Mathematics - 2015-01-10 (page 1 of 3)

robjohn

00:15

I haven't seen Chris's sis since this moring (morning here)

Hmm... she was on the main site an hour ago.

Jasper Loy

00:36

@robjohn Well, maybe I wasn't here, so she thought there was no point coming, lol.

robjohn

@JasperLoy That must be it.

Jasper Loy

I am trying to find the insect which bit me ten times or so, either mosquito or something else, in my room.

Hello @Alizter remember not to read Lie Groups, just study for your A level exams and get into Cambridge.

Alizter

@JasperLoy I have been studying

Jasper Loy

@Alizter Good for you. I have been thinking of how to solve my mental problems.

Alizter

@JasperLoy Start reading one of your books. You know, just for fun. You don't have to do the exercises. Just read through. You can come back to exercises later

5

anorton

00:51

@Alizter That is really good advice... I find that if I ever set out to work every exercise in a book, I'll never finish. But, if I set out to just read through a book, I'll certainly finish. :)

user97303

01:26

hey @JasperLoy

user97303

does anyone have a copy of "Algebraic Graph Theory", by Gordon Royle and Chris Godsil?

Kevin Driscoll

02:19

I was about to watch a video about a board game. But then it seemed too complicated for the moment. So, instead I decided to watch a video about Ultrafilter by Terry Tao........ how does this make any sense????

Jasper Loy

03:20

Hello @KajHansen how are your classes?

Kevin Driscoll

05:36

Whats the name of the NOT baby Rudin analysis book? If it Real and Complex Analysis?

Kaj Hansen

Yes @KevinDriscoll

@JasperLoy, not bad so far. I'm in for a very busy semester though.

@MaryStar, I've been busy this evening. Did you figure out your problem?

Jasper Loy

@KajHansen It's weird that you replied two hours later, after I went out and returned.

Kaj Hansen

I should be doing work, but I think I'm going to relax with some video games...

@JasperLoy, is there a gym near you?

Jasper Loy

@KajHansen Not too near, but yes. Why do you ask?

Don't tell me you are planning to live with me.

Kevin Driscoll

@KajHansen What are you going to play?

Kaj Hansen

05:45

@JasperLoy It's one of those rare times when I'm actually feeling not-depressed after going hard in the gym a bit ago. Though it's mostly temporary, it helps.

@KevinDriscoll, an indie game called Binding of Isaac. Might play some Skyrim later.

And even later maybe I'll get around to TeX'ing my topology homework.

Jasper Loy

@KajHansen I am going to give myself another 6 more years to enter grad school. I set the upper bound at age 40.

Kaj Hansen

Life is too short to wait @JasperLoy. Plus you get less creative as you age.

Kevin Driscoll

@KajHansen Ah, excellent game. You playing Rebirth?

Jasper Loy

@KajHansen Sure, but healing takes time. I might never get well. And the creativity part is bullshit to me.

Kaj Hansen

Indeed! I'm nearly platinum god @KevinDriscoll. I've beaten all the challenges and all bosses using all characters on hard.

I just need to actually find all the items I've unlocked now.

@JasperLoy, something weird happens when we age. I definitely remember having a more extensive imagination when my age was in the single digits.

Kevin Driscoll

05:49

Aaaaah, excellent. You're farther than I am. I was very close to Platinum God in the original. I think I just had a handufl of items and some no damage achievements to do....... then Rebirth came out finally

Kaj Hansen

I don't know what I'm talking about of course :P

Jasper Loy

@KajHansen One reason I got mentally ill is because I am too imaginative.

Kaj Hansen

Do you remember a time when you were healthy @JasperLoy ? My problems started in 2010. I miss my old self.

I feel like I'm getting a little better as of late.

Jasper Loy

@KajHansen There is no clear time when I became mad. But if you want me to give a time, I would say 16 years ago, when I was 18.

Kaj Hansen

@KevinDriscoll, where are you in your education?

Jasper Loy

05:51

I need to kill that fucking mosquito. It has bitten me ten times.

Kevin Driscoll

@KajHansen Working on my PhD

Kaj Hansen

Oh nice. I'm just a third year undergrad myself. Where are you at?

Kevin Driscoll

Georgia Tech! About an hour from Ted Shifrin.

I'm a physicist though

Kaj Hansen

Nice! I see Ted several times a week. I've had him for three courses now.

I have countless friends attending GA Tech.

Kevin Driscoll

Ah, another Bulldog. Sorry we can't be friends anymore :-(

(but Im a grad student so it doesn't really matter to me!)

Now if you went to UNC...... THEN we'd have problem (#GoDuke)

Kaj Hansen

05:56

haha! I don't really get into rivalries myself.

I just stick to my math and my weightlifting :P

Kevin Driscoll

As close as UGA and GaTech are, they'll never be as close as Duke and UNC.

Kaj Hansen

Yeah? How far apart are we talking? @KevinDriscoll

Kevin Driscoll

@KajHansen 8 miles. 20 minutes by bus, every half hour on the half hour. Less by car.

Kaj Hansen

Woah. You weren't kidding!

So do you know JimmyK on here? I was talking to him the other day.

Kevin Driscoll

No I cant say that I do

how would I?

Kaj Hansen

06:05

He's also a grad student at GA Tech

In math of course

Kevin Driscoll

Ah ok do you know who he works for? Ive had very limited interaction with the math department. Theyre reasonably far from each other on campus.

(which is a change for me, at Duke they're in the same building!)

Kaj Hansen

Can't say that I do. Are you studying physics/statistics?

Kevin Driscoll

Just physics. Cold atom theory.

Kaj Hansen

Oh very cool. My dad is a physical chemist. He's doing work in that area.

He's studying supercooled liquids, particularly those that tend to form networks of hydrogen bonds, like water.

Kevin Driscoll

Some of the things those people do, I can't even begin to imagine. They have the craziest ideas sometimes, those chemists!

Kaj Hansen

06:09

Well, I actually don't know what cold atom theory entails. It's probably a lot more different than I realize.

He was a math major at UW back in the day. He took real analysis from Rudin himself.

Kevin Driscoll

I suppose he has a Rudin number of 1 then

and his students a Rudin numbe rof 2, and so on

Kaj Hansen

LOL, and I guess that makes me an Erdos 2 :P

My combinatorics professor last year was an Erdos 1.

Kevin Driscoll

I think Mike Miller is working on his Erdos-Bacon number

which is the sum of the steps you are removed from both Erdos AND Kevin Bacon

Kaj Hansen

hahaha

Kevin Driscoll

If you don't mind me asking, what ethnicity are you @KajHansen? I like to mentally keep track of the background of people I talk to. If you were named John or something, I'd just assume you were Caucasian European, but Kaj is unique to me.

Kaj Hansen

06:17

I am Caucasian European. My ancestors were vikings, and both sides of my family are from the Norway/Denmark area. "Kaj" means "Earth" in some Old Norse dialect (as in earth, fire, water, air).

So the "j" in my name is pronounced as it is in all those cool Norse words like fjord or the names of Skyrim characters.

Kevin Driscoll

Ah! Okay, that's pretty interesting. Shattering my stereotype that all Scandanavian people are named Magnus, Jon, and Sven or something similar.

Kaj Hansen

I.e. Kaj = Sky without the "S"

My last name is in the traditional Norwegian style though. "sen" = "son of". So you get a lot of surnames like Christiansen, Arnoldsen, etc.

Kevin Driscoll

No relation to the Japanese word, kai, I imagine

Kaj Hansen

None at all. I get asked that a lot though, lol

Kevin Driscoll

AH WOW! And now I know. I had noticed the patter, e.g. Magnus carlsen

Kaj Hansen

06:22

Yep!

Sometimes you see that sort of thing but it's "son" instead. That indicates a Swedish background instead of Norwegian.

Kevin Driscoll

Is that a recent change? Like, I know Norwegian has like 2 ways of writing; the old way and the new way that they updated to be inline with phonetic pronunciation

Kaj Hansen

I'm not sure. I don't actually speak any of those languages.

Kevin Driscoll

Gotcha. I roomed with 2 Swedish guys a couple years ago so I learned all of their prejudicial feelings about people or Norwegian decent

Kaj Hansen

I'm actually not familiar with that either. I'm not first generation or anything like that.

My family's been in America for 3 or so generations.

Kevin Driscoll

Ya, I figured. What about Lutefisk, anyone ever make you eat that stuff?

Kaj Hansen

06:32

My grandmother makes it sometimes, LOL

Kevin Driscoll

My step-grandfather was from Minnesota but his family was form Norway as well

Loved that stuff.......but oh man was it TERRIBLE

Kaj Hansen

hahahaha

Kevin Driscoll

Also, I saw your pic with Ron Paul on your Facebook (creepy Facebook stalker, RIGHT HERE!!!), and it makes me jealous

I dont always agree with him, but I am a libertarian

Don't tell the mods; we might not be allowed in the chat at the same time anymore

Kaj Hansen

LOL. I'm fine with people looking me up on Facebook. I added the link for a reason.

That was in South Carolina at Wofford College. I saw him in a debate live too that day with the other Republican candidates.

I'm not a Republican though. I'm definitely on the libertarian side of things.

Kevin Driscoll

Oh yea, I've been to Wofford once or twice. I'm from Greenville so its not too far.

makes sense that he'd be debating in SC given its primary position

Kaj Hansen

06:41

What sorts of music do you listen to?

Kevin Driscoll

Did you ever identify as Republican? I stopped being a Republican around the time the President was elected. I wonder now if there will be people who will not be going basically their whole adult lives self-identifying as libertarians

You know, I don't take time out to listen to music all that often, despite the fact that I play the guitar. But I go through phases. Usually I want to hear some catchy top 40 song from the radio so I go listen to it on youtube. But then I end up in some bizarre string of songs hopping through youtube.

Kaj Hansen

Kind of? But by the time I was sufficiently well-informed and interested in politics, I was a libertarian.

I'm actually not interested in politics very much anymore. I just focus on math at this point.

Kevin Driscoll

That goes through Kanye West and Killer Mike but also Depeche Mode and A-Ha AND Metallica AND John Coltrane....... and well you sorta see how this goes

About the only things I don't listen to are country (but I like bluegrass) and anything labeled metal from the past 5 years or so

Kaj Hansen

Yeah, I feel you. I'm very, very much into music myself. I have a fairly extensive library....all sorts of genres. It's great because it allows me to connect to a lot of people as sort of an icebreaker.

Kevin Driscoll

Yea, I love that aspect. In my own experience I know its easy to mischaracterize whole groups of people if you don't have any of them in your social circle. And my social circle is VERY white (as you might expect from my academic post), so I really appreciate that hip hop exists as a thing I can usually talk about with someone who's black. Once you get over that initial hump its a lot easier to see other stuff to talk about.

Kaj Hansen

06:52

Yeah, that's a good point. I don't listen as much to modern hip-hop, but I listen to a ton of 90's hip hop.

It's a very natural topic of discussion when you're hanging out with someone for the first time, so there are huge advantages like you pointed out.

skullpatrol

:D

Kaj Hansen

LOL @skullpatrol

I don't think I've ever asked...where are you in your education @Skull ?

Kevin Driscoll

Ya to be fair, I've sworn off a lot of Top 40 hip hop these days. I basically listen to Kanye now and some slightly more underground stuff that I heard about recently like Run The Jewels. I love that 90s flow. People spend a lot of time now, I think, on writing catchy hooks and choruses and such and relatively less time on treating their voice like a percussion instrument that can add interesting rhythm to the song.

skullpatrol

And then there is humor.

Kaj Hansen

Like I said, I don't listen to much modern hip hop, but I do like Kanye's Beautiful Dark Twisted Fantasy.

Lupe Fiasco is decent for recent stuff too.

skullpatrol

06:58

I'm some where between a banana and a mango @KajHansen ;-)

Kaj Hansen

Of course :P

Kevin Driscoll

MBDTF is part of my "work mix." If I really need to focus and get lots done I have a playlist for that and its a big part of it.

Kaj Hansen

It's not so much study music for me. But hip-hop is a go-to genre when I'm getting myself psyched up on the way to a test or on the way to the gym.

Kevin Driscoll

How do you feel about Yeezus?

Kaj Hansen

Eh, not as familiar as I probably should be. Any favorite tracks I can YouTube?

Jasper Loy

07:06

@skullpatrol I just called my best friend. He said some things to me which showed he did not understand me anymore. So I think I have just lost my best friend. I won't talk to him anymore.

Kevin Driscoll

It's a pretty weird Kanye album. My favorite track is Bound 2, I think

Kaj Hansen

Interesting music video

Kevin Driscoll

Oh ya I forgot about that.....absolutely ridiculous

Kaj Hansen

hahaha

Jasper Loy

@KajHansen Aren't you sleeping?

Kaj Hansen

07:10

@JasperLoy, it's not that late here

Kevin Driscoll

2 AM is basically early for a Friday

Really, @KajHansen should be out gettin shwasty-faced

Kaj Hansen

LOL, you're probably right. Maybe I'd actually find a girl for once.

Kevin Driscoll

If you keep asking , you'll find one with probability approaching 1. Only question is if you want to find a girl at a bara.

or dorm party

Do they have parties in the dorms at UGA?

Its basically not allowed here.... which I find perplexing

Kaj Hansen

I greatly prefer dorm/house parties.

Balarka Sen

@KajHansen Hi.

Kaj Hansen

07:18

Yeah, it's allowed. Though where I live is technically a dorm, it very much feels like an apartment.

Large bedrooms, living rooms, kitchen, etc.

Mostly upperclassmen.

Jasper Loy

Over the years, my friends have left me one by one, due to my mental illness. Now I have only a few left.

Kaj Hansen

Hey there @BalarkaSen

Jasper Loy

I am still looking for that mosquito.

I don't want ten bites to become twenty.

The bites are extremely big and itchy.

Kevin Driscoll

Unfortunately for you, the number of bites grows GEOMETRICALLY

Jasper Loy

WTF

If it bites again I will need to think of a solution.

Balarka Sen

07:25

@KajHansen Think of an infinite group. Quick.

Kaj Hansen

$\mathbb{Z}$.

Balarka Sen

@KajHansen Draw the Cayley graph.

Jasper Loy

Why are you ordering him?

Kaj Hansen

Ugh, I have to go back and review how to draw them.

Balarka Sen

@KajHansen if $G$ is a group, the cayley graph of $G$ with generating set $S$ is the graphs with nodes being elements of $G$ and edges joined between nodes $a$, $b$ if and only if $a^{-1}b \in S$

Kaj Hansen

07:27

Ok, so $\{1\}$ is a generating set.

Balarka Sen

$\{1, -1\}$, sure.

Kaj Hansen

Woah, woah, let's be careful @BalarkaSen

There's a difference between generating set and the set of generators.

Balarka Sen

OK, yeah, I meant the latter :P

Sorry.

Kaj Hansen

Ok cool

So for one, we get this giant "line" of nodes/edges. (n, n+1) is an edge for all $n \in \mathbb{Z}$.

Balarka Sen

Exactly.

Kaj Hansen

07:30

I guess that's all you get, haha

Balarka Sen

It's the real line with integer marked nodes.

@Kaj Now choose a node from your graph.

Kaj Hansen

Ok, I'll choose $2$.

Balarka Sen

Sure. Now oh noes I forgot one thing. Nodes $a$ and $a\cdot s$ ($s \in S$) in the Cayley graph are joined and directed. OK?

There is an arrowed edge going from $a$ to $a \cdot s$

Kaj Hansen

Ok, so there is an arrowed edge going from any element to any other element in our case?

Balarka Sen

Right.

Actually the direction is irrelevant. Hehe. My bad.

@KajHansen, so you chose $2$.

Kaj Hansen

07:34

Sure

Balarka Sen

You see, there are two "rays" going from $2$. One begins at $2$ and goes towards infinity and begins at $2$ and goes the opposite ways towards negative infinity.

Kaj Hansen

Certainly. I see that.

Balarka Sen

Informally, the "boundary" of this Cayley graph is $\{\infty, -\infty\}$

Now let us look at another example before going into the definition of a boundary.

Kaj Hansen

Ok, what would be a good one to look at?

Balarka Sen

Consider the free group on two generators $\langle x, y \rangle$ and the generating set $\{x, y\}$

Kaj Hansen

07:37

Not familiar with free groups unfortunately

Balarka Sen

@KajHansen Well, they're just group consisting of elements being finite words (say $xyxxxyxyxyxyyxx$)

Just two generators with no relation.

Kaj Hansen

Ok, that's simple enough (no pun intended)

So would $\mathbb{Z}$ be a free group? I.e. $n = 1111111111...$ (n times)

Balarka Sen

Actually, no because $1$ and $-1$ in $\Bbb Z$ commutes ;)

That's a relation.

Kaj Hansen

Hmm. Let me ask you another question: Do finite free groups exist?

Just trying to get a feel for these.

Balarka Sen

Yes, no problem, asking questions are good. I won't look at Cayley graphs of free groups since you're not familiar with those. No, free groups are not finite.

Kaj Hansen

07:41

Ex: Would the dihedral group be a free group, with generators Flip and Rotation?

Balarka Sen

Dihedral groups have a relation.

If $r$ is rotation and $s$ is flip, then $srs = r^{-1}$

You can check that.

Kaj Hansen

I'm already familiar with that. Just probing what you mean by "relation" and I think I understand now.

Balarka Sen

Yeah, so your group of $x$ and $y$ have no relation.

You can just build your words and go on and on and on and it won't reduce to anything simpler.

Kaj Hansen

Cool cool. Are there any examples of free groups that can be expressed in a form I already know?

Balarka Sen

Hehe. Actually $\Bbb Z$ is a free group. My bad.

It's a free group on a single generator.

Kaj Hansen

07:44

-_-

Cool. I guess there simply cannot be a relation when there's only 1 generator

Balarka Sen

@KajHansen I do that everytime. Decline true statements as false with weird logic and then decline my logic and apologize. Please bear with me :P

Kaj Hansen

lol ok

Balarka Sen

@KajHansen Yeah. Infinite cyclic groups are free groups on single generator. (Of course, all of them are isomorphic to Z)

Kaj Hansen

Yeah. Kind of a boring example, but oh well

Balarka Sen

Free groups with n > 1 generators are complicated ;)

Complicated in the sense that they are not of the same isomorphism type of any group you know of.

Kaj Hansen

07:49

I figured that'd be the case

I'm taking topology right now btw. I figured you'd like that

Balarka Sen

But they are important since any group appears as quotient of free groups.

@KajHansen Cool beans.

Metric spaces and whatnots, @Kaj?

Or are you onto general topological spaces already?

Kaj Hansen

Yeah. It looks like we'll be covering a lot of material. We started with metric spaces though.

Balarka Sen

Interesting. Cayley graphs can also be equipped with a metric. :)

Kaj Hansen

Graphs in general can, lol

Balarka Sen

Yes, true.

But the metric on cayley graph induces a metric on the group.

so any finitely generated group is a metric space

Kaj Hansen

07:55

Oh cool. And so it's also a topological space.

Balarka Sen

indeed.

Kaj Hansen

Here's the problem set I'll be working on next week: math.uga.edu/~pete/4200HW_one.pdf

Balarka Sen

in fact the groups are metric spaces is the single revolutionary idea that was used by Gromov to formalize geometric group theory.

@KajHansen let me look.

Pete's notes are always great ;)

eh, that's not topology @Kaj

Kaj Hansen

Yeah, we're stuck in metric space stuff for the first week or two @BalarkaSen

Balarka Sen

But that's analysis.

Kaj Hansen

08:01

Sure, but we'll get to the juicy topology soon enough.

Balarka Sen

I suppose you guys are revising analysis, @Kaj?

Kevin Driscoll

Ive never taken a topology class, but Ive seen several video lecture series and lecture notes and such

and they all seem to go over some analysis stuff to start with

Kaj Hansen

Yeah, in a sense. Here's the syllabus if you're interested: math.uga.edu/~pete/4200_syllabus_spring_2015.pdf

Balarka Sen

I've never studied analysis but I still have been able to study topology @Kevin

Maybe it depends on the book. I used Simmons, and it never touched much of analysis (because it's a book on analysis :P) yet introduced topology in a nice fashion.

Everything's nice except the first two weeks @Kaj :P

Topology over R is epsilons all over. Ugh.

Kaj Hansen

I have faith in Pete. He certainly knows what he's doing, and I enjoy his lectures.

Balarka Sen

08:08

Yeah Pete is a great guy.

I just have an allergy for R :P

I hate epsilons so much that I always think of continuity by the open-set definition :P

(Of course, that's the right way to think about it for general top. spaces, but I do that in C and R too :P)

Kaj Hansen

I think about both definitions depending on what's most convenient at the time.

Balarka Sen

Maybe I'd have to study analysis some times soon.

You have quite an open mind, @Kaj.

Kaj Hansen

What makes you say that @BalarkaSen ?

Balarka Sen

You're studying topology (plus analysis) after studying PDEs and like combinatorics, even though your primary interest is algebra. That's rare.

Kaj Hansen

I studied ODEs, not PDEs. I'll be taking a course in PDEs next year though :P

Balarka Sen

08:20

Or maybe not rare, but just that I am a narrow-minded prat.

Kaj Hansen

I like to think I'm getting a lot of breadth, but I feel like I don't have much depth either. I'm trying to fix that though.

Kevin Driscoll

Grad school is the place to really fix that, honestly

Balarka Sen

You have more depth than the grad schoolers I study algebraic topology with, @Kaj

So shuddap.

Kaj Hansen

I'm planning on grad school. I just hope a university will take me. My GPA isn't the best, unfortunately.

I have something like a 3.3 right now, mostly because I tend to overload myself with math courses every semester.

Balarka Sen

What's GPA?

Kaj Hansen

08:24

@BalarkaSen, you are assigned grades at the end of each semester. A, B, C, D, or F. If you get an A, then you earn 4 points. If you get a B, 3 points, and so forth.

Your GPA is the average number of points you've earned.

Balarka Sen

3.3 should be ok, then, @Kaj.

Kaj Hansen

There's a bit of simplification there. Basically, my "average grade" over all my university coursework is exactly a B+

Maybe. I've been under the impression that grad schools are mostly interested in applicants with >3.5 GPA. I don't really know what I'm talking about though.

Balarka Sen

You'll be fine. Don't worry about it.

Kevin Driscoll

I can't speak for math, but in physics the cutoff is normally closer to 3.0, My GPA was in that 3.3 area and I had a 780 on the physics GRE. Was good enough to get into a couple tier 2 grad schools.

And that was with really only 2 semesters of research experience

Kaj Hansen

I'm fine with a grad school on the same level as UGA or GA Tech. I don't have any internal NEED to get into MIT, e.g.

I accept that I'm not the best mathematician, and I lack the superhuman creativity to be truly great. But I know that I enjoy math, and it's what I want to do for a living.

Kevin Driscoll

08:41

Ok so Im trying to do this analysis thing

so if I want to prove $f(x) = x^2$ is continuous

How do these things usually go?

I need to fix $\epsilon > 0$

and then consider a point, say, $x-\delta$ and then demand that $\delta$ be such that $x^2 - (x-\delta)^2 < \epsilon$?

Kaj Hansen

@BalarkaSen, I actually wonder if "preimage of open sets is open" would be easier in this case.

Yes @KevinDriscoll. To say a function is continuous is to say that it is continuous at every point in its domain.

So you not only choose an $\epsilon > 0$, but also some $x \in \mathbb{R}$.

Kevin Driscoll

Indeed, so this specific strategy will work?

ah ya, of course, I kinda did that implicitly

Kaj Hansen

It will. I've never actually considered this though, so I'm interested in how it'll turn out.

So let's play around. Expanding that and simplifying, we'll get...

Kevin Driscoll

How would you have done the problem? The open sets way?

Kaj Hansen

This way is fine. I've never actually considered the open sets way for a problem like this.

So expanding and simplifying, we'll get $2\delta x - \delta^2$

Kevin Driscoll

08:48

@KajHansen WLOG take $\delta > 0$ then $2 x \delta > \delta(2 x - \delta)$

Kaj Hansen

And it's kind of obvious that, making $\delta$ small enough, you can force that to be as small as you like. That is, $\displaystyle \lim_{\delta \rightarrow 0} 2\delta x - \delta^2 = 0$. But you can crunch it out even further to get an appropriate $\delta$ in terms of $\varepsilon$.

Kevin Driscoll

So if $\delta < \epsilon/2 x$ then we cna satisfy the condition

Kaj Hansen

Let me check that right quick

That works, yep

Kevin Driscoll

Oh, there should be an absolute value

I suppose just to be careful

in case x is negative

Kaj Hansen

Oh sure. You'd have absolute value bars around the original thing.

I.e. we want to find a $\delta$ so that $|x^2 - (x-\delta )^2| < \varepsilon$

Just to belabor things, you want to consider all points $\delta$ distance from $x$, which includes everything between $x$ and $x + \delta$ as well. You're basically done though.

Kevin Driscoll

09:02

Well, I made $\delta$ arbitrary to take care of that

Im not sure that suffices though

Kaj Hansen

Remember $\delta$ is strictly positive.

Kevin Driscoll

OH, I see what you mean. Yea.

Kaj Hansen

Ultimately it'll come out the exact same way as the original.

Kevin Driscoll

I guess one could fix that by considering the distance between $x-\delta$ and $x+\delta$, requiring that to be less than $\epsilon$ and then using the triangle inequality

or rather the distance between the images of those points

Mike Miller

wow you folks are still up

Kaj Hansen

09:06

Your $\delta < \varepsilon / 2x$ works. Just see what you get evaluating $|x^2 - (x + \delta)^2|$ with that.

It comes right out.

I'll probably go to sleep in a bit lol

Kevin Driscoll

So then to show that this functions is not uniformly continuous, suppose that we have a $\delta$ which works for $x>0$, for $x+n$, $\lvert (x+n)^2 - (x+n + \delta)^2 \rvert = \delta(2n + 2x +\delta)$. So, making $n > \epsilon/2 \delta$ yields a problem. @KajHansen

Kaj Hansen

Yeah. That's essentially the idea @KevinDriscoll. I'm not sure if your specific choice of $n$ is correct, but I suspect $|(x+n)^2 - (x+n+\delta)^2|$ grows without bound as $n \rightarrow \infty$, which indicates that you'll eventually be outside of that $\varepsilon$ range.

For uniform continuity, you assume that you have a $\delta$ that works for all $x$, not just a specific one though. Everything else is fine.

Actually, it's a lot simpler!

Kevin Driscoll

@KajHansen Well, I assumed you gave me a $\delta$ for some $x$ and then demonstrated an $x$ for which is does not work.

Which to me sounded easier than doing it by contradiction

Kaj Hansen

Given the $\varepsilon > 0$, suppose you have such a $\delta$ that works for all $x$, and then consider $|x^2 - (x+\delta)^2|$ as $x$ gets large.

Kevin Driscoll

and I think my $n$ works because in $n > \epsilon/2\delta$ then $\delta(2n + 2x + \delta) > \epsilon + \delta(2x + \delta) > \epsilon$ so you're in trouble

Kaj Hansen

09:19

I suppose that's basically correct.

The big difference between your thing and mine is that you have some extra terms to keep track of when you square. I think it does work though.

Kevin Driscoll

oh ya I suppose you could do it your way and then just pick an $x$ rather than picking an $n$

its the same thing

Kaj Hansen

mhmm

Kevin Driscoll

Ya I guess my physics background get sin the way

For some reason I wanted to keep around the stuff I had before because I already knew some things about it

so I wanted to keep $x$ and $\delta$ and then introduce a new thing

But here because most things are inequalities, its totally unnecessary. You don't need to keep around EXACTLY what you had before.

Alright, gonna go to sleep

glad I cna at least do the simple problems

Kaj Hansen

Good luck!

Balarka Sen

09:38

@KajHansen not at all. it's a bad way to think about continuity in R.

Chris's sis

10:46

Greetings

Mary Star

@KajHansen I have written my idea at the exercise, but I don't know if it is correct...

Nick

11:10

@DonLarynx I meant someone like Sharona from Monk. I don't know what @Jasper's OCD motif is but in any case, a human being can be able to help him cope with the stress emotionally. Also, hand him wet napkins after a handshake :D

It's just an idea to help him integrate him back into being an earning member of society.

If he isn't open to this, we can't blame him. It's hard to trust people IRL as much I idealize.

He's one of those resilient people who try to own their own burdens, no matter how heavy it is. And he's dreaming of just one day shedding all his load as opposed to trying to ease his burdens one step at a time.

I hope he does okay in the end. Good people deserve good things.

Get well soon, @Jasper

@Chris'ssis: Greetings to you too :D

Chris's sis

@Nick Hello :-)

Nick

@Chris'ssis What are you upto?

Chris's sis

@Nick I'm creating some stuff and working on my book (that if I'm lucky enough I'll be able to publish it by the end of the year).

Nick

@Chris'ssis Oooh. Found a publisher yet?

Chris's sis

@Nick Springer

@Nick It also depends on the papers I manage to publish. I still wait for other very nice papers to be published. I don't even mention the problems I proposed to the mathematical magazines (around the world).

Nick

11:28

@Chris'ssis You mean Springer Science+Business Media the guys behind Encyclopaedia of Mathematics !?

Chris's sis

@Nick link.springer.com

Nick

@Chris'ssis You don't mention the problems because you want the content of your book to be unique?

Chris's sis

@Nick Not really. I mentioned some already. I think the first problem in my book will be the first problem Ramanujan submitted to JIMS done in a way that will amaze the audience. :-)

Nick

@Chris'ssis I think no matter the contents of your documents, the typesetting, stylization and overall presentation will matter much to the reader. Maybe, not the intended reader but I speak on behalf of readers in general. Who might be editing and/or verifying the contents of your book? Have you left all these details to the publisher?

Chris's sis

@Nick I intend each problem in the book has something special, that "wow" element that makes you love it!

Jasper Loy

11:34

@Nick Thank you.

Chris's sis

@Nick Sure, I take into acount all these details. Well, you're right, that's very important, and I already talk with some guys about it, trying to find the best way of doing things. Besides that, the publisher will give you support on this side.

@Nick The hardest part is to make the selection of the problems ... (I have over 10,000 problems ...)

You modify the list again and again and again ... (trying to justify to yourself that it's better this way than the previous way ...)

I'll probably have around 300 problems ... (each having that wow element)

evinda

11:52

Hello!!!

evinda

12:30

Hi @DanielFischer!!! Could I ask you something? In order to show the commutativity in respect for the addition we show that $n'+m=m+n'$ without the assumption that $m+n=n+m$. When we want to prove that $n'm=mn'$ do we necessarily have to consider that $mn=nm$ ?

Daniel Fischer

@evinda I don't know if it's necessary, but it is not entirely unlikely that it will be convenient.

evinda

@DanielFischer I tried to prove it without the assumption that mn=nm and I got the following:
We want to show that $n' \cdot m=m \cdot n'$.

For $m=0$: $n' \cdot 0 \overset{\text{definition}}{=}0 \overset{\text{previous sentence}}{=}0 \cdot n' \checkmark$

We suppose a $m$ such that $n' \cdot m=m \cdot n'$ for all $n \in \omega$.

We want to show that $n' \cdot m'=m' \cdot n'$.

$n' \cdot m'=n' \cdot m+n'=m \cdot n'+n'$

@DanielFischer Is there a way to continue without using this assumption?

Daniel Fischer

@evinda I don't know, would have to think about that. But is there any reason to not use the assumption as part of the induction hypothesis?

evinda

@DanielFischer I thought that we firstly show that $0 \cdot m=0$, then that $n' \cdot m=m \cdot n'$ and finally that $m \cdot n=n \cdot m$.

Daniel Fischer

Sorry, I don't see what overall strategy you try to use.

evinda

12:40

@DanielFischer At the induction hypothesis don't we assume only that the proposition we want to show is true for a natural number $n$ and then at the induction step we want to show that it is true for n'?

robjohn

@Chris'ssis: I see that you answered this question almost 2 years ago. I saw the recent duplicate and added a different approach to the old question.

Daniel Fischer

@evinda Yes, that's what we do, but what, precisely, is the proposition you want to show?

evinda

@DanielFischer I want to show that $n' \cdot m=m \cdot n'$.

Chris's sis

@robjohn Yeah, DCT works nice as well.

Daniel Fischer

That's not precise, @evinda. For one specific $m$, for all $m$?

evinda

12:46

@DanielFischer We are given the definition:
For each pair of natural numbers $m \in \omega, n \in \omega$ we define the multiplication between $m,n$ (as a function $\cdot: \omega \times \omega \to \omega$) like that:
$m \cdot 0=0 \\ m \cdot n'=m \cdot n+m$

So in order to show the commutativity we have to show that $0 \cdot m=0$, $n' \cdot m=m \cdot n'$ and that $m \cdot n=n \cdot m$, right?

So in order to show that $n' \cdot m=m \cdot n'$ we show that it holds for $m=0$, then we suppose that it holds for a fixed $m, \forall n \in \omega$ and we want to show that it holds for $m'$.

robjohn

@Chris'ssis my first thought was that on $[0,1]$, $nx^{n-1}$ is an approximation to $\delta(x-1)$, but before I even noticed that the question was a duplicate, I started thinking about the substitution $x\mapsto x^{1/n}$. $nx^{n-1}$ seemed to be asking for it. I am surprised that no one else had used it in an answer.

Daniel Fischer

@evinda You need to specify the quantifiers. Each proposition you want to prove should contain some $\forall$. Write down (for yourself) precisely which proposition you want to prove in each step of the total proof of commutativity. The strategy should probably contain a nested induction (inductively, for every $m$, you prove by induction that [for every $n$ something involving $m$ and $n$ holds]).

evinda

@DanielFischer We want to show that $n' \cdot m=m \cdot n'$.

For $m=0: n' \cdot 0 \overset{\text{definition}}{=}0 \overset{\text{we proves that } 0 \cdot m=0 }{=} 0 \cdot n'$

We suppose a $m$ such that $n' \cdot m=m \cdot n', \forall n \in \omega$

We want to show that $n' \cdot m'=m' \cdot n'$.

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$Mathematics$ Mathematics