thoughts on floyd?

correct answer :)

I thought that would get a star.

@FernandoMartin Arctic Monkeys?

star begging?

That's the name of a song by David Bowie in his first, relatively unknown because it sucked , album

meh/10

Though I really like one song by them

Fernando we're musical brothers

Fluorescent Adolescent

fistbumps @Mike

@Mike Partner's in musical taste crime.

do you use last.fm?

I think I've already asked that before

I don't

seems like a hassle to set up

Pedro is a shoe brand here.

how about zeppelin?

who doesn't like zeppelin

some people

floyd to is not for everyone

I don't particularly like Pink Floyd. Led Zeppelin is OK.

see^

just OK

times are a changing...

I am going to bed, good night. It's 8 am here.

later

the times they are a-telling, and the changing isn't free

@skullpatrol No, I just think blind reverence for Pink Floyd, Led Zeppelin, or Bob Dylan is a-chanin'. I don't think any are particularly bad, I just don't think they're mind blowing.

fair enough

@AlexYoucis What if skull is blind, huh? Then he doesn't have a choice but blind reverence.

Dick.

@Mike Then he wouldn't be offended by my text inconsideration.

Text to speech.

@AlexYoucis I honestly think The Division Bell is a hell of an album.

That's fine. It's just not my thing?

@Mike Not computer braille?

@AlexYoucis Just sharing. =P

honestly I think everyone should model their lives off their musical superstars

@PedroTamaroff Haha, no problem. To each their own.

I plan to live like Ian Curtis, for instance

@AlexYoucis YAS. Let me see if I can find a local band you might like.

living like a rock star?

@Mike Suicide at 23?

@FernandoMartin DAWG.

We have incompatible musical tastes @Pedro

Ouch

My mom

@FernandoMartin We're musical magnet poles.

@Daniel I'm glad someone else got it.

@FernandoMartin youtube.com/watch?v=8DNEB3-1u5s

BBNG is great

@FernandoMartin Have you ever heard of Halógena?

@FernandoMartin Exxxccellent.

Their cover of The World is Yours is about as good as the original

@Pedro No, where are they from/what do they play?

I like Bajofondo's Fandango

@FernandoMartin It used to be a metal band here in Argentina. The drummer used to play tennis with me sometimes. Now he does pop-rock.

Their band is called Juegos de Azar.

youtube.com/user/juegosdeazar

@FernandoMartin You a Gesaffelstein fan?

Let's see

I liked the previous band, this one not so much.

Not Halógena, but Factor Inicial.

never heard of them @Alex

@FernandoMartin Here

Do you guys know Ill Niño

@user127001 The storm?

Band

@FernandoMartin FB a sec.

ok

The last message was posted 1 hour ago.

Is [a,b) open or closed?

It's complement puzzles me.

Complement $(-\infty, a) \cup [b, \infty)$

The left set is clearly open. The right one though . . .

I would say it is closed?

@r9m. Did you add me on facebook btw?

Oh no. Getting my definitions mixed up. The complement is $(-\infty,a) \cup [b,\infty)$. The left set is open but the right one needs a little more work. The complement of the right set is $(-\infty,b)$. This set is clearly open. Therefore $[b,\infty)$ is closed. Thus the whole complement of $[a,b)$ is not open. Hmmm, but this does not help me.

I see my answer lies here: proofwiki.org/wiki/…

@eXtremiity ok, you've proven that $[a,b)$ is not open.

now prove that it's not closed (almost there)

@Mike. I am trying to find the continuity of a function. I have been given the set and metric for its domain, but only the set for its codomain (i.e. no metric).

Should I be taking the hint that the metric of the codomain is irrelevant to the functions continuity?

wat

can you explicitly tell me what your problem is?

I really want to get this on my own. But there are just little things that are bothering me.

the metric on $C(\Bbb T)$ is implicitly the sup metric

......

Oh wow.

$d(f,g) = \sup_{t \in \Bbb T}\{|f(t)-g(t)|\}$

Would have never figured that out.

if they never talked about that, that's weird.

Oh, I have done many examples with the sup metric.

anyway, that metric makes sense for $C(X)$ for any compact metric space $X$.

I just never knew I was meant to assume $C(\mathbb{T})$ implied the sup metric.

ah, well that's how we metrize $C(X)$ for any compact $X$.

I see. Well that clears a couple of things.

But, that metric will tell me the supremum of the values of |$f(x)$- $g(z)$|. My function $H$ maps numbers from $[0,1]$ to functions in $C(\mathbb{T})$. Does it the metric still apply?

Yep

What's the definition of continuity?

anyone here up for an eigenvalue inequality question?

(It maps elements of one metric space to elements of another :) )

There are a couple. But I think the best one to use in this case is that every open, $U$ in the codomain, $f^{-1}(U)$ is also open.

I'm never up for eigenvalue inequality questions

yikes

I'd use the one specifically for metric spaces

For all $\varepsilon >0$ there exists a $\delta >0$ such that if $d_{X}(x,x_{0})<\delta$ then $d_{Y}(f(x),f(x_{0}))<\varepsilon$?

that's what I'd advise

Thank you. Appreciate it. I am annoyed that I did not implicitly know the metric on $C(\mathbb{T})$ is the sup metric.

Far out.

No worries, man.

Why isn't is $d_{1}$ though?

You mentioned compact set.

what's $d_1$?

$d_{p}(f,g) = (\int_{a}^{b} |f(t)-g(t)|^{p}dt)^\frac{1}{p}$.

well, first you'd have to define what you mean by integrating on the circle.

anyway that sounds gross.

Integrating around the circle. Well, from complex analysis that translates to contour integration?

sure, but now you see how it gets messy

there's a lot of reason to want to integrate things on $\Bbb T$. now is not, however, the time.

And when $p = \infty$ that's when you get the sup metric.

Ok, so I can definitely see how it can get rather messy.

Sup makes it nicer. Is that why I assume sup metric?

no, it's because the sup metric is the metric one cares about for $C(X)$ when $X$ is compact :/

it was a mistake that the author didn't mention it and did mention that $I$ has its usual metric

but it's what's meant

what book is this?

The questions are from my lecturer. The information I get are from his lecture notes.

what does [Ass3] denote

I have a set of questions to do from my problem sheet. Q1 (Ass3) implies that this is one question that I must present to my lecturer to mark. I.e. (Assignment 3). There's only 2 questions from every 20 quesstions.

ah

gotcha

anyway

i can't give you a good reason, other than that the sup norm is the natural thing to use here, and i legitimately can't imagine it's anything else

so trust me on this one :)

I definitely will. I was wondering for hours how I could have attained a metric for the codomain. Now things should become easier.

You've helped a lot @Mike. Appreciate it. I'm going to have a break now.

Thanks again:)

np

If I have a smooth manifold $M$ and functions $\gamma: \mathbb{R} \rightarrow M$ and $f: M \rightarrow \mathbb{R}$, how do I take the derivative of $(f \circ \gamma)(x)$?

$f \circ \gamma$ is just a function $\Bbb R \rightarrow \Bbb R$...

1 hour later…

@user1876508 you mean the chain rule?

no, chain rule doesn't make sense here

we take derivatives of functions to (or on) manifolds precisely by turning them into functions between euclidean spaces

what command clears the scilab console? tried clear but it does nothing

$f'(\gamma)$ is a linear transformation on the tangent space $T_\gamma$ acting on the tangent vector $\gamma'$. that's what the chain rule means to me. to actually compute any of that of course one needs coordinates and then we're differentiating things in a Euclidean setting, which $f\circ\gamma$ was to begin with.

hi, all

hello

I have a group of points (x,y), and suppose the two variable meet a relation as y = (x - a) * b. so a and b are unknow coefficient which is to be estimated. how to do that? A least square estimation is fit for this problem?

@PeterSheldrick Have any idea?

yeah, it's probably better to write it as y = m*x+c with m=b and c=-a*b

that's a line

how to code it in Matlab?

define the matrix A with two columns, first column: is [x1;x2;x3;...;xn] and second ones(n,1) then [m;c] = (A^-1*A)*A'*[y1;...;yn]

here's some matlab code of mine that does linear least squares fitting of polynomials and as a bonus also absolute value fitting pastebin.com/RyPYUt21

A/b is the same as inv(A'*A)*A'*b in matlab

so it's not (A^-1*A)*A'*[y1;...;yn] but inv(A'*A)*A'*[y1;...;yn]

A invalid linker?

the link works for me

hi

eyup

In fact, I can write it like this, [y1;y2;...;yn] = [x1,1;x2,1;...;xn,1] * [m;c]. And then get [m;c] = Y / X, right?

@PeterSheldrick

@Pro-grammer It seems that you example i.e. c3 conflict with general formula cn

which one can be sure?

@DarkHorse how doyou mean?

obviously, you can get c3 is {x1,x2},{x2,x3},{x3,x4} when deduced from cn, but that's not in your example above

I doubt that one of them is fault

i think i worked it out using mod clock

@DarkHorse

mod clock?

yes

"mod n"

after 12, it loops back to 1. Formally, we say two numbers are "equivalent mod n" if they have the same remainder when divided by n.

okay, I got it

thanks though :)

:D

@eXtremiity ya :P

@robjohn Assuming $5=3$ can you prove my name is Rob?

hi hi

ooo that mod n stuff that's modular arithmetic.

@Sawarnik Depends on what you mean by $5$ and $3$. Depends on what you mean by $=$ too.

You can do the same with $\infnty$ = 1/0. @Sawarnik :-)

Is there a way to denote the calculation $1+2+3+\dots+n$? and What is the term for a factorial type operation, but with summation instead of products? ask basically the same thing. The question, which is 2 days old, already has more views that the other one, which is 2 years old.

such is the way of the world

@MartinSleziak Happens all the time.

@r9m Hey.

@Sawarnik hello

Yay for descriptive question titles: math.stackexchange.com/questions/741752/…

@r9m Prove that if a regular heptagon ABCDEFG has side 1, then 1/AC+1/AD=1.

@Sawarnik regular polygon .. what does ptolemy say ? :D

@r9m Hee, you are quite sharp :)

@(removed) whats F ?

@r9m I was asking if an acute triangle ABC has heights CE, BF then how will you find angle FEB?

what is F ? another altitude ?

@Sawarnik EBCF is a cyclic Quad

@r9m maybe not.

what are angles $BEC$ and $BFC$ ?

90.

so that means BC is the diameter of the circle passing through E and F

Oh ho. Yup.

so FEB + angle C = 180

Thanks.

(y)

Today is the big match.

@Sawarnik t20?

@boywholived Yup!

@boywholived How do you know?

@Sawarnik Any guesses?.

@boywholived You are an Indian.

why not sri-lankan?.

@boywholived Chances are less. 1.2 billion is more likely than some few millions.

Why not chineese?.

@boywholived Then you wouldn't know what is t20.

@Sawarnik I will google search

@boywholived Google won't help with "Today is the big match."

@boywholived Whom are you supporting?

@Sawarnik I know your location hence google searching "today big match india" will give you t20.

@Sawarnik I don't support anybody.

@boywholived No, my sixth sense knows you knew before.

@boywholived Then you are an Indian :p

@Sawarnik I don't believe in sixth sense. I believe in logic.

@Sawarnik You can't prove it. And unless you don't, I am not.

@boywholived Probability.

"Impossible to predict my exact location and momentum." Sorry, but I have definite evidence.

@Sawarnik What is it then?.

I can't predict your momentum but your location can be pinned down to one country.

@Sawarnik How?.

@boywholived First of you knew the details about the match, which is highly suggestive. And "You can't prove it. And unless you don't, I am not." means you are resigning to the fact although I are correct, but I can't prove it.

@boywholived :p

. . .

@Sawarnik. You are bending the facts against me. I told you a google search today big match india will give you the answer. (Try it yourself). And if I reveal my location some time for fun is gone, So I write sentences to suggest your affirmation is true. Note: I write those purposefully. Now let me have my chance of :p

@boywholived Sure, I have not proved it but logic suggests I am true. You arguments looks like they have been deliberately made up.

@boywholived Although this is doubtful.

Ok, you win.

@Sawarnik I am not fighting to win.

@boywholived Ya, of course, you are fighting to lose :p

@Sawarnik I don't mind losing.

rawr

@usukidoll meow

my class is gonna fail the midterm on MOnday D:

you know something is not right when the last two assignments are 3 out of 10 s X(

will you too?

@ccorn, Hi!

dude nobody passed the first midterm

so lol at the second one. I bet everyone is going to get the modular arthimetic questions right

@usukidoll wat!

and that's it because everyone bombed on the relations and the cardinality homeworks

which are the 3 out of 10s that I told you about

everyone scored the same thing

yes my prof sucks

nice class.

sucks more than my calc iv professor. I could sure use his grading style now

do whatever pass regardless XD

at least when you ask for help....my calc iv professor would help you and get you involved . He would make sure YOU KNOW what you're doing

but this recent prof just takes the problems that I have questions on and does it himself

@Sush Hi

like I don't exist

super rolls eyes

@Sush Hi. Why is your name Sush?

@Sawarnik, Hello

though for some odd reason the course did help me understand how to tackle the linear algebra proofs

^^'

Please help me here

I know the easy proofs... but for the ones that uses like 3-4 definitions lol need more time X(

PLEASE

this member is new at proofs points to self

so I don't have a big background when it comes to it X(

but if it's pure computation...go ahead ^^

I can't understand why did we add $\binom{12}1$ here

hi

hi

rawr

@ccorn, please help me!

any probability people in?

i know some.

I love differential equations ^^

I get stuck at some point.

Is it true that the set of condensation points of $E\subseteq\mathbb R$ is perfect?

@Sush Well, the $-\binom{13}{1}\binom{12}{1}$ accounts for patterns ...X...YZ... or ...XY...Z... where the ... may be empty. But this assumes that the position of the XY pair is unique. This fails for the situation ...XYZ... = ...(XY)Z... = ...X(YZ)... That is, this pattern has been subtracted twice and must be added back once.

@ccorn, THANK You so much! Sir. Your explanations are always complete!

I have a question relating to max flow min cut

@Sawarnik do you have any idea how to play 'Eights!' or 'Threes!' on PC?

@ccorn,added your reply as a comment for other readers.

Hi, anyone here willing to help me understand quadratic forms? Please?

@Sush Thanks, but now I note that my wording has been a bit ambiguous --- well, I hope it helps :-)

@ccorn, only your explanation has helped!

@Sawarnik Here is a list: $\{\text{anon},\text{Chris's sis},\text{Rob},\text{skullpatrol},\text{Sawarnik}\}$. Your name has index $5$, but if $5=3$, then your name has index $3$.

@Hawk What are they?

@robjohn Ooh, nice.

@Sush If you re-read Sabyasachi's answer, you will recognize the last paragraph as (basically) the same as what I have said. Somehow you have understood me better, but I do not know why :-)

@Sawarnik mathematical games...i saw one play...but cannot play on PC

@Hawk You could play the related 2048.

@Sawarnik, are you iitian?

@Sush Wat! NO.

I m just a kid.

Going to Class 9th.

OK

:)

@Sawarnik Can you give me a link of that game?

gabrielecirulli.github.io/2048

@Sawarnik thanks, actually this was the game I was searching for!!!

:) There are many variants, and many questions on SO over that game.

@Sawarnik right you are!

@Hawk I didn't had nightmares btw.

Here is a nice one: Sometimes you want to draw an ordinary triangle with interior angles no greater than 90 degrees. Here, ordinary means the following: Its angles shall be neither too small (i.e. close to zero), nor too close to each other, nor too close to 90 degrees. Let's say "too close" for a difference of less than 15 degrees. Turns out that we can now pin down the angles exactly. What are they?

@ccorn, will you please help me here?

He says: Since 24 can not have more than 4 prime factor,so the number can can not have more than 4 prime factors

But 24 has only 2 prime factors!

How did he know that the number will not have more than 4 prime factor?

@Sush only 2 distinct prime factors. $24=2\cdot 2\cdot 2\cdot 3$, that's 4 prime factors, with 3 of them equal.

@ccorn, Oh, Ok.

How did he conclude that "so the number can can not have more than 4 prime factors"

Sorry,@ccorn, got it!

@ccorn, so he should have written "so the number can can not have more than 4 'distinct' prime factors"

Right?

@Sush right, but the question asker apparently has english problems anyway. He probably has had the right thought and just did not take the pains to express it accurately.

@ccorn, thank you!