Mathematics - 2017-02-23 (page 3 of 4)

Mateen Ulhaq

11:13 AM

@user8469759 I believe so. Alternatively, you can visualize it as a surface: f(x, y) = |(x, y)|

user84215

11:44 AM

Hello

user84215

can we have an uncountable set of real numbers that none of its element is limit point ?

Alessandro Codenotti

No, intersect with [-n,n] and use Bolzano-Weierstrass

user84215

ok by bolzano it has limit point. But is it necessary this limit point belong to set ?

BAYMAX

wow nice pictures .. where are you coding it?

Alessandro Codenotti

@aminliverpool ah, I misread, the answer is still yes, but I don't remember a proof of the top of my head

user84215

12:00 PM

from where do you answer "yes" ?

Alessandro Codenotti

It's a known fact

Every uncountable subset of R contains one of its limit points, you can probably find it on mse

anonymous

@DHMO You there?

user84215

@AlessandroCodenotti I need an uncountable set of real numbers that none of its element is limit point. So by your statement, it never occurs.

Alessandro Codenotti

Here for example math.stackexchange.com/questions/310113/…

@aminliverpool right

anonymous

$$\lim_{x->0} \frac{1}{\ln(1+x)}-\frac{1}{\ln(x+\sqrt(1+x^2)}$$ @DHMO Try finding this limit!

Semiclassical

12:35 PM

Probably overkill, but I'd substitute $x=\sinh t$ so that the second term simplifies

user8469759

12:49 PM

Is any of you familiar with splines functions? and related properties

(BSplines)

Balarka Sen

@AkivaWeinberger I see you and Zach talked about some interesting problem above. Want to restate?

@MateenUlhaq It's the "right" notion of a metric on a smooth surface/manifold.

The dot product on each tangent space gives you the ability to define arclengths of paths on the smooth surface.

That's why it's also known as a Riemannian metric.

BAYMAX

Guys, any1 aware of video lecture regarding calculus of variations, any university lectures??

Akiva Weinberger

@BalarkaSen The complement of finitely many arcs (injective images of $[0,1]$) in $\Bbb R^2$ is path-connected. Is the same true for infinitely many arcs? If not, what about a countable infinity of arcs?

Balarka Sen

Disjoint arcs, you mean?

Akiva Weinberger

Yeah.

Alessandro Codenotti

12:59 PM

Pairwise disjoint arcs?

Akiva Weinberger

Yeah.

Alessandro Codenotti

Ah, Balarka already asked

Akiva Weinberger

Also, it matters that it's "path-connected" and not "connected". I think that it's not too hard to show that, for countably many arcs, the complement must be connected.

Alessandro Codenotti

My internet is being very slow today

I don't think that's true even for countably many arcs

Balarka Sen

I don't know how to prove the finitely many version for topologically embedded arcs though. For smooth arcs you can probably do it very easily because locally the arcs look like the x-axis in R^2 (immersion theorem).

Ah, I guess the same argument generalizes by Schoenflies (embedded arcs are ambient isotopic to a bit of the x-axis).

Akiva Weinberger

1:08 PM

Re my "it's not too hard to show" comment above: I take that back.

user84215

is it possible to have an uncountable set of real numbers that none of its element is a limit point respect to its elements ?

user84215

why ?

Akiva Weinberger

Isn't that the same as the question here: math.stackexchange.com/questions/310113/…

which Alessandro linked to above

user84215

no

Akiva Weinberger

Oh, wait, "limit point" is different from "accumulation point", isn't it

SteamyRoot

1:14 PM

is it...?

Akiva Weinberger

Actually, I have no idea

user84215

it is not important. I want none of its element is limit point respect to "its element" ?

SteamyRoot

In some contexts they're definitely the same...

Oh well

Akiva Weinberger

@aminliverpool With respect to the set's elements, you mean?

Balarka Sen

@aminliverpool What does that even mean

Akiva Weinberger

1:17 PM

In any case, you can't have an uncountable set made up of only isolated points.

SteamyRoot

Maybe it would be helpful if you gave your definition of a limit point and such

user84215

that is not respect to the bigger set that is real numbers

Balarka Sen

What does "respect to" mean???

user84215

a limit point of a set S in a topological space X is a point x (which is in X, but not necessarily in S) that can be "approximated" by points of S in the sense that every neighbourhood of x with respect to the topology on X also contains a point of S other than x itself.

Akiva Weinberger

So you want X to be the set S, rather than the set of real numbers?

user84215

1:20 PM

yes

Akiva Weinberger

That's the same as letting X be the set of real numbers and only looking for limit points in S.

user84215

no

Akiva Weinberger

Why

Do you know what a subspace topology is

Balarka Sen

@aminliverpool You're just looking at S with the subspace topology from X.

user84215

because maybe the sequence used to define limit points dont lie in S

Akiva Weinberger

1:23 PM

> can be "approximated" by points of S

user84215

ok

Akiva Weinberger

@aminliverpool Also, see where it says "also contains a point of S other than x itself"

@aminliverpool Note also that your definition doesn't explicitly mention sequences

But, nah, the sequence would have to lie in S.

Alessandro Codenotti

(And that not all limit points of S are the limit of a sequence in S)

In a generic topological space I mean, sequences are fine in R

Akiva Weinberger

@AlessandroCodenotti In this case they're the same, though.

Yeah.

user84215

What is the conclusion ?

Akiva Weinberger

1:29 PM

@aminliverpool The only way for it to be false is if you had an uncountable set made of only isolated points. This can be shown to be impossible

so every uncountable set must have a limit point.

(It's impossible for an uncountable set to be made of only isolated points, because then you could define a surjective function from a subset of rationals to the set, contradicting its uncountability.

The function is defined by mapping every rational q to the largest element of the set less than or equal to the q, if it exists. (To be cont'd)

user84215

I want the elements are not limit point respect to set S itself.

Akiva Weinberger

@aminliverpool I know.

@aminliverpool But do you agree that the only way for it to happen is if every element is an isolated point (respect to S)?

user84215

yes

Akiva Weinberger

(Cont'd) This is surjective, because every point is isolated, meaning it has an open set around it containing no other elements of S. Thus, the rationals in the open set greater than that point get mapped to that point. QED.)

DJanssens

Hi there, I'm having some troubles coming up with a function that does the following: I have a range of values [0 to 4.000.000] and I need to split it up in 40 points. Linearly this would be simple each point would be a multiple of 4.000.000/40, Which is 0, 100.000, 200.000, ... However I need this to be somewhat more exponential, where the values start going up with higher numbers at the start, but at a lower pace at the end.

So it would become something like 0, 500.000, 900.000, 1.300.000 (this is just a random example :D)

user84215

1:44 PM

I have problem in your argument. next day I will continue the discusion

FormlessCloud

Hi guys

Akiva Weinberger

@DJanssens Maybe 4000000^(n/40)

FormlessCloud

I have a question that I would like to post here

on math.stackexchange

Akiva Weinberger

so that the 1st will be the fortieth root, the 20th will be the square root, and the last will be four million @DJanssens

FormlessCloud

but I have already asked it on ux.stackexchange

do you think that if I rearrange it would it be a good fit for this site?

2

Q: What is the 8.8 mm R Criterion Curved Design?

Can you exlplain me what is the 8.8 mm R Criterion Curved Design? Apparently it is a criteria used by the UMi Touch phone. Why it uses 8.8 mm curves (what does this even mean? Do they mean that the radius of the fillets is 8.8 mm?) and what is the relation with the golden ratio? What is a cutting...

basically I would like to ask why using 8.8 mm ensure the golden ratio

and why using 45 degress ensures a cutting rate of 0.333% whatever that is

thanks

Akiva Weinberger

1:47 PM

A lot of the stuff about the golden ratio being "more pleasing to the eye" is BS

FormlessCloud

lol thanks

DJanssens

@AkivaWeinberger that's more or less what I needed, I think I need to turn it around though to 4000000^((40-n)/40) since I need the start to be bigger compared to the tail.

Akiva Weinberger

Oh, I see

FormlessCloud

so do you guys think that the company that designed that phone is just saying random things?

Akiva Weinberger

I dunno

I mean, they didn't make up the stuff about the golden ratio. It's been around for a long time. It's still BS, though

DJanssens

1:55 PM

@AkivaWeinberger That's indeed more or less what I needed, however it seems that the exponential function is a bit too powerfull, is there a way to influence the steepness of it :D

FormlessCloud

mmm okay thanks. I may try to ask a similar question on this site maybe

MathematicsStudent1122

This is sort of a random question, but does anyone know any awesome math-related pick up lines (ie., if you're trying to get laid)? No, I'm actually serious.

2

FormlessCloud

focusing more on the mathematical aspect

Akiva Weinberger

@DJanssens

How about that? No exponential function, but you can vary the steepness by changing n.

(Click on the image to make it bigger)

The app is desmos.com, by the way

desmos.com/calculator

DJanssens

the calculator looks cool @AkivaWeinberger but it doesn't seem to plot it :/

Akiva Weinberger

2:05 PM

@DJanssens Change the scale? I put in 4000000 so it probably is out of the frame

2

SteamyRoot

@MathematicsStudent1122 You could probably get some inspiration by listening to "Finite simple group of order 2" by the Klein Four Group

DJanssens

@AkivaWeinberger that's pretty awesome! thanks for the site and tip

JamalS

The $\int \frac{dx}{x}$ question today has had many views, and it got me wondering, is there an algebraic geometry motivated answer to it? I recall a section on Abelian integrals in Griffith's text...

Akiva Weinberger

What question

JamalS

41

Q: The deep reason why integral of 1/x is a transcendental function (log)

In general, the indefinite integral of $x^n$ has power $n+1$. This is the standard power rule. Why does it "break" for $n=-1$? In other words, the derivative rule $$\frac{d}{dx} x^{n} = nx^{n-1}$$ fails to hold for $n=0$. Is there some deep reason for this discontinuity?

calculus real-analysis logarithms indefinite-integrals

I'm just wondering if there's an algebraic geometry motivated explanation (which albeit would go over the OP's head)

(See for example, chapter 2 of Griffiths)

Akiva Weinberger

2:18 PM

@JamalS I wonder if you can connect it with the sinc function. $\sin(\pi x)/(\pi x)$ (extended to $0$ by continuity) is equal to $0$ for all integers except for $x=0$, where it equals $1$.

@MathematicsStudent1122 Something about laying tangent to someone's curves

3

JamalS

Or blowing up a singularity on them...

s.harp

2:52 PM

Are there open sets in $\Bbb R^2$ that are simply connected but not contractible?

Balarka Sen

@s.harp It actually has to be homeomorphic to R^2. This is the big conclusion of the uniformization theorem.

If you identify R^2 with C you moreover get that it has to be biholomorphic to a disk

user8469759

does anyone have a good grasp in graph theory?

s.harp

does it have to be the same as the disk or the same as the disk OR $\Bbb C$?

maybe im an idiot, but i thought they were not conformally equivalent

Balarka Sen

disk if your open set is proper. If you include all of C what you said is right, yes.

No, you're correct

s.harp

gotcha

Astyx

3:07 PM

Hi chat

Sha Vuklia

Hello

Astyx

What's up ? @ShaVulkia

Sha Vuklia

@Astyx Analysis homework :P You?

Astyx

Swimming pool in 20 minutes :)

Sha Vuklia

Competitive swimming? :P

Astyx

3:11 PM

No, thankfully not .. I'm bad at swimming

Sha Vuklia

Haha, well that sounds good! Physical exercise is good :)

Astyx

Just to do some sports in the week

Sha Vuklia

I mean

the fact that you're going to do that for fun

:P

Yea I should do that too actually. But I have to choose between physical exercise or more sleep... And I've chosen sleep for now :P

Astyx

Sleep is cool too :p

Sha Vuklia

XD

quick question:

Let $(V,\rVert \lVert_V)$ be a $k$-dimensional linear space over $\mathbb R$. Choose a norm $\lVert \rVert$ for $\mathbb R^k$. Prove using Bolzano-Weierstrass that
$$
\inf_{\{\vec x\in\mathbb R^k:\lVert \vec x\rVert=1\}}\lVert\Phi(\vec x)\rVert_V>0.
$$
I've chosen the Euclidean norm on the $k$-dimensional Euclidean space $\mathbb R^k$. So we have
\begin{align}
\lVert\vec x\rVert=\sqrt{x_1^2+\dots+x_k^2}.
\end{align}
Consider a sequence $(\vec x_n)$ in $\mathbb R^k$ with $\Vert \vec x_n\Vert=1$ for each $n\in\mathbb N$. It's obvious that this sequence is bounden. By Bolzano-Weierstrass, we k…

My teacher told me to use contradiction here

So I have to assume that
$\inf_{\{\vec x\in\mathbb R^k:\lVert \vec x\rVert=1\}}\lVert\Phi(\vec x)\rVert_V=0.$

Astyx

3:17 PM

What's $\Phi$ ?

Sha Vuklia

However, I don't know how to use this convergent subsequence.

oh right

it's a linear map:

$\Phi:\mathbb R^k\to V$

and it's a bijection

Astyx

Oh right

If it wasn't a bijection that wouldn't be true

Sha Vuklia

so basically the coordinate function :P

Or at least, the coordinate function could be an example

Astyx

Well the linear map is continuous

So you have $||\Phi(x)||=0$, where $x$ is a limit point

Topologicalife

Hi. A stupid question: I have $\displaystyle\lim_{x \to{+}\infty}{(x^4+7x+2)^c-x} = L$ where $0\neq L \in \mathbb{R}$. I have to compute $c$.

Sha Vuklia

3:19 PM

wait, why is it continuous?

Astyx

Because all linear maps are continuous in finite dimensionnal spaces

Sha Vuklia

okay

Topologicalife

How do I do compute c? I was thinking on simply define $c$ as $c = \dfrac{\log(x+k)}{\log(x^4 +7x+2)}$

Astyx

That means $x=0$ because it's continuous

And from there I think you can find the contradiction :)

Sha Vuklia

But how did you get $||\Phi(x)||=0$ ?

oh wait

how do you know that $\vec 0$ is the limit point?

because why does the subsequence have to converge to the infimum?

Astyx

3:22 PM

It doesn't

Suppose you have a sequence such that $\Phi(x_n)$ goes to 0

Sha Vuklia

Ohh

right

Astyx

Then as you said it's closed and bounded so it has a limit point

(finite dimension, once again)

Sha Vuklia

because if no sequence goes to o, then we couldn't have had the infimum equal to 0 in the first place

Astyx

Yup

Sha Vuklia

aha, okay

yea that helps!

thanks, I'm gonna write it out neatly now:)

Astyx

3:23 PM

Glad to help

@Topologicalife What's $k$ ? Do you mean $L$ ?

Topologicalife

Yeah.

Little Rookie

Hello all, i have a question pertaining to behavior of 2nd order linear homogeneous ODE with constant coefficients. Please help if you can. Thanks :)

http://math.stackexchange.com/questions/2157986/behaviour-of-2nd-order-homogeneous-linear-ode-with-constant-coefficients

Astyx

$x^{4} + 7x + 2 \sim x^4$ as $x\to \infty$

Topologicalife

Yeah.

Astyx

Or even $x^{4} + 7x + 2 = x^4 + O(x)$

Topologicalife

3:27 PM

But yeah, and?

I don't get your point.

You can define $c = 1/4$ but then the limit is $0$.

Astyx

Which is $x^4(1+O(x^{-3}))$

Then use Taylor expansions at 0

So you get $x^{4c}(1+O(x^{-3}))$

Well if the LHS converges then $c={1\over 4}$

Necessarily

Topologicalife

Yeah. But then my limit is zero.

And remember:

11 mins ago, by Topologicalife

Hi. A stupid question: I have $\displaystyle\lim_{x \to{+}\infty}{(x^4+7x+2)^c-x} = L$ where $0\neq L \in \mathbb{R}$. I have to compute $c$.

Astyx

So you have a contradiction

Topologicalife

Indeed.

That's what is confusing me.

I think there doesn't exist $c$ such that.

Astyx

Well that's what we just proved

Akiva Weinberger

3:31 PM

@BalarkaSen Here's another question: Must the complement of countably many compact line segments (not curved!), pairwise disjoint, be path-connected? :P

Astyx

I gotta go now, I'll be back later

Sha Vuklia

Have fun at the pool!

Akiva Weinberger

(I still have no idea what happens when we change "path-connected" to "connected")

Topologicalife

Thanks @Astyx

Our ways are pretty the same.

Frank Science

I don't know where I gest stuck with.

I think that the full subcategory of torsion abelian groups is a Grothendieck abelian category.

Alessandro Codenotti

3:37 PM

What happens if I choose $\{q\}\times [0,1]$ with $q\in\Bbb Q\cap(0,1)$ in $[0,1]^2$ and then cover the other squares in the plane in a similar fashion but turning 90 degrees between adjacent squares?

Frank Science

The generator is given by $\oplus_n\mathbb Z/n$.

Sha Vuklia

4:05 PM

Let $(V,\rVert \lVert_V)$ be a $k$-dimensional linear space over $\mathbb R$. Let $\Phi:\mathbb R^k\to V$. Choose a norm $\lVert \rVert$ for $\mathbb R^k$. Prove using Bolzano-Weierstrass that
$$
\inf_{\{\vec x\in\mathbb R^k:\lVert \vec x\rVert=1\}}\lVert\Phi(\vec x)\rVert_V>0.
$$
I've chosen the Euclidean norm on the $k$-dimensional Euclidean space $\mathbb R^k$. So we have
\begin{align}
\lVert\vec x\rVert=\sqrt{x_1^2+\dots+x_k^2}.
\end{align}

I have one question. Assume $\inf_{\{\vec x\in\mathbb R^k:\lVert \vec x\rVert=1\}}\lVert\Phi(\vec x)\rVert_V=0$. Consider a sequence $(\vec x^{(n)}…

Balarka Sen

@Alessandro That's a pretty good idea. In fact think about $[-1, 1] \times \{-1, 1\}$ with $\{-1-1/n\}_{n \in \Bbb N} \times [-1, 1]$ and $\{1+1/n\}_{n \in \Bbb N} \times [-1, 1]$. It looks like two line segments of equal length parallel to the x-axis and a bunch of lines accumulating to the line connecting their endpoints, at both ends.

Stuff in between $[-1, 1] \times \{-1\}$ and $[-1, 1] \times \{1\}$ should be disconnected from the world.

So literally take a square and replace a pair of two opposite sides by a bunch of lines accumulating to them, I mean

Everything is disjoint but if you try to get outside from the "interior" of the square you're gonna hit the lines which accumulate to the boundary sides

Sha Vuklia

Never mind, I got it!

Akiva Weinberger

4:20 PM

@BalarkaSen @AlessandroCodenotti Nice!

Ramanujan

Hello.. Is there anyone who can help me with geometry circle problem!?

Mithlesh Upadhyay

-1

Q: Proving Partial Order

Testing .... Making sure this works. Tick tock

proof-writing proof-theory

Akiva Weinberger

@MithleshUpadhyay I rolled back the edit

Alessandro Codenotti

4:38 PM

@BalarkaSen neat! So even segments can beheave weirdly

Ahmed99

Find five different positive integers such that the sum of any three terms divides the sum of the other two. Any hints?

Evinda

Hello

Is anyone of you familiar with cryptography?

MartianCactus

in this

angle POQ = 30

and angle PQO is 90

so shouldnt PQ be one third of PO instead of being half of it?

user8469759

5:04 PM

@Evinda I'm

Alessandro Codenotti

@Balarka if you're interested I found a reference for a simple construction of an uncountable subset of R containing no perfect set

Evinda

@user8469759 I want to design an encryption system of type Feistel that has not the weakness that the pure message can be found, given the encypted message and also a pair of a pure message and its cryptogramm. The system should have the following parameters: size of block: 8 bits, length of key: 6 bits, number of rounds : 3.

The encyption algorithm computes the following:

$L_i=R_{i-1} \\ R_i=L_{i-1}+F(k_i, R_{i-1})$.

I have thought to pick $F(k,R)=AR+B+k$, where A is a 24x24 matrix, $k=(k_1,k_2,k_3), k_i \in \mathbb{F}_2^4$ and we would have to find specific A and B such that the orig…

Astyx

Hi again chat

user8469759

what do you mean by "has not the weakness that pure message can be found"?

Evinda

5:19 PM

@user8469759 I am looking at an exercise where it is the second question.

It is stated as follows:

Alice communicates with Bill using an encryption system of type Feistel. Specifically, each block has length 128 bits, the algorithm has 4 rounds and the common, secret key of Alice and Bill is $(k_1, k_2, k_3, k_4)$, where $k_i \in \mathbb{F}_2^{64}, i=1, \dots, 4$ is the key of the i-th round. The original message is divided into two parts, of 64 bits each , the left and the right , $L_0$ and $R_0$. Next, the encryption algorithm computes the following

Sha Vuklia

@Astyx! I need your help XD

Astyx

Do you now ? :p

Sha Vuklia

Yes, it's still that same question! Let me just type it in math jax

@Astyx
I've proven that for some norm $\lVert\rVert_V$, there exist $c,C>0:$
$$
c\Vert \vec x\Vert\leq\Vert \Phi(\vec x)\Vert_V\leq C\Vert \vec x\Vert.
$$
Now I have to show that any two norms on $V$ are equivalent. My problem is that I've shown equivalence between two norms, that are not meant for the same vector space. I need something of the form:
$$
c\Vert \Phi (\vec x)\Vert_{V_2}\leq\Vert \Phi(\vec x)\Vert_{V_1}\leq C\Vert \vec x\Vert_{V_2}.
$$

How can I do that?

Astyx

Let $N_1$ and $N_2$ be two norms on $V$

Then you just proved $c||\Phi^{-1}|| \le N_1 \le C||\Phi^{-1}||$

And $c||\Phi^{-1}|| \le N_2 \le C||\Phi^{-1}||$

Sha Vuklia

Let me see what you did

I really don't get it. How are we going to use this inverse? I know that $\Phi$ is a continuous bijection, so both $\Phi$ and its inverse are strictly increasing or decreasing

but... I don't see it @Astyx

Astyx

5:39 PM

What ?

A bijection isn't necessarilly increasing or decresing

As a matter of fact increase qnd decrease doesn't make sense in vector spaces

Sha Vuklia

oh wait, that's only on $\mathbb R$

Astyx

It is

Rewrite the inequalities I wrote so as to have $||\Phi^{-1}||$ in the center

Sha Vuklia

The way I interpret you, that's impossible? @Astyx

I can only move one $|| \Phi^{-1}||$ to the middle

wait I'll show you

Astyx

You have $c||\Phi{-1}|| \le N_1$ and $N_1\le C||\Phi{-1}||$

Sha Vuklia

5:54 PM

And how can I combine $c$ and $C$? @Astyx

Astyx

$\Bbb R$ is a field :)

Sha Vuklia

oooooooohhhhhh

WAIT.

$\frac{N_2}{C}\leq \Vert\Phi^{-1}\Vert\leq\frac{N_1}{c}$ @Astyx

Astyx

Right :)

Now you get the same for $N_2$ (different c and C though)

Using this you can compare $N_2$ and $N_1$

Sha Vuklia

okay let me try!

Ahh, I got it. If we let $c_1, C_1$ belong to $N_1$, and $c_2,C_2$ to $N_2$, then we get
$
\frac{c_1}{C_2}N_2\leq N_1\leq\frac{C_1}{c_2}N_2
$

Thanks so much! @Astyx

Astyx

Right

There another way to do this

Sha Vuklia

6:03 PM

yea?

Astyx

You can show that $||\Phi^{-1}||$ is a norm

And what you did is show that it's equivalent to any other norm

So all norms are equivalent

because equivalence between norms is (thankfully) an equivalence relation

Sha Vuklia

but $\Vert \Phi^{-1}\Vert$ is a norm on $\mathbb R^k$, right?

Astyx

No

It takes values in $V$, right ?

Sha Vuklia

Does that mean that $V$ is the domain or the codomain?

because it's the inverse of $\Phi$, and we know that $\Phi:\mathbb R^k\to V$

Astyx

Domain

Sha Vuklia

6:07 PM

yea okay, but the domain doesn't matter?

Astyx

Not sure what that means, English is not my native language :p

Sha Vuklia

we take the norm of the assigned values by $\Phi^{-1}$

So $\Phi^{-1}(v)=\vec x$

so if we take the norm of $\Phi^{-1}(v)$, we take the norm of some $\vec x$

Astyx

Yes

Sha Vuklia

So it's a norm on $\mathbb R^k$, right

Astyx

No, $||\cdot||$ is a norm on $\Bbb R^k$

$||\Phi^{-1}(\cdot)||$ takes vectors of $V$ and returns a positive real

So it's a norm on $V$

Sha Vuklia

6:12 PM

Oh right, like that

Simple

Given a matrix $A=\begin{pmatrix}-3 & 5 \\ -2 & 2\end{pmatrix}$ and $X'=AX$, we can determine the direction portrait by looking at $\text{tr}(A)\pm\sqrt{\text{tr}(A)^2-4\det(A)}$

is there a way to draw the exact phase portrait with these information?

Astyx

I'll be back, bye

Sha Vuklia

Bye bye!

Akiva Weinberger

6:36 PM

@BalarkaSen @AlessandroCodenotti Easier construction (though unbounded): $\bigcup_{n=1}^\infty(\{1/n\}\times[-n,n])$. That is, a bunch of vertical line segments that approximate the line $x=0$.

Balarka's wins if we want it to be bounded, or if we want them all to have unit (or bounded) length

Sirmimer

Let $X$ be a set and define $f : X → X$. Let $A ⊆ X$. Prove that $f^{-1}[A]=A⇔f(A)⊆A$ and $f^{-1}[A]=A⇔f^{-1}[A]⊆A$. How do I go about proving "<=" in both statements?

Alessandro Codenotti

7:29 PM

@Akiva @Balarka there's this question somewhat related to our discussion, apparently removing countably many lines from $\Bbb R^3$ can't mess it up too badly, weird

Vrouvrou

7:40 PM

Hello, i have that $z$ is a limit point of $\phi_{t}(y)$ that is $z=\lim_{t\rightarrow+\infty} \phi_{t}(y)$ and we have that $f(\phi_{t}(y))$ is decreasing

can i deduce that $f(z)\leq f(y)$

Jing Weng

7:56 PM

Does the harmonic series approach positive infinity?

greedIsGoodAha

8:07 PM

yes. try using the comparison test to convince yourself

Akiva Weinberger

8:22 PM

@JingWeng Yes. Details can be found here:

https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

@Sirmimer That doesn't seem to make sense. Let $X=\Bbb R$, and $f(x)=x/2$. Then $f([0,1])=[0,\frac12]\subseteq[0,1]$, but $f^{-1}[[0,1]]=[0,2]\ne[0,1]$.

Zach Hauk

8:47 PM

i'm here

Ran Wang

Hi Zach

New to this site and is still figuring out how to use it...

Zach Hauk

hi @Ran

you'll get the hang of it pretty fast :P

Ran Wang

Well. Answered and asked a few questions. I feel the people here are exceptionally nicer than.....well, most of online forums I know

Zach Hauk

well, it's nice to hear you feel that way :P

Ran Wang

Yep. Math makes people nice~

Frank Science

8:52 PM

That's due to the difference of functionality of MSE and online forums.

Ran Wang

Agreed. But for example, I come from China and there is a site in China that serves similar purposes. Still people are very scary and always condemn the people for asking "simple and stupid" questions

By the way, I have a question to ask. I want to ask a question about reference for Mackey topology in general. Especially that when properties in weak topologies (such as compactness) implies that in Mackey topology. Is this too broad to ask as a question in MSE?

Frank Science

I think that it's okay, although I don't know what you're talking about.

with a tag reference-request

You needn't be too serious about their comments of simplicity and stupidity. It just indicates that they don't want to answer, or that the question itself is below the threshold of the forum. There is little meaning to say a question is stupid without reference to the questioner.

Jing Weng

9:24 PM

Is it always true that \sum |r_n| = \sum r+_n -\ sumr-_n ?

where r+_n = |r_n|+r_n / 2 and r-_n = |r_n | -r_n /2

Zach Hauk

9:48 PM

hey @Akiva

Akiva Weinberger

Hey

Zach Hauk

im studying spectral theorem now

Akiva Weinberger

I don't think I know that one

Zach Hauk

it says that the eigenvectors of a symmetric linear map form an orthonormal basis for $\Bbb R^n$

Akiva Weinberger

Ah, right

Zach Hauk

9:50 PM

(and, thus, that the eigenvalues are all real)

Akiva Weinberger

I wonder why it's called that

"Spectral"

Zach Hauk

well it doesn't have to do with prime ideals :P

@AkivaWeinberger if i make it in but can't go this year, am i still an "alum"?

Akiva Weinberger

I don't think so

I don't know if that means you have to retake the quiz, though

Zach Hauk

:/

Akiva Weinberger

@ZachHauk You should email Marisa or someone like that

Zach Hauk

9:52 PM

that's what i mean

Akiva Weinberger

Someone who works there

Zach Hauk

Marisa replied to me about financial aid

she said it's very flexible, and we would only have to pay as much as we can afford

Astyx

How does one prove that the isometry group is isomorphic to the group of symetries of the unit sphere ?

Zach Hauk

isometries of what metric space...

Astyx

Finite dimensional vector space

Zach Hauk

9:52 PM

like, any metric?

or euclidean norm

Astyx

Any I'd say

Zach Hauk

ah

Astyx

It seems quite intuitive

Akiva Weinberger

What about translations

Do you want origin-persevering isometries?

Astyx

Yes

Akiva Weinberger

9:54 PM

If you preserve the origin, then the unit sphere maps to itself

Astyx

Yes

Akiva Weinberger

so it bijects to the group of symmetries of the sphere that way

Astyx

Exactly, how do you show it's a bijection ?

Akiva Weinberger

Find an inverse

Astyx

The injective property is quite straightforward

Akiva Weinberger

9:56 PM

If $f$ is a symmetry of the sphere, how to you get the symmetry $g$ of the space corresponding to it?

Astyx

You use homogeneity of the norm I guess

Akiva Weinberger

Make $g(\alpha\hat u)=\alpha f(\hat u)$ or something

Extend $f$ linearly

Astyx

Not really linearly ?

Akiva Weinberger

Oh, right

Extend it… multiplication-by-scalars-y?

Astyx

Homogeneously

Akiva Weinberger

9:58 PM

That.

Astyx

Or homogeneous-of-degree-1ly to be more precise

Zach Hauk

oh i taught a kid induction today

Akiva Weinberger

Now show that it's really an inverse, and that it's a group homomorphism

Zach Hauk

and i also figured out how bad of a teacher i am

Akiva Weinberger

@ZachHauk Cool

@ZachHauk Less cool

Astyx

9:59 PM

Yeah that's the tricky part

Zach Hauk

i always flub words and don't know how to phrase things

Astyx

Except for the group part

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$Mathematics$ Mathematics