shifrin

No, if you set up that integral, @Pallas, as you showed me, there is no *2.

Pedro Tamaroff

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.

Ted Shifrin

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.

Pedro Tamaroff

4:47 AM

@TedShifrin Yes.

Ted Shifrin

Ugh, it keeps giving me timeouts.

Brody

What is a monic polynomial? Leading coefficient equal to unity?

Ted Shifrin

Yes, @Brody. Google is your friend.

Brody

Definitely. A certain sentence on a Wiki article seems off, but it's probably my own poor understanding.

Ted Shifrin

What sentence?

Brody

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)"

Ted Shifrin

Yeah, that is very misleading.

Brody

One does not preclude the other, no?

Ted Shifrin

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.

Brody

Neat.

Pedro Tamaroff

Second definition, second definition!

Ted Shifrin

4:53 AM

I always do first definition @Pedro.

Brody

Thanks for clearing that up @TedShifrin. The original phrasing confused me a bit.

Ted Shifrin

When $n$ is even, both hold fine.

Pedro Tamaroff

Ted. No.

Ted Shifrin

Too many minus signs for students to mess up with $-A$.

I'm making this on pedagogical grounds. I don't really care otherwise.

Pedro Tamaroff

People have to learn how to deal with minuses.

Brody

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.

Ted Shifrin

It isn't that important, @Pedro. Chill out.

Brody

That is, $\det(\mathbf{A}-\lambda I_n)$ as our text notates it

Ted Shifrin

I know you're thinking of module theory stuff, @Pedro, but it really is not a big deal.

Brody

whoops

Pedro Tamaroff

@TedShifrin I'm thinking about orientations.

Ted Shifrin

4:56 AM

You're probably only doing $2\times 2$ also, @Brody.

Brody

yep, for the most part

Ted Shifrin

How could the sign on the characteristic polynomial matter for orientations, @Pedro?!

Pedro Tamaroff

@TedShifrin I saying I'm thinking about something else.

Not about polynomials.

Ted Shifrin

LOL, oh. Good. Brat.

I'm tired of all the delays and timeouts. I need to ice my neck and shoulder. Bubye all.

Joshua Lamusga

Adios.

Brody

4:57 AM

I appreciate the responses. Cya

Forever Mozart

huh

Stan Shunpike

@TedShifrin You sound like an athlete

heading to the training room

Brody

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.

Forever Mozart

5:17 AM

you know whats better than radiation?

knowledge

Axoren

Darn, I was thinking cupcakes.

Forever Mozart

they are good too

chernobyl is freaky stuff

they said the radiation could kill you in 20 minutes

but I wonder how

Brody

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

Jessy Cat

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

I 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: $\forall t_{1}, t_{2} \in [0,A]$...

Forever Mozart

5:55 AM

cool

Tobias Kildetoft

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?

Tobias Kildetoft

@AlexClark It should be fine as long as you ran it with 768,768

user147690

I did

Tobias Kildetoft

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?

Tobias Kildetoft

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

Tobias Kildetoft

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 $A_5\times S_3$

Mike Miller

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?

Itachí Uchiha

Is |x| a continuous function?

Tobias Kildetoft

@ItachíUchiha Yes

bolbteppa

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?

Itachí Uchiha

@TobiasKildetoft Thanks

Forever Mozart

7:36 AM

@ItachíUchiha can you prove it?

PVAL

@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.

Balarka Sen

8:42 AM

holy crap. it's 115F in my hometown.

Hi @AlexClark

Tobias Kildetoft

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

Paradox 101

9:21 AM

Hi, the direct product of $\mathbb Z_{12}$ with itself isn't a cyclic group right?

Tobias Kildetoft

@Paradox101 Right

Balarka Sen

Hey @EricStucky.

Eric Stucky

eya

Does Stokes' theorem work if the form's rank is not one less than the manifold dimension?

Balarka Sen

Dunno Stokes.

Eric Stucky

ah :/

We haven't officially learned it in topology but it comes up on all the past prelims so...

Paradox 101

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?

Balarka Sen

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.

Tobias Kildetoft

@Paradox101 In general, this can be tricky

Paradox 101

@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

Tobias Kildetoft

@Paradox101 depends on the concrete case.

Paradox 101

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

Balarka Sen

9:30 AM

@ForeverMozart Hey. You.

Forever Mozart

hi

Balarka Sen

You should tell me what Barratt and Milnor says.

Tobias Kildetoft

@Paradox101 So the subgroup is of order $6$?

Forever Mozart

who is that?

Paradox 101

@TobiasKildetoft yes

Tobias Kildetoft

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 $12$

Kari

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.)

Paradox 101

@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 $G$ $12$?

Tobias Kildetoft

@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

bolbteppa

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?

Balarka Sen

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

Paradox 101

@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!

Balarka Sen

I am not going to reply with the two letter statement starting with an 'F'.

Tobias Kildetoft

@Paradox101 Exactly

Balarka Sen

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...

Tobias Kildetoft

@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

Balarka Sen

@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"

Forever Mozart

@BalarkaSen did you read my paper?

Balarka Sen

Nope, @Forever. Link me?

Forever Mozart

9:48 AM

do you understand the Lelek fan now?

Balarka Sen

Yeah

I know what it means, at least.

Forever Mozart

oh ok

the link is in my special room

I explain it in the first 2 pages

Balarka Sen

Nice. What did you prove there?

Forever Mozart

in the first 2 pages I just show how to construct the fan

cause there were no good sources

Balarka Sen

ah, ok.

Forever Mozart

9:53 AM

the rest is just me rambling, trying to prove more stuff

Balarka Sen

Weren't you trying to prove a conjecture by Erdos or something like that?

Forever Mozart

yep

its tough

but I have at least one piece of information that Erdos did not have

which may be useful

Balarka Sen

what's that?

Forever Mozart

There is a closed non-disconnecting subset of the buckethandle continuum which meets every composant

1991 result

4 years after Erdos died

Balarka Sen

mmkay.

Forever Mozart

9:56 AM

I can maybe use that

its so frustrating

so many examples just barely fail

which means maybe its not possible

Balarka Sen

maybe there are no counterexamples

so you need to prove it

Forever Mozart

:(

:((((((((((((((

:|

:'(

:D

Balarka Sen

I need to grab hold of someone who can tell me how to do resolution of singularities for plane curves.

Forever Mozart

:Ë

:Œ

mustache

Balarka Sen

10:12 AM

ugghhh. I can't find a single useful reference on the internet.

Everyone's just crazy about normalization.

Forever Mozart

normalize

Algebra 2015

HI

may I ask here 3 short physics questions ?

Eric Stucky

"The question isn't who is going to let me, it's who is going to stop me." —Ayn Rand

Forever Mozart

10:30 AM

that is very aggressive

bolbteppa

Made it nicer math.stackexchange.com/questions/1749434/…

user147690

10:57 AM

@Algebra2015 Usually you won't get an answer I would imagine. People aren't that big on physics questions here

Tobias Kildetoft

@AlexClark Yay, the conjecture still holds for orders other than 768

user147690

@TobiasKildetoft Oh good, so we are just waiting on me haha

Tobias Kildetoft

yep

user147690

Good good :D

Tobias Kildetoft

I wonder how much time could have been shaved off if I had been better at the whole computational group theory thing

Balarka Sen

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

Balarka Sen

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

Tobias Kildetoft

@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?

Tobias Kildetoft

@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

Tobias Kildetoft

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

Tobias Kildetoft

Do the same with the bibtex file

Abhishek Bhatia

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. :)

Oliver

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 $x>0$

Albas

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

Jessy Cat

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

I 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: $\forall t_{1}, t_{2} \in [0,A]$...

analysis compactness lipschitz-functions arzela-ascoli equicontinuity

Evinda

2:26 PM

Hi @robjohn
How can we find the weak derivative of $u(x)=|x|^{-\gamma}, 0< \gamma< \frac{n-p}{p}$ ?

Mike Miller

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.

ramsay

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

ramsay

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

Martin Sleziak

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 slope, tangent-line, rate-of-convergence, convergence-rate, rounding-error, rounding-unit, machine-precision, 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.

Owatch

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?

Mike Miller

@PVAL I just got funding for the thing in Rice so I guess I'll see you there.

Owatch

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?

Mike Miller

4:23 PM

@BalarkaSen Torus?

Danu

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!

Owatch

I'm going to assume the paper I'm reading is wrong..

Mike Miller

@Danu hooray!

Danu

Indeed! Now I just need to remember how to diffgeo...

Evinda

I want to write the following in latex: