Mathematics - 2012-08-10 (page 2 of 3)

Frank Science

11:01

For example, you can define $1/0$ well and say that it's just a notation, and then, state the critical properties of $1/0$ and keep manipulating it.

skullpatrol

One of your "critical properties" must be that it is not a real number.

Frank Science

But it could be well behaved just like real numbers.

Incidentally, some mathematicians manipulate logic.

Did you hear something called intuitionistic logic?

skullpatrol

It could also not be well behaved...

Only the real numbers behave like the real numbers :-D

Nimza

Guys, who does know some parametric representation of surface $z_1^4 = 1+z_2$?

Frank Science

$z_1=t$ and $z_2=t^4-1$.

Nimza

11:10

@FrankScience $z_1, z_2 \in \mathbb{C}$

Frank Science

@Nimza It's also OK when $t\in\Bbb C$.

Nimza

@FrankScience so nice! Thanks!

Frank Science

@Nimza Whether it is nice depends on the problem you're working with.

skullpatrol

As long as t doesn't represent time.

Nimza

@FrankScience now I want to integrate some functions on surface, but I think this parametrisation is ok for me

skullpatrol

11:13

Time being a real number.

Jonas Teuwen

@FrankScience Brouwer? Our national maths hero.

BenjaLim

@JonasTeuwen no

Jonas Teuwen

What?

@BenjaLim What?

skullpatrol

@JonasTeuwen Is admission to the University of Amsterdam difficult?

Jonas Teuwen

@skullpatrol Depends on your degree. If you have a European one: fine.

If you have a reasonable US one: fine as well.

user19161

12:17

@jonas Remember to see a doctor about your tongue today.

user19161

@matt I hope you are feeling better today.

Nimza

Who does know how a projection of $z_1^4 = z_2^2 + 1$ on $\mathbb{R}^3$ should look like? I expected something like torus, but in Matlab received some self-interlacing surface without holes

N3buchadnezzar

13:19

Heya

Is there any way to describe the red function using a single expression? I can easilly get it using a piecewise definition, but in this particular case this is not wanted.
( I am asking if the red function can be expressed on the form f( x ) = ... )

f(x) = -abs(x) - abs(3 - x) + 7

That did it

Martin Sleziak

You probably meant $4-|x|-|x-3|$.

Doh! $10-|x|-|x-3|$

@N3buchadnezzar The constant 7 does not work (try plugging in $x=0$).

N3buchadnezzar

@MartinSleziak Yes I know

robjohn

$y=7+ax(3-x)$

Martin Sleziak

And you probably need 5/4 somewhere.

robjohn

Oh, I see that it is capped off

Martin Sleziak

13:29

Now the problem seems to be which of the lines is red. :-)

Jayesh Badwaik

@N3buchadnezzar Would you mind using using rectangle function?

Martin Sleziak

What robjohn wrote was the orange one.

robjohn

@MartinSleziak it is the family of parabolas that are not truncated.

N3buchadnezzar

@MartinSleziak Why 5/4 ?

Martin Sleziak

Is the red one piecewise linear?

N3buchadnezzar

13:30

@MartinSleziak Yeah

Martin Sleziak

7/4 then?

Jayesh Badwaik

@N3buchadnezzar If you want to take fourier transforms or something similar that requires only piecewise continuous, then you can use a rectangular function to desribe it.

Martin Sleziak

$|x|+|x-3|$ has slope 2.

The red line has slope -7/2 in the right part.

robjohn

Try $y=7+\frac{7}{20}(x(3-x)-|x(3-x)|)$

It doesn't cross at $x=-2$, so some of the numbers I am assuming from the picture are not right.

Matt

@JasperLoy Yeah, thanks, sort of.

N3buchadnezzar

13:37

I just made it like this in pgfplots

@robjohn \addplot[color=red,thick] coordinates {(-2,0) (0,7) (3,7) (5,0)};

@robjohn And the function is linear not quadratic.

robjohn

@N3buchadnezzar Oh... then I am assuming things wrong :-)

Nimza

How do I post my image here?

N3buchadnezzar

Making things linear usually makes things easier.

Jayesh Badwaik

@Nimza use the upload button on the right, besides the send button.

Nimza

@JayeshBadwaik oh, thanks. Didn't see it

Martin Sleziak

13:42

Or using [text](link) format; see faq.

Nimza

Does it look like $w^2 = z^4 - 1$?

N3buchadnezzar

$$ f(x) = - \frac{7}{4}\left( |x| + |3-x|\right) + \frac{49}{4} $$

Finally

robjohn

$\frac78(11-|3-2x|-|3-|3-2x||)$

@N3buchadnezzar They both work :-)

N3buchadnezzar

@robjohn Yeah, I saw. Now I was trying to figure out why ;)

user19161

@N3buchadnezzar Is this pgfplots?

N3buchadnezzar

13:56

@JasperLoy Indeed

user19161

@Nimza Is this also pgfplots?

N3buchadnezzar

I would assume mathematica.

Nimza

@JasperLoy what is "pgf"? It is an orthoprojection on 3d plane

user19161

@Nimza It is a graphics program in TeX.

Nimza

@JasperLoy a, no... it is Matlab

user19161

13:58

@Nimza OK, the three M's are not free though.

user19161

Matlab, Mathematica, Maple

Nimza

@JasperLoy yeah :(

user19161

I guess I would only use it if I am in a university that subscribes.

Jonas Teuwen

@JasperLoy It is not free, I agree, but for some things they really are much better (Mathematica) :-(.

And I usually don't have much time to write a decent integrator for X.

N3buchadnezzar

Everything is free for Pirates, Yarr!

user19161

14:00

@JonasTeuwen Aha! But anyway I would not want to get too involved in computational stuff. Seen a doctor yet?

Jonas Teuwen

Kickass, bro.

@JasperLoy I have seen one.

And three cute girls doing some tests.

user19161

@JonasTeuwen Er, no flirting. You already have a girlfriend.

Jonas Teuwen

@JasperLoy I didn't. But they are cute anyway.

user19161

@JonasTeuwen OK, good good. I wonder if I will ever have a girlfriend.

Jonas Teuwen

@JasperLoy If you work well on everything, I suppose: yes.

But don't see it as a holy grail because often they bitch quite a lot.

But anyways... I'm off for coffee!

Matt

14:02

See ya!

Jonas Teuwen

See ya!

user19161

@skull I saw the starred message. I dislike Lang's books too. Many of them seem to be copy and paste of each other. And they leave some important things to the exercises. Lang wrote his books while on vacation.

N3buchadnezzar

Is there a function that is undefined for every x between oh lets say x=0 and x=3?'

$$f(x) = \begin{cases}
x^2 & \text{if} \ \ x<0 \cap x>3
\end{cases}$$
is not valid, no piecewise functions.

user19161

@N3buchadnezzar Can it be undefined in some parts outside that as well?

N3buchadnezzar

@JasperLoy Yeah, are you thinking about fourier series?

Martin Sleziak

14:08

Do you mean something like functions used in Batman equation?

user19161

$\frac{1}{\min (0,x)}$

N3buchadnezzar

What I am really looking for is a function that has $f(0)=7, f(3)=7 ,f'(1)=-1$ and has no minima or maxima on $x \in [0,3]$

user19161

@N3buchadnezzar That's quite a different question!

N3buchadnezzar

I know! I am sorry...

Nimza

14:24

@MichaelBoratko is it a soap membrane on your avatar?

Michael Boratko

@Nimza Yes, a computer generated one.

bugman123.com/MinimalSurfaces/index.html

There's a lot of really great math art on that site, done by Paul Nylander.

N3buchadnezzar

AH

I did it!

Nimza

@MichaelBoratko wow, thanks. I'm mesmerized by holes

N3buchadnezzar

$$f(x) = \frac{-x^2/3 + 8x - 14}{x-2}$$

Michael Boratko

@Nimza No problem

Nimza

14:32

@MichaelBoratko Are you working in this direction?

Michael Boratko

I am very interested in minimal surfaces, yes

Not quite enough background yet to really get my hands dirty

Matt

@JonasTeuwen No wait! I have a question about an integral.

Nimza

@MichaelBoratko it is interesting, is it possible to find genus by boundary without drawing a surface? As I know in inverse Dirichlet-Neumann problem for Riemann surfaces it is impossible now

Michael Boratko

@Nimza I don't know, sounds like an interesting question though

Jayesh Badwaik

14:48

@Matt Try me if it is not too advanced.

Matt

@JayeshBadwaik I have an operator $T_n : X \to \mathbb R$, defined as $T_n f = D_n \ast f (0)$, the convolution with the Dirichlet kernel evaluated at zero. I want to compute the operator norm of it.

Jayesh Badwaik

So basically you want to find the supremum of $\left|\left| D_{n} \ast f(0) \right|\right|$ correct?

Matt

Yes.

Jayesh Badwaik

@Matt so, you have an integral
\begin{equation}
\int_{-\pi}^{\pi} f(x)D_{n}(-x) dx
\end{equation}

Matt

Almost: $$ \int_0^1 f(x) D_n(x) dx$$

Just about constants. Doesn't really matter.

Jayesh Badwaik

15:02

Yeah. Just a moment.

Matt

Here, from my notes.

I am trying to go from penultimate to ultimate line.

OH!!

Have it.

Obvious.

Shouldn't hatch chickens before they're counted...

But I think I'm very close.

Jayesh Badwaik

15:29

@Matt I guess it has something more to do with the conditions on $f$, rather than the integral itself.

@JasperLoy I just saw some of your answers on english. LOL

Matt

@JayeshBadwaik I think one can use Hölder and then $L^2 \subset L^1$.

Now I need to show that first to see why, forgot why.

(if the measure space is finite)

Somehow I want $L^1 \subset L^2$ implies $\|f\|_2 \leq \|f\|_1$.

skullpatrol

Matt

Unforgivable. I really need to finally remember the proof.

Jayesh Badwaik

@Matt It is the holder inequality proof I guess. $\|f\|_2 \leq \|f\|_1$

Matt

How do you get $\|f\|_2 \leq \|f\|_1$ from Hölder?

I am misreading you, aren't I.

Jayesh Badwaik

15:37

@Matt Yup. Just a min, will be back.

@Matt Holder's ineqality says if $\frac{1}{q} + \frac{1}{p} = 1$ then

Sorry, I am wrong, I doing it opposite.

$\|f\|_2 \geq \|f\|_1$. Awfully sorry.

user19161

@JayeshBadwaik Which one made you LOL?

Jayesh Badwaik

@JasperLoy possessiveness of the s/o

user19161

@JayeshBadwaik OIC, well, not really funny.

Matt

It's wrong. I looked it up: we have $$ \|f\|_1 \leq \|f\|_2$$
Not the other way around.

Jayesh Badwaik

15:52

@Matt yeah, but I am wondering why do you need to do that? Are there any conditions given on $f$?

the only thing you need to do is $\|f\|_{\infty} \leq 1$

Matt

16:08

I posted it as a question.

skullpatrol

@Matt How many questions have you posted this month?

Jayesh Badwaik

@Matt what do you mean by "The operator norm is the sup over f with norm equal to 1."? Are you saying that norm of $f$ is $1$?

Matt

No, I mean $$ \|T\| := \sup_{\|f\|=1} \|Tf\|$$

anon

@Matt really?

Matt

17:08

@anon Well according to my lecture notes:

anon

what is X here?

Matt

A finite measure space. This only holds for finite measure spaces.

Gawd I haz teh hunger. Need to eat, how inconvenient : (

bbl

anon

I mean, for domain of $f$ of cardinality $n$ and $f\equiv1$ identically we have $\| f\|_1=n$ and $\|f\|_2=\sqrt{n}$

so you are missing a constant

Matt

But the constant is there. The measure of the space. In my question on main, the space has measure one, so the constant is one : )

Really have to go and eat something now, I'm starved.

Jonas Teuwen

@Matt Teh Hungur!

skullpatrol

17:20

Jonas Teuwen

@Matt Continuous inclusion. Jensen's theorem.

Or Hölder.

skullpatrol

Now that is my kind of math teacher ;-D

She's the one that gets the job.

Jonas Teuwen

I prefer the math teachers to know math well, not the ones that look good.

skullpatrol

Yes, of course.
They should also be able to communicate well, in my opinion.

I find there is nothing worst than listening to something you don't understand
through an accent you don't understand.

skullpatrol

17:52

@robjohn I seem to remember you talking about this :-D

Old John

18:05

@JonasTeuwen I definitely didn't fall into your second category :)

skullpatrol

@OldJohn Do you mind if I post one last one, Sir?

Old John

@skullpatrol Go ahead!

skullpatrol

Old John

@skullpatrol Nice

skullpatrol

:-D

hi @HenryT.Horton @HenningMakholm

Henry T. Horton

18:24

Skullpatrol? Is that you?

skullpatrol

Yep.

Matt

@JonasTeuwen Was having a good laugh today. Went to see the school counsellor. : D

breeden

hi

skullpatrol

hi

breeden

hmm, can i access this channel through an IRC server?

Matt

18:34

Him: "Well you certainly waited way too long to seek for help with your anxiety problem. I can write something saying that they should give you an extension if you do a therapy with me."

Me: "Sure I'll do anything. Though I don't think it's curable."

Him: "But if you don't believe in it it won't work!"

Me: "What? So you're saying that everything you can do is entirely based on placebo effect?"

Him: ... *blank face*

anon

this chatroom is on an SE network, not an irc server.

Matt

Ok, feel free to delete it.

(irc server is before my time, I don't really know what it means.)

anon

..what?

Martin Sleziak

@Matt anon was talking to breeden (see above)

Matt

Oops, missed that. I thought he was referring to my conversation with Jonas.

anon

18:39

irc an old (but still thriving), comparatively minimal protocol for text-based chat, and there are many servers out there dedicated to it

well, visually minimalistic anyway

Martin Sleziak

Internet Relay Chat was it?

Matt

I think I will have to check it out at some point too.

Martin Sleziak

"IRC is just a multiplayer notepad"

skullpatrol

Internet Relay Chat

Internet Relay Chat (IRC) is a protocol for real-time Internet text messaging (chat) or synchronous conferencing. It is mainly designed for group communication in discussion forums, called channels, but also allows one-to-one communication via private message as well as chat and data transfer, including file sharing. IRC was created in 1988. Client software is available for every major operating system that supports Internet access. As of April 2011, the top 100 IRC networks served more than half a million users at a time, with hundreds of thousands of channels History IRC was cre...

Matt

Is 4chan an example?

anon

18:41

try mibbit (channel #math on the EFNET server, for example)

the mibbit site itself is a middle entity, not part of irc. otherwise you can get a chat client like say IceChat, and you connect to servers and join channels and so forth with text commands

breeden

multiplayer notepad :)

skullpatrol

So 4chan is not an example.

Matt

I'm there. Another one off my to do list : )

bbl

skullpatrol

later

Henning Makholm

Isn't there somewhere better to redirect this meta question? The currently alleged duplicate seems to be based on misunderstanding one or both of the questions.

user19161

19:21

@skullpatrol I never used IRC.

skullpatrol

@JasperLoy Me neither.

user19161

@skullpatrol In fact SE chat rooms are the only ones I ever used. So nerdy.

anon

:o

skullpatrol

@JasperLoy Me too ;-)

user19161

@skullpatrol You need to wait for her to turn around and see her face first.

skullpatrol

19:24

@JasperLoy Yes.

Henry T. Horton

Nobody look at my faaaaace!

skullpatrol

Because you, Sir, are a monster :O

user19161

@skullpatrol I used to look much hotter than the girl in the pic.

user19161

@HenryT.Horton My eyes!

skullpatrol

@JasperLoy "Used to"?

user19161

19:27

@skullpatrol Well, various things have taken a toll on me.

Henry T. Horton

@JasperLoy Aren't you a guy? Why are you comparing your looks to a girl?

user19161

@HenryT.Horton In the same way you can compare algebra to analysis.

Henry T. Horton

Well it doesn't matter now. You only used to look hotter. Now you fall below my standards.

skullpatrol

hi @OldJohn

user19161

@HenryT.Horton Well, one day I will be restored to glory.

Old John

19:30

@skullpatrol Hi folks

user19161

@OldJohn Refreshed after the spa!

Old John

@JasperLoy Oh yes - but rather shattered after the travelling

skullpatrol

@HenryT.Horton I used to think you were a nice guy.

Henry T. Horton

@JasperLoy That would be quite a feat of modern surgical technique

user19161

@OldJohn Sp $1-1=0$. :-)

Henry T. Horton

19:31

@skullpatrol I'm one of the worst people to ever exist.

Old John

@JasperLoy Yeah - feels a bit like that :(

user19161

@HenryT.Horton No, it just takes a few weeks of workout, very simple. In fact, maybe I already am hotter than her.

user19161

@skullpatrol You must take it with a sense of humour, but maybe that was part of it as well. :-)

Peter Tamaroff

Guys.

Henry T. Horton

@JasperLoy Post a pic of yourself in the same pose and outfit

user19161

19:33

@PeterTamaroff Hello Pedro! How was the talk?

skullpatrol

@JasperLoy Her bod is happening, no doubt about it ;-D

Peter Tamaroff

@JasperLoy Good. I liked it.

user19161

@PeterTamaroff OK. Did you ask for his autograph? :-)

Henry T. Horton

What was the talk about

Peter Tamaroff

@JasperLoy I took a picture, not an autograph =P

skullpatrol

19:34

@HenryT.Horton The talk around here is about how mean you are.

user19161

@PeterTamaroff Oh OK. Want to show it here?

Peter Tamaroff

@JasperLoy I have to wait for it. The photographer should send it by email...

user19161

@PeterTamaroff Ah, I thought you used your handphone to take a digitial one.

Peter Tamaroff

@JasperLoy I have a crappy handphone =D

user19161

@PeterTamaroff Mine has no camera, the way I want it.

Peter Tamaroff

19:36

I mean. It is a decent phone. The camera is crappy.

user19161

But when my current one spoils it would be hard to find a phone without a camera.

user19161

They are always making things worse aren't they?

Peter Tamaroff

Could someone give me an example of a quotient/identification map?

@JasperLoy Things aren't made to last.

user19161

I had to search the whole island to get a pair of specs that have plain black plastic frame and a nosepad.

skullpatrol

@JasperLoy I don't even have a cell phone.

user19161

19:38

And it's also hard to find a T-shirt with one colour and no words or designs at all.

user19161

@skullpatrol Did you watch the movie Cellular?

skullpatrol

No, should I?

user19161

@skullpatrol It's very exciting...

skullpatrol

@JasperLoy I don't watch many movies either.

user19161

@skullpatrol OK. Maybe that's because you actually have a life. :-)

Henry T. Horton

19:41

Movies cut into time I could be spending doing nothing on the computer

skullpatrol

This room is better than any movie, in my opinion :-D

user19161

@skullpatrol Of course, that's because I am here for you bro.

skullpatrol

@JasperLoy You, Sir, are definitely one of the good guys.

Henry T. Horton

We're the good guys, Michael

user19161

@skullpatrol Yeah, I met many many bad guys in my life, not just trolls but demons.

Peter Tamaroff

19:45

Peoples!

Henry T. Horton

Humans

Peter Tamaroff

Could anyone care about the math for a second?

Henry T. Horton

That's all I care about. Enough with this extraneous conversation.

Old John

@PeterTamaroff Fire away, young man

Peter Tamaroff

@OldJohn =D Hello there.

Old John

19:47

@PeterTamaroff Hola

Henry T. Horton

Was Old John watching the whole time? Did he hear me singing to myself?

skullpatrol

BANG, BANG....

Peter Tamaroff

Now I'm reading about identifications (quotient) maps. And although the author doesn't mention it, I think he is doing the following:

Henry T. Horton

He's got a gun!

Old John

I'm always watching ... and listening ...

Peter Tamaroff

19:48

Let $p:E\to B$ be an identification.

Old John

@PeterTamaroff Not Mendelson, I hope :)

Henry T. Horton

Mendelson is for homo[morphism]s

Peter Tamaroff

And let $G:E\to Y$ be such that for each $x,x'\in E$, $p(x)=p(x')\Rightarrow G(x)=G(x')$

Henry T. Horton

I thought you were doing Spivak.

Peter Tamaroff

@HenryT.Horton I decided not to mix things up. When I end this, I start another. Else I'll get stuff mixed up. Plus, I like topology.

skullpatrol

19:50

yo @N3buchadnezzar

N3buchadnezzar

Yo

Peter Tamaroff

Now, we define an equivalece relation.

$x\sim x'\iff p(x)=p(x')$

Old John

@PeterTamaroff should that be iff ?

Peter Tamaroff

@OldJohn Yes, yes.

Old John

OK

Peter Tamaroff

19:52

Now what we do is $Y/\sim$

And for each equivalence class in $Y/\sim$ and $b\in B$ define $g(b)=G(\hat b)$

correct?

Wait, nah. Let me rewrite.

OK.

Let $\sim:E\to E$ such that $x\sim x' \iff p(x)=p(x')$

Note that $x\sim x \Rightarrow G(x)=G(x')$

Now for each $b\in B$, define $\beta =[p^{-1}(\{b\})]$, the equivalence class of $b$.

Let $p:E\to B$ be an identification and $G:E\to Y$ be continuous, and $p(x)=p(x')\Rightarrow G(x)=G(x')$ (this should've gone first)

$[\cdot]$ mean equivalence class of.

REWRITE: Now for each $b∈B$, and $x\in p^{−1}({b})$ set $\beta =\hat x$, the equivalence class of $x$.

N3buchadnezzar

Heya

Let us say I take the euclidian inner product between two vectors, what does the number i recieve tell me?

Old John

20:08

@N3buchadnezzar It can tell you if the vectors are orthogonal (if the inner product is zero)

Henning Makholm

And if not, then whether the angle between them is acute or obtuse.

MeAndMath

Hola!

N3buchadnezzar

Thanks

Henry T. Horton

Holas

MeAndMath

¿cómo estás?

Old John

20:16

salam, hale soma chetouri?

Henry T. Horton

Soy cansado... y tu?

N3buchadnezzar

Pablo no englis...

MeAndMath

@HenryT.Horton estoy muy bien!

MeAndMath

20:30

bueno que és viernes...

Mariano Suárez-Alvarez

exacto! :-)

MeAndMath

@PeterTamaroff Hi Peter!

@MarianoSuárez-Alvarez Gracias a Dios!

Mariano Suárez-Alvarez

@PeterTamaroff, you should have stayed for lunch!

MeAndMath

@PeterTamaroff What lin.alg book do you use?

Transcript for

$Mathematics$ Mathematics