Mathematics - 2015-07-23 (page 1 of 2)

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12:42 AM

reset up the driver

or update it

3 hours later…

3:47 AM

I have a question about BCL (en.wikipedia.org/wiki/Binary_combinatory_logic)

Don't both rewrite rules reduce the size of the string? How can you end up with a string larger than the one you started with?

looking at that page naively, i'd tend to agree: the first rule decreases the number of characters by 4, the second by 1. so no rewrite could create a string larger than the previous one

out of curiousity, is there a reason why that should be possible? i've no insight into such things

evening @ted

3 hours later…

6:36 AM

Anybody know how to find the Taylor series of the Lambert W function?

I know a way to do it but I'm curious if there are any "standard" ways (I used a kind of odd method)

7:06 AM

@MikeMiller OK, so what about this reformulation : $S^1 \to M$ be an embedding into some $3$-manifold. Let's assume that the embedded circle is not only nullhomotopic, but bounds a "nice" disk, that is, a disk with the self-intersections inside the interior of the disk. Is it necessary that it bounds an embedded disk, then?

Doesn't seem as if there's some nontrivial knot boundind a "nice" disk inside $S^3$ : look at the overpasses and underpasses. Surely the knot itself intersects the interior of any disk it bounds at least once.

7:48 AM

@DanielFischer: I know some Mobius transformations can be used as isometries from the upper half plane model to the disk model of hyperbolic space. How do I find an isometry from the disk model to itself sending $(0,0) \to (a,0)$? @BalarkaSen @MikeMiller

@Huy In complex notation, $$z \mapsto \frac{z + a}{1 + \overline{a}\cdot z}.$$

@DanielFischer: How does one find it with reasoning and not just by trying out?

I mean it's clear to add $a$ in the numerator, but why must I have a $\bar{a}$ in the denominator?

@Huy How much complex analysis do you know? From the reflection principle, it follows that a holomorphic automorphism of the disk extends to an automorphism of the sphere, hence is a Möbius transformation. Then you can see which Möbius transformations are automorphisms of the unit disk.

@DanielFischer: Just some basics, but forgot most of them unfortunately.

@DanielFischer: Nobody told me I needed to revise complex analysis to study Riemannian geometry. :(

The reflection principle tells you that if $b$ is a zero of an automorphism, then $1/\overline{b}$ is a pole, So you get $z - 1/\overline{b}$ in the denominator. Scaling it so that the unit circle is mapped to itself gets you $1-\overline{b}\cdot z$.

7:56 AM

Hm.

I see.

@Huy You don't need to, it's just simpler if you can use complex analysis.

8:13 AM

@DanielFischer: If we define an isometry as a diffeomorphism $f: (M,g) \to (N,h)$ with $f^*(h) = g$ and on the other hand an isometric immersion as an immersion such that $f^*(h) = g$, then an isometric immersion is not necessarily an isometry, right?

(just a bit confused by the notation)

@Huy Right. An isometric immersion can be into a higher-dimensional manifold, or its image can be a proper open subset of the codomain if the dimensions are equal, and an isometric immersion need not be injective.

Good, I was worried I made a mistake copying the definitions, but then all is good.

8:27 AM

"Indecomposable is a weaker notion than simple module". Surely, it should be the other way round?

@Boni No, simple implies indecomposable, so it's the stronger notion.

@DanielFischer Right, right. The combination of the negations in the definition of "simple" and in the indecomposible confused me and made me negate the wrong things in my head. Thank you for checking.

8:52 AM

In representation theory, I had learnt that there are representations of groups and these induce CG-modules. We were taught to think of these modules as C-vector spaces with a G-action on them. At that point, I had been thinking that the scalar of the vector space is just C, with some compatibility axioms from G-actions.

However, looking at group rings, it looks like the base ring is much larger and more complicated than that.

Can someone give me a hint as to how I can link these two interpretations together?

@Boni C[G] is comprised of C-linear combinations of elements of G. you wanna know how they act on a rep V? well, you know how each elt of G acts on V, so just extend linearly. thus is born an action of C[G] on V out of the action of G on V.

say you have two elts g,h of G and somewhere in the midst of things you have an expression gv+hv, and perhaps for whatever reason the shorthand (g+h)v for it (with the obvious interpretation using the distributivity property) is suggestive. well, if we consider the group algebra C[G], then this is no longer mere shorthand, it's literal fact that gv+hv=(g+h)v.

if you know any category theory, what this means is there there is an equivalence of categories, of (complex representations of G) with (modules over the complex group algebra C[G]).

@anon That sounds like the first definition I had learnt in the representation theory course. The group G acts on V as automorphisms. Compatibility axioms enforces the linearity you have mentioned.

if your definition was a group homomorphism G->GL(V), i.e. a linear group action, then you probably didn't have anything at that stage to interpret an expression like 'g+h' (an element of C[G]).

I think I should be more specific about my confusion. The Wikipedia page mentions that the elements in group rings RG are maps G --> R with similar compatibility. This looks (to me) nothing like group homomorphisms.

that's a set-theoretic trick to define formal linear combinations of things

9:04 AM

@anon You are right. I did learn them as G --> GL(V).

I guess that is why I am having trouble linking them in my head.

a formal linear combination of things is determined by a collection of coefficients, one for each thing being combined. thus, the coefficients may be described as a scalar function on the set of things you're combining.

Oh. I see. That is quite neat.

think of a function $f:X\to\Bbb C$ as a formal linear combination $\sum_{x\in X}f(x)x$

the utility of this viewpoint is when you want to generalize to infinite $X$

Reminds me of how I am supposed to see sequences as functions.

then you can put conditions on the function, like continuous, smooth, rapid decay, compact support, finite L^2 norm, etc.

@Boni exactly

9:08 AM

Right. Looks like group ring requires finite support according to Wikipedia.

yes. the 'correct' generalization of C[G] for infinite G, though, is L^2(G) for locally compact G

(now we're getting into abstract harmonic analysis)

I guess that makes sense. We want some sort of "convergence" to make sense of the sum.

@anon I think I can see how it is defined now. The representation of coefficients as a function was the missing bit for me. Thank you.

@Boni I'd recommend getting used to the formal linear combination viewpoint before trying to translate between that viewpoint and the functional one.

@anon Good idea. I would like to stick to representations in formal sums as well. I just wanted to make sense of the definition by Wikipedia when I came across it.

Hey @anon.

9:23 AM

hey

D'you know the answer to the question I asked above?

It's probably something very silly.

goes fishing

I don't know what that expression means.

means I started going up through the transcript

oh, I see.

here is le question.

9:29 AM

maybe there's a "flattening" flow on the space of disks the circle bounds which yields an extremum with no self-intersections. sounds pretty complicated. but your claim seems plausible.

that's pretty high-faluting. you want to choose an embedded disk from the space of all nullcobordisms of the embedded circle. the space would be pretty wacky, it's not clear if you can do much calculus with it.

dunno what a nullcobordism is

a cobordism with the nullmanifold. it's equivalent to the "space of all disks bounded by the circle", just a -probably?- meaningful way to state it.

hey, could I define my chain complexes in the following way? Let $C_n({\cal M})$ be the $\Bbb Z$-linear combinations of closed, compact, oriented, $n$-dimensional submanifolds of $\cal M$, subject to the following two relations: if $\cal N=A\cup B$ with $\cal A,B$ intersecting only at their boundary (and inheriting their orientation), then $\cal N=A+B$, and if $\cal C,C'$ are the same submanifold but with opposite orientation, then $\cal C'=-C$.

I am not going to check through the details, but yes! Homology of that chain complex would give you bordism homology.

@anon you want a submanifold with boundary, not just a submanifold.

the boundary operators $C_n \to C_{n-1}$ are defined by $M \mapsto \partial M$.

erm, wait, I don't understand your equivalence relation.

9:39 AM

@BalarkaSen say I put two squares side-by-side in R^2 with the same orientation. the sum of those two squares is the resulting rectangle.

the idea I guess is that chains are domains of integration, cells are boundaryless, and boundaries are, well, boundaries. but domains of integration with "multiplicity." for instance, in complex analysis, one can add/subtract contours this way. from what I can tell, this is what de rham cohomology describes.

at first glance, I thought this was bordism homology, but I am not sure anymore. I guess I don't understand your first relation yet. let me re-look.

you could just say $\cal A+B=(A\cup B)+(A\cap B)$ (when $\cal A,B$ restrict to the same orientation on $\cal A\cap B$) I think

no, I guess, this is something rather exotic.

one may take the topological boundary and linearly extend to get $\partial$, and thus define $H_n({\cal M})$. presumably if one homotopes (or at least smoothly isotopes, just to be safe) an oriented submanifold $\cal N_1$ to a new one $\cal N_2$, then the region bounded between them (so, we're assuming a nice picture here) has boundary $\cal N_1-N_2$, so $\cal N_1=N_2$ in $H_n$. Putting a directed CW complex on $\cal M$, presumably any oriented submanifold can be homotoped to a chain.

this should mean the "anonhomology" can be discretized (via complexes) to get the usual definition more or less, with the utility of finitizing all of the relevant computations

also, since every compact manifold can be triangulated (simplexicated?), we could probably relate them to singular homology as well

@Mike see above discussion

2 hours later…

12:03 PM

Hi@TedShifrin...

Chris's sis the artist

How is it going?

All is fine it seems.

@Balarka Was my rest of the proof fine .. except the first statement which I corrected after reading the definition

@Balarka this is my proof for it being an equivalence relation
http://chat.stackexchange.com/transcript/message/22941161#22941161

Hello,Soham

Chris's sis the artist

12:25 PM

0

Q: Calculating in closed form an integral in Airy function

Can we hope for a nice closed form for the integral below? $$\int_0^1 \frac{\displaystyle \text{Ai}\left(-\frac{t}{2^{2/3} \sqrt[3]{3-3 t}}\right)^2+\text{Bi}\left(-\frac{t}{2^{2/3} \sqrt[3]{3-3 t}}\right)^2}{ \sqrt[6]{1- t}} \, dt$$ What tools do you recommend me to use? Some information abou...

calculus real-analysis integration definite-integrals special-functions

More than half of the unanswered questions are topology related or topology question :p

Soham Chowdhury

@Huy what are those? they look like tensors or whatever.

@Rememberme allo

@SohamChowdhury: They are tensors. They are Einstein's field equations.

Went to school ?@Soham

Soham Chowdhury

wow.

I imagine it feels nice to understand.

Are "physics tensors"' the same as those in algebra?

yes, Rem.

12:34 PM

Einstein's field equations ... wow (though I don't know anything about 'em)@Huy

You know there are people in my class who claim they have general relativity thoroughly @Soham...

@SohamChowdhury: I don't know much algebra but I think the answer is positive. In physics, you need to know much less about tensors though to use them, it suffices to think about them as some multilinear forms (or "multidimensional matrices"). When I learnt about tensors in linear algebra, they were introduced through the tensor product and its universal property, which was a bit abstract to imagine at that point.

Soham Chowdhury

I think I should ask @Ted or @Semiclassical.

Yes, I've seen the "matrix" thing, @Huy. I wonder . . .

@SohamChowdhury: The LHS of the equation consists of Ricci curvature and Einstein tensor, the RHS is the stress energy tensor. The LHS basically describes how curved spacetime is, the RHS gives information about the energy/mass in the space you are studying. The equation itself thus says that the curvature of spacetime depends on how much stuff there is, and that the way how things move depends on the curvature.

@Soham We define kernels in abstract algebra as the map $\phi : G\to H$ and $\operatorname{ker}(\phi)$ as the elements in $G$ which are map to the identity subgroup of $H$..
I was thinking is there any similar way to define kernels in topology such that the process of construction of both the kernels is somewhat similar

Soham Chowdhury

interesting.

@Rem, use the universal and see.

12:42 PM

Universal?@Soham

Soham Chowdhury

Universal property

In various branches of mathematics, a useful construction is often viewed as the “most efficient solution” to a certain problem. The definition of a universal property uses the language of category theory to make this notion precise and to study it abstractly. This article gives a general treatment of universal properties. To understand the concept, it is useful to study several examples first, of which there are many: all free objects, direct product and direct sum, free group, free lattice, Grothendieck group, Dedekind-MacNeille completion, product topology, Stone–Čech compactification, tensor...

(btw just for the sake of completeness, the LHS actually just is the Einstein tensor. The Einstein tensor itself consists of Ricci tensor and scalar curvature of course)

Soham Chowdhury

thanks, @Huy.

@Rem, learn about the u.p. that ker satisfies in rings/groups and see what analog it has in topology.

@SohamChowdhury: If you want a more or less compact introduction, I can send you the lecture notes of the lecture I took last fall.

Soham Chowdhury

Please do.

that's very kind of you. :)

12:46 PM

@Huy well I haven't in my whole life read anything about general relativity .. But is general relativity all about the famous Einstein equation (E=mc^2)

@Rememberme: Not really, no. That famous equation was actually found when Einstein was studying special relativity, before general relativity.

So what is general relativity all about?

@Rememberme: Gravity. :P

Earth's gravity ? But wasn't it already been touched and done with by newton?@Huy

@SohamChowdhury: Do you know SR already?

Soham Chowdhury

12:49 PM

like Lorentz transformations and stuff?

@Rememberme: Newton studied things in terms of "forces", as you probably know from Newton's laws. There is a very short list of fundamental forces, and gravity was thought to be one of them. However, Einstein's theory of general relativity says that gravity is actually not a force, but a geometric property intrinsic to spacetime.

@SohamChowdhury: For example, yeah.

Soham Chowdhury

user image

:)

Soham Chowdhury

@Huy superficially.

@SohamChowdhury: itp.phys.ethz.ch/education/hs14/GR/gr

Can you access that PDF?

Soham Chowdhury

12:54 PM

yes.

ETHZ is cool. But a) Switzerland is expensive and b) they ask for a C1 in German, and I only have a B1. :'(

@SohamChowdhury: If you want exercises (including solutions) at some point, just ping me. Also a short PDF on SR if you think you don't remember all of it: itp.phys.ethz.ch/education/hs14/GR/sr

@SohamChowdhury: Do the C1? I doubt it's very difficult, and there's a lot of foreigners here who hardly speak any German.

@Huy you are in Zurich?

@SohamChowdhury: Also there are scholarships if money is a problem, and if you have excellent grades (which I wouldn't doubt for a second), you'll surely get the scholarships.

@Rememberme: I am currently at the border of Zurich, but I study in Zurich, yes.

My bro went there for an internship I guess

Cool!

12:57 PM

I want to go there but .. how is my question @Huy

Soham Chowdhury

it is quite hard, @Huy. that's near-native fluency.

@Rememberme: Apply, or ask Soham how to apply.

Well we both have 1 year left before we start applying for these places .. if we manage to get good marks

Soham Chowdhury

I was in Germany last year. It was sort of . . . saddening to see ~8 yo kids speak flawless German. :P

haha

I dont even know German .. so i guess I cant enter in Zurich ...

Soham Chowdhury

1:00 PM

ETHZ teaches math properly. I remember some first-year undergrad on /r/math who said they were doing tensor products in algebra.

If you can't get in due to German for your BSc, you can consider entering for your MSc. Iirc there are no language restrictions there.

Soham Chowdhury

That GR pdf is beautiful. Adjoints!

@Huy yes, I know that.

@SohamChowdhury: We did have tensor products in linear algebra, that's what I was just saying before. :P

Soham Chowdhury

First-years! :')

can you study pure math there?

The pdf also has bilinear forms .. I saw one :p

Soham Chowdhury

1:01 PM

yes, indeed.

@SohamChowdhury: Yes, of course. I study pure math. :D

Soham Chowdhury

GR?

Well, almost pure, hehe.

@Huy what are the costs there .. any idea ?

@SohamChowdhury: You can choose from almost any math-related lecture, but I was interested in some physics, that's why.

Soham Chowdhury

ah, cool.

1:03 PM

nice!@huy

Soham Chowdhury

Switzerland is expensive, @Rem.

@Rememberme: Living costs are really the biggest problem. Uni costs about $1200 per year.

damn costly!!

Compared to top UK or US unis, it's really nothing.

Soham Chowdhury

Yes. What are the living costs like?

1:04 PM

I know look at Caltek...

@Huy that is already 75,600 bucks in India

Soham Chowdhury

Caltech*

@SohamChowdhury: Roughly $1000 per month, or a bit more, depending on the way you live.

@SohamChowdhury: I know few people who can get by with less.

Soham Chowdhury

:(

That is a deal-breaker, indeed.

I can live anyhow ... just I want to study maths there

That's where your scholarship helps! :)

1:06 PM

What are the range of the scholarships you can get

Soham Chowdhury

I think you have to give some kind of test there, in person?

dcholarships in this country dont suffice to buy a book

@Rememberme: I don't know about the scholarships to support people who can't afford to live there, but a friend of mine got an excellence scholarship because of his great grades. He received about 10k USD per semester (so 20k USD per year) for his MSc, and didn't have to pay tuition fees.

Thats amazing !!

Yeah, with 20k you can live very decently as a student. Plus you can further get a job as a TA and earn some more to buy many books. :D

1:09 PM

:)

@SohamChowdhury: I have no idea about a test. I just know a friend who had amazing grades and told me he got that scholarship. He also got a personal office at the ETH to work/study in.

@Huy ETHZ comes under SAT?

even in internet availability of free books is decreasing

Soham Chowdhury

@Rememberme SAT?

I have no idea how SAT exactly works and what it exactly is, except that it's some standardized test.

Soham Chowdhury

1:10 PM

No, ETHZ does not accept SATs afaik.

Sad...

i remember five years ago, e-books were more spread than these years

Soham Chowdhury

@Huy yeah, I get it. It's just that actually flying out for a test you might not do well enough in is a bit hard.

@SohamChowdhury: I can imagine.

Yes its really hard for especially us Indians @Huy

1:13 PM

@SohamChowdhury: I'm grateful I grew up here, I just went to a public high school and then could go to the ETH without further problems. It's a lot harder for people from elsewhere, unfortunately.

you can just get free papers where free books are barely found except if you use peer to peer servers and thats dangerous

And we have really less good places to do maths in India.. though we have ISI and dont have any idea how this math powerhouse is

Hello , Help needed

Is there anybody

Total cost of studying in Zurich (without travel expenses is ):
15,62,826 inr@Soham

:D

Soham Chowdhury

1:17 PM

a year?

Thats damn costly!!

Soham Chowdhury

that's around a third of any American univ.

Yes @Soham

A year

@SohamChowdhury: And probably not including living costs in a US univ. While food etc. is cheaper, it's still not free. :(

:(

Soham Chowdhury

1:18 PM

ah.

@Soham you have idea about costs in ISI?

If any of you guys come to Zurich at some point, I'll promise to invite you to some great food.

Sure !!

Food's my best forte :p

I really want to learn to cook more things when I'm done with uni.

I love to eat

1:21 PM

I do too. :D

Okay to compare costs of ETHZ and MIT

Abstract algebra : Claim D3(reflections) isomorphic to S3(permutations) , I COULD NOT find a homomorphism preserved in a bijection. Is there any hint ?

Chris's sis the artist

@robjohn amazing results come one after another, there is a flood of them here. :-)

MIT $63,250$ ETHZ $24,748$ ... my god!!@Huy

Yeah, the US unis' tuition fees are crazy.

1:26 PM

@Chris'ssistheartist Great! I have to check my computer in for repair. There is something wrong with its wireless.

How do I get better at impossible ODEs?

@MickLH: Drop the word "impossible".

I should be clear, but I wasnt ;)

Chris's sis the artist

@robjohn I see. I also have problems with this one, it seems HDD spins so much that everything blocks here once in a while.

I have an ODE that's stacked up to like 8th order with radicals inside trig and all manners of horror

1:28 PM

@Chris'ssistheartist: SSD.

Is there really nothing to do except numerical solutions?

Chris's sis the artist

@Huy Yeah, that would be a good solution. I intend to buy a new computer anyway.

@robjohn did u see my proposition ?

See, I know you guys are holding out... chris sis could just solve all the specific cases that are interesting and make me look ridiculous for asking

"Can't you see these elegant solutions, kid? All you had to do was manipulate through like 7 infinite series and use recursive induction... did you even try?"

2

@Chris'ssistheartist since you keep passing these enormous series to your chipset its likely eventual that it would crash out

Chris's sis the artist

1:33 PM

@Agawa001 :-)))))))) I think i7, 5th generation would be great. :-)

@robjohn did u try updating your driver ?

Chris's sis the artist

(although I'm afraid not to buy it and then see the 6th one being lunched)

@Chris'ssistheartist ah, cool, i still run an i3

Chris's sis the artist

@Agawa001 Someone suggested me to try FX-8350, but ...

fast ram might be worthwhile if you're stressing memory bandwidth with heavy computations

Chris's sis the artist

1:35 PM

DDR4

@Chris'ssistheartist it doesnt help me for my researches, i struggle hard to replace it

Chris's sis the artist

@Agawa001 What do you wanna get instead?

the clocks and timing of individual vendors and even between their products can vary even within a grade like DDR4

i5 with 4G of ram

And the clock speeds aren't always apples to apples when you account for latency figures and burst modes

1:39 PM

i had a computer when i was 11 and look at me now deteriorating, fate irony

@Rememberme hey!

1:53 PM

in the answer to this question math.stackexchange.com/questions/1352291/… what does $\binom n{k,l,n-k-l}$ mean

2:06 PM

@Lembik its called multinomial

$\frac{n!}{k!l!(n-k-l)!}$

thanks @Agawa001 !

@Agawa001 say I changed the probability of getting a -1 in v, can you see what would have to change?

well, sorrry, ididt quiet read your question carefully, i just replied regarding what i m informed , excuse me

Hi @Agawa001, are you still here ?

If you take the example you made yesterday and drag the "B" point at the bottom of the circle, the "point3" will stay on top of the circle instead of following it. Is there a workaround to this ?
http://tube.geogebra.org/m/1431445

@A_V i think people sleep at night, unless you are nocturnal

@A_V yes i know

i made that quickly without so many constraints

2:23 PM

Well I wish sleep got me better at mathematics

a thorough work need to be surrounded from all sides, so that must be some inequalities i wasnt aware of

$cos(\theta)$ might be negative, thats one prediction of system failure

things can be fixed if you use $|cos|$

and $|x-x_0|$ for segment length

okay, so the absolute value of cos in the Y formula ?

that might bring it to work

@TedShifrin Have a safe trip! Let us know how things are in SD!

@Agawa001 It's annoying that the notation usually changes. I think the trinomial should be $\binom{n}{k,l}$

The absolute value of what exactly ?

I'm lost again :(

Should I use $|x-x_0|$ while calculating the radius ?

2:32 PM

@robjohn yea it sould be like this, thats what i thought about

@A_V wait

This is where I figure I should had spent those 1000s of hours learning useful stuff like maths instead of playing video games

Did you at least play good video games?

@Agawa001 I don't think it is the driver. I am able to use the wireless for a while after restarting and even longer if the computer has been off longer before restarting. It seems like a hardware problem. I am taking it in this morning for a more extensive diagnostic.

Michael Albanese

A smooth invertible function may not have a smooth inverse (consider $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x^3$). Does it necessarily have a continuous inverse?

@Agawa001 of course, if we adopt the trinomial $\binom{n}{k,l}$ then that equals the quadrinomial $\binom{n}{k,l,n-k-l}$

2:36 PM

@Huy Yes only the best

@robjohn what kind of computer is it?

...just wondering.

Thanks, @robjohn. I plan to stop in LA the week(end?) after Labor Day on my way back from SF.

@MichaelAlbanese: yes.

Would anyone be so kind to help me understand the answer to math.stackexchange.com/questions/1352291/… please

Michael Albanese

Hi Ted. Why?

I'm sure I should already know why.

what for example, would change, if the probabilties for v were P(v_i= 0 ) = 1/2 , P(v_i= -1) = P(v_i= 1) = 1/4?

I don't understand some of the explanation given.

2:48 PM

Well, you know it for $\Bbb R$ ....

oh what does XOR mean in the explanation?

hi @TedShifrin

Enjoy your flight professor @TedShifrin

hi @robjohn

Hi Lembik

Exclusive or?

ok but what is being exclusive OR'ed with what?

the sentence doesn't parse for me

2:50 PM

I haven't looked.

"If $v$ has $k$ entries $+1$ and $l$ entries $-1$, the probability of its product with a row $r$ of $M$ being $0$ is $2^{-(k+l)}$ times the number of ways of selecting $k$ of the $k+l$ non-zero entries in $v$ and choosing $r$ to be $1$ when (entry is selected) XOR (entry is $+1$). "

hey guys someone can help me

with one "simple" problem?

@MichaelA: I guess it depends on your domain ... Manifold? Otherwise

the standard counterexample stillholds.

Michael Albanese

Yeah, manifold.

Hi @TedShifrin

How goes packing/moving?

Michael Albanese

2:53 PM

What counterexample are you referring to?

Is it valid to write $a \neq b = c$ to mean $a \neq b$ and $b=c$ or is it ambiguous/incorrect?

$[0,1) \to S^1$

Valid.

About to board the plane, Balarka.

ah. Hope everything goes well.

2:55 PM

d(n)=15-2,5*cos(╥(n-31)/360)

Hello. Can somebody tell how I change my username to any name I like ?
Does everbody has this option ?

this about days

Michael Albanese

Uh, manifolds with boundary.

when take this on my calculator is one rect

@MichselA: I think smoothness is a red herring. It should be an exercise in point set, assuming manifolds (without boundary).

2:57 PM

and i have one question ask to me when the result is 12,3 but for me is always 12,5

with boundary, I.gave you the counterexample.

Can someone help me ?

Michael Albanese

Sorry, I'm not sure I understand what you are getting at. What is the point-set statement you seem to be referring to?

@skillpatrol It is an old MacBook Pro, circa 2010.

d(n)=15-2,5*cos(╥(n-31)/360)
can you help me with that function?

3:00 PM

Ok, not point set. I guess it might be invariance of domain. I have to board the plane. Outta here ...

Michael Albanese

No worries. Have a safe flight.

;_;

I have a theorem with proof that given a paracompact smooth manifold $M^n$, for any finite $d$ we can realize a rank $d$ real vector bundle over $M$. The proof starts by taking a smooth map $u: M \to G(N,d)$. How do I figure out $N$?

@robjohn time for an iPad Air 2 :-)

Circa 2015

@TedShifrin Let me know when.

3:04 PM

@MikeMiller

@robjohn math.stackexchange.com/questions/1281770/… .. is my answer alright now?

0

Q: Can you help me with a function with cos and radians?

I have this function d(n)=15-2,5*cos(╥(n-31)/360) this function talks about the difference between when day start and day over . when i change n for 5 result for me is 12,5 when i change for 10 result is 12,5 when i go insert the function on my calculator the function create one graph straight ...

functions trigonometry

can you help me pleasE?

please*

What does it mean to find a linear, homogeneous system that characterizes the column space of a matrix?

3:25 PM

@Huy: I don't understand. You can always realize a smooth vector bundle. Pick $M \times \mathbb R^n \to M$ the projection map.

@MikeMiller: Nontrivial, I guess?

But that's not true, every vector bundle over a contractible whatever is trivial.

So you don't know what kind of construction that was supposed to be?

No. WHere are you reading it?

Notes from lecture.

Apparently the construction simply yields the pullback of the tautological bundle of the Grassmanian.

3:28 PM

@Huy: OK, well in any case they actually are building a vector bundle. $N$ doesn't matter.

As long as $N \geq d$, $\text{Gr}(N,d)$ is a thing that makes sense, and the construction will work fine.

Ok, thanks.

@BalarkaSen: I'm going for a hike, I'll think later. The answer is probably still no.

@anon: I'm going for a hike. I'll think about this when I get back.

I was apparently very popular last night.

You wish!

tsk tsk

3:50 PM

@robjohn try to fix or embed the antenna of your hardware, maybe it cant capture the traffic because its kinda loosy

@Agawa001 I saw you played a bit with your live example, did you find an answer to this circle equation ?

im still working

lol did it extend to public view ?

I can see the extra points you added, how does geogebra works? Where's the equation behind the graph ?

@A_V its an absolute value issue (i felt it before)

Only the Y axis seems to suffer

to I tried adding ABS() functions everywhere in my code and it did not happen to work

so I tried *

3:56 PM

when the mobile point go beyond the circle axis, the other point changes direction

i know why and i m fixing it

the point is, you must split cases

there 3 or 4 cases you must take count of

I can make conditions, that wouldn't be hard

different equations for different angles ?

fixed

what about now ?

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