Mathematics - 2013-09-22 (page 1 of 4)

Charlie

00:00

Hi... @anon ....

anon

hello charlie

user87637

I should take my ten holy books to an island and do nothing but study next year.

Charlie

@anon ....how are you?

anon

alright

user87637

If I take one month to read one book, I can finish the ten in a year.

user87637

00:03

If I take two months per book, I will take two years.

user87637

Anyway, the first five books should be more than adequate for the GRE.

user87637

And the next five should be adequate for the prelims.

Karl Kronenfeld

It's not uniform though. Maybe one will take one month and another will take three.

Charlie

@anon ....okay

user87637

@KarlKronenfeld Yeah, that's right.

user87637

00:07

anon is trying to remain anonymous so that he can retain his godly aura.

Charlie

@JasperLoy I missed your anon lines

user87637

anon must be working on the millennium problems now, lol.

user87637

Let us not distract him.

Charlie

Not anon, though

....anon......

user87637

I miss nobody, except...that's my secret.

Charlie

00:15

@JasperLoy I annoyed you all this time .... You must be sick of me

Hello, mr.Splaver

cyberskull

00:34

All hail! @robjohn

Hi @Charlie

robjohn

@cyberskull You may rise...

cyberskull

@robjohn thank you, how are you?

robjohn

@cyberskull pretty good. Been doing manual labor all day. Now I have to take the dog to the park. It will be nice to get back to doing math.

cyberskull

@robjohn enjoy your walk.

Charlie

@cyberskull hi skullucky

cyberskull

00:40

@Charlie :D

@Charlie what a memory!

Charlie

@cyberskull I don't forget

@cyberskull ;)

cyberskull

@Charlie How does the old saying go...if you only knew half the things I have forgotten...

Charlie

@cyberskull hehehehe

cyberskull

@Charlie :D

Charlie

@cyberskull I liked the "still waters run deep" stuff

cyberskull

00:46

@Charlie Yes, me too.

@Charlie How are you doing?

Charlie

@cyberskull getting better

cyberskull

@Charlie good, good

Charlie

@cyberskull I can't work with all this heat :(

cyberskull

@Charlie Do whatever you can, it will pass...

Hi @TedShifrin wazzup?

Charlie

@cyberskull :/

cyberskull

00:52

@Charlie :-|

Charlie

:(

cyberskull

:P

gag

from the heat

or :'''( sweating

Charlie

':( sweating

cyberskull

much better :D

@TedShifrin have you heard the old saying?

robjohn

@Charlie Just yesterday, you were commenting on the nice weather. It has changed?

cyberskull

00:56

12 mins ago, by cyberskull

@Charlie How does the old saying go...if you only knew half the things I have forgotten...

Charlie

@robjohn it changed brutally

robjohn

@Charlie sounds unpleasant

cyberskull

7 hours ago, by Charlie

TWO days ago was 17

Charlie

@robjohn incredibly unpleasant

cyberskull

I gotta run see you guys later @Charlie @robjohn et al

Charlie

00:59

@cyberskull bye!

robjohn

@cyberskull later

Twink

hi

pal

i was banned for 24 hours with no explanation

MSE mods are so unfair

Gustavo Bandeira

@Charlie Don.

anon

ah, eternal recurrence

Peter Tamaroff

@Twink Mariano explained it very well.

Charlie

01:07

@GustavoBandeira oi

Peter Tamaroff

@anon LAWL.

Twink

what did he explain?

Peter Tamaroff

@Twink, before you say anything, please spend a second reflecting on whether you might remove it in the future: if you will, then just don't say it
otherwise you become noise
and noise will be chased away

Twink

omg I stopped removing my comments

then, I just removed one or two, as EVERYOBODY do

Peter Tamaroff

@Twink Not really, there are few people that remove stuff here.

But you're the only one that keeps coming back, trolling, bothering and whatnot.

Twink

01:10

I did it because I wrote something but then I included it in my last comment by editing it, and I removed the other one

Gustavo Bandeira

@Charlie hhh

Twink

what is wrong with you guys? don't be so unfair

Charlie

@Twink you were being quite obnoxious yesterday

Gustavo Bandeira

@Charlie What did he do?

Peter Tamaroff

@TedShifrin Hey.

anon

01:13

@PeterTamaroff His comments were not flagged until finally he decided charlie wasn't fun to play with anymore and started randomly accusing her of wishing he was dead. Apparently he also accuses cerebus in ELU of threatening to gay-rape him. I feel somewhat left out only having been accused of ~~diffamating~~ defaming him.

Ted Shifrin

Hi, @Peter. I finished using up my red pen :)

Peter Tamaroff

Proceeds to ignore for a pleasant evening.

@TedShifrin Good grades?

Twink

@PeterTamaroff is my derivate correct? math.stackexchange.com/questions/500788/…

Ted Shifrin

For advanced grad courses I grade more holistically. Some fine, some not.

Charlie

@GustavoBandeira read what anon said

Peter Tamaroff

01:14

@TedShifrin Holistically? What does that mean?

Twink

can anyone tell me if my derivate is correct? math.stackexchange.com/questions/500788/…

Ted Shifrin

Lots of comments but no numerical grades on the papers.

Twink

@TedShifrin can you tell me if my derivate is correct? math.stackexchange.com/questions/500788/…

anon

@Twink the derivative of $(1-x^2)^k$ is $-2kx(1-x^2)^{k-1}$, there is no $k-1$ outside of the exponent

also there is no need to ask the question three times in a row in quick succession.

Twink

yes you're right

Gustavo Bandeira

01:17

@Charlie He was trolling you?

Twink

thanks @anon

I know you hate me so thanks for helping me

Peter Tamaroff

@TedShifrin Ah, I see.

Twink

and thanks for making fun of my grammar mistake diffamating

Peter Tamaroff

@TedShifrin Can you throw me a bone on the lines problem? Ugh, I don't know if I should be doing this! =D

Ted Shifrin

@Twink: You should pay careful attention to your algebra and calculus, or more advanced math will be utter hell. More time on thought about math, less on making scenes here, please.

@Peter: Doing what, exactly?

Twink

01:20

yes it was because I was making some integrals and I got confused

but when @anon told me I realized of my mistake

Peter Tamaroff

@TedShifrin Asking for further help without more tries. What I told you the last time was the furthest I got.

Ted Shifrin

Even old professionals have to double- and triple-check their work.

Twink

it was nit a scene I was just asking

everyone has the right to make mistakes

don't be so incomprehensive

Ted Shifrin

I know, @Twink. I wasn't referring to this moment.

Peter Tamaroff

I am trying to see if I can enunciate the problem in terms of the existence of certain function. I am not sure how to approach it. For example, should I aim to give one concrete example which I know will give the greatest amount of lines possible?

Twink

01:22

which moment where you refering to, then?

@anon yesterday I asked you and Charlie to watch one video

it was about a prank call

and in that video they say the phrase "why did you tell me that you wanna kill me?"

Ted Shifrin

@Peter, I imagine it's not easy to visualize from a formal algebraic viewpoint. My hint (again) is to ponder what we learned from that example of the set of lines meeting THREE lines.

Twink

that's why i said that, but you obviously didn't watch the video

Peter Tamaroff

@TedShifrin What we learned. Hmm.

Twink

and you took it personal and as usually, you started to make some absurd accusations

Ted Shifrin

LOL.

Peter Tamaroff

01:25

@TedShifrin I am thinking about reverse engineering $z=yx$.

But I first need to think what is the analog in $\Bbb C^3$.

Ted Shifrin

What is remarkable about that surface? Don't worry about $\Bbb C$. That's only relevant to the algebra at the end.

Peter Tamaroff

@TedShifrin Doubly ruled. You said that.

Twink

@TedShifrin can you tell Peter to unignore me please?

tell him I'll be a buen chico

Ted Shifrin

Oh, darn, I shouldn't give things away to you ... @twink: I think Peter and others have had it with your antics and argumentative nature.

Peter Tamaroff

@TedShifrin Yes, that was my point. Giveaways.

Twink

01:31

I'm sorry

I have the right to a second chance

Ted Shifrin

@twink: One of your personas also acted like a total twelve-year old to me, so my patience is almost done, as well, and I'm the openly gay person.

Twink

I'm an openly gay person too

Ted Shifrin

Yes, @Peter, I should be reprimanded. So ... Doubly ruled ... Hmmm ...

user60887

hey guys could you check if my proof is correct. Here is the link. math.stackexchange.com/questions/500120/…

Ted Shifrin

Yes, @twink, I know. And let it go for now.

Peter Tamaroff

01:34

@TedShifrin Continues to think. To the Whiteboard!

user60887

i know it maybe a bit long.

Ted Shifrin

@user60887: I don't have the patience to read it,sorry. Have you considered induction?

Peter Tamaroff

@user60887 Induct, induct, induct.

Ted Shifrin

@Peter: only a plane is triply ruled!

user60887

its ok. I have used induction where m,n>0

Peter Tamaroff

01:35

@TedShifrin Heh, I have edited my comment!

Twink

but why did you mention being an openly gay person? @TedShifrin

Ted Shifrin

Really? With all those steps?

user60887

but i dont think i can induct when the integers are negative. I got like 9 cases already but idk if those cases are correct

Ted Shifrin

To emphasize how tiresome your antics and behavior have been, @twink.

@user. The negative case is one step once you've done the positive, because you argue $(a^k)^{-1}=\dots$.

Peter Tamaroff

@user60887 Remember that $a^{-n}:=(a^{-1})^n$

Twink

01:38

I don't understand

Ted Shifrin

Now who's a giveaway? :D@Peter

Peter Tamaroff

@TedShifrin Hey! That's a definition!

Ted Shifrin

Go think, @twink. Be an adulr.

Peter Tamaroff

@TedShifrin Come to the Dark Side. We have ignore buttons.

Ted Shifrin

@Peter: ponders :)

Not on my iPad. :(

user60887

01:40

oh yes i already used the fact that $a^{-n}:=(a^{-1})^n$ already in the negative cases

Ted Shifrin

Then there should only be one case, I think, for the negatives, once the positives are nailed.

Peter Tamaroff

@TedShifrin Yes, look for "menu".

Ted Shifrin

Interesting nontrivial theorem, @Peter: Other than a plane, every doubly ruled surface is projectively equivalent to $z=xy$. Amazing, eh?

2

No clicks or menus on the iPad.

user60887

Actually i got 9 altogether. Well I was thinking I still have to prove when m=n=0, m>0 and n<0, n<0 and m<0, m>0 and n<0, m>n, etc

Ted Shifrin

Oh, wait.

Peter Tamaroff

01:43

@TedShifrin Projectively equivalent means...? I would try to guess, but I am a little tired.

Ted Shifrin

Equiv under the group of projective transformations ... You don't know them yet ... Hence my shipping you info on projective stuff.

Peter Tamaroff

@TedShifrin =D

Twink

@TedShifrin can you please tell Peter what I told you, please?

you removed a comment

you will be banned

Ted Shifrin

Hmm, I figured out how to edit and remove on my iPad.

Twink

@TedShifrin can you help me with something?

Jonathan told me how to find the supremum math.stackexchange.com/questions/500788/…

of that set

I need to calculate the limit of that supremum when $k \to \infty$

but I don't know how! :S

I posted that question because I thought that whould be a good way to solve one problem that I have

but finding that limit took me to the same place

because I don't know how to do it :S

Ted Shifrin

01:51

You do limits as you did in basic calculus, @twink.

Twink

I need to calculate $\lim_{k\to \infty} \frac{k^2(2k)^k}{(1+2k)^{k+1/2}}$

I tried but I don't know how

it's a very difficult limit :(

Ted Shifrin

Focus on the net exponent on $k$.

Twink

the limit should be $0$

Ted Shifrin

@Peter: Aren't you entitled to a break after all this conferencing?

Why? @Twink

Twink

because I know the answer

Peter Tamaroff

01:55

@TedShifrin Meh, I am allright.

Twink

I think I can do this comparing that fraction with another one

$0<\frac{k^2(2k)^k}{(1+2k)^{k+1/2}}<\frac{k^2(2k)^k}{(2k)^{k+1/2}}$

Ted Shifrin

@twink: Not if the algebra is correct. i've not checked it.

Yes, so?

Twink

$0<\frac{k^2(2k)^k}{(1+2k)^{k+1/2}}<\frac{k^2(2k)^k}{(2k)^{k+1/2}}=\frac{k^2}{(2‌k)^{1/2}}$

Ted Shifrin

What power of $2k$ are you left with?

You need to learn exponent rules, @twink! Sigh.

Twink

why? :(

Ted Shifrin

01:59

What is $2^8/2^3$?

Twink

$2^5$

Ted Shifrin

Good. Now do the other one right.

Twink

$0<\frac{k^2(2k)^k}{(1+2k)^{k+1/2}}<\frac{k^2(2k)^k}{(2k)^{k+1/2}}=\frac{k^2}{(2‌k)^{1/2}}=(\frac{k^4}{2k})^{1/2}$

Peter Tamaroff

@Twink That is of no use to you.

Twink

$0<\frac{k^2(2k)^k}{(1+2k)^{k+1/2}}<\frac{k^2(2k)^k}{(2k)^{k+1/2}}=\frac{1+k^2}{(‌2k)^{1/2}}=(\frac{k^4}{2k})^{1/2}$

Peter Tamaroff

02:01

In fact, the whole point of this is the function is unbounded at that point.

Ted Shifrin

NO. Fix the term after the second $<$.

What point? The point is moving.

Twink

I don't understand :(

$a+1>a$

Peter Tamaroff

@TedShifrin Sure. The points.

Twink

so $\frac{1}{a+1} < \frac{1}{a}$

Ted Shifrin

Do the algebra correctly on the next thing. The first inequality is valid.

Twink

02:04

$0<\frac{k^2(2k)^k}{(1+2k)^{k+1/2}}<\frac{k^2(2k)^k}{(2k)^{k+1/2}}=\frac{k^2}{(2‌k)^{1/2}}=(\frac{k^4}{2k})^{1/2}$

$\frac{k^2(2k)^k}{(1+2k)^{k+1/2}}<\frac{k^2(2k)^k}{(2k)^{k+1/2}}=\frac{k^2}{(2k)‌^{1/2}}=(\frac{k^4}{2k})^{1/2}=(\frac{k^3}{2})^{1/2} \to \infty$

Ted Shifrin

I've told you twice your equality is wrong.

Use correct laws of exponents

Twink

It was supposed to converge to 0 :(

what is the incorrect law I used?

Ted Shifrin

It does not, but you are doing it wrong. You need to pay attention to algebra. Explain how you did $2^8/2^3$.

Twink

The $(2k)^k$ cancel

Ted Shifrin

There is no $(2k)^k$ in the denominator. You are NOT thinking.

Twink

02:08

Yes because $(2k)^{k+1/2}=(2k)^k (2k)^{1/2}$

Ted Shifrin

OK ... I need to go back to the original. Hold on.

Twink

¬¬

Peter Tamaroff

$${{{\left( {\frac{{1 + 2k}}{{2k}}} \right)}^{2k}}}\to \sqrt e$$

$$\frac{{{k^2}}}{{\sqrt {1 + 2k} }}\to\infty$$

Therefore, all to $\infty$.

As I said, the whole point of this seems to be exhibiting a function that converges poinwisely to zero but not uniformly.

Karl Kronenfeld

That's a lot of braces.

Twink

you thought it was $(2k)^{\frac{k+1}{2}}$?

Peter Tamaroff

02:11

@KarlKronenfeld My TeX has fucked up teeth, yes.

Twink

you made me feel like a moron

Peter Tamaroff

...and here comes the drama.

Twink

lol

yes Peter

it's the point

to exhibit that function

but it's supposed to converge to 0

pointwise

Ted Shifrin

I apologize, @Twink. I thought about it in my head and screwed it up. As I said, even we old professionals screw up if we don't write things down carefully. So the correct way to write this is: $f_k(x_k) = \frac{k^2}{\sqrt{2k+1}}\left(\frac{2k}{2k+1}\right)^k$.

Twink

but I'm trying to prove it doesn't converge uniformly to 0

using this theorem:

Peter Tamaroff

02:13

@Twink It does converge to zero pointwise.

Ted Shifrin

The first term goes to $\infty$, but let's think carefully about the second term. It is surprising.

Twink

$f_k$ converges to $f$ uniformly if and only if $\lim_{k \to \infty} \sup_{x \in I}|f_k(x)-f(x)|=0 $

Ted Shifrin

Gee thanks, @Peter :P

Peter Tamaroff

@TedShifrin $\text{converges}_\infty$.

Ted Shifrin

HA: I got my thanks ahead of your comment :P

Twink

02:14

first I tried to find the supremum

Peter Tamaroff

Some people use a $f_k\to f$ for pw and $f_k \stackrel{\to}{\to} f$ for unif.

Twink

and I posted this question

Ted Shifrin

@Twink: The second term in the product is something you have to be careful with!!!

Twink

1

Q: How to find the supremum of this set?

Let $k \in \Bbb N$ be fixed and $S=\{k^2x(1-x^2)^k\ : x \in [0,1]\}$ I'm trying to find $\sup S$ and at which $x \in [0,1]$ one obtains this supremum. This $\sup$ is not at $x=0, 1$ because $k^2x(1-x^2)^k=0$, for those values of $x$, and for example, for $x=1/2$, $k^2x(1-x^2)^k>0$.

real-analysis

now that I have the supremum, I wanted to prove that it...

OOH!!

Ted Shifrin

So now you're ignoring me, @Twink?

Twink

02:15

it must not converge to 0 :D

no Ted let me read

Ted Shifrin

OK ... Do you see why I've written it the way I have, breaking the product up to see what's going on?

Charlie

Goodnight @peter

Twink

$f_k(x_k)$?

Ted Shifrin

@Peter, your double arrow doesn't look right :P

Try $\rightrightarrows$ :P

Twink

the second term converges to 0

Ted Shifrin

02:17

Look carefully at $\left(\frac{2k}{2k+1}\right)^k$.

But how fast? The first term in the product goes to $\infty$ so you have to be careful here.

Twink

hmm

Ted Shifrin

What about $\left(\frac{n+1}n\right)^n$?

Twink

to $e$

Ted Shifrin

Good! Now work it all out carefully.

Twink

$(\frac{1}{1+\frac{1}{2k}})^k$

something like $1/e$

?

Ted Shifrin

02:21

Something like, yes :)

Better to write it as $\frac1{\left(1+\frac1{2k}\right)^k}$ ...

Peter Tamaroff

@TedShifrin Ah, didn't know the code.

Ted Shifrin

@Peter, I never use that symbol, so I didn't either :P

Boy, I posted what turned out to be a question I should have figured out on MathOverflow and then I flub this limit w/@Twink. Time to hang up my hat and quit math.

Twink

but $(\frac{1}{1+\frac{1}{2k}})^k>(\frac{1}{1+\frac{1}{k}})^k$

Ted Shifrin

True, but you should know exactly what it is, @Twink. If the exponent in the denominator were $2k$, what would the limit be?

Twink

$1/e$

but it's not $2k$

it's $k$

Ted Shifrin

02:25

OK, so how do we correct because the exponent is only $2k$? I teach my students this is called the method of wishful thinking. So make it be $2k$ and then correct what you do.

Twink

does it congerge to $\frac{1}{\sqrt e}$?

Ted Shifrin

Yuppers. So, since the first term in the product goes to $\infty$, the sups do approach $\infty$. That $k^2$ messes it up. Without it, it'd go to $0$, but no luck.

Twink

thanks :(

i'll try to think faster :(

Ted Shifrin

I apologize for the algebra mishap. But do be super-careful. This stuff requires careful attention to algebra and calculus.

And it's very late. So get some sleep.

Peter Tamaroff

@TedShifrin Ted.

I don't see how to "set up" the problem.

Ted Shifrin

02:33

Are you dreaming lines, @PeterTamaroff? :D

Peter Tamaroff

@TedShifrin I probably will.

Ted Shifrin

Don't go back to square one. Start with where we were with all the lines meeting the set of three (particular) skew lines.

Peter Tamaroff

@TedShifrin Yes, we had $z=xy$.

Ted Shifrin

Right, and our three lines all lie in that surface. Now out of space comes a general fourth line ...

You still awake, @Charlie?

Peter Tamaroff

@TedShifrin Aha.

Ted Shifrin

02:35

grins @ "Aha."

Think I should discuss this problem when I give my talk for undergraduates at a regional meeting in a few months?

Peter Tamaroff

@TedShifrin It is mind twisting.

Ted Shifrin

Might do a bunch of tidbits of classical geometry ... including the Fermat problem I love. I know about three or four proofs for it. Given three noncollinear points $A$, $B$, $C$, find the point $X$ that minimizes $|AX|+|BX|+|CX|$. Oops, more nightmares for @Peter :D

user60887

srry but did anyone get a chance to look over my proof? i had 9 cases to show $(a^m)^n=a^{mn}$ but idk if i went overkill

Ted Shifrin

@user60887: I'm not going to read it until you distill it down as @Peter and I suggested.

Peter Tamaroff

@TedShifrin The points where?

Ted Shifrin

02:38

in the Euclidean plane :)

This is classically phrased as the "electric company problem." They want to minimize the length of cable they lay from their office to three cities ... :P

Anthony Carapetis

@TedShifrin: that reminds me of soap films

Ted Shifrin

Ah, very apt reminder.

You're ready for geometric measure theory now? :D

user60887

So what cases do I need to show then? When m>0 and n>0? and m<0 and n<0?

Ted Shifrin

This is a one-dimensional problem of a soap film, yes.

Anthony Carapetis

Hahaha, I tried learning a little GMT for the paper I was reading for my honours project last year

Ted Shifrin

02:40

Polish off the positive case efficiently, @user60887. Then deal with the negatives in one or two sentences total.

Peter Tamaroff

@TedShifrin I am thinking triangles.

Ted Shifrin

Very cool, @AnthonyCarapetis. I'll avoid puns about integral currents.

Peter Tamaroff

Mediants.

Anthony Carapetis

hehe

user60887

ok ill do that right now

Ted Shifrin

02:41

Or @PeterTamaroff might decide to electrocute me.

Peter Tamaroff

@TedShifrin ?

Anthony Carapetis

There's a convenient "rectifiable" pun lying around here somewhere...

Ted Shifrin

There are all sorts of amazing coincidences about points in triangles. Incenter, circumcenter, orthocenter ... this point is a different one.

Only if @Peter lies prostrate @AnthonyCarapetis :P

Peter Tamaroff

@TedShifrin Won't Google "Fermat Point".

Ted Shifrin

I'll spank you if you do, @Peter. I should never give you anything google-able.

Peter Tamaroff

02:42

@TedShifrin STAHP!

The PUN-O-METER is about to blow.

Ted Shifrin

It's a beautiful multivariable calculus problem, @Peter, but there are solutions that are purely geometric based on versions of the triangle inequality.

Well, you won't wish for my return so soon again, @Peter :D

If you read a certain section of Chapter 3 of my book, it would give you geometric insight to the multivariable solution, @Peter. :D

Peter Tamaroff

@TedShifrin Of Fermat's Point?

Ted Shifrin

@Anthony, what was your Honors thesis?

How to find it using multivariable calculus, yes.

It's a key conceptual point to me, but probably not to too many other calculus teachers.

conceptual point = interesting example, in this case

Anthony Carapetis

@TedShifrin: I spent the year reading emis.de/journals/NYJM/JDG/p/2001/59-3-1.pdf then wrote about how hard it was

Ted Shifrin

LOL ... not my idea of the outline of a thesis :P

Anthony Carapetis

02:46

:P

Ted Shifrin

Ah, I know both those authors. Do you have a lot of background in differential geometry and PDEs?

Twink

@TedShifrin :-( math.stackexchange.com/questions/500788/…

Should I calculate $f''$ to check that $f''(\frac{1}{\sqrt{1+2k}})<0$ to be sure it's a maximum? :-(

Anthony Carapetis

@TedShifrin: a little, I'm increasing it as fast as possible :p

Fernando Martin

Evening

Ted Shifrin

No need, @Twink. The function $f_k$ is $0$ at the endpoints and has only one critical point in the open interval. What do you conclude?

Anthony Carapetis

02:48

when I started the project I knew everything in do carmo and that's about it, so there was a lot of new ground on the PDE side of things

Ted Shifrin

Awesome, @Anthony. Where are you now?

baby doCarmo, I presume

hi, @FernandoMartin.

Anthony Carapetis

@TedShifrin: nah, Riemannian doCarmo

Ted Shifrin

Wow, I've succeeded in shutting @Peter up with not one, but two problems :P

Anthony Carapetis

I'm doing a PhD with Ben Andrews at ANU now

Twink

it has only one crtical point because it's a quadratic polynomial function?

Ted Shifrin

02:49

Damn impressive for an undergrad, @Anthony. I still don't know everything in big doCarmo :P

Anthony Carapetis

:)

Ted Shifrin

No, @Twink, it's definitely not a quadratic. But you computed the derivative and solved for critical points, right?

Peter Tamaroff

@TedShifrin Let $y$ be a point in the plane. Set $f(y)=|x_1-y|+|x_2-y|+|x_3-y|$. Then $$f(y)\geqslant \frac{|x_1-x_2|+|x_2-x_3|+|x_3-x_1|}2$$

Ted Shifrin

A lot of you guys doing geometry in Australia. Neato.

Peter Tamaroff

Here the bars are Euclidean dist, obviously.

Twink

02:51

yes but

Ted Shifrin

I have never thought about that, @Peter, but it seems ok. Doesn't help, though :P

Twink

I don't know if $f'$ has more zeros

Peter Tamaroff

@TedShifrin Bleh.

Twink

I only found one

user60887

ok ive fixed it

Twink

02:51

the one that is in the answer

user60887

0

Q: Prove $(a^m)^n=a^{mn}$ for all $a\in G$ and $m,n\in\mathbb{Z}$

I have to prove $(a^m)^n=a^{mn}$ for all $a\in G$ and $m,n\in\mathbb{Z}$ where $G$ is a group. Is it enough to just expand $(a^m)^n=(a^m***a^m)$- $n$ times. And then from here we can expand it a bit more to there there are $mn$ amount of $a's$? Or do I need to break it up into cases. I felt if I...

homework abstract-algebra

Peter Tamaroff

@FernandoMartin Heya.

Shahab

hey everyone

Ted Shifrin

@Twink, well, you know how to find all the critical points and be sure there's only one in the open interval.

Fernando Martin

@PeterTamaroff How's it going?

Twink

02:51

I like Banchoff of differential geometry

Peter Tamaroff

@FernandoMartin Thinking.

Twink

but how do you know it has only one critical point? :S

just looking at the function :S

Ted Shifrin

@user60887: I think it's a bit too terse. You should put in the step for $(a^m)^{k+1}$. Then you need to handle the case of two negatives or one negative. You also got tangled up in negatives a bit. Make it clear what you're using and assuming before you get started. Explain why $a^{-n} = (a^n)^{-1}$, for example.

I know what it looks like, @Twink, but you don't need to know. Do what I told you to do TWICE. Damn, you hate to take advice.

Banchoff and I are mathematical brothers :P

user60887

oh ok besides that the proof as a whole is correct?

Ted Shifrin

Yes, the outline is much cleaner and more readable. Do include a case for one negative as well as two negatives.

You do have a mistake as it stands with your $-$ and $-1$'s.

Twink

02:56

you know Banchoff in person?

Ted Shifrin

Yes, @Twink, very well, actually.

user60887

oh ok thanks alot for you help. srry if i was being a bit annoyinh

Twink

I like his book

more than Do Carmo's

Ted Shifrin

I like my own better than either for undergraduates :P

But I think doCarmo isn't a good undergraduate book unless students have had some rigorous analysis background. And our students haven't, except for one or two.

@user60887: You needn't apologize. But, darn, so many of you young'uns hate to take advice and/or listen :P

@Peter: Did anyone ever tell you you're easily distracted/waylaid? :P

Twink

you have your own book?

Ted Shifrin

02:59

Yes, @Twink, I've written 4. The diff geo book is free in .pdf form on my website.

Twink

wow I want to be like you when I grow up :$

Fernando Martin

@TedShifrin: Which other books have you written?

Ted Shifrin

Well, tennis comes early in the morning, so it's bedtime for this bonzo :P

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$Mathematics$ Mathematics