hi

is a <=> b the same as b <=> a

yep

hi

hi

here was an interesting question I saw:What is an example of: Complementary adjacent angles formed by two intersecting lines.

i don't think it is possible with only two lines

anyone know?

This should be simple if it's possible, but how can I find the intersection of $y=x^3$ and $y=x^2 + x$? I factor an $x$ out of the second equation, then divide both sides by $x$ to obtain $x^2 = x + 1$ when I set them equal to each other. I get stuck there.

@JoshuaLamusga Did you try solving $x^2 = x+1$?

@user19405892 No. For some reason I forgot that finding the roots of that was the same as finding the intersection. And I have a calc II test tomorrow. I hope my brain isn't as fried then. (I've been cram-studying for 4 hours)

Just remember that $x = 0$ is a solution as well and you can solve $x^2 = x+1$ as normal.

Thanks. I've got a solids of revolution test tomorrow and I'm trying to find the hardest problems I can so I'll be extra prepared... do you happen to know any good ones?

Yes i do

Here is one;

In a space, let $B$ be a sphere (including the inside) with radius of 1. Line $l$ intersects with $B$, the length of the common part is the line segment with the length of $\sqrt{3}$. Find the volume of the solid generated by rotating $B$ about $l$.

I've got a picture of a sphere with a straight line drawn from any angle and intersecting its center. Is that the right mental image?

If the straight line intersects the center, that means that the segment is a diameter which is 2, but it is √3.

So it's a chord that intersects the sphere. I have to find a line rotated about a sphere. I'm not sure if I've got that level of knowledge yet; I've been using the disc/washer/shell method. I've heard of a cylinder method, but I don't know if that would help. I've been rotating about lines parallel to the x or y-axis, which would be a cylindrical body, I suppose. I guess I'd need something like $\dfrac{4 pi}{3}\int$stuff?$^3$. Is that close-ish?

Yeah you're getting closer. This problem may be a little advanced but I think doing it will give you a better intuition for revolutions of solids

I meant "since $g \circ f$ is surjective" rather. Is it correct?

@JoshuaLamasgu As a hint, find the distance between the center of $B$ and $l$.

(I'm actually taking a short break. I'll work on it when it's over.)

To find the distance from the center of the sphere, I was going to evaluate the semicircle $y = sqrt(1 - x^2)$ (from a 2D perspective) and solve for $sqrt(3)/2$. I got +/- $\dfrac{1}{2}$.

I just learned that when you said the sphere included the inside that it was a ball rather than just the surface area. Hmm. This is an interesting problem, but I might have to learn some more before I can tackle it.

Hello all! Analysis n00b here - couldn't the least upper bound property just be rephrased as "all continuous subsets must have a least element"?

That seems a whole lot simpler to me.

No.

no

least upper bound means exactly what you think it means

Hi

Hi

Hi

I proved something Mike.

It may be trivial but it isn't written down anywhere that I can find

Grats

I don't care if it's trivial. Is it interesting?

It's nice sounding

That's what matters

Sometimes

The contact manifold (the contact structure coming from the Milnor fiber) $\Sigma (p,q,r)$ for p,q,r coprime is never Legendrian surgery on a knot.

@JoshuaLamusga yep you are right, it's 1/2

congrats @PVAL

I like that a lot actually. If it's trivoal why is it true?

You just compute the homotopy class of the plane field associated to the trace and the Milnor fiber.

Why does that help? What's the obstruction to realizing such s thing by conscg surgery?

conscg ?

Contact

Is the least upper bound property equivalent to the axiom of choice?

No.

There's an integer invariant of plane fields $c_1^2-2\chi(X)-3\sigma(X)$ where $X$ is an almost complex filling of the plane field.

Computing this mod 8 (with X the Milnor fiber and the trace of a knot) deals with this.

@MikeMiller Cheers. Just wanted to make sure, since so many things seem to be equivalent to the axiom of choice.

Ah and if it was obtained as a surgery you could calculate that (considering the surgery as a filling) and it's wrong.

Ya.

This is for Legendrian surgery though (-1 contact surgery).

I can ask Ko if he knew this on Fri if you want.

I'd imagine if Bob didn't know this hed know who does.

Fair enough

Do you think this can be turned into a general obstruction?

Since both the homotopy classes of plane fields stuff and the Legendrian surgery stuff is him.

What does general mean?

This invariant is due to Bob in his 1998? Annals paper.

I just mean something you can apply to nice effect on more manifolds. But wow, annals, it seems so simple.

It's more complicated if the 3-dimensional $c_1$ isn't torsion.

I mean this is the paper where he essentially developed Legendrian surgery and put Stein structures and tight contact structures on everything he could find.

@Mike

@MikeMiller The Milnor fiber has some nice facts about it that make this easier. I can write a similar proof for contact manifolds filled by a manifold with these facts, but $\Sigma (p,q,r)$ is pretty general (no idea if its true if you allow the contact structure to vary).

It's false for $\Sigma (2,2,2)$

Which is $\Bbb RP^3$ and is filled by the -2 trace on the unknot.

I don't have another Brieskorn example where its false.

Is there a more or less natural contact structure on the bigger seifert fibered homology apheres?

@MikeMiller not really. There's those coming from reverse-slam dunking the rational pieces (when the base has framing -2).

Those aren't unique.

Yikes

I think I am modifying my candidacy (in close to one month now) to focus around this..

Even if it is known, its a nice application of the things I wanted to talk about.

It seems your candidacy is a bigger deal than ours are

It isn't. It's just some people present nice results during it, and I'd like to not feel so inferior to those people.

Apparently someone had to give a 3-hour uncondensed version of their talk to their adviser as a practice talk.

Ya fair enough

each cycle is a function for instance (1,2,3) sends 1 to 2 ,2 to 3 , 3 to 1 and fixes 4 and 5. The multiplication is just composition. See what $\tau(i)$ is and what $\sigma(tau(i)$ is for each i between 1 and 5.

@PVAL My reasoning was that $\sigma = (1,2,3)(4,5),~ \tau = (1,5)(2,4)(3)$. So we wish to compute $\sigma \tau = (1,2,3)(4,5)(1,5)(2,4)$ since we can drop the $(3)$. Starting with the smallest number $1$ we have the first cycle sends $1$ to $2$, second and third fix $2$ and fourth sends $2$ to $4$. Next the first cycle fixes $4$, second sends $4$ to $5$, third sends $5$ to $1$ and fourth fixes $1$. So we close the bracket $(1,4)$ since we already have $1$. Is this wrong?

@PVAL Next, we pick the smallest number that isn't already in the disjoint cycle we have constructed. That's $2$. The first cycle sends to $2$ to $3$, second and third and fourth fix $3$. So we have $(2,3 \ldots)$. Next, the first cycle sends $3$ to $1$, second fixes $1$ , third sends $1$ to $5$. So we have $(2,3,5)$. Oops, so the answer is $(1,4)(2,3,5)$?

I gave you a method that works. Sorry, but I am not checking your computation for you.

@PVAL I'm not asking you to check the computation. Just whether the answer I got using your method is correct (and I'll happily go back to the drawing board if it isn't).

Basically whether $(1,4)(2,3,5)$ and $(1,4,3)(2,5)$ are the same thing. (I'm a novice).

@GGG you need to compose from the right

and no, they are not the same. Cycle decomposition (i.e. with disjoint cycles) is unique

@TobiasKildetoft Thank you very much! I was going insane on this!

Hooray! Tobias saves another person's sanity :-)

@skullpetrol All in a days work...

...and we do appreciate it!

@MikeMiller: you said yesterday it's pretty obvious that the Laplacian is self-adjoint. did you really mean symmetric? I'm trying to show it's self-adjoint right now (in the manifold setting) and I'm comparing with what we've done in functional analysis (that was over $\mathbb{R}^n$) and it was quite a lengthy thing to do.

@MikeMiller: the proof we did required first knowing that the domain of the closure of the Laplacian is $H^2 \cap H_0^1$, we also need to know that under certain conditions $\| u \|_{H^{k+2}} \leq C \| f \|_{H^k}$ where $u \in H_0^1$ is the weak solution of the Dirichlet problem, we even required the trace operator. am I completely misunderstanding something here?

Hello Herr @DanielFischer

Hello!!

When we have that $f′$ is concave up on an interval what information do we get for the graph of $f$ on that interval?

can anyone tell me what does that exactly mean by "function g is the convex envelop of the function f on a unit l_inf ball". I understand the part of "a function being the convex envelop of the other function" but what is the "on a $l_inf$ ball" means.

hi

hi

i need help with a calc problem

In a space, let $B$ be a sphere (including the inside) with radius of 1. Line $l$ intersects with $B$, the length of the common part is the line segment with the length of $\sqrt{3}$. Find the volume of the solid generated by rotating $B$ about $l$.

Any ideas about how to find the volume of this thing (I know what the figure looks like)

Is the chat dead?

Well I guess it is.

how do we use the washer method if the axis of rotation is inside the figure?

Wouldn't that be like dividing by zero?

@AnastasiaDunbar What do you mean?

Is the figure sphere?

the one that results by rotation isn't a sphere: it looks like a torus smooshed toghether

@user19405892 you should find an example of such a line, then show that (probably) every other line with this length is a rotation of this line, then calculate the volume with your sepcific example

how do you rotate a sector of a sphere?

@s.harp

huh, don't know actually

if we can do that i can just subtract that from the area of the cylinder with height 2 and radius 2 to get the volume we need

I need to rotate the sector about the √3 line

Hi @AkivaWeinberger.

anybody know how to rotate a sector of a sphere

When we have that $G$ is a group and $H$ is a subgroup of $G$, why does it stand that $|G/H|=[G:H]$ ?

@Huy: I think we're talking about self-adjointness in two different ways. I mean formally self-adjoint: in the $L^2$ metric on $C^\infty$ functions, we get something like $\langle \Delta f, g\rangle = \langle f, \Delta g\rangle$ rather trivially. When you're talking about self-adjointness I suspect you're talking about it with respect to the Sobolev inner products.

@MikeMiller: no. if you have a densely defined unbounded operator, you can define a densely defined adjoint operator. now an operator is self-adjoint, if it is its own adjoint AND the domain on which the adjoint is defined coincides with the domain on which it was originally defined. in general, the adjoint of a symmetric operator has a larger domain than the original one.

@MikeMiller: so what we did in functional analysis is was first to prove that one can extend the Laplacian defined on $C_c^\infty \subset L^2$ to $H_0^1 \cap H^2$ and then we had to prove that the adjoint of this extension again has domain $H_0^1 \cap H^2$

yes, I think formally self-adjoint is what we call symmetric

Hello @AkivaWeinberger !! Do you have an idea about my question above?

Hey

Any linear algebra expert here?

I'm having so much trouble with coordinate changes

@Huy: OK, then that's what I was saying.

Symmetric seems like a strange choice of words but I can survive.

I do agree that your second condition - getting the domain to be right - is difficult.

hi

what is an Complementary adjacent angles formed by two intersecting lines.

Does anyone have any online multivariable calculus notes/sources?

anyone know about my question?

@MaryStar are you talking about lagranges theorem?

Yes. $[G:H]$ is the number of different cosets. Is $|G/H|$ also the number of different cosets? @user19405892

@MaryStar It is the quotient of the order of the groups

You can prove lagranges theorem by defining an equivalence relation on G that partitions G into left cosets

Does it always stand that $|G/H|=[G:H]$ or only when $H$ is a normal subgroup of $G$ ? @user19405892

H is a subgroup, not necessarily normal

that is lagranges theorem, yes

G must also be finite

Ah ok... Thank you!! :-) @user19405892

I have also an other question... Which are the possible groups that have order $6$ ? One such group is $S_3$, right? @user19405892

that's right

Are there also other groups that have order $6$ ? @user19405892

Actually there is

the cyclic group Z6

those are the only 2

up to isomorphism

@Evinda No. Consider that argument for a circle $x^2+y^2=1$. The points $(1,0),(0,1)$ are in the line, but the line passing through those points isn't contained in the circle.

In fact no line is contained in the circle.

Hey @PedroTamaroff.

Hey.

Then we would have $l(t)=(1-t,t)$, right? Doesn't this satisfy the equation for $t=0$ and $t=\frac{3}{2}$? Or am I wrong? @PedroTamaroff

What's up, @PedroTamaroff?

The line is in $\Bbb R^3$, @Evinda.

Give it some more thought, perhaps draw the surface given by that equation.

So you meant that I should just consider z=0? @PedroTamaroff

@Evinda Does $t^2+(1-t)^2=1$ hold for any value of $t$?

Think about it a bit more. Make a drawing, perhaps.

Do you know what's the general shape of $x^2+y^2=1+z^2$?

It's a hyperboloid, isn't it? @PedroTamaroff

Aha.

Now consider its cross sections planes parallel to the $xz$ plane.

Cross sections with the $xy$ plane give circles, of course.

And cross sections with the $xz$ plane give hiperbolas, but there is a degenerate case where it gives two lines.

In fact you can look at the circle $x^2+y^2=1,z=0$.

The hyperboloid contains many lines passing through that circle.

@PedroTamaroff So you mean that we just look at the cross section with the xy plane?

Hi

hi

@Evinda Did you read what I wrote? =/

Does this stand only for groups of order $6$ ? Or do we have in general, for an even number $n$, that all the groups of order $n$ are the cyclic $\mathbb{Z}_n$ and $S_{\frac{n}{2}}$ ? @user19405892

@MaryStar No for example the klein four group has order 4 but it is a direct product of two copies of the cyclic group Z2

and also there is like the dihedral group

Ah ok... Why do we not have the dihedral group also in the case of groups of order $6$ ? @user19405892

you can tell there have to be 2 groups by the classification of groups of order a product of two distinct primes

@user19405892 I haven't understood that... Could you explain it further to me?

What does the "|" in x | y represent?

"304 | 19"...is that division?

Hi -- out of curiosity, does anybody have insight on this question? math.stackexchange.com/questions/1672134/…

I ave not been able to prove the statement (and half suspect it to be false, but couldn't come up with a counter example either)

(in short, the question asks if uniform convergence to f on the reals of $f(x+1/n)$ (for f continuous) implies uniform continuity of f)

@DemCodeLines it means that $19$ divides $304$.

*have

Does the fact that $2i+bj$ cannot be orthogonal to $k$ have something to do with the basis {1,j,k} ?

@MaryStar sorry ill be back in 30 min

@user19405892 Ok... No problem...

Thinking about it again, $2i+bj$ is orthogonal to k for any b. This holds since i and j are both orthogonal to k , so also any linear combination of them, right? @DanielFischer

Does anyone have an idea why the equality holds?

$e_i \cdot e_i=1$, $e_i \cdot e_j =0$ for $i \ne j$.

@PVAL So it is equal to $(a_2 b_3-a_3 b_2)^2 e_1 \cdot e_1+ (a_2 b_3-a_3 b_2)(a_3 b_1-a_1 b_3) e_1 \cdot e_2+ (a_2 b_3-a_3 b_2) (a_1 b_2-a_2 b_1) e_1 \cdot e_3+ \dots$, right?

hi

@MaryStar are you asking why is the symmetric group_3 and the cyclic group Z6 the only two groups up to isomorphism with order 6?

@L33ter sorry for taking so long to reply but yeah I know a fair amount. What's the problem you're working on?

@MaryStar Basically the rule of classification of groups of order $pq$ a product of two distinct primes is this: If $p$ does not divide $q-1$, then there is only one isomorphism class of groups of order $pq$, namely, the cyclic group.

and

If $p$ divides $q-1$, then there are two possibilities: the cyclic group of order $pq$ and the semidirect product $\mathbb{Z_q} \cross \mathbb{Z_q}$ where $\mathbb{Z_q}$ is thought of as the additive group of integers mod $q$ and $\mathbb{Z_p}$ is identified with the subgroup of order $p$ in $\mathbb{Z_q}^*$, which is cyclic of order $q-1$.

Why do we take the $\theta$ to lie on $[0, \pi]$ ?

@Evinda the angle between the vectors is between 0 and pi

@Evinda What was the preceding problem?

Why is the angle between the vectors between 0 and pi? @user19405892

that is how angles are defined between vectors is the smallest angle between the two

hi

hi

hi