Mathematics - 2017-03-23 (page 3 of 4)

@SteamyRoot

34 mins ago, by SteamyRoot

But the idea is that, since in every neighbourhood of such point, the ratio is bounded, hence it's a removable singularity

@Brody I'm okay with it. It makes people less vulnerable to "one bad quiz".

I think some profs have done it as: "If you do better on the final than your lowest quiz score, then that lowest quiz score gets replaced by your score on the final."

Which I think is fine as well.

@SteamyRoot Yeah. The fact that I use Mathematica a lot can be chalked up to my having access to a free copy of it through the university.

If I didn't, I doubt I'd pay for it.

Hi@Semiclassical

3:15 PM

hi @baymax

You know about casimir functions?

I know about Casimirs in algebra.

Lie Algebra ?

Yeah.

Yea that one only!

I was thinking in that paper that to know about the dynamics

A plot of Hamiltonian and Caimir was taken ?

3:18 PM

In quantum mechanics, a casimir is an observable which commutes with all other observables. (So it's in the center of the relevant algebra.)

Yes

At the level of classical mechanics, I suppose it'd be something which has zero bracket with all coordinates/momenta.

That is intersection of the surface of H and C can give us information about the dynamics ?

Yeah

I suppose the point is that if C has zero bracket with all coordinates, then it's necessarily an invariant of the motion.

At the same time, H should also be an invariant of the motion.

as H is conserved right

3:21 PM

If these are independent quantities, these will represent two surfaces in phase space, and the system's trajectory in phase space should lie on both of them.

the system's trajectory in phase space should lie on both of them. ?

Let me put it differently. Suppose I have a surface of constant energy in a 2n-dimensional phase space.

ok

That'll correspond to a codimension-1 surface in that 2n-dimensional space.

oh..codimension?

3:24 PM

Yeah. A plane in 3D space has dimension 2 and codimension 1.

A line has dimension 1, but in 2D the codimension is 1 while in 3D it's 2.

A plane in 3D space has dimension 2 and codimension 1.\

is it like this

3-2

Right.

=1

ok

ha interesting!

In other words, by requiring H to be constant I've in effect reduced the number of variables for my trajectory by 1.

If I know 2n-1 of them, I can infer the last one from H.

However, that still gives me a lot of freedom for what my trajectory is. If I started out with 2*2=4 dimensional phase space, knowing the total energy still leaves me in a 3 dimensional subspace.

So it doesn't uniquely prescribe the system's trajectory.

still stuck at this -"That'll correspond to a codimension-1 surface in that 2n-dimensional space."

3:29 PM

Suppose you've got two particles of mass m=1 subject to some interaction potential. Then your Hamiltonian would be H=p1^2/2+p2^2/2+V(q1,q2).

Yes

If I know the initial energy E, I know that H=E for all times. So my trajectory is restricted to that level set H=E.

a surface in 2n dimension space has codimension 2n -2 ?

as a surface is a 2dimensional

Ah. I didn't mean it like that.

Suppose V(q1,q2)=(q1-q2)^2 (a Hooke's law force).

ok

3:32 PM

Suppose I know what the total energy is. If I know what q1,q2,p1 are, then I can find p2 from this H.

(at least, I can find only finitely many possible p2. Could be plus/minus.)

yes

So my hypersurface H=E isn't 4D. It's only 3D.

I've got 1 constraint, so the hypersurface it produces is a 3D subset of 4D space.

(If they say 'surface', they probably mean 'hypersurface' i.e. not necessarily 2D.)

I need to know a hypersurface

One more way to say it. Suppose I'm only in 3D space, and I give the condition that x^2+y^2+z^2=1. Then that's a 2D surface---the constraint has dropped the dimension by 1.

If I'm in 2D space and I have x^2+y^2=1, then that's a circle i.e. a 1D subset.

x^2+y^2+z^2=1 - a sphere (3D) ?

3:38 PM

If I'm in an n-dimensional space and I have a constraint equation, then in general I'll have an (n-1)-dimensional subset (hypersurface).

Right. If you zoom in on the sphere, it looks like a plane up close. Hence, 2D.

oh.ok

So if you have a 2n-dimensional phase space, then the requirement that H is constant limits you to a 2n-1 dimensional subset of that space.

Nice

However, that's not enough to tell you what the trajectory of your system in that phase space is. Such a trajectory is a function of time, so it's 1D.

Going back to the sphere picture, there's a lot of curves you can draw on it.

So just knowing that you're on the sphere doesn't give you a unique trajectory.

On the other hand, suppose someone tells you that you always stay at the same height on the sphere i.e. z=constant.

Then the trajectory is unique: You're stuck on a given line of latitude for all times.

interesting

Such a trajectory is a function of time, so it's 1D.

?

3:42 PM

Right.

how we arrived at that

where are other coordinates

like x,y,z

x=x(t), y=y(t), z=z(t)...

okk

If I change t, then all variables change at once.

If I wanted to have a different trajectory, I'd need to (for instance) change my initial x,y,z.

Yes

3:44 PM

But once I pick the initial condition, the time evolution of the coordinates/momenta is prescribed by Hamilton's equations.

Hence they're unique functions of time, and the only parameter that matters in the trajectory is the time itself. Hence, 1D.

ok

So conservation of H limits me to a 2n-1 dimensional subspace, but my trajectory can in principle be any 1D curve in that subspace.

Yes

If I want to limit that trajectory further, I have to look for other conserved quantities of the system.

Trajectory - a 1D curve in a 2n -1 dimensional subspace.

3:46 PM

Right.

Going back to the two-particle example, energy conservation restricts me to some 4-1=3D subset of phase space in which my 1D curve must live.

yes

this 3 is the codimension?

no, 1 is the codimension. 3 is the dimension.

4 because of p1,p2,q1,q2

Right.

oh,sorry ok

3:49 PM

There exist open simply connected subsets of $\Bbb R^3$ that are not homeomorphic to the complement of finitely many points, fun fact

However, we can say more. Note that if H=p1^2+p2^2+(q2-q1)^2, then the Hamiltonian is invariant under (q1,q2)->(q1+Q,q2+Q).

In other words, the Hamiltonian doesn't change if I shift the coordinates by an overall constant.

nice there

so H is invariant under coordinate shifts.

As a consequence, we have that $\frac{dp_1}{dt}=-\frac{\partial H}{\partial q_1}=2(q_2-q_1)=\frac{\partial H}{\partial q_2}=-\frac{dp_2}{dt}$

Or, rearranging, $\frac{d}{dt}(p_1+p_2)=0$.

In other words, the total momentum $P=p_1+p_2$ of the system is a conserved quantity!

So that's another constraint that my trajectory must satisfy: It lives on a codimension-1 hypersurface P=const.

yup

Now, P and H are distinct quantities. So therefore these two hypersurfaces are distinct.

Hence if both quantities are conserved, my trajectory must lie in the intersection of two such hypersurfaces.

3:55 PM

Yes

And the intersection of two codimension-1 hypersurfaces is usually a single codimension-2 hypersurface.

Wait, how is the complement of the Whitehead manifold simply connected, let alone contractible?

I don't see any ways to nullhomotope a loop through the middle

So now I'm down to a 4-2=2 subspace of the full 4D phase space.

So our trajectory is a single codimension-2 hypersurface?

yes

No. It's still a 1D trajectory. But we've narrowed down the portion of the 4D phase space in which it can live.

3:58 PM

ok

Which if we think in terms of physics is hardly a shock: Suppose my system starts from rest. Then once I pick the initial positions of the two particles, the trajectory thereafter is completely determinate.

yes

it is easy that it could be determined

So my system trajectory will in general depend on the two initial conditions starting from rest.

yes

Hence I've got a 2D space of possible initial conditions starting from rest, each of which determines a 1D trajectory through the full 4D phase space.

Now, let's go to your example.

How big is your phase space?

4:01 PM

yup

(how many coordinates/momenta)

oh .. its Chuas system

so 3

3 positions?

x,y,z

yes

That'll be a 6D phase space, then.

4:03 PM

$\dot{x},\dot{y},\dot{z}$

Knowing the Hamiltonian restricts you to 5D; knowing the Casimir furthermore drops you to 4D.

yes

If you then assume that the solution starts from rest, then you're restricted to a 1D space of possible initial conditions.

actually 1D space ?

Right.

To be more definitive:

You start from rest, and you pick some specific starting y,z.

4:06 PM

ok

Hmm, I think I may have spoke too soon..

I'm not sure exactly how they're using it.

I feel like they've got too few constraints to be able to visualize anything.

hindawi.com/journals/jam/2013/764108

A 4D subspace is still pretty big, even if it's a big restriction on the full 6D phase space.

ok

Oh, I see. Their Hamiltonian and Casimir only depend on x,y,z.

4:10 PM

yes

In that case, the dimension counting is different. You start with x,y,z; knowing the Hamiltonian restricts you to a 2D subset, and knowing the casimir further restricts you to a 1D subset.

e.g. the intersection between the cylinder and the plane.

So in that case, you know that the trajectory must lie on that circle.

yup

You don't know how it moves on that circle, of course: It could move at a uniform rate, or it could stop/start/reverse.

nice

But it can never move off of said circle.

4:11 PM

yeah

Hence you've constrained the dynamics quite strongly.

since the dynamics follow both H and C so we get trajectory in the intersection of them that is intersection of H and C and both are surfaces ?

Right.

In this case, they're surfaces in the usual 2D sense.

yeah

cylinder is 3d

?

2D.

It lives in 3d space, but if you zoom in on any part of it it's a plane.

4:15 PM

Hey all :-)

hi

actually i am thinking why are we zooming it?

Yo, @Semiclassical

To see that it's 2D not 3D.

yo yo @Kari

4:16 PM

hey hey, @BAYMAX

oh

A plane in 3-space is certainly a 2D object, and if we zoom in on the cylinder we see that it looks the same as a plane. Hence the cylinder is also 2D.

Got a quick question on why two different defs of continuity are equivalent. Wanna hit me with some insight, @Semiclassical?

Gonna have to pass.

Perception changing?

4:17 PM

Cool :-)

ok,cylinder lives in 3D but is 2D , so funny :)

Hi guys

Which defs? @kari

Eh, same with a sphere.

anyone familiar here with multi-variable calculus, gradient, Jacobian?

4:18 PM

okay, leaving for now.

yo its a plane

ok bye@Semiclassical

nice day!

Q: Sequential continuity on metric spaces

Ah, I seem to have found an answer on stack exchange, @Alessandro

1

Please give me a hint for proving this statement: Let $(X,d)$ and $(Y,d')$ be metric spaces, $f$ a function from $X$ to $Y$. If $f^{-1}(B) $ is closed in $X$ for all closed subset $B$ of $Y$, then $f$ is sequentially continuous. Note that $f$ is sequentially continuous if and only if for a...

@BAYMAX some 2D surfaces live in 4D space

yeah

like f(x,y)=0 in g(x,y,z,w) = 0

@AlessandroCodenotti

or is it silly mistake there?

Not sure what you mean. I was thinking about $\Bbb P^2(\Bbb R)$ for example

You don't even need a metric space, sequential continuity and continuity are the same in first countable spaces @kari

4:23 PM

That's a bit general for my taste at the moment

ok

Basically in metric spaces you use the balls with radius $1/n$ to construct the sequences needed to show the equivalence, in first countable spaces you use a countable local basis instead but the proof is the same

That makes sense

$1/n$ is countable so the basis idea followss

(metric implies first countable because of the 1/n balls by the way)

Is there some kind of diagram with all the inclusions?

Like normed $\subset$ metric $\subset$ Hausdorff etc.

4:27 PM

There are some huge ones in "counterexamples in topology" including plenty of little known properties that it's not really worth remembering

The important thing to know is that metric implies first countable and at least $T_4$ as far as separation axioms are concerned (probably more, I'm not sure)

Hm, yeah, metric spaces should also be $T_5$ and $T_6$

I don't think any of this is super important to remember :P

It's very helpful in the grand scheme

The diagrams (inclusions and implications) are really helpful

There's plenty of them. Like compact+Hausdorff implies $T_3$ and $T_4$

4:46 PM

Any one wants to commit?

current statues
19/100 committers with 200+ rep

need 81 more committers
Please help
Sorry for the inconvenience I am causing

where to commit?

link

I don't know any Maple

59

Maple

Proposed Q&A site for students, teachers and researchers who are using Maple for symbolic and numerical computation.

Currently in commitment.

Good syrup though

@BAYMAX Thanks for your interest

@AkivaWeinberger Just help us out and commit

4:55 PM

All the best@MapleSE-Area51Proposal

@BAYMAX Thank you dear

@MapleSE-Area51Proposal What does "commitment" mean

@AkivaWeinberger literally "dedicated to a cause"

But what does it mean for this? Does it mean I have to regularly participate in the Maple site once it's created?

Or does it just mean "I think this should exist, even if I won't use it"

You don't need to participate

5:01 PM

The whole point of the committing is to ensure the site has an active community once it's released...

We need 100 SE users with rep more than 200 to reach beta site

The problem is that the commitment made by mostly new users but to reach beta we need 100 SE users with rep more than 200

Out of 60 committers 41 are new users

So any of you guyz having rep more than 200, please commit.

Was trying to solve an improper integral $\int_{-\infty}^{\infty}\frac{\cos(3x)}{(x^{2} + 1)^{2}} dx$

By Cauchy's Residue Theorem

Got the singular points $i , -i$

Seriously:
1) do you really have to spam this everywhere and every few hours?
2) committing is supposed to be something serious: if you commit to a proposal, you are saying that you want to be active on it when it goes to beta.
3) These "new users" you speak of are all 51 reputation; with not a single other SE. This smells of someone mass-creating accounts to boost the proposal.

I'm all for creating new SE sites if there's an audience (even if I don't like what it's about)

But this isn't the way to do it.

but I was thinking that why we take the contour semicircle also there are some missing residues and we are not taking them ?

Dear, I am advertising it here if someone interested then he/she can commit

5:15 PM

Yes if this is the way then it is wrong . I hope you are doing the right thing.

If someone is interested, then they should commit.

Asking someone to commit and telling them they don't need to participate afterwards, however, invalidates the whole purpose of commitment.

If they dont want then how can you force them?

Do you think I am getting any benefit from it? I am trying to create for others. I myself is well versed in both Maple and Mathematica, So I don't need any site for myself.

I'm not saying you get a benefit from this or that you should force people to participate

I'm saying you shouldn't ask people to commit if they're not interested in participating.

From my experience, there are negligible number of SE users who are interested/know about maple.

Julius

Hi. Any suggestions on how I could show that the maximum likelihood estimator, when exists, coincides with a method of moments estimator only for exponential family?

5:20 PM

Point taken.

eurocoder

At the very start of these MIT lecture notes the author integrates 1/(y-1) and gets log|y-1| - Why is it not just log(y -1) without the absolute value sign? I've seen this in many places, why do they use the absolute value sign here?

Oops, the lecture notes are here: The lectures are: ocw.mit.edu/courses/mathematics/…

@eurocoder What will be log(y-1) if y<=1?

5:48 PM

hi

anyone has some familiarity with multivariable calculus?

@nbro I'm kinda ish aight with it, not as much computation but eh

just some partial derivatives... no?

:D

Show me and I'll try

Balarka Sen

@skullpetrol Ah. I wouldn't know.

I know essentially the topics, but I've not applied them much, so when I have to do it I'm basically unsure on how to proceed

anyway

I've asked this question here: math.stackexchange.com/questions/2200017/… regarding how to find the Jacobian matrix of a multivariable function

the problem is that the function isn't that is to take the partial derivatives of

*isn't that easy

5:56 PM

Alright, I'll check it out once I have breakfast

ok, thanks!

anyway, apparently we should be able to find the Jacobian with the help of the gradient that you can find at the end of the question... I used it in the result I'm suggesting, but I think I used it wrongly...

the mentioned gradient is with respect to j1 (and j2), which is actually defined in terms of the angles $\phi_1$ and $\theta_1$, but, still, I don't think I can do what I did in the suggested Jacobian...

why? essentially because the first row should be the row of partial derivatives with respect to phi1, phi2, theta1 and theta2, so we should have 4 columns (right?)...

but what I did results in the subtraction of two 3D vectors...

which isn't clearly correct...

but, again, somehow I should be able to use the gradient to find this Jacobian...

Anheuser

6:20 PM

Has anyone ever heard of an Amish mathematician?

Come across one in their studies?

HSM has downvoted me and thrown me out.

And math.stack moved it to HSM.

I've shown that for $X\subset \mathbb R^n$ it holds that: $X$ is compact iff $X$ is closed and bounded. Now I would like to show that this equivalence holds for any set in an $n$-dimensional vectorspace $V$. I was thinking of using an isomorphism between $V$ and $\mathbb R^n$, but I'm not sure how. Could someone help me with this?

Whoa, Sha Vuklia, I just realized that I've been misreading your name as Sha Vulka the whole time

I'm really sorry for the elementary question, guys, but can anyone link me to a good resource on rearranging formulas? I need to rearrange delta = Y/X^1/t so that t is the subject. I really need to go back to basics, here.

@Akiva Hahahah :P It's from Bosnian "vuk" (=wolf), so I like to think of it as "the wolf-like" :P

Cool!

6:30 PM

Correction: So that X is the formula. I'll see what I can find on YouTube, Khan Academy, those kinds of resources.

@jserv Try raising both sides to the t-eth power

*So that X is the subject. God, it's been a long day. Thanks @AkivaWeinberger , I'm dimly remembering the procedure of applying things to both sides, but I'm struggling to go through it in a sensible way with that formula

skull petrol

That's a good place to start @jserv :-)

Khan academy.

@jserv This page and this page have decent reviews of the rules for exponents

Cool, thank you @skullpetrol . And that's extremely helpful @AkivaWeinberger , much appreciated.

6:44 PM

hello everyone

7:10 PM

Which wild sphere do you like better, Alexander's horned sphere or this thing

@Sha So the way to do it is to choose a basis for the finite dimensional space, and then do a coordinate mapping. That is an isomorphism, which in finite dimensions is also a homeomorphism.

7:23 PM

Hi chat

Mike Miller

@AkivaWeinberger Take the bad side of the horned sphere; call that the horned ball $B$. As you know, it's not simply connected. Now take its double: $D(B) = B \sqcup B/(\partial B \sim \partial B)$. That is, glue two copies of it together along the boundary. Then this is still homeomorphic to $S^3$, which means $S^3$ has a wild sphere whose complement is the horned ball on both sides.

@Daminark Yea, I thought of that too, but how to use that isomorphism?

and hi @Astyx

let me think

Hello @ShaVuklia

@MikeMiller I'll take your word for it that that's still homeomorphic to $S^3$$—though it's not hard to construct such a sphere anyway. Take a horned sphere, and take its inversion, and then connect sum them.

(Its "inversion" is essentially the same as "the bad side" I guess)

@Daminark We want to show that for each continuous function $L$ it holds that $L(X)$ is compact, if $X\subset V$ is compact, right? Now how can I use the coordinate function? First of all, it's a specific continuous function. Second, I don't even know how to show that it holds for the coordinate function :P whatever "it" may be

Mike Miller

7:34 PM

@AkivaWeinberger I don't think I buy your claim that it's not hard to construct such a sphere. Seems to me yours still has a good side and now an even worse side.

Basically, you're getting what's also a homeomorphism, so a set is compact in one iff its image is

Hello

hi chat

Also it's worth noting that norms on a finite dimensional space are equivalent

@Daminark ah ok I guess. I think I'm lacking knowledge, because I'm unfamiliar with the theorem that a set is compact iff its image is, if we're dealing with a homemorphism

7:40 PM

So the image of a compact set under a continuous function is also compact

Do you see why that's true?

yes, I've proven that once

@MikeMiller It has one set of horns facing out and one set of horns facing in

OK, so if a function $f$ is a homeomorphism, say $V$ is compact, then $f(V)$ is compact by that. Now, if $f(W)$ is compact, $W = f^{-1}(f(W))$ is also compact, since $f^{-1}$ is continuous.

Mike Miller

Hm

oooooooooooooooo...oooohhhhh

7:43 PM

You're dealing with 2 different vector spaces, you can't necessarily give the identity

oh, I thought what I said was really stupid :P

ahh right

Forever Mozart

whats going on here

Suppose that $p,q$ are two distinct primes , $n=pq$ and d=gcd(p-1,q-1).
I want to show that if $a^{n-1} \equiv 1 \pmod{n}$ for some $a$ then $a^d \equiv 1 \pmod{n}$.

That's what I have tried:

$$
a^{pq-1}\equiv 1 \pmod n \Rightarrow a^{pq-p+p-1}\equiv 1 \pmod n \Rightarrow a^{p(q-1)+(p-1)}\equiv 1 \pmod n \Rightarrow a^{Ad+Bd}\equiv 1 \pmod n \Rightarrow a^{d(A+B)}\equiv 1 \pmod n \Rightarrow \left (a^{d}\right )^{A+B}\equiv 1 \pmod n$$

But from this, we cannot deduce that $a^d \equiv 1 \pmod{n}$, can we?

our coordinate function is as close as we get get to the identity function, from an intuitive perspective at least

That's why you pick a basis $b_1,\ldots,b_n$ in $V$, and then map $(x_1,\ldots,x_n)$ to $\sum_{i=1}^n x_ib_i$

7:44 PM

right right right

got it!!

thanks, you are a-ma-zing

No problem

Do convince yourself that all norms on a finite dimensional space are equivalent though

hahahaha

dude

I've spent two days proving that XD

I can dream the proof

(not that you could have known that :P)

does it make sense to talk about 'corresponding' sets of the natural filtration of two different random processes?

@Evinda yeah, that doesn't look like a sound conclusion.

for example, the set $ \{X_{t_2}<0\}$ and the set $\{Y_{t_2}<0\}$?

7:46 PM

@Semiclassical What else could we do?

can I call them 'corresponding' sets?

Not sure. My impulse is to use Bezout's theorem i.e. there's integers x,y such that x(p-1)+y(q-1)=d.

Lmao, that's good

We had it on a pset once, I remember

But I'm not confident that helps.

I mean, the task was to prove that if you have some Banach space $X$, then any finite dimensional subspace was closed

7:49 PM

@Daminark Wait but... did we prove what I wanted to prove now? That any compact set is closed and bounded?

Well, that compact sets are closed and bounded holds in every metric space

:l

To prove the reverse is what requires finite dimensions

I did so @Semiclassical so maybe this will not help :/

ah like that

7:50 PM

This also feels like a Fermat's little theorem problem, though.

Because then once you have an isomorphism/homeomorphism between $V$ and $\mathbb{R}^n$, your function say is $f$

Actually, hmm. Maybe the thing to do is view it mod p and mod q first, then see what that implies about mod pq.

If $W\subset V$ is closed and bounded, clearly $f(W)$ is closed and bounded

Which means $f(W)$ is compact, so that $W$ is compact

oh right

i.e. something something Chinese remainder theorem.

7:52 PM

wow okay

thanks!

No problem!

sorry I don't know why I'm so slow today, but really, thanks a lot

It's fine, don't worry about it!

@Semiclassical We have that $a^d \equiv (a^{q-1})^y \mod{p}$ and $a^d \equiv (a^{p-1})^x \mod{q}$, right?

Is there a use of Fermat's little theorem in there? If so, I think I agree.

7:55 PM

well I guess the chat is in number theory mode right now :|

(Well I'm not doing much, so rather what are you doing)

What are we doing ?

@Astyx You want to join the squad of people who would like to know number theory soon but for now are doing analysis?

Suppose that $d=gcd(p-1,q-1)$ for distinct primes $p,q$. Show that $a^{pq-1}\equiv 1$ mod $pq$ implies $a^d\equiv 1$ mod $pq$.

I know more about number theory than about analysis I think

Which isn't that meaningful anyway, but hey

Yes, I have used the fact that $a^{p-1} \equiv 1 \mod{p}$ and $a^{q-1} \equiv 1 \mod{q}$. @Semiclassical

7:57 PM

Okay.

How can we continue?

I'm not entirely sure it's valid, if I'm honest. The main problem is that I forget how negative exponents work in this business.

I mean from what I've seen of number theory it seems like a pretty dope subject, but I won't be doing that class until next year

And you have to think like that, because in general d=x(p-1)+y(q-1) would have x or y negative.

I mean there's an IBL number theory which I prob could do