Mathematics - 2012-09-17 (page 1 of 3)

user19161

00:00

@Link What do you mean by velocity cut into half?

Link

actually

intial velcoity

one sec

i'll show you a graph

user19161

@link Have you been kidnapped?

user19161

LOL

Link

no

k

here is the question:

How would graph A change if its accelration was kept the same but its intial velcoity was cut in half?

the line is graph A

user19161

@Link Acceleration is the rate of change of velocity with respect to time.

user19161

00:09

So it is the gradient of the graph.

Link

okay

I know that

gradient?

user19161

So start from 3 instead of 6.

Link

oh

user19161

And the slope should be the same.

Link

thats it?

user19161

00:10

Slope=gradient.

Link

wow

thanks

user19161

Wait.

user19161

As you can see, even the acceleration in the first graph changes.

user19161

So what does the question really mean?

Link

huh?

user19161

00:12

Acceleration changes from negative value to zero (flattening out of the graph).

Link

yes

user19161

So I ain't sure what the question is.

Link

it's what I said

how would the line change

thats it

user19161

No, I mean it is not clear whether they want both graphs to have the same acceleration at all points.

user19161

I guess if you ask this question, you don't really know what I am asking here.

Link

00:15

yea

wait

thats what it wants

user19161

Anyway I have said everything I can say here.

Link

to have the same accelration

but thanks

its a simple high school class

so thats probably it

user19161

Often school teachers don't know what they are talking about

Link

calc is a lot easier -.-"

user19161

@Link Calculus?

Link

00:17

yep

ayways

if the slope of velcoity vs time graph is the same throughout

then the slope of an accelration vs time graph is just = 0 right?

Peter Tamaroff

@anon

user19161

@Link If slope of VT graph is the same, that means acceleration is constant, which means that slope of AT graph is zero, yes.

user19161

From displacement to velocity to acceleration, one just takes derivatives.

user19161

In the other direction, one takes integrals.

user19161

Derivatives correspond to slopes and integrals to areas.

Link

00:22

okay

that just helped me so much

love the calculus way of descbing it

Peter Tamaroff

@MarianoSuárez-Alvarez Hola!

N3buchadnezzar

^^

Link

well bye guys and gals, and thanks for the help @WillHunting

anon

huh?

@PeterTamaroff what's up

Peter Tamaroff

@anon I'm stuck

anon

00:33

I assume commas represent decimal points

Peter Tamaroff

@anon You assume well.

N3buchadnezzar

@anon Any quick pointers why $\sin x / x$ is analytic?

anon

@PeterTamaroff If $f(a+h)\ge10^{-n}$ and $h<10^{-n}$ then $f(a+h)/h>1\not\to0$.

@N3buchadnezzar Using only the side-over-hypotenuse definition?

You clearly need only check $x=0$, and it should be equivalent to showing the limit at 0 exists. Which is to say, the derivative of sin() exists at 0.

N3buchadnezzar

@anon No, by any means neccecary. And I know the problem is at the origo.

anon

But I assume you don't automatically have the Taylor expansion or the formula as a linear combination of complex exponentials. So essentially you start with only the triangle definition.

N3buchadnezzar

00:40

The problem is indeed a bit backwards. The goal is to prove that $\lim_{x \to 0} \sin x / x = 1$ by using the taylor series. And in order to justify switching the limit and the sum, we need the function to be analytic.

Jayesh Badwaik

00:59

@anon About that semigroups we were talking about.

user19161

@jayesh I saw your removed message. It implies continuity at the point but not on an interval containing the point.

user19161

I am going to sleep now. Have fun boys and girls!

Jayesh Badwaik

@WillHunting You are correct, I just found a counter-example to it.

math.tamu.edu/~tvogel/gallery/node4.html

Similarly, the ruler function may be differentiable at 0.

leo

hello :-)

Jayesh Badwaik

@leo hi.

leo

01:10

@PeterTamaroff Imk

Peter Tamaroff

@leo I think I have proven $2.$ one way.

leo

@PeterTamaroff I guess "$\implies$"

Peter Tamaroff

@leo Well. Suppose that $\lim_{x\to a}f(x)=L$.

leo

yep

Peter Tamaroff

Then for every $\epsilon >0$, there is a $\delta >0$, such that, for all $x$, $0<|x-a|<\delta\implies |f(x)-L|<\epsilon$

leo

01:13

yep

Peter Tamaroff

Let $\{x_n\}$ be any sequence with $x_n\to a$.

Then for every $\delta>0$ there exists an $N$ for which $n\geq N$ implies $|x_n-a|<\delta$

But for this $N$, we'll have then $$|f(x_n)-L|<\epsilon$$, from where the sequence $f_n=f(x_n)$ converges to $L$.

Am I left or right?

leo

@PeterTamaroff not only for that $N$ but for all $n\geq N$

and the everything is right

Peter Tamaroff

@leo Yes, that is what I meant.

leo

the "$\impliedby$" is true because it can't be false

@PeterTamaroff ok fine

Peter Tamaroff

@leo Proof by contradiction?

leo

01:17

@PeterTamaroff indeed :-)

proof by tradition

:6177004 Tomae's function?

Jayesh Badwaik

@leo Yes, I was thinking. but then realized, f(0) = 1 right?

Peter Tamaroff

@leo So I should say "supongamos ahora que para cada secuencia $\{x-n\}$ con $x_n\to a$, la secuencia $\{f_n\}$ con $f_n=f(x_n)$, converge a $L$, pero que $\lim\limits_{x\to a} f(x)\neq L$"?

leo

@JayeshBadwaik I think is not even well defined at $0$, because $0$ is rational but $0=0/q$ with $q$ any natural number

@PeterTamaroff exacto!

Jayesh Badwaik

@leo Okay, hmm. Yup, hence my confusion. If we define it as $0$ at $x=0$, then it is differentiable.

Peter Tamaroff

@leo La idea es mostrar que la suposicion $\lim f\neq L$ es absurda con lo que debe ser $\lim f=L$?

Es casi trivial, no?

leo

01:29

@JayeshBadwaik no. No matter how you define $f$ at $0$ you get a discontinuity at that point, so it can't be differentiable at $0$. She resist :-)

Peter Tamaroff

@leo O sea, creo que seria mas prolijo probar $A\implies B$ y $\neg A\implies \neg B$, no?

leo

@PeterTamaroff correcto

@PeterTamaroff así es

@PeterTamaroff pero me gusta más la prueba por contradicción en este caso

Peter Tamaroff

@leo OK.

revolviendo el tacho de basura

desdoblando bollito de papel

leo

@PeterTamaroff good

anon

@JayeshBadwaik eh?

leo

01:40

@PeterTamaroff lo digo porque en esa forma todo empieza: "suponga que para toda sucesión $x_n\to a$ la sucesión $f_n$ asociada converge a $L$. Suponga que $\lim_{x\to a}f(x)$ no existe o no es $L$"

Peter Tamaroff

@leo Creo que la contradicción es tan obvia que no me doy cuenta.

leo

si $\lim_{x\to a}f(x)=K\neq L$ por la dirección que ya probaste se sigue que todas $f_n$ convergen a $K$ y eso es contradictorio

Jayesh Badwaik

@anon You remember I asked you about when can semigroups be isomorphic and stuff. Got this which was related.Hence, shared that with you.

Peter Tamaroff

@leo Asi de directo? XD

leo

@PeterTamaroff si $\lim_{x\to a}f(x)$ no existe entonces se puede cocinar una sucesión $a_n$ tal que la $f_n$ asociada no converja a $L$

@PeterTamaroff si. Es solo usar la dirección que ya tienes

Peter Tamaroff

01:45

@leo La contradiccion seria en realidad que las secuencias convergen a dos limites distintos, que es absurdo, no?

leo

@PeterTamaroff eso estaba a punto de decir

piensa la otra. Voy a comer :D, ya vuelvo

leo

02:01

@PeterTamaroff ya volví

Peter Tamaroff

@leo No soy muy buen cocinero aparentemente

leo

@PeterTamaroff Supongamos que $\lim_{x\to a}f(x)$ no existe. Entonces para cada número real $K$ existe un $\epsilon\gt 0$ tal que para cada $\delta\gt 0$, hay un $t$ con $|t-a|\lt\delta$ pero $|f(t)-K|\gt\epsilon$

Peter Tamaroff

@leo Si.

@leo Pero esa $l$ debe ser una $K$! =)

leo

ok

la cosa es para cada número real $K$. $L$ es un número real entonces apliquemos eso a $L$

Peter Tamaroff

@leo OK.

leo

02:10

entonces existe un $\eta\gt 0$ tal que para todo $\delta\gt 0$ bla, bla,...

la cosa es para cada $\delta\gt 0$, en particular $1/n\gt 0$ para cada $n\in \Bbb N$

Entonces para cada $n\in\Bbb N$ existe un $t_n$ tal que $|t_n-a|\lt\frac{1}{n}$ pero $|f(t_n)-L|\gt\eta$

Peter Tamaroff

@leo Perfecto. Cuando vi el $1/n$ ya me di cuenta... =D

leo

ok

esa $(t_n)$ es la queríamos cocinar

Peter Tamaroff

@leo Pero entonces $\{t_n\}$ es una secuencia con $t_n\to a$ pero $f_n\not \to L$, que es absurdo.

leo

@PeterTamaroff sí, contradicción. Por lo tanto debe ser que $\lim_{x\to a}f(x)$ existe y es $L$

Peter Tamaroff

@leo B-E-autiful!

leo

02:17

recuerda como empezamos:

37 mins ago, by leo

@PeterTamaroff lo digo porque en esa forma todo empieza: "suponga que para toda sucesión $x_n\to a$ la sucesión $f_n$ asociada converge a $L$. Suponga que $\lim_{x\to a}f(x)$ no existe o no es $L$"

Mohamed Ahmed Nabil

why is i^2 = -1?

Peter Tamaroff

@MohamedAhmedNabil What do you mean by "is"?

leo

@MohamedAhmedNabil why is $1\neq 0$?

Mohamed Ahmed Nabil

@PeterTamaroff can anyone give me the proof to

i^2=-1 ?

Peter Tamaroff

@MohamedAhmedNabil That is not a theorem. And if it is, then the definition of $i$ is $i=(0,1)$, and you have defined complex numbers as ordered pairs, and multiplication of complex numbers accordingly.

leo

02:20

@MohamedAhmedNabil There is no proof. It is an axiom.

Mohamed Ahmed Nabil

@PeterTamaroff why are you writeing $ everywhere?

Peter Tamaroff

@MohamedAhmedNabil I guess you don't have CHATJax

@MohamedAhmedNabil Here

Mohamed Ahmed Nabil

@PeterTamaroff ty

leo

@MohamedAhmedNabil Or you can start with more primitive axioms and deduce it from it as Peter suggest

@PeterTamaroff hay mucho de eso que acabamos de hacer en análisis. De negar una hipótesis del tipo "para cada $\delta\gt 0$" se consiguen cosas para cada $n\in \Bbb N$ y así.

Son bonitas construcciones

Peter Tamaroff

@leo Si. Justo ayer me daba cuenta lo detallado que hay que ser a negar la existencia de un limite, al negar la continuidad o mas complicado aún, la continuidad uniforme!!!

@leo Si =)

@leo Es mas. Estoy intentando probar que $x\sin x$ no es uniformemente continua en $[0,\infty)$.

leo

02:26

@PeterTamaroff esa o $x\sin (1/x)$ ?

Peter Tamaroff

@leo $x\sin x$.

Porque (supongo) probe que si $f$ y $g$ son unif continuas y acotadas en $A$ entonces $f\cdot g$ es u.c. en $A$. Quiero probar ahora que la hipotesis de que ambas sean acotadas es necesaria.

La idea es que si $x=n\pi+\pi/2$ e $y=n\pi$ entonces $|x\sin x-y\sin y|=n\pi+\pi/2$

Y $|x-y|=\pi/2$

leo

correcto

Peter Tamaroff

Es decir, tomamos una raiz y un valor pico.

Entonces, la negacino de u.c. sería
"Para *cierto* $\epsilon >0$, se tiene que *para todo* $\delta >0$ existe *algun* $x$ y *algun* $y$ para el cual $|x-y|<\delta$ pero $|f(x)-f(y)|>\epsilon$??"

leo

@PeterTamaroff eso es correcto

Peter Tamaroff

@leo OK. Ahora tengo que aplicarla.

Mi problema es que lo mio funciona para todo $\delta >\pi/2$!

leo

02:36

@PeterTamaroff si ese es el problema.

Peter Tamaroff

@leo Lo tendré que ver mañana, ya es tarde...

leo

@PeterTamaroff pero pa probar que la acotación de ambas es necesaria, $x\sin(1/x)$ te sirve, puedes hacer algo parecido. Es que esa tuya no estoy seguro.

Gustavo Bandeira

@PeterTamaroff ENGLISH!!!

leo

@GustavoBandeira but you speak potugues right?

Gustavo Bandeira

02:52

@leo I'm kidding, when I speak portugues, people say: "Speak english!", was just trolling, haha

leo

@PeterTamaroff mentira, la que yo digo no sirve. La idea es encontrar $f,\ g$ unif. continuas, $f$ acotada y $g$ no acotada, tal que $fg$ no es unif. continua, cierto?

@GustavoBandeira oh I see :-), but you already understand something of what is above

don't you?

@PeterTamaroff si ese es el caso $x\sin (1/x)$ no sirve porque $\sin(1/x)$ no es unif. cont.

Mariano Suárez-Alvarez

05:42

Hola, Oi, Hello

Parth Kohli

08:16

Hello there.

Gustavo Bandeira

@leo Yep. I understand everything.

@MarianoSuárez-Alvarez Hello.

@ParthKohli Hello.

Parth Kohli

Hello again, Gustavo.

Gustavo Bandeira

What's up?

Parth Kohli

I just came back; took an exam.

Gustavo Bandeira

08:32

I just woke up.

Math exam?

Parth Kohli

Yes :)

Gustavo Bandeira

Nice

Parth Kohli

Does MathOverflow belong to the StackExchange network?

Gustavo Bandeira

I need coffee.

Still not.

But I've heard they were packing things, for being integrated in the SE network.

Parth Kohli

Very well.

Gustavo Bandeira

08:39

It will allow the migration of questions.

Parth Kohli

That is good.

Which of them has a higher level Mathematics?

Gustavo Bandeira

MO

Parth Kohli

OK.

Both of them have a similar idea.

Gustavo Bandeira

Yep.

Parth Kohli

08:56

Is it weird that Arturo Magidin, who has already left this site, was seen on MathOverflow 4 hours back?

Gustavo Bandeira

He doesn't like us.

Parth Kohli

Really? Why so?

Gustavo Bandeira

Nope, I'm kidding.

I have no idea of his reasons to do so.

Parth Kohli

May I know what accept rate means?

Gustavo Bandeira

Yep.

Is the rate of accepted questions.

Parth Kohli

09:03

Accepted? In what sense?

Gustavo Bandeira

You have 100 questions, and you accepted a answer in 80 of them, therefore: 80% accept rate.

Parth Kohli

Is there an "accept" option?

Gustavo Bandeira

yep.

Parth Kohli

Oh yes, there is.

Shall I press it if I found a good answer?

Gustavo Bandeira

Yes.

Parth Kohli

09:07

Thank you very much.

Gustavo Bandeira

You're welcome.

Gustavo Bandeira

09:28

6

Q: Grasping mathematics

First, I'm not trying to make this sound like a "poor-me" story. I understand fully that every decision I've made leading to this is my fault. I am genuinely looking for advice. So, I am a high school student who is a sophomore and feel I have developed an interest in mathematics a bit late. As ...

My history is the same.

Parth Kohli

The title is catchy, definitely need to check it out. :) @GustavoBandeira

Noah

Hello

Morning

Gustavo Bandeira

@Noah Hello. =)

Parth Kohli

Hello, @Noah!

Noah

Hows it going

anything new?

Gustavo Bandeira

09:32

Not much, I just made a new question.

And I need coffee.

Parth Kohli

Is there a space where you may see all your starred questions?

Noah

Hope you have come up with a solution to your books @ParthKohli

@GustavoBandeira Coffe is good, but I like tea better

Gustavo Bandeira

There should be a way to do that, but I'm unaware of. @ParthKohli

Noah

@GustavoBandeira what's the question about?

Gustavo Bandeira

0

Q: What are the dangers of visually exposing of mathematics?

I've heard several times (such as this one) that it's dangerous to learn/prove/teach mathematics through images. I've also read somewhere that showing mathematics through images helps one's intuition because we understand better through images, due to our long date use of the vision sense. I can...

geometry soft-question learning

Noah

09:37

That's an interesting question

@GustavoBandeira I guess you got an answer

@GustavoBandeira How do you write/compose equations in the question box on this website?

Gustavo Bandeira

@Noah Equations in the middle of $$

Parth Kohli

Write between $ and $.

Noah

@GustavoBandeira but do we need another editor?

Parth Kohli

@Noah Do you know about MathJax?

Gustavo Bandeira

@Noah For the chat, you need mathjax.

45

A: Should chat have TeX support?

This bookmark processes the current page with MathJax. It has been modified from the bookmark on this page to handle $$...$$ and to handle \\[...\\] properly, to include AMS additions, and to update automatically. The COPY TO CLIPBOARD link on pastebin.com should copy the bookmark to your clipbo...

Noah

09:44

Okay

in the website how do I write say 2 to the power of 2?

Parth Kohli

2^2 between dollar signs.

Be careful, because two to the power 34 is 2^{34}.

Noah

ouch. I dont see that char on my keyboard

Parth Kohli

Remember: if you write 2 to the power 34 as 2^34, it renders it as the following:

Noah

Okay, one more thing. I remember a similar website but it wasnt hosted by stack exchange. Do youo guys happen to know that?

It was about mathematics, I guess they started a similar initiative when stack overflow was the only website at this community...

Zhen Lin

MathOverflow?

user19161

09:51

@Noah Are you thinking of math overflow?

Parth Kohli

MathOverflow, yes.

Noah

Right

That's teh one... so what happened to that?

Zhen Lin

still around

Parth Kohli

It uses StackExchange 1.0

user19161

Chat is loading so badly that I did not see Zhen's message when I replied.

Parth Kohli

09:53

A lot of users have migrated from Math. Stack Exchange to MathOverflow.

Noah

Okay, but what's the difference? Which is better?

user19161

@ParthKohli Migrated? No, the two sites are quite different.

Parth Kohli

Yes, of course they are.

user19161

MSE is for all math, MO is for research math.

Parth Kohli

What I am saying is, some users have left Math.SE, and they now contribute to MO.

user19161

09:54

So ignorant people like me only hang around on MSE and not on MO.

Noah

Omm, okay... Jasper

Parth Kohli

Yes, Will.

user19161

@ParthKohli Not really, maybe they are on both sites.

Noah

But is it owned by the same people?

user19161

Sometimes people take a break.

Parth Kohli

09:55

No, Noah.

user19161

@Noah I get the impression that it is the same people but under a different licence, and that they will all be the same again soon.

Parth Kohli

@WillHunting how about an example, say, Arturo Magidin? (Incidentally, I was discussing the same thing with @GustavoBandeira)

Noah

You mean JOel and Jeff? @WillHunting

user19161

@ParthKohli Like I said, people come and go all the time. They are not married to the site you know.

user19161

@Noah I am not too sure myself of the details.

user19161

09:58

Of course, users like to speculate when a great contributor has seemingly left.

Noah

Route

user19161

But they are just speculations, no more no less.

Noah

@WillHunting specs are bad

no more no less

Gustavo Bandeira

People go to MO because of digievolution.

Noah

People go there when they retire...

See the list of users... Most of them are retired professors

Gustavo Bandeira

10:05

So... What ya doing here JohnSenior?

Noah

See the list of users... Most of them are retired professors

@GustavoBandeira He maybe one of us ;)

Gustavo Bandeira

48

Computer Music

Proposed Q&A site for anyone engaged in the theory, technique, practice, performance, and study of computer music as it has developed from the early Music N days of Max Matthews to modern signal flow languages and beyond.

Currently in commitment.

This is nice.

Noah

How many votes are needed for a website to move to the beta phase?

Gustavo Bandeira

Dunno.

Noah

@GustavoBandeira and poeple who propose sites here, do they get some sort of share at the company?

Gustavo Bandeira

10:13

@Noah I guess that proposers with proeficiency become moderators.

Noah

@GustavoBandeira so no monetry value?

Gustavo Bandeira

I guess no

Noah

omm, okay

Gustavo Bandeira

I want coffee.

Noah

And I dont see any ads on these sites, not sure why is that

Gustavo Bandeira

10:14

=(

Noah

Okay...

Gustavo Bandeira

Maybe someone rich decided to give this to us.

Noah

Sip it slowly...

Stir it up...

And try to add some creamer

Gustavo Bandeira

Yep.

Gustavo Bandeira

10:27

@ParthKohli The classes on the course I suggested you have began.

Parth Kohli

10:46

@Gustavo thank you very much for the reminder!

What should I start with?

Gustavo Bandeira

12:33

@ParthKohli Sorry, I'm back.

robjohn

13:15

Am I alone in thinking that this is not in the best taste?

His followup comment has gotten two upvotes.

The reference to other sites for support without actually mentioning those sites provides no support (and is essentially passive-aggressive). Actually mentioning those sites would be unacceptable.

Of course, being compared to something described so poorly is almost damning with faint praise.

Matt

@robjohn The teddy is gone.

robjohn

@Matt has he said so?

Matt

Quite a blow.

@robjohn ...

robjohn

@Matt I just looked at his profile.

Jayesh Badwaik

@robjohn Yup, it is uncalled for. For two reasons.

First the reason you mentioned, about not naming and then trying to get support is one.

Second is even if the rest of the stackexchange network was completely awful and MSE was the least awful in them, it would be no excuse to defend MSE. MSE is MSE, what happens here should be maintained to an appropriate standard irrespective of anything else that is happening somewhere else.

Peter Tamaroff

13:26

Say fo shizzle to my nerdizzles

robjohn

@PeterTamaroff hey there!

@JayeshBadwaik I don't understand the upvotes his response has gotten. Oh, well, there are many things i don't understand.

Jayesh Badwaik

@robjohn Hmmm, probably due to perceived wittiness. "Not naming policy" in isolation sounds nobler than when taken in the context, people who upvoted failed to see it like that.

Peter Tamaroff

Link to the fuzz?

Jayesh Badwaik

fuzz?

Parth Kohli

Hello, g'evening to all the Indian fellas.

Peter Tamaroff

13:38

Tb is gone?

@parth what about non indians?

Parth Kohli

Well, time zones are different, so I can only say g'day folks!

Jayesh Badwaik

I guess it is high time to coin a phrase which means "Appropriate Greetings for your time zone which mean Hello".

3

And "Appropriate Greetings for your time zeon which mean Goodbye. "

Parth Kohli

What does a pencil next to one's chat message mean? Editing?

Jayesh Badwaik

Edited post probably.

Parth Kohli

Thank you.

user19161

13:52

Did anyone see Jonas? Is he returning to chat?

Jayesh Badwaik

@WillHunting No, he hasn't come back since that starred message he posted.

user19161

By the way, why was Jonas auto-suspended from chat?

user19161

What was the thing he said?

user19161

@JayeshBadwaik Maybe he is flying somewhere now.

Jayesh Badwaik

No idea.

user19161

13:56

@parth I see you have removed your picture.

Gustavo Bandeira

@WillHunting I thought that was because he said: "We should make gay themed jokes"

But he deleted the saying.

Jayesh Badwaik

@WillHunting About the continuity thing, I thought I was correct, but I was wrong.

user19161

@GustavoBandeira It might have been auto-deleted.

Jayesh Badwaik

Or may be this.

user19161

@JayeshBadwaik I doubt it.

Gustavo Bandeira

13:59

@JayeshBadwaik I guess not, Is that offensive?

Transcript for

$Mathematics$ Mathematics