Mathematics - 2019-06-01

The Testosterone Fanatic

12:09 AM

man I wish I could mention it in my resume that I have access to messaging in this chatroom

Ted Shifrin

Why is that a point that makes you a good hire?

The Testosterone Fanatic

I gained some points here

which is required to get here

just a trivial achievement but an achievement nonetheless

never mind

nbro

Your nick though :P

Do you guys want see my approximation of the diffusion equation using central differences?

I is the function (in my case, an image)

This is basically an approximation of the RHS of the equation above

Ryan Unger

@TedShifrin cool

anakhro

12:30 AM

Hey cool cats.

@RyanUnger if your gf used local coordinates, is that a turn on or turn off?

BananaCats Category Theory App

1:09 AM

What do you call a mathematician that regularly cuts trees in his lumberjack hobby?

A log-a-rhythm

:D

anakhro

puns are the bane of humour

anakhro

1:25 AM

I still love you, though

BananaCats Category Theory App

I don't know homological joke theory yet

anakhro

The best math jokes are anecdotes.

Ryan Unger

2:17 AM

@anakhro what's wrong with local coordinates

BAYMAX

3:17 AM

Hi chat1

I was trying to do the coordinate change and rotation too

here - autodraw.com/share/052UF41XP17O

I am trying to obtain the matrix

anakhro

3:48 AM

@RyanUnger choices, choices, everything is about choices.

Ryan Unger

5:13 AM

@anakhro that's the wonderful thing about analysis

it's very flexible

there's lots of room for stuff

Subhasis Biswas

@BalarkaSen, even the PREFACE written by Artin to his abstract algebra book is on a whole new level

skullpatrol

5:38 AM

That is the hallmark of a classic textbook.

it forces the reader to sit up and take notice immediately

s.harp

6:12 AM

@TedShifrin I think this example cannot work, because isometries are uniquely determined by their derivative at a point. So if $f_n$ is the identity on some ball the only way to extend it by an isometry is to let it be the identity on everything (that is connected to the ball)

Subhasis Biswas

@skullpatrol which one?

s.harp

(of course $M$ should be connected otherwise there is no trouble finding counter examples :) )

skullpatrol

@SubhasisBiswas all classic textbooks make the reader aware of what to expect in the preface

Subhasis Biswas

@skullpatrol although I have a very little experience with classic texts, but I can confirm that both Rudin and Artin has the best prefaces to their texts. Not tasteless like other ones.

I am going through it right now. The book.

skullpatrol

Yup, those are classics.

Silent

6:31 AM

Let $(x_n)$ be a convergent sequence of real numbers. Denote $X=\{x_n:n\in \Bbb N\}$ and let $f:X\to X$ be a set map. Then $(f(x_n))$ converges if $f$ is injetive. How to show this?

NoLand'sMan

Hello

I didnt know chat rooms existed

skullpetrol

7:04 AM

Welcome :-)

Subhasis Biswas

can anyone find this in his hard-copy version?

or is this just a mistake in the pdf version

page 5

flowian

8:26 AM

@Silent Suppose $(f(x_n))$ does not converge and is injective. This implies that $(x_n)$ does not converge.

This of course cannot be true, as we supposed that $(x_n)$ converges

FYI. I used negation of $a \implies b$. Which is $a \land \neg b$.
Here $a$ is '$f$ is injective' , $b$ is '$(x_n)$ converges'.

Luyw

11:00 AM

hey guys

is the use of infinitesimals justified?

i have just read in thompson's calculus made easy a proof of the derivative of a constant being 0

for $y=x^3+5$ he does $y+dy=(x+dx)^3+5$ then substract $y$

well, the book is centered around them

sum rule $y=u+v \implies y+dy=u+du+v+dv \implies dy=du+dv$

so yeah, generally, is this kind of use of infinitesimals okay? are there ways such use is wrong?

and is the concept of orders of smallness taken seriously?

$dx\neq 0$ but $(dx)^2=0$

Leaky Nun

11:37 AM

Non-standard analysis

The history of calculus is fraught with philosophical debates about the meaning and logical validity of fluxions or infinitesimal numbers. The standard way to resolve these debates is to define the operations of calculus using epsilon–delta procedures rather than infinitesimals. Non-standard analysis instead reformulates the calculus using a logically rigorous notion of infinitesimal numbers. Non-standard analysis was originated in the early 1960s by the mathematician Abraham Robinson. He wrote: [...] the idea of infinitely small or infinitesimal quantities seems to appeal naturally to our intuition...

2

@Luyw

Luyw

what's the opinion of the "standard analysis" though?

s.harp

12:03 PM

@Luyw that the derivative of a constant function is zero is "easy" to prove in any setting

when you talk about infinitesimals without being precise about what they are supposed to be mathematically, then you are basically just motivating statements, and not proving them

Albas

If $\nabla$ is a connection on some manifold $M$, then we define for a vector field $Y$, $\nabla Y$ to be a $(1,1)$ tensor field such that when contracted with $X$ it gives the covariant derivative $\nabla_X Y$. I don't get the meaning of this definitio. How does acting $\nabla$ of a vector field give you a $(1,1)$ tensor field? And what does contracted with $X$ even mean?

skullpetrol

@Luyw standard analysis does not use infinitesimals

An "infinitesimal-number" just like an "infinite-number" are not "real-numbers."

Think about the question: is there a real-number in-between 0.999... and 1.000...

If there was it would have to be greater than 0.999... and less than 1.000...

What real-number can you add to 0.999... to make it larger but less than 1.000...?

Luyw

12:30 PM

thank you @skullpetrol

i see what you mean

skullpetrol

np

gateprep

1:40 PM

IS there anyone active in the chatroom

Subhasis Biswas

Hello

Here is a question

user193319

0

Q: Every Vertex Bijection Induces a Graph Isomorphism/Symmetry

First some setup: Definition 1.9 A graph $\Gamma$ consists of a set $V(\Gamma)$ of vertices and a set $E(\Gamma)$ of edges, each edge being associated to an unordered pair of vertices by a function "Ends": $\text{ENDS}(e) = \{v,w\}$ where $v,w \in V$. In this case we call $v$ and $w$ the ends...

Subhasis Biswas

2:02 PM

Let $G$ be the finite group generated by two distinct elements $x,y$ of order $2$. What is the order of $xy$?

2 i guess?

If they commute

otherwise they are isomorphic to $D_{2n}$...?

(the Dihedral group of symmetries of n gon)

ÍgjøgnumMeg

$D_{2n}$ isn't generated by two distinct elements of order 2

Subhasis Biswas

$xy$ should also be in the generating set, right?

and $o(xy)$ determines the order of the dihedral group?

found the answer

i was referring to this

5

Q: Prove that a group generated by two elements of order $2$, $x$ and $y$, is isomorphic to $D_{2n}$, where $n = |xy|.$

I am completely stuck at the question Let $G$ be a finite group and let $x$ and $y$ be distinct elements of order 2 in $G$ that generate $G$. Prove that $G \cong D_{2n}$, where $n = |xy|.$ I have proved that Let $x$ and $y$ be elements of order 2 in any group $G$. If $t = xy$ then $tx =...

abstract-algebra group-theory

ÍgjøgnumMeg

2:28 PM

lool

fair enough

user193319

In ÍgjøgnumMeg defense, "otherwise they are isomorphic to $D_{2n}$..." is quite vague and, in fact, false to say: elements aren't isomorphic to a group, groups are isomorphic to groups.

Astyx

What's an example of a Banach algebra with non trivial radical ?

Subhasis Biswas

@ÍgjøgnumMeg :p :p :p

@user193319 -_-

Leaky Nun

@ÍgjøgnumMeg it actually is

Astyx

He figured

Alessandro Codenotti

2:41 PM

@Astyx $M_2(\Bbb C)$ should work

Leaky Nun

oh

Astyx

@AlessandroCodenotti Oh neat

Alessandro Codenotti

Because it has nilpotent elements

Astyx

For some reason I had come to the conclusion that $M_n(\Bbb C)$ wouldn't work because I forgot that nilpotent matrices exist

Leaky Nun

doesn't mean they're in the radical right...

Alessandro Codenotti

2:42 PM

(this is even a $C^\ast$-algebra)

Leaky Nun

I think the upper-triangular matrices would work instead

at least that's what modular representation theory taught me...

Alessandro Codenotti

Isn't the nilradical contained in the radical?

Leaky Nun

nilradical doesn't exist for non-commutative rings

$\begin{pmatrix}0&1\\0&0\end{pmatrix}$ and $\begin{pmatrix}0&0\\1&0\end{pmatrix}$ are both nilpotent but their sum isn't

Alessandro Codenotti

Ah right, $M_2(\Bbb C)$ actually has trivial radical

Leaky Nun

yeah it's even simple right

Alessandro Codenotti

2:49 PM

Is there a simple example of a commutative Banach algebra $A$ and an $x\in A$ such that the ideal $(x^2)$ is closed? Because then the quotient will be an example of a Banach algebra with nontrivial radical

user193319

What about continuous complex valued functions?

Over some locally compact Hausdorff space?

Astyx

Then the radical is trivial, because it is the set of functions that are zero everywhere

user193319

Hmm...well, that rules out all commutative $C^*$-algebras.

Astyx

I don't get why $M_2(\Bbb C)$ doesn't work

Actually maybe I do

Martin Sleziak

> As an algebra, a unital commutative Banach algebra is semisimple (i.e., its Jacobson radical is zero) if and only if its Gelfand representation has trivial kernel. An important example of such an algebra is a commutative C*-algebra.

Wikipedia seems to confirm that commutative C*-algebras don't give such an example. (Although I do not know what Gelfand representation is.)

Astyx

2:56 PM

It's the map $g:A\to C(\hat A)$ such that for a characteristic $\chi$ $g(a)(\chi) = \chi(a)$

(where $\hat A$ is the set of characters on $A$)

Alessandro Codenotti

I think they're called characters of $A$?

Astyx

You're right

Alessandro Codenotti

But yeah for commutative $C^\ast$-algebras the Gelfand representation is an injective isometry, so it has trivial kernel

Ajay Mishra

3:16 PM

Do the method of checking differentiability via the continuity of partial derivative in the vicinity of region, fails? or I am getting it wrong?

Astyx

4:00 PM

Does $\Bbb C[X]/(X^2)$ work ?

@AlessandroCodenotti

Astyx

4:19 PM

Where $\Vert aX+b\Vert = |a|+|b|$

This way $\Vert PQ\Vert \le \Vert P\Vert\Vert Q\Vert$

Alessandro Codenotti

Is it complete?

Astyx

I think, since it looks like $\Bbb C^2$

But I need someone to tell me it is to be completely sure :p

Is $l^1(\Bbb N)$ with convolution a Banach algebra ?

Alessandro Codenotti

4:49 PM

If you use $\Bbb Z$ instead you can construct the group C*-algebra en.m.wikipedia.org/wiki/Group_algebra

Ted Shifrin

@s.harp Yeah, I was totally stupid. The derivative won't be an isometry at every point of those disks.

Alessandro Codenotti

Hi @Ted

Subhasis Biswas

5:22 PM

hi bois

I have come up with a question. Perhaps a very stupid one

Let us consider a metric space $(X,d)$ . Define a function (sort of like a linear-functional) $f_{x_0}:X \to \mathbb{R}^+ \cup \{0\}$ such that $f_{x_0}(x)=d(x_0,x)$.

My question is, if $(X,d)$ is connected, will $f_{x_0}$ be continuous?

Alessandro Codenotti

5:37 PM

What do you think?

Hint: connectedness is irrelevant

Subhasis Biswas

$(X,d)$ is open in itself. Now, $\mathbb{R}^+ \cup \{0\}$ is also open. So, $f_{x_0}^{-1}(\mathbb{R}^+ \cup \{0\})=X$, which is open. So continuous.

@AlessandroCodenotti

Subhasis Biswas

5:53 PM

nah my argument seems terribly wrong

take $X= [1,2] \cup [3,4]$ with usual metric of $\mathbb{R}$. Take $f_2(x)=d(2,x)$

@AlessandroCodenotti it should be continuous and connectedness should not be irrelevant

Mathphile

We may write \begin{align} \int_0^{\pi /2} {\frac{{dx}}{{\left( {1 + x^2 } \right)\left( {1 + \tan x} \right)}}} &= \int_0^{\pi /2} {\frac{{\cos x}}{{\left( {1 + x^2 } \right)\left( {\cos x + \sin x} \right)}} \cdot \frac{{\cos x - \sin x}}{{\cos x - \sin x}}dx} \\ &= \int_0^{\pi /2} {\frac{{\cos ^2 x - \sin x\cos x}}{{\left( {1 + x^2 } \right)\cos 2x}}dx} \\ &= \frac{1}{2}\int_0^{\pi /2} {\frac{{\sec 2x}}{{1 + x^2 }}dx} + \frac{1}{2}\int_0^{\pi /2} {\frac{1}{{1 + x^2 }}dx} - \frac{1}{2}\int_0^{\pi /2} {\frac{{\tan 2x}}{{1 + x^2 }}dx} \end{align} — mwomath Jul 9 '14 at 12:00

can someone help me understand the third step?

how did he divide into three integrals?

Mathphile

6:10 PM

anyone there?

Cannon444

Hello everyone!

I was wondering if anyone knows where I can post questions that would be answered as a solution rather than a formula on how to solve the question?

Is there a separate place in Stack Exchange or another website that deals with this?

Subhasis Biswas

@AlessandroCodenotti I was trying to go with the argument: Let the function (which is real valued) be discontinuous at a point $c$. Partition all the elements of $X$ into two subsets. One with $d(x_0,x)\leq c$ and one with $d(x_0,x)>c$

In either of the sets, a significant amount of points would go missing. Which will help us to construct two non empty open sets whose intersection is empty and whose union gives us $X$

Let $\lim_{x \to c^+} (=m) \neq \lim_{x \to c^-} (=n)$. Take a point in between $m$ and $n$, and consider a neighbourhood around it with diameter less than $|m-n|$. That region would be empty.

@BalarkaSen, what do you think?

Alessandro Codenotti

@SubhasisBiswas you need to check the preimage of every open set

Subhasis Biswas

@AlessandroCodenotti oh ok. I missed it.

now, what is your opinion?

Alessandro Codenotti

Or you can check the preimages of a basis, like the open intervals for example

Subhasis Biswas

6:24 PM

why the connectedness should be irrelevant here?

Alessandro Codenotti

Check the preimages of the open intervals as I suggested

@SubhasisBiswas $c$ should be in $X$ for $f$ to be discontinuous at it, but you're treating it like an element of $\Bbb R$ immediately afterward

Cannon444

Hey fellas, do you know where I can post questions to get answers as solutions rather than answers as formulas that answer the questions? Is there a different website for this?

Subhasis Biswas

@AlessandroCodenotti yes. I should not have done it. What I had in mind was: If discontinuous at $c \in X$, we get a corresponding value of $f_{x_0}$ in $\mathbb{R}$.

Alessandro Codenotti

Still, there is no reason why the set with $\leq$ should be open though

Subhasis Biswas

@AlessandroCodenotti is there any guarantee that the set would either closed or open?

or even both?

Alessandro Codenotti

6:31 PM

Hmm I don't think so

Start from the easy case, what's the preimage of $[0,a)$ through $f$?

Ted Shifrin

belated hi to demonic @Alessandro

Subhasis Biswas

@AlessandroCodenotti an open set

Alessandro Codenotti

Which open set?

Ted Shifrin

@SubhasisBiswas: Exercise for you: The metric function $d\colon X\times X\to\Bbb R$ is always continuous.

Subhasis Biswas

@TedShifrin ok. I have to do this one.

Ted Shifrin

6:35 PM

It answers your current question immediately.

And you should understand why.

Alessandro Codenotti

It's close to what we're trying to do now

Subhasis Biswas

@TedShifrin yep. I got it

@AlessandroCodenotti is this one a bit stronger?

No.

then? equivalent?

Au contraire.

6:37 PM

There's a mantra for doing anything involving metric spaces

Subhasis Biswas

@AkivaWeinberger what is it?

Akiva Weinberger

which is, "When in doubt, use the triangle inequality"

Ted Shifrin

That sounds like sound advice.

There shouldn't even be doubt :P

hi, DogAteMy

Alessandro Codenotti

Just look at the preimage of an interval, I've been trying to say this for a while now!

Subhasis Biswas

Ok @TedShifrin. I am going to try this. Help me out if I get stuck

@AlessandroCodenotti It will take some time for me.

Ted Shifrin

6:39 PM

I don't like using preimages for things like this.

The $\delta$-$\epsilon$ definition is perfectly fine for metric spaces.

Ultradark

7:02 PM

Can integral curves be non parametric curves?

Vladimir Reshetnikov

Is there a theorem connecting coefficients in the asymptotic expansion of the Hilbert transform of a function f with the moments of the function f itself?

Balarka Sen

@TedShifrin Sequences > $\delta-\epsilon$ > preimages in metric spaces

Alessandro Codenotti

7:17 PM

filters>sequences>...

Balarka Sen

This is cool. I was wondering if you could do this a few days ago.

Rather, I was wondering if there's an adjoint to the forgetful functor LRS -> RS

Mathphile

can someone help me understand how $\int \frac{x^{2n}}{1+x^2} dx$ can be expressed by the hypergeometric function as shown on wolfram alpha wolframalpha.com/input/?i=int+x%5E(2n)%2F(1%2Bx%5E2)?

Ultradark

Hey $\to$ chat

Mathphile

any help guys?

Ultradark

7:34 PM

Mathphile maybe you should wait more time before you send a follow up message to the chat. Ask your question, and then be patient. Maybe move on, go outside and integrate some flower petals in your backyard.

Balarka Sen

@SubhasisBiswas It's good to understand why exactly $D_{2n}$ is generated by two elements of order 2; two pairs of reflections about different axes generate all the rotations!

Ultradark

The chat answers when the chat wants to answer. You cannot make the chat do anything. And asking a question multiple times will anger the chat. The chat will get so angry that the chat will explode and then nobody can chat, so please.

Semiclassical

@Mathphile the answer isn't all that interesting. If you expand $x^{2n}/(1+x^2)$ around zero, you get $x^{2n}-x^{2n+2}+x^{2n+4}-\cdots$. Then integrating this term by term gives $$\frac{x^{2n+1}}{2n+1}-\frac{x^{2n+3}}{2n+3}+\frac{x^{2n+5}}{2n+5}-\cdots$$

Mathphile

@Ultradark okay sorry if I annoyed anyone

Ultradark

You didn't. I'm just saying

Balarka Sen

7:42 PM

The crucial point is if $|x| = |y| = 2$ then $\langle x, y \rangle = \langle x, xy \rangle$. So in fact it's a group which has a normal cyclic subgroup of index $2$ generated by $xy$, so is a semidirect product $\langle x \rangle \ltimes \langle xy \rangle$.

Semiclassical

That can be rewritten as $$\frac{x^{2n+1}}{2n+1}\left( 1+\frac{2n+1}{2n+3}(-x^2)+\frac{2n+1}{2n+5}(-x^2)^2-\frac{2n+1}{2n+7}(-x^2)^6+\cdots\right)$$

Presumably, the portion within parentheses is just ${_2}F_1(1,n+\frac12;n+\frac32;-x^2)$

Ultradark

I'm listening to a song that's 12 minutes long

Semiclassical

(note that the first interesting coefficient is of the form $(n+\frac12)/(n+\frac32)$, same as the parameters in the hypergeometric function)

Balarka Sen

In fact, I don't think finiteness of the group is necessary. If the group is not finite $xy$ is forced to be of infinite order (semidirect product of finite groups is finite), which implies it has to be $\Bbb Z_2 \ltimes \Bbb Z$ or $\Bbb Z_2 * \Bbb Z_2$ - they are isomorphic. This is the "infinite Dihedral group"

Mathphile

@Semiclassical yes

Semiclassical

7:45 PM

So yeah. The indefinite integral has that form, but it's not much more revealing than "it's a way to write the series expansion"

Mathphile

the problem is that i don't know much about hypergeometric functions and i was hoping someone could explain me how to express the integral in that form

Semiclassical

You mean like I just did?

Mathphile

@Semiclassical but this helps

Semiclassical

ah

for reference, here's the series expansion for the hypergeometric function (for |z|<1):

$${}_{2}F_{1}(a,b;c;z)=\sum _{n=0}^{\infty }{\frac {(a)_{n}(b)_{n}}{(c)_{n}}}{\frac {z^{n}}{n!}}$$

where $(a)_n = a(a+1)(a+2)\cdots (a+n-1)$ is known as the Pochammer symbol

Balarka Sen

Honestly I don't know the point of these functions. They just look cool

Semiclassical

7:49 PM

in particular, when $a=1$ (as in your example) you get $(a)_n = (1)_n = 1(2)(3)\cdots(n)=n!$

Balarka Sen

It always feels nice to write ${}_n F_m$

Semiclassical

hence $\displaystyle {}_{2}F_{1}(1,b;c;z)=\sum _{n=0}^{\infty }{\frac {n! (b)_{n}}{(c)_{n}}}{\frac {z^{n}}{n!}}=\sum _{n=0}^{\infty }{\frac {(b)_{n}}{(c)_{n}}}z^n$

Mathphile

@Semiclassical okay this really helps

Semiclassical

oh, bad move on my part: we're already using $n$ for the indefinite integral, so we can't reuse it as a summation index

so replace $n$ with $k$ in my last two series expansions

Mathphile

okay

Semiclassical

7:54 PM

And then $b=n+1/2,c=n+3/2$ gives (doing the replacement I indicated) $$\frac{(b)_k}{(c)_k} = \frac{(n+1/2)(n+3/2)\cdots(k+n-1/2)}{(n+3/2)(n+5/2)\cdots(k+n+1/2)}=\frac{n+1/2}{k+n+1/2} = \frac{2n+1}{2n+1+2k}$$

which matches the series expansion I gave earlier for the integral

Mathphile

yes

Thank you @Semiclassical

Semiclassical

So it does indeed match. But I'm not sure this really tells you anything interesting; if you've got a series expansion with simple-looking coefficients, you can usually get it in terms of a hypergeometric function

Moreover, when you pick some value of $n$--say, integer or half integer---you can typically simplify further, as the WA page you linked initially indicates

So I'd say the lesson is really: "the series expansion of this indefinite integral has simple coefficients"

Mathphile

@Semiclassical wasn't really looking for anything interesting. I was just trying to figure out how this hypergeometric function works which you have made me understand

Semiclassical

kk

Mathphile

thanks again :)

Ultradark

7:59 PM

has anyone here ever read a harry potter book

Semiclassical

@BalarkaSen yeah, even having worked with them they're sorta wtf to me

Though the unifying feature there is that they're solutions to linear second-order ODEs

Where things get weird af is nonlinear odes

Painleve functions etc

the strange thing is that there's parallels in the story of painleve functions (solutions to certain nonlinear odes) and the hierarchy of hypergeometric/bessel/airy special functions (solutions to certain linear odes)

never really wrapped my head around that

c.f. en.wikipedia.org/wiki/Painlev%C3%A9_transcendents#Degenerations

Ultradark

nice

Semiclassical

(something something singularity analysis, i guess)

Ultradark

8:16 PM

I don't want to read a textbook, and I also don't want to read a popular math book. Any suggestions?

Subhasis Biswas

8:31 PM

@BalarkaSen I will try to learn further. But not rn. I am very slowly getting used to it

@TedShifrin $d: X \times X \to \mathbb{R}$. What is the metric of $X \times X$?

I am inclined to consider the metric $d^* : X \times X \times X \times X \to \mathbb{R}$ defined by $d^*((x,y),(a,b)) =|d(x,y)-d(a,b)|$

Now, with respect to $d^*$ metric on the product metric space i will now try to prove the continuity:

Pick an arbitrary $\epsilon >0$ and consider the inequality $|d(x,y)-d(a,b)| < \epsilon$. Now, the metric function will be continuous at $(a,b)$ if we can find an open ball $B((a,b),\delta)$ in $X \times X$ such that, $(x,y) \in B((a,b),\delta) \implies |d(x,y)-d(a,b)| < \epsilon$. We choose $\delta = \epsilon$.

@Ultradark This is beyond science...

Astyx

Hi chat

Subhasis Biswas

8:49 PM

okay, the metric $d^*$ should be $d^* : (X \times X) \times (X \times X) \to \mathbb{R}$

Alessandro Codenotti

9:46 PM

@SubhasisBiswas that cannot work, think about the relationship between the metrics in $\Bbb R^2$ and $\Bbb R$

What would be the distance between (1,3) and (6,8) with your definition?

Subhasis Biswas

Let $(X,d)$ be an infinite metric space such that $d(x,y) \in \mathbb{Q}, \ \forall x, y \in X$ with usual metric of reals.

A) Does such a metric space exist?

B) Is this metric space connected?

Attempt: Yes, such a metric space does exist. Take $\mathbb{Z}$ with the usual metric of $\mathbb{R}$.

B) Pick any two $p, q \in X$. Let $|p-q|=r$ where $r$ is a rational. Let $r=2$ (WLOG). So, $p =q+2$ or $p=q-2$. (we choose any one of the possible values). Now, the metric space would be connected, only if $X$ contains all the values between $q$ and $q+2$. But, $q + \sqrt {2} \notin X$

@AlessandroCodenotti let me check

@AlessandroCodenotti so, you are asking me to check the value of $d^*$?

Yes

damn. that's zero

Indeed

but the points aren't identical

that cannot work

Alessandro Codenotti

9:51 PM

So think about how you calculate distances in $\Bbb R^2$ normally instead

Subhasis Biswas

@AlessandroCodenotti euclidean distance. So, I can define the metric to be $\sqrt{(d(a,b))^2+(d(x,y))^2}$

can't we?

okay. I will try to work that out myself and then I will get it checked. don't post this

answer

Alessandro Codenotti

Right idea, wrong formula

Subhasis Biswas

@AlessandroCodenotti :(

Alessandro Codenotti

What's the distance between (0,0) and (1,1) with your formula?

Subhasis Biswas

I will redo it. I will also go through the preimages as you asked me to do.

But right now I can't keep my head

can you kindly check the new problem I posted?

it has been bugging me

Ultradark

9:56 PM

Alessandro are you one of the owners of this room?

@SubhasisBiswas do you wanna check my most recent question and I'll look at yours?

Subhasis Biswas

@AlessandroCodenotti do you mean $f_2(x)$?

@AlessandroCodenotti new one?

@Ultradark man. I probably don't know much of math to look at your question. An honest reply. Trust me. I am here to learn. One day I might be capable enough

Alessandro Codenotti

@SubhasisBiswas $f_{x_0}$ for any $x_0$

@SubhasisBiswas Yes

@Ultradark Yes

Subhasis Biswas

@AlessandroCodenotti again zero. Now I'll try this : $\sqrt{(d(x,a))^2+(d(y,b))^2}$

Alessandro Codenotti

@SubhasisBiswas Yeah that's how you compute distances in $\Bbb R^2$

Subhasis Biswas

@AlessandroCodenotti :D

thank you so much Sir. I kept missing it

Ultradark

10:01 PM

@SubhasisBiswas I am mostly here to learn too. I am slowly starting to learn stuff

But I also realize most of the folks in this chat are in the top 1% in the world in math

so I'm not going to compare myself to them. Yet.

Alessandro Codenotti

Note that if $X$ and $Y$ are metric spaces there's infinitely many metrics $X\times Y\to[0,\infty)$ inducing the product topology

Subhasis Biswas

@AlessandroCodenotti does that mean we can choose our own metric accordingly?

Now, can we have a metric space where $d(x,y) = $rational for every $x,y$ in $X$?

for example $(\mathbb{Q},d)$

Alessandro Codenotti

If $(X,d_X)$ and $(Y,d_Y)$ are two metric spaces any function $X\times Y\to[0,\infty)$ of the form $d((x_1,y_1),(x_2,y_2))=(d_X(x_1,x_2)^p+d_Y(y_1,y_2)^p)^{1/p}$ is a metric that induces the product topology, for $p\in (1,\infty)$

@SubhasisBiswas Right

Subhasis Biswas

@AlessandroCodenotti $L^p$ norm?

Alessandro Codenotti

It's the $p$-norm of the vector of distances

It works for a product of finitely many spaces as well

(The product of countably many metrizable spaces is also metrizable, but that's a little harder)

Ultradark

10:11 PM

@SubhasisBiswas or @AlessandroCodenotti do you know if superellipses are solutions of a differential equation or a set of independent differential equations?

Alessandro Codenotti

I don't even know what a superellipse is

Ultradark

superellipses are the equations for different p-norms of $L^p$ space

Subhasis Biswas

does it always have to be disconnected when the space is infinite? my argument would be like this:

Take any $x, y$ in $X$ (metric is $|x-y|$). Fix $y$ and suitable $r \in \mathbb{Q}$. Now, $x=y+7$ (or $y-7$, let's ignore that) (WLOG, $r=7$). So, for $X$ to be connected, all values between $x$ and $y$. But pick $y+ \sqrt{2}$, which does not belong to $X$.

Ultradark

For example $x^n+y^n=1$ for different values of $n$

Subhasis Biswas

@Ultradark more general version of an ellipse?

Ultradark

10:14 PM

Superellipse

A superellipse, also known as a Lamé curve after Gabriel Lamé, is a closed curve resembling the ellipse, retaining the geometric features of semi-major axis and semi-minor axis, and symmetry about them, but a different overall shape. In the Cartesian coordinate system, the set of all points (x, y) on the curve satisfy the equation | x a | n + |...

Subhasis Biswas

@AlessandroCodenotti more generally, after choosing $r$, we can choose $y+r/\sqrt{2}$. Which is not going to be in the set

Alessandro Codenotti

@SubhasisBiswas You're not really justifying why for $X$ to be connected all intermediate values between $x$ and $y$ must be obtained

Subhasis Biswas

@AlessandroCodenotti assuming that's an established theorem... I encountered that proof in Rudin.

Alessandro Codenotti

(That fact is indeed true, but it would need some justification)

Subhasis Biswas

@AlessandroCodenotti can we argue that w.r.t the usual metric of R, the fact holds for any $X$?

especially since $d: \mathbb{R} \times \mathbb{R}$, $X$ has to retain some fundamental properties of $\mathbb{R}$

Alessandro Codenotti

10:18 PM

Sure, but that needs a few things, the image of a connected space through a continuous function is connected, product of connected spaces is connected, and the distance is a continuous function from the product

The distance has $X\times X$ as domain

Subhasis Biswas

@AlessandroCodenotti oh...

@AlessandroCodenotti okay. let me try

Bob

anybody here good with statistics?

Jeff

Question: Why doesn't the linear algebra textbook give this proof of $0\bm v = \bm 0$?

Proof: $0 \bm v = 0 (v_1, v_2, ..., v_n) = (0v_1, 0v_2, ..., 0v_n) = (0, 0, ..., 0) = \bm 0$

It gives a different, longer, proof.

Ultradark

Just ask your question and someone will answer if they are interested @Bob

Bob

could somebody look at my post:

0

Q: A problem Dealing with Sampling with replacement

Below is a problem from the Schaum book: "Probability and Statistics". I started the problem but I am confident that I am on the wrong approach. I am hoping somebody will tell me where I went wrong. Problem: An urn holds $60$ red marbles and $40$ white marbles. Two sets of $30$ marbles are drawn ...

probability statistics

Alessandro Codenotti

10:24 PM

@Jeff What's \bm supposed to be? It's not rendering

Jeff

i thought that was how to bold the next character (so that it's understood to be a vector).

Bob

\bm is the bold character in LaTex but I think you need an include file

like tools.ins

Alessandro Codenotti

Seems fine to me, what's the proof given in the book?

Well I guess the proof in the book works regardless of how scalar multiplication is defined in your vector space, while yours is specific to $\Bbb R^n$

Subhasis Biswas

I was thinking about going like this: $d: X \times X \to \mathbb{R}^+\cup \{0\}$ is a continuous function. Take $f:d(X \times X) \to \mathbb{R}$ such that $f(x)=x$, which is indeed continuous.

Jeff

@AlessandroCodenotti Ah, is that the secret?

Subhasis Biswas

10:34 PM

Nah...I am heading in the wrong direction

Alessandro Codenotti

No need to write it

I think my point above is the reason they have a longer one

Jeff

To answer your question, @AlessandroCodenotti, the proof in the book is:
0v = 0
0v=(0+0)v
0v=0v+0v
0v+(-0v)=0v+0v+(-0v)
...

Alessandro Codenotti

Yeah, that kind of manipulation works in every vector space

Anyway I'm leaving now, it's well past midnight here

Jeff

@AlessandroCodenotti Goodnight. TY

Subhasis Biswas

@AlessandroCodenotti

wait

a bit

It's 4:08 here and I haven't slept yet.

Suppose that for some $p, q \in X$ , there is some $r$, such that $p<r<q$ and $r\notin X$.

one question here. Does $X$ have to have an ordering?

johnny09

10:54 PM

why the particular solution for nonhomogeneous differential equation depends only on one variable?

for example for the heat equation $u_{xx}=u_t-F$, we seek $u(x,t)=u_s(x)+X(x)T(t)$

Subhasis Biswas

11:07 PM

Supposing that $|p_0-q_1|=r_1$ , $|p_0-q_2|=r_2$ , $|p_0-q_3|=r_3$ such that $r_1<r_2<r_3$ (fixed) and any $p_0, q_1, q_2, q_3$ (consider the values like a set) satisfying the equalities are such that $p_0, q_1, q_3 \in X$ but $q_2 \notin X$.

Now, consider the set $\{x \in X: |x-p_0|< r_2\}=A$ and $\{x \in X: |x-p_0|>r_2 \}=B$. Evidently, $A \cup B =X$. Now, $A \cap B = \emptyset$. And $A, B$ are non-empty. Both $A$ and $B$ are open. So, $X$ must be disconnected.

Ultradark

11:28 PM

mic check mic check...

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$Mathematics$ Mathematics