Mathematics - 2019-03-23

jacobian

12:01 AM

that's what two dimensional means though

might be asking if the embedding is unique in some sense

Semiclassical

maybe, but that isn't graph theory

Ultradark

I'm talking about graphs that look like they are curved

for example a mesh graph that looks like it's the outer shell of a hemisphere

jacobian

what's the question though

a graph will have a lot of different embeddings into a lot of different spaces

Ultradark

Semiclassical

oh hey, a golf ball

Ultradark

12:07 AM

This golf ball graph (with nodes at intersections) looks positively curved.

Semiclassical

possibly what you want to look up is Euler's formula (and more generally the Euler characteristic)

some examples for motivation:

A cube has 6 faces, 12 edges, and 8 vertices

Ultradark

okay

Semiclassical

A tetrahedron has 4 faces, 6 edges, and 4 vertices

An octahedron has 8 faces (hence the name), 12 edges, and 6 vertices

A common fact between these: The number of faces, minus the number of edges, plus the number of vertices, is 2.

Math geek

2

Q: $(X,\mathscr T)$ is normal and each closed subsets of $X$ is a $G_{\delta}$ set.Then, $(X,\mathscr T)$ perfectly normal.

Prove the following result without using Urysohn's lemma. $(X,\mathscr T)$ is normal and every closed subset of $X$ is a $G_{\delta}$ set.Then, $(X,\mathscr T)$ perfectly normal. My effort: I have proved using Urysohn's lemma. How do I prove without the use of Urysohn's lemma? Let $A$ a...

general-topology

jacobian

the spherical mesh is not two dimensional @Ultradark

but you probably meant graphs corresponding to piecewise-linear structure of 2 dimensional manifolds

Semiclassical

12:11 AM

more generally, it turns out that this formula holds for any (convex) polyhedron in three dimensions

jacobian

or something like that

Semiclassical

That includes your golf ball

jacobian

in which case the stuff Semiclassical is saying is what you want, euler characteristic

Ultradark

okay

Semiclassical

So if you were to count up all the faces, edges, and vertices (including those on the back side of that picture which you can't see)

you'd still have faces - edges + vertices = 2

That 2 is known as the Euler characteristic of the sphere

By contrast, the plane has Euler characteristic 1.

And a donut turns out to have Euler characteristic 0.

Now, that's not exactly telling you about positive curvature as such. It's more telling you that your graph could be drawn on a sphere without messing things up

There are notions of graph curvature, but I know basically nothing of them

(I saw the lead author of this give a talk on the subject of graph curvature, for example: arxiv.org/abs/1502.04512)

Leaky Nun

12:26 AM

so for every $f \in L^2(\Bbb R/\Bbb Z, \Bbb C)$ there is $a_n \in \Bbb C$ with $f = \sum_{n \in \Bbb Z} a_n e^{2 \pi iz}$?

this sounds quite dodgy if like the function isn't continuous

not every $f \in L^2$ is equivalent to a continuous function right

MatheinBoulomenos

@LeakyNun the Fourier series only converges in $L^2$, not pointwise (and even if it converges pointwise, there is no reason why the limit should automatically be continuous)

I don't see what is dodgy about it

Leaky Nun

because when I was trying to prove it... I explicitly used continuity

and there's also the result that if it is jump discontinuous then it converges to the average of the two one-sided limits

is it easy to prove that it converges in $L^2$?

MatheinBoulomenos

it's just Hilbert space theory

Leaky Nun

I think a pathway would be to show that they converge pointwise for continuous functions and then use the fact that smooth functions are dense in L^2?

although that "fact" seems difficult to prove

maybe change smooth to continuous

oh I see how to prove it

we know that simple functions are dense in L^2

MatheinBoulomenos

you're overcomplicating this

Leaky Nun

12:35 AM

simple functions are piecewise constant

MatheinBoulomenos

there is no density argument needed

Leaky Nun

I mean I don't even know how to prove that L^2 is complete

Semiclassical

Plancheral iirc

MatheinBoulomenos

you just show that $(e^{2 \pi n i z})_{n \in \Bbb Z}$ is an orthonormal basis for $L^2(\Bbb R/\Bbb Z)$

Leaky Nun

and how would I show that?

MatheinBoulomenos

12:36 AM

orthonormality is a computation

and the fact that it has dense span is Stone-Weierstraß

Leaky Nun

doesn't Stone-Weierstraß just say that the polynomials are dense in the continuous functions?

MatheinBoulomenos

that's Weierstraß

Leaky Nun

what does Stone-Weierstraß say?

MatheinBoulomenos

If $X$ is a compact Hausdorff space, then a unital *-subalgebra $A$ of $C(X,\Bbb C)$ is dense iff it separates points

Ultradark

@Semiclassical does the outside part of the shape count as one of the faces

Semiclassical

12:42 AM

that's too vague for me to answer

MatheinBoulomenos

okay you also need that $C(S^1, \Bbb C)$ is dense in $L^2(S^1, \Bbb C)$

Ultradark

a square has four vertices and four edges and 1 face

Semiclassical

right

Ultradark

i was wondering if the faces would be two

counting the outside of the square

but I think I'm thinking of something else

Semiclassical

depends on how you think of it, really

if you think of it as a polygon in the plane, you wouldn't

on the other hand, suppose you imagine drawing a square on a sphere

then clearly the inside and 'outside' both count as faces

that'll change the face count by one, so you'll go from the Euler characteristic of the sphere being 2 to the Euler characteristic of the plane by 1

Ultradark

12:56 AM

okay thanks

user193319

If $a$ and $b$ are two group elements with finite order which commute, what can we say about the order of $ab$?

Semiclassical

well, what can you say about $(ab)^k$ given that they commute?

user193319

$(ab)^k = a^k b^k$.

Isn't the order $\displaystyle \frac{|a| \cdot |b|}{gcd(|a|,|b|)}$?

Semiclassical

It is. There's a simpler way to write that tho

MatheinBoulomenos

no, the order is not always that

consider the case that $b=a^{-1}$

Semiclassical

12:58 AM

oof

yeah, and that's true even if $b,a$ had infinite order

this Q&A on the main site covers it rather well:

38

Q: Examples and further results about the order of the product of two elements in a group

Let $G$ be a group and let $a,b$ be two elements of $G$. What can we say about the order of their product $ab$? Wikipedia says "not much": There is no general formula relating the order of a product $ab$ to the orders of $a$ and $b$. In fact, it is possible that both $a$ and $b$ have finite ...

abstract-algebra group-theory examples-counterexamples

Leaky Nun

1:15 AM

desmos.com/calculator/xa8a9xp7cy

Lagrange interpolation

@Semiclassical so we have a whole course about proving existence (and uniqueness) of ODEs and analyzing behaviour of equilibriums

Semiclassical

oof

Tanner Swett

2:20 AM

Hey everyone. I'm trying to tackle a control problem using the Laplace transform. I dunno how to figure out whether I'm doing this all correctly or not.

The problem is: There's a damped harmonic oscillator. It's driven by a P controller which is trying to drive the position of the oscillator to 0 as quickly as possible. However, the controller has a delay of 1 second.

So, the differential equation describing this system is reasonably simple: $f''(t) = c f' (t) + k f(t) + K_p f(t - 1)$.

I took the Laplace transform of both sides and got $s^2 F(s) - s f(0) - f'(0) = c (F(s) - f(0)) + k F(s) + K_p e^{-s} F(s)$. Solve that, and you get $F(s) = (s f(0) + f'(0) - c f(0)) / (s^2 - c - k - K_p e^{-s})$.

Now, suppose that $K_p = 0$ and $- c - k = 1$ (since $c$ and $k$ are both going to be negative in realistic situations). Then the denominator becomes $s^2 + 1$, meaning that $F(s)$ has poles at $i$ and $-i$.

So... what can I conclude there? Those poles have a real part of 0, so does that mean they represent oscillations which don't decay? That doesn't make sense; as long as $c$ and $k$ are negative, and $K_p$ is zero, all oscillations should decay.

Or am I totally misinterpreting what the Laplace transform means?

Tanner Swett

2:49 AM

Ah, I see that I made a mistake. I transformed $c f'(t)$ into $c (F(s) - f(0))$, but I'm missing an $s$ there; it should be $c (s F(s) - f(0))$, so the denominator should be $s^2 - c s - k - K_p e^{-s}$.

Now if $K_p = 0$ and $c = k = -1$, the behavior is what I was expecting: the poles both have a negative real part, indicating that the oscillations decay. If I make $K_p$ too large (positive or negative), one of the poles comes to have a positive real part, indicating an unstable oscillation mode.

Neato.

Sorry for the monologue.

Tanner Swett

5:31 AM

I went ahead and posted a question related to all this, if anyone's interested: math.stackexchange.com/questions/3158979/…

Rudi_Birnbaum

11:32 AM

Hi people!

math.stackexchange.com/questions/3159130/… any suggestiones welcome!

Jose Enrique Calderon

12:54 PM

Hello All..

Can someone direct me to how a post may be transfer or copy from one groupr from another? I want to transfer the following post to Mathematica exchange. math.stackexchange.com/questions/2369651/…

Or to copy in both becuase it is relevant in bothe areas.

Manolis Lyviakis

2:01 PM

anyone knows graph theory?

id like to talk about this covering graph

is an example of nielsen scheier theorem

why the arrows go like this

Alessandro Codenotti

Because $H'=aH$, so there is an arrow from $H$ to $H'$ and similarly with all the others

That's how the Schreier coset graph is defined

Manolis Lyviakis

2:17 PM

so Hb=H'

and a^-1H'=H

$a^{-1} H'=H$

Alessandro Codenotti

no, the arrow on $a^{-1}$ is pointing toward $H'$

Manolis Lyviakis

so

why is it crossing itself

Alessandro Codenotti

I don't know why they chose to draw it this way, but it makes no difference, all that matters for the edges of a graph is where they start and where they end

Manolis Lyviakis

ohh ok

now suppose i find a spanning tree of that

say the 2 egdes of a,b

how can i shrink it?

can i join 2 egdes?

i mean how do i get the buquet

Alessandro Codenotti

@ManolisLyviakis That's not a spanning tree, one edge is enough

Manolis Lyviakis

2:25 PM

oh yeah right

so i cant shrink one egde with 2 vertices

ohhh i get it

i shrink it so the 2 points join

and the remaining

Alessandro Codenotti

If you choose just one edge and contract it the two vertices get identified and you obtain a bouquet of $3$ circles

Manolis Lyviakis

is 2 circles

yeyeye

thanks

user10478

5:24 PM

For a vector field with circulation but no curl due to an undefined point, i.e., $\begin{bmatrix}\frac{-y}{x^2 + y^2} \\ \frac{x}{x^2 + y^2}\end{bmatrix}$, is there always (ever?) a different, otherwise identical vector field, where the point is defined and has curl equal to the circulation it induces?

Mary Star

6:03 PM

Hello!!

In a box there are 10 balls, 3 blue and 7 red. We take 9 balls with replacement. Which is the probability that we get exactly 5 red balls?

This is equal to $$\binom{9}{5}\cdot \left (\frac{7}{10}\right )^5\cdot \left (\frac{3}{10}\right )^4$$ since we use here the formula $$P(X=k)=\binom{n}{k}\cdot p^k\cdot p^{n-k}$$ right?

user10478

8:38 PM

Does anyone understand this comment? I feel like it's a joke over my head that will be ruined if I reply asking for clarification. math.stackexchange.com/questions/3159678/…