Mathematics - 2015-03-25 (page 1 of 5)

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12:00 AM

Hello @columbus8myhw

@columbus8myhw $g^2 = f\cdot h$

Ohh

infinitesimal simplicio

5

A: Sporadic MathJax failure

Peter from the MathJax team here. Our CDN provider has been experiencing attacks on some of its DNS servers (see also this thread on the MathJax User Group). This attack leads to erratic DNS resolution failures. We're sorry for the inconvenience this is causing users and we'll send out an updat...

What an inconvenience

I wish people would just stop attacking other people

@ᴇʏᴇs Yeah… no.

12:04 AM

@robjohn, since $b-a>0, c>0$, so to have $b-a=cm$, we can't have $m\le 0$. So, If $c>b-a$, then $cm>b-a$. Thus $cm\le b-a<b$. Am i right?

By the way, the maxima and minima of $\dfrac{\sin x}x$ are precisely where $\dfrac{\sin x}x$ intersects $\cos x$. This can be proven easily with calculus, but I wonder if there's a geometric proof.

Well, since $m\gt0$ and $m\in\mathbb{Z}$, we must have that $m\ge1$. That means that $c\le b-a\lt b$

@robjohn, thank you very much.

@ᴇʏᴇs Hi Bart.

Hi @JasperLoy

@JasperLoy How are you

12:09 AM

@ᴇʏᴇs Not too good. I did some things these few days which unintentionally made my OCD a bit worse. Now I need to try to fix it.

@JasperLoy :(

12:41 AM

@teadawg1337 I have test tomorrow.

Hype.

@robjohn Could you take a look at my question?

0

Q: Differential equation Laguerre $xy''+(1-x)y'+ay=0, a \in \mathbb{R}$

The differential equation Laguerre $xy''+(1-x)y'+ay=0, a \in \mathbb{R}$ is given. Show that the equation has $0$ as its singular regular point . Find a solution of the differential equation of the form $x^m \sum_{n=0}^{\infty} a_n x^n (x>0) (m \in \mathbb{R})$ Show that if $a=n$, where $n \in ...

differential-equations

infinitesimal simplicio

How are you going to review @Owatch?

I did the best thing I could. I went out and used my moneys to buy many drugs to take away anxiety!

Just kidding, I'll do some practise problems.

Probably tomorrow since I have work for other classes to do tonight.

infinitesimal simplicio

12:57 AM

Try to find questions that combine methods

I am being tested on five sections.

Two problems each, worth eleven points each.

Per section.

infinitesimal simplicio

ok

So I will do problems from each section.

I work really slow though, so I may only do three in the four hours before class.

Maybe 5 at most.

infinitesimal simplicio

Good luck my friend :-)

I performed poorly on an exam today :(

1:09 AM

Happens to the best of us

infinitesimal simplicio

:(

Does it really happen to the best

I cannot do badly.

I must do well.

I remember the time I got a 5% on an exam

I rarely do very well.

1:11 AM

It wasn't because I didn't study either

Then what was it?

The thing about actuarial math... is that no matter how well you know the theory, the only way to really learn it is to do $\infty$ many problems

For me, pure math came much easier.

I had studied everything except one thing I didn't get, and with my bad luck that one thing was like half the exam problems lol

I always make mistakes on some simple questions and get zero points

Hi @meer2kat

1:22 AM

@ᴇʏᴇs sup

For those who used my "interesting questions" bookmarklet (maybe @MikeMiller?): it was broken for a couple of months. Fixed version

(SE changed the format of time of posts on the front page, which I did not account for.)

@ᴇʏᴇs doing anything fun?

@meer2kat Studying I guess

@meer2kat Complex analysis is kind of fun, some parts

@ᴇʏᴇs cool!

@meer2kat Number theory not so much

1:27 AM

@ᴇʏᴇs yeah

@meer2kat What about you

@ᴇʏᴇs reading up on modular arithmetic

@meer2kat I have trouble with modular arithmetic

@ᴇʏᴇs i'm enjoying it so far though in class we only covered basic operations so far4

Thanks for the fix, @Woodface. I hadn't used it since I changed computers but I'll bookmark the fixed one.

1:34 AM

Hi @meer2kat @ᴇʏᴇs

Hi @JasperLoy

@JasperLoy Are you doing okay

Committing to a challenge

Should anyone care about the hyper reals?

Hi @Committingtoachallenge

Committing to a challenge

Hi @ᴇʏᴇs

@Committingtoachallenge Yes

Committing to a challenge

1:35 AM

Why should we care about the hyperreals?

@ᴇʏᴇs Not too good, like I said. I messed up some things these few days, now I have more OCD themes to resolve.

@JasperLoy you'll do fine

and hi

@meer2kat Thanks. How did you come across that video?

Committing to a challenge

They seem a little ridiculous to me

@Committingtoachallenge Don't you like non-standard analysis

1:37 AM

@meer2kat Oh by the way you could just have posted that video in this room, no need to create another room.

Committing to a challenge

@ᴇʏᴇs I don't like infinite numbers

@Committingtoachallenge Racist

Committing to a challenge

Why would anyone care about them if they aren't possible under any application?

@JasperLoy Remember I used to create new rooms just to chat with you lol

@Committingtoachallenge Why does anyone care about set theory

Committing to a challenge

@ᴇʏᴇs Because it is integral for the construction of much of math

And that 'much of math' has many applications and makes sense

Whereas there is no idea of an infinite number

1:40 AM

@Committingtoachallenge Why do you choose to construct math with set theory

Committing to a challenge

Well it has to be constructed somehow right?

@Committingtoachallenge Why not type theory

Committing to a challenge

Well sure, I don't know much of that

But if it works then fine

But what are hyper reals being used to bridge to?

They work

Committing to a challenge

They might work, but they aren't inherent in reality, there is no infinity in reality

There is no infinite number

1:42 AM

About about higher order logic

Committing to a challenge

What?

@ᴇʏᴇs: Sets are interesting (or really, logic is interesting). That suffices for me.

Committing to a challenge

@Mike do you care for hyper reals?

I never learned any nonstandard analysis. I think I might be amused.

Committing to a challenge

They have infinite numbers in hyper reals...

1:44 AM

@MikeMiller @Committingtoachallenge @ᴇʏᴇs @meer2kat I am going to take a nap, good night, pray for me.

@JasperLoy i watch a lot of spoken word. i didn't realize that i could post links

Committing to a challenge

@JasperLoy Good night, I shall pray for you by lighting candles

@Woodface Thanks! The bookmarklets are really very helpful.

Committing to a challenge

@PeterWoolfitt He usually doesn't respond to us lowly humans

@Committingtoachallenge I don't mind. Hopefully he will at least read the comment - it's nice to get positive feedback.

1:58 AM

well this made sense until we hit exponents

2:13 AM

YAY MATHJAX IS BACK!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1111111111111

:D

well that was fun. i just went through the entire chapter in about ten minutes

Committing to a challenge

What chapter from what book?

2:39 AM

oh from my textbook from class. it's just over modular arithmetic

Committing to a challenge

@robjohn I thought he quit during the Bill D situation. Isn't that the case?

wow it got quiet

Committing to a challenge

@meer2kat Ahh $\Bbb Z / n \Bbb Z$ is definitely very important

i'll learn more in depth about notation tomorrow

Committing to a challenge

@meer2kat $\Bbb Z / n\Bbb Z$ is just integers mod $n$

2:54 AM

@Committingtoachallenge yeah

3:07 AM

@Committingtoachallenge It was before that.

@meer2kat What is spoken word

Committing to a challenge

@robjohn Oh okay, maybe he just got bored

@Committingtoachallenge It's more than that. When people ask him to go to chat, he says he won't.

Hey, everyone

Hi @Axoren

Committing to a challenge

3:16 AM

@robjohn Well I suppose assuming that is his real name, the people he knows in his private life could read all of his transcript, and maybe it could reflect badly on him

Without looking up a procedure or giving me an answer, just tell me what your first approach to solving this problem would be: $$\left[\begin{array} \\ 7 & 10 \\ 15 & 22 \end{array}\right]^{\frac 1 2}$$

Committing to a challenge

Is that a half?

Yeah, the square root of that matrix.

Another way to put it is the matrix $B$ such that $BB$ is that other matrix.

Obviously, $B$ is 2 by 2.

Committing to a challenge

Yep

I thought of a bunch of ways to solve it, but they're all disgusting.

Committing to a challenge

3:19 AM

I'll think about it after I finish my functional assignment

Looks interesting, I have never seen such a thing before for some reason

And there's an actual identity for it, and I know I would have never gotten it.

Not specifically for THAT matrix

But that's the matrix someone presented to me.

When you're done with your assignment, try a blind attempt at it.

If I had to speculate, I bet he got pinged all the time from chat of people asking him questions that he didn't want to do, and should have been just asked on main so that he can choose which problems that he wants to do without people bothering him. @Committingtoachallenge

Committing to a challenge

@Paul That is another fair idea, but I find MSE is usually pretty dramatized. Probably excessively so(But it is subjective considering this website means more to some than others)

Who are we talking about

Committing to a challenge

Asaf

3:23 AM

I could also believe drama

Wait, he left the site or something?

I've been gone a while because of school.

I haven't seen @DavidWheeler

Committing to a challenge

He left chat almost two years ago, but he was a big member of chat(and a founder)

No, he just use to be the owner of the chat room but he now refuses to use the chatroom

Committing to a challenge

David got bored because of me I believe haha

3:24 AM

I've only been around since Rob and Anon owned it.

Committing to a challenge

He helped me out for a bit and then got bored of my questions, my bad, he was extremely helpful

Maybe he got overwhelmed because a lot of people were asking him questions lols

Committing to a challenge

The end of David

It was late, so don't blame my questions ahaha

@Committingtoachallenge That is why we can't have nice people in the chat room...

:)

Committing to a challenge

I am a monster

3:26 AM

Do you think you will be able to do all those books in 2 years

?

Committing to a challenge

Well I am doing very poorly in that regard, but I am covering the content fully which is good

And the challenge has been successful motivation

I am somewhere around a fifth or where I should be

I guess in the end it is the journey that matters

Committing to a challenge

Yep

But I will get to speed up a heap when my semester ends

Thats what they all say

Committing to a challenge

Well I worked through my christmas break 2h a day, so it wasn't tooooo bad

and that was with heaps of trouble in personal life simultaneously

3:29 AM

Were you still behind schedule, even if you discount any falling behind when you were in school?

Committing to a challenge

@PaulPlummer Yep, half where I should have been, but I was homeless for 20 or so days there, so it isn't toooo bad

@ᴇʏᴇs poetry, but spoken

@meer2kat Oh poetry

infinitesimal simplicio

...of logical ideas?

@Committingtoachallenge oh wow. It is definitely good that it is motivating to you and that it seems to be working (even if going slower than your ideals). I am a little to erratic, and get bored if I follow a plan like that (I have tried), and I tended to underestimate the density of material, 10 pages doesn't sound like a lot, and maybe could be read easily but to understand and use the material takes a lot more time.

3:37 AM

I always get so depressed when I compare myself to other students

Committing to a challenge

Yes, I often had times where I was doing 2 pages a day xD

There is a high school student that has been taking PhD level courses :(

infinitesimal simplicio

:-O

Committing to a challenge

A little out of date, but here are my page numbers roughly

Zorich is a painfully boring textbook, I am tempted to find a replacement and start some replacement from scratch...

@Committingtoachallenge Also, maybe one of the other reasons I tend to not do that, is that if I do fall behind, I am like "I mine as well not do anything since I am so far behind". I am not familiar with Zorich, what does he cover?

Committing to a challenge

3:39 AM

I have replaced ‘Dugundji – Topology’ with ‘Munkres – Topology’

@PaulPlummer Zorich - mathematical analysis I. It is pretty much calculus + real analysis + 10000 applications(random number I chose)

Ah so it has some stress on applications,

That is why you are doing it together with Rudin?

Committing to a challenge

Yep, apparently it is a great way to learn real calculus, and I will give it a chance to get better

(real in the sense that you will remember it in a few years, rather than normal calculus that you forget within 6 months, as I have experienced)

@Committingtoachallenge Maybe not a bad idea. It seems like there will be a lot of redundancy in your study, not sure if that was by design. Is you main interest towards analysis?

This is why I hate latex.

I have to write a point cloud instead of drawing one.

What's a point cloud

3:53 AM

A graphical representation of points that can only be described as a cloud.

Disconnected vertexes with some axes for reference.

Are you drawing an atom or something

user image

Committing to a challenge

@PaulPlummer Where is the redundancy? I think only a few chapters will be redundant + all of the intro stuff for most of the texts + zorich/Rudin conflict

A doughnut atom

beautiful

3:55 AM

That's an example of a point cloud with specific distributions.

I hope that's not what you have to draw in LaTeX

I have to draw something with a much lower resolution.

Still, like 20 points and shaded areas.

On a graph.

Sounds annoying

Committing to a challenge

Main interest was unknown until now haha @Paul

@Committingtoachallenge What is it now

Committing to a challenge

3:56 AM

@ᴇʏᴇs It seems to be algebra, but it might change

@ᴇʏᴇs Baking.

Committing to a challenge

@ᴇʏᴇs I am finding algebra amazing

Damn, not fast enough, lol

Baking sounds better than algebra

You can't eat algebra.

3:56 AM

does anyone know how to draw a 3-D gradient or vector filed in LaTeX?

@Simple Our LaTeX questions probably belong on the TeX chatroom

@Committingtoachallenge My professor is kind of desperate to solve the Monster group

@ᴇʏᴇs Solve it?

Yes, that's what he calls it

How would one solve a group? Find all the elements of them and their interactions?

3:58 AM

@Axoren I asked this question few hours ago, but no one answer it - -

I guess everything there is to know

@Committingtoachallenge Maybe it isn't as much as I am thinking, just looks like a lot of analysis and related fields. Getting interested in algebra, anything in particular/specific, or more of a feeling that the stuff you have been studying is more up your alley?

Committing to a challenge

@PaulPlummer I have been able to sit down for 8 hours on algebra a few times and not get even remotely bored, so that's a good indicator haha

Definetly!

@ᴇʏᴇs He's an ambitious man. Hopefully a wise man in a cave somewhere has bestowed upon him a sword and shield.

Committing to a challenge

3:59 AM

I am learning a variety of things in class for it, so I guess I will see if it stays that good

I personally want to play the Ring Game on a really large ring.

Committing to a challenge

Learning lie and commutative algebras atm

@Axoren He did spend a good chunk of his life devoted to it but he's depressed he might not be able to understand it before he passes

I had a feeling when you asked about $\mathfrak{sp}$

Committing to a challenge

yep xD

4:00 AM

That's my goal. I want to play a Ring Game the same size as Chess.

And be good at it.

Committing to a challenge

That being said I am doing real analysis sort of stuff right now for my functional analysis class

0.0

Committing to a challenge

I seem to be really bad at analysis

Which might be why I don't like it, so I won't commit to algebra until after this semester(taking complex and functional analysis)

Yeah.

Committing to a challenge

I hate hyper real stuff hahaha

4:05 AM

But it leads to a simple proof of why, say, $x^2$ isn't uniformly continuous.

Actually, who wants to make a Ring Video Game, after this: math.stackexchange.com/questions/158924/the-ring-game-on-kx-y-z

I tend to think it is a mistake to narrow really early, especially since one of the reasons I like math so much is the connections, although "focusing" on algebra isn't really focusing just deciding what you like.

Let $H$ be infinite, so that $\frac1H$ would be infinitesimal.

Committing to a challenge

infinite 'numbers' are yuck

$H$ and $H+\frac1H$ are infinitely close to each other. If $x^2$ was uniformly continuous, we'd have $H^2$ and $(H+\frac1H)^2$ being infinitely close to each other.

However, those expand to $H^2$ and $H^2+2+\frac1{H^2}$, which are not infinitely close to each other. Thus, $x^2$ is not uniformly continuous.

@Committingtoachallenge They might be yuck, but you can make this all rigorous.

4:08 AM

@columbus8myhw Thanks

(This specific approach requires the Axiom of Choice, though, the way the hyperreals are normally defined...)

Committing to a challenge

I don't understand the axiom of choice, but I haven't tried to in a long time. I will check it out again tonight

(it's only 2pm)

Uh, well, first you need to know how the hyperreals themselves are constructed in set theory, which is complicated.

So the connection to the Axiom of Choice will be far from obvious.

@Committingtoachallenge You're from Queensland, right?

I need to go, sorry.

Committing to a challenge

Yep

@columbus8myhw Have fun!

There's an old joke: the axiom of choice is clearly true, the well-ordering theorem is clearly false, and who can say about zorn's lemma?

Committing to a challenge

4:13 AM

How can I prove that $f(x) = \left\{\begin{array}{cc} x,&x\in\Bbb Q\\-x,&\in\Bbb I\end{array} \right.$

That in isn't continuous for any $x\ne 0$

I am trying to show $\exists \epsilon \gt 0, \forall \delta \gt 0| |x-c|\lt \delta$ and $|f(x)-f(c)|\gt\epsilon$

And I have $|f(x)-f(c)| = |x-c|\gt \epsilon$

Assuming both are rational as case 1

Pick your $x$ first. THEN pick $\epsilon$

Committing to a challenge

Well I want $x$ to be non-zero

And I have $\delta \gt |x-c| \gt \epsilon$.

Or I prove it for one $x$ and use something to extend it to all other cases?

$$f_0(x) = x^x \\ f(x) = \sup \{x, f_0(x), f_0(f_0(x)), f_0(f_0(f_0(x))), ...\} \\ \mathcal{C} = \left\{ f(\alpha) \mid \alpha \text{ is an ordinal}\right\}$$

I argue that for any ordinal $\alpha \in \mathcal{C}$, and any other ordinal $\beta$, that if $\beta \ge \alpha$, then $\beta \in \mathcal{C}$.

Committing to a challenge

@DavidWheeler I somewhat get this joke after google searches

Ugh, wait. Should that be $\sup$? Ugh, it's just the right-limit of that sequence.

Committing to a challenge

4:20 AM

@DavidWheeler So the order of quantifiers is $\exists \epsilon,\forall \delta, \exists x(\epsilon,\delta)$?

@Axoren There are a couple of similar animations at the end of this answer

@robjohn Thanks Rob, but I wanted to do it in LaTeX. Our homework is required to be typeset in LaTeX.

No animations, no image imports.

So, I think that $\mathcal{C}$ is the class of the almost-largest ordinals. The largest being $\delta = \sup \mathcal{C}$.

@Axoren Oh. You'd need a special graphics package to do that kind of animation.

Ugh, wrong letter.

Committing to a challenge

@Kaj can you help me with my easy problem(not for me) above?

4:23 AM

Oh yeah, your quantifiers are weird

I think I've finally got it. The largest ordinal ever possible is that delta.

Committing to a challenge

Oh above that haha

Oh wait

"Is not continuous"

Is $\mathbb{I}$ the irrationals?

Committing to a challenge

Yep

So you can get an irrational arbitrarily close to any rational

Start by choosing any $x$ and take $\varepsilon = f(x)$

Committing to a challenge

4:26 AM

Choosing any $x$ rational or not?

I think that'd work, let me think. And doesn't matter.

Committing to a challenge

I don't understand taking any $x$

Shouldn't $x$ remain arbitrary?

Well, we can show that this function is discontinuous at any $x \neq 0$ @Committingtoachallenge. To do this, choose an arbitrary such $x$, and let's find an $\epsilon$ for which there isn't a corresponding $\delta$.

Yeah, it is arbitrary.

Oh, I missed the question and read something else.

If we can get it to work for one such $x$, then you can apply the same proof over and over for any given $x$ since we'll be working in general terms.

Committing to a challenge

4:28 AM

Ahhh okay

So we are extending one thing to the rest okay

So I think choosing an $x$ and taking $\varepsilon = f(x)$ will work

So let's assume $x$ is rational WLOG. Then for any $\delta >0$, we can find $y \in [x-\delta, x+\delta]$ such that $y \in \mathbb{I}$.

Committing to a challenge

$|2-c|\lt \delta$ and $|f(2)-f(c)|\lt \epsilon = f(2)$ means

$|2-c|\lt \delta$ and $|2-f(c)|\lt 2$

Now depending on whether $c$ is rational or not we get two cases

Is this going in the right direction?

Yeah, so let's see what happens if $x=2$.

Which is what it looks like you're doing.

Committing to a challenge

So thats a contradiction so I screwed up

No no. So if $x = 2$, we can find an irrational number, call it $y$, within any delta distance of $2$.

And $f(y) < 0$.

Which means $|f(2) - f(y)| > f(2)$ right?

This is problematic. You can't get that distance on the function values less than $\varepsilon = f(2)$

Committing to a challenge

4:34 AM

Oh I inversed my inequality sign

$|2-c|\lt \delta$ and $|2-f(c)|\gt 2$

Anyone have time for a quick notation question? The last part I will get once I understand what the hell is happening in the first part: i.imgur.com/6sktbNX.png

Yeah, exactly, for some $c \in \mathbb{I}$.

Can a collection of sets be inside itself if it's a collection and not a set?

Does that notation mean something else when used with two collections?

god that's painful to read, @Axoren

Yeah, it is.

But I understand the last part perfectly.

I just don't know what the first part means.

Committing to a challenge

4:38 AM

$|2-c|\lt \delta$ and $|2+c|\gt 2$

$c$ can depend on delta?

Well, it needs to lie within delta distance of your chosen $x$

@Axoren I am guessing it should be more like $\{c_\infty \cap c_\in \mid c_\infty \in \mathcal{C}_\infty, c_\in \in \mathcal{ C}_\in \}$. I am saying this because if these are collections of sets, but then they contain themselves they are a set, but they can't be a set because they contain themselves.

@PaulPlummer That's what I'm thinking, too. I'm also kind of curious what the $\wedge$ would mean between two collections.

Well I think they are defining what it is

Yeah, but it's not used later.

4:45 AM

Where are you getting this? What is $\Pi$?

Is it a function that involves $\wedge$?

It's a well defined function on collections.

In our context, it's called a growth function.

Would it make sense if $\mathcal{C}= \mathcal{C}_\infty \wedge \mathcal{C}_\in$ but they just forgot to put that in?

I'm trying to think of a fast way to explain it.

@PaulPlummer That's what I'm thinking.

But that means that $\mathcal{C} = {\mathcal{C}_\infty}_{\mathcal{C}_\in}$

By a previous definition.

Lol

Well if that makes sense, then I would guess that is the case, but it is very strange. This is a strange book or article

It's homework.

4:49 AM

Oh, then I would ask for clarification, sometimes things like this happen

The symbols $\infty$ and $\in$ are meaningless placeholders used for distinguishing purposes only.

As opposed to using $1$ and $2$ to imply some order.

Last homework assignment, he used the prime symbol as an element of a set and everyone started to panic.

$\prime$

I think it's a good idea, just not with symbols we use everywhere else.

What class is this? Obscure confusing notation 301

Lol, no. Machine learning.

I think I understand the higher-level meaning of the proof I'm expected to make.

Is this stuff standard notation or does the professor just want to get people comfortable with all these symbols are just symbols with no inherit meaning

There isn't a standard notation for a lot of this stuff.

I think $\Pi$ is the only standard notation used in that question.

4:54 AM

So the prof is just bad at making up notation...

Again, I don't think it's that bad.

I just think he could have used different symbols to avoid the implication of order.

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