Center 1, 0

I think there is some weird algebra trick at play here

and I dont like it

I guess $x^2 + y^2 = r^2$

can I write it as $r^2 cos^2 \theta + r^2 sin^2 \theta = r^2$?

@JasperLoy Center of the circle

@Jordan yes, you can factor out the $r^2$ and use $\cos^2\theta+\sin^2\theta=1$

I am hopped up on caffeine

I am going to start doing my math tests in pen, the person grading it will see the confidence in my choice of pen and probably not even check my work

what the heck is y = 2

x = 0?

@Jordan How are you doing, pal?

@PeterTamaroff I just found out I got a 62% in my calc test, but I think I am going to keep taking it and go for the C. Not sure how colleges will look at that but I want to keep attempting to learn

@Jordan Welcome to the math world!

But yeah, I am doing okay. Way less stress now that I don't care about my grade as much

@Jordan Great for you. You're welcome, by the way! =P

Welcome for what?

@Jordan Just kidding, for the help.

I always appreciate help, but you act like I am leaving, I am not!

@Jordan Leaving? Leaving what/where?

I was just saying you are talking like I am about to die or go away :P

@Jordan Hahaha, how did you get that?

Maybe I just read too much into things

There are so many ways to manipulate polar coordinate stuff I a;lways overlook the way I am supposed to do the problem

@Jordan Always think about the implications of your statements.

I have no idea how to even approach this next problem

I dont think I know of any methods to do it

$y = 1 + 3x$

@Jordan You want to write that in polar coords?

yes

$$r\sin \theta = 3r\cos \theta + 1$$

can you solve for $r$?

what is that from?

$$y = r\sin \theta $$

$$x = r\cos \theta $$

oh

I see

How was I suppose to know to do that? I think it is the only way to solve this

@Jordan It is.

there are so many options, how do I know this is the right one?

@Jordan There aren't many, really. Since you have $x$ and $y$ and want $r$ and $\theta$, why not try substituting with the relation between polar coords and rectangular coords?

because I wanted to try and to solve x^2 + y^2 = r^2 first

Even if I found this on my own I wouldn't have gotten an answer

I can't factor it

I get $cos^{-1} \frac{rsin \theta+1}{3} = \theta$

So that isn't helpful

this is way too frustrating

I need to take a break

@Jordan Just a sec!

Remember you need to solve for $r$!

Honestly this just seems fucking pointless, I can't even do these basic easy problems after 45 minutes of trying and everyone else can get them in 10 seconds

@Jordan Usually the problem is the new symbols confuse us. It is normal. If you had $r S = 3 r C+1$ and I asked you to solve for $r$ you probably would've done it in 10 secs too.

This is why I fail every math class the first time

ouch

I learn nothing doing my homework really

I can't get any question done by myself, so I ask it here and then I don't learn anything

@Jordan Then try to do them by yourself. Try till you do it, even if it takes a lot.

I pretty much gave up in my math class and now I am a week behind

@Jordan I don't think so. I'm pretty sure that you learn something by asking here

@leo He got -29 in one of his questions!

@PeterTamaroff I do but I waste hours on the problems and usually never get the answer

@PeterTamaroff That must be the record!

@Jordan Then never try to study math. I have being stuck in a problem for weeks!

@leo I guarantee that problem was way more complicated than anything I have ever seen. I also bet there is probably no one in the world that would solve it in 10 seconds or so

but if you can solve it at the end, what you feel is the best thing you have ever feel

I think that's because people work in math

@Jordan probably it would be easy for people that work on the field everyday

There is no satisfaction in solving a problem out of a book like this really

@Jordan You need to solve this ones to get to the bigger party Jordan. You're doing well, really.

@Jordan just keep going. Math is not easy, but is not impossible. The only way to learn math is solve tons of problems. It doesn't matter if other people think that the problems you work on are to easy. It does't matter

and who doesn't love big parties? amirite?

And that's true. You are doing well!

@anon can you explain how do you feel when you solve a hard problem?

@robjohn doesn't look all that different actually.

same way I feel when I beat a difficult level in a video game

@anon i don't like big parties. i'm not very social.

I was joking about Peter's typo. I don't either.

@anon oh i see it now. =p

@anon I actually meant to write "Party". I'm having fun!

well have a good night. See you soon

@anon It's like a mental "LEVEL UP!"

@PeterTamaroff nerd

I don't even like LAN parties

i spent the whole of today reading up on categories, sheaves, and presheaves. now i will go for schemes.

I have a Makoto Kako type question.

Rudin says "let $A \subset \Bbb R$, and assume $\beta$ (a cut) is an upper bound of $A$ (i.e. any $\alpha \in A$ is contained in $\beta$)

Now

"Define $\gamma$ to be the union of all $\alpha \in A$."

What should the cardinal of that union be?

I suppose it is countable since it is a union of sets of rational numbers, which are countable, right?

(I mentioned Makoto Kako because the Axiom of Choice came to mind =P)

A function is zero over an entire open interval, can we say anything about its Fourier series coefficients?

@RajeshD Since they are determined by an integral involving $f$ I assume they should all be zero in that open interval.

@PeterTamaroff Fourier series coefficients? they vary with frequency not in time/$x$ for $f(x)$.

$a_k$, $k = 0,1,2,3,...$

@RajeshD Aren't they $a_k = \int_{T-\pi}^{T+\pi} f(x) \cos(kx) dx$ or something the sort (I'm making that up but twas something similar)?

yes ofcourse

they vary with $k$ right?

Ah ! the limits of intergration should be over 1 period of $f$

@RajeshD Sure, but if $f(x) =0$, the integral is clearly zero.

$f(x)$ is not totally zero, but only in any open interval contained in one period

@RajeshD Could you rephrase that¿

Let $(a,b) \subset (0,T]$, $f(x) = 0 \forall x \in (a,b)$

render

@PeterTamaroff

@RajeshD Oh, OK.

Then just remove tht part from the integral and evaluate the coefficients. Why do you want to know if anything can be said about the coefficients?

I have Fourier coefficients of a function given, I want to say that since these coeff's do not satisfy a so and so condition, the corresponding function cannot vanish over an entire pen interval.

One idea I got is that such $f$ can be expressed as a sum of two orthogonal functions.

each function having one segment on the left and right of the open interval and zero elsewhere

magma is super-cool.

compact operator!

Whoa I didn't realize today is sunday!

That is because today is Saturday

@jordan : Its Sunday morning 8:46 am here

@anon Could you give me a hand?

okay

@anon The inverse additive of a cut if confusing me a little.

Say I take a cut $\alpha \in \Bbb R$

Then Rudin seeks to prove that $\beta$ is a cut, defined as the set of $p$ such that there is an $r$ such that $-r-p \notin \alpha$. He says "in other words, some rational number smaller than $-p$ fails to be in $\alpha$"

I follow the proof that it is a cut.

It is OK.

Now he seeks to prove $\alpha + \beta = 0^*$ where $0^* = \{r \in \Bbb Q : r <0 \}$

He says

"If $r \in \alpha$ and $s \in \beta$, then $-s \notin \alpha$."

"there is an $r$ such that" do you mean "there is a rational $r<-p$ such that"?

@anon Any $r>0$

some $r>0$?

@anon Sorry, some, yes.

@anon I understand assuming $a+b=0$ that $-s$ can't be in $a$, or $s+(-s)=0$ which is not in $\bf 0$, but beforehand I can't understand why he asserts that.

OH!

Got it.

... glad to help.

If $s \in \beta$, then there is an $r>0$ such that $-r-s \notin \alpha$. But $-s > -s-r$ so that it can't be in $\alpha$ either, right (by def. of a cut)?

@anon But you did!

@anon ...still, I can't visualize the inverse additive cut of $\alpha$.

it's just the complement of the negation of the elements of $\alpha$

(in $\Bbb Q$)

want me to mspaint it? :P

@anon The complement of the negation. (Just as a sidenote, I think saying something is "just" takes some credit away from that concept, which I personally don't like. The other day a guy said "It is just because a polynomial of degree $n$ has $n$ complex roots and I was like "JUST!?!?!?!?")

@anon Hhehe, OK!

Hi, guys.

@FrankScience Sup?

I'm intrigued by your name.

@PeterTamaroff $\sup$?

@FrankScience What' s up?

what'(s up)

Fine. How do you do?

but then what'$\sup$

@FrankScience Good. Studying.

@PeterTamaroff right. it's a kneejerk thing. everything's trivial after you figure it out!

Dedekind's cut?

the blue is the complement of the purple

yes

well, sort of the complement. it doesn't actually include -a (in case a is rational).

@anon Hahaha well, Ben has a quote by his tutor which is "That is trivial. I can prove it in two lines."

notice that if you negate the elements of b, you will have the "complement" of the elements of a, so it makes sense to say that every element of the set of negations of elements of b has some rational less than it but greater not in a

I have professor that used to say "los patitos!" to refer to something easy to prove

Dedekind's cut is the fundamental chapter of my calculus book, but I'm not sure I'm familiar with it.

ducks?

little ducks

@anon It seems! Why ducks, @leo?

@leo Duckies!

maybe because duck rhymes with ... luck

@PeterTamaroff don't know :-)

@leo I laughed and cried inside when my phsyics teacher called parabolas "happy" and "sad" instead of convex and concave.

@PeterTamaroff that used to happen here as well

@PeterTamaroff what would $x=y^2$ be? drunk?

@anon HAHAHHA, well, "vertical" parabolas.

Let me try to rebuild the theory of dedekind's cut.

@FrankScience It will take some time!

I think he say so because here "los patitos!" sometimes is used to refer to cheap things.

@PeterTamaroff I think it makes sense.

@leo Oh, OK.

@PeterTamaroff I can't thing in some Argentine equivalent right now

A partition $\alpha|\beta$ of $\Bbb Q$ is Dedekind's cut if and only if $a<b$ whenever $a\in\alpha$ and $b\in\beta$.

Trying to memorize the proof of Jordan-Holder / Schreier refinement via Zassenhaus / butterfly via isomorphism / lattice theorems is like trying to memorize the layout of Dry Dry Desert and Forever Forest combined in my childhood.

@anon fyi: Schreier refinement proves crucial for unique factorization in noncommutative rings. See Cohn's AMM survery on unique factroization, where he calls it the "lattice method" iirc.

Oh, I was wrong.

$\alpha$ and $\beta$ should be non-empty.

@leo We usually worry when something is too expensive, not too cheap, maybe that's why!

@FrankScience And should be $= \Bbb Q$

@PeterTamaroff I've mentioned that $\alpha|\beta$ is a partition of $\Bbb Q$.

Lemma1 $\alpha,\beta$ is infinity.

@BillDubuque I have no academic institution, so it looks like I'll have to wait. :/ Also, not sure how much I'm interested in UF in noncomm rings atm; I haven't attempted fully digesting the more basic comm theory!

@FrankScience Is infinity?

What do you understand by infinity?

What is "\alpha, \beta"? Is $,$ a kind of operation?

@PeterTamaroff It seems that anon is discussing some theory, so I should stop.

@FrankScience Oh, don't worry about that

We can handle parallel talking!

@PeterTamaroff $\alpha,\beta$ are infinity sets.

@FrankScience What do you understand by an "infinity set"?

@PeterTamaroff I only want to show that we can always delete finity elements from them.

@PeterTamaroff how do you call the sort of cheap things you buy and then get broken so quickly?

"infinity" is a noun, "infinite" is the adjective

@anon Yeah, I was just sticking to his nomenclature.

@leo Hm, "baratija" comes to mind, but it is definitely not used here!

@PeterTamaroff Should dedekind's cut be settled without set theory?

@FrankScience No, but I don't follow when you talk about "infinite sets".

Any Dedekind cut is countably infinite, since $\Bbb Q$ is countably infinite.

@PeterTamaroff $A$ is infinite if and only if there exists $B\subsetneq A$ such that there's a bijection between $A$ and $B$, but I don't want such definition.

@FrankScience What other do you suggest?

@PeterTamaroff I have no plan, so let lemma1 vanish. We would build up it if necessary.

@PeterTamaroff Now I want to define equivalence relation of Dedekind's cut.

For each Dedekind's cut $\alpha|\beta$, $\alpha$ has maximum element, or $\beta$ has minimum element, or neither has max/min.

$\alpha|\beta=\alpha'|\beta'$ if and only if $\alpha=\alpha'$ and $\beta=\beta'$ when max/min from $\alpha$,$\alpha'$,$\beta$,$\beta'$ is deleted.

we should prove that $u=u$ and $u=v\,\Rightarrow\,v=u$ and $u=v,v=w\,\Rightarrow\,u=w$.

You're using too many simbols you don't define. It is almost 2 am so I have to get some sleep, I woke at 5:50 am today! We can talk about this tomorrow, OK?

@anon By the way, is it hard to prove that any two fields with the lub property are ismorphic?

@PeterTamaroff what symbol is not defined?

@FrankScience $\alpha ^\prime$

@PeterTamaroff you mean ordered fields? haven't thought about it.

$\alpha \mid \beta$

@PeterTamaroff $\alpha|\beta$ is only a tuple of $\alpha$ and $\beta$

@anon Rudin says that any two ordered fields with the lub property are isomorphic.

@PeterTamaroff $\alpha'|\beta'$ is Dedekind's cut.

@FrankScience And I assume $\alpha$ and $\beta$ are cuts?

@PeterTamaroff $\alpha|\beta$ is cut.

@FrankScience OK. We'll continue tomorrow! I'm going to fall asleep!

@PeterTamaroff well, I can see that any two LUB fields contain $\Bbb R$

@PeterTamaroff For example, $\alpha=\{x:x\in\Bbb Q,x<1\}$, $\beta=\{x:x\in\Bbb Q,x\ge1\}$

but I dunno how to show there aren't bigger ordered fields with LUB

I finally find that my calculus is very bad. I can't build up Dedekind's cut immediately.

@anon does $\Bbb R$ is the maximal ordered field with LUB?

What about $\Bbb R^*$?

you don't really need cuts to do and understand calculus well

Eh? Maybe the construction of real numbers is an important part of calculus.

@leo according to Rudin I guess

@anon cool

@FrankScience less so than you may think. it was never even covered in my calc classes.

besides, equivalence classes of Cauchy sequences are another route, and may be more pedagogically useful because they can be generalized and prepare you for some convergence topics

@anon I heard that there're many fundamental theorems about real numbers which are equivalent, for example, minimum upbound property, Borel's lemma, and so on.

@anon My elder schoolmates told me that, I might as well prove all the pairs. For example, if theorems $T_1,T_2,\ldots,T_m$ are those equivalent theorems, I should try to prove $T_j\,\Rightarrow\,T_k$ for all $j\neq k$, then I'll touch calculus more closely.

Do you think the author of this post is aware that he sounds like a crank?

And should I tell him?

his most recent blog entry is notable

Uh oh.

looks like he got published in integers which is a pretty solid journal

I don't mean to say that he is a crank, only that he sounds like one, and should take a week off and calm down.

yeah i wouldn't know about that

@Eugene Is it good to explain some proof to the others while learning mathematics?

@FrankScience huh?

@Eugene I find I have never understood the calculus. It might be because I seldom explain to others.

@FrankScience well that is up to you. some believe explaining it helps yourself understand it and others don't. i believe reading and doing a buttload of exercises is the way to go

@Eugene Well, a lot of exercises are necessary.

if you're feeling bored read this terrytao.wordpress.com/career-advice

@Eugene In the noon I suddenly realized that I have never understood Dedekind cut, which I thought I knew.

@Eugene wordpress is unaccessible here.

@FrankScience well i don't really either. it's all voodoo anyway

@Eugene There's no doubt that a lot of exercises are necessary.

@FrankScience are you in high school?

@Eugene Now I'm free home.

as in are you still a high school student?

No

well in my experience the research process is not all that elegant. you basically muck around for awhile and eventually something that makes sense comes out.

@Eugene What is elegant and research process is not elegant?

@FrankScience most of the slick proofs are from really old things

@Eugene slick?

Clever.

@Eugene You meant that these proofs are pretty but useless?

he didn't say anything about pretty proofs being useless

anon: He said "slick" which is unclear to a not-native-speaker-of-English.

This is a very difficult word.

slick: Superficially attractive or plausible but lacking depth or soundness

"slick" has special meaning to mathematicians :)

frank: I think in this case Eugene just meant "clever" and perhaps that they prove the result by a method that may not work for other results.

Anon has it.

graceful would be a good synonym

You meant that a slick proof is a proof which seems from Mars?

The people cannot understand why the thought comes out?

that's sometimes symptomatic of slick proofs, but I don't think it's inherent in the meaning of the word "slick" itself

It was these proof that I often touched wih when I was in high-school.

I consider these thought would never come out from my brain.

sorry i went to brush my teeth. listerine tastes godawful

yes slick means clever proof. those are elegant.

Well.

Yesterday I mentioned a limit which anon pointed out that it's from Ramanujan.

$\sqrt{1+2\sqrt{1+3\sqrt\cdots}}$.

@FrankScience and?

@Eugene Did you touch it when you're in high school?

@FrankScience no

@Eugene I saw it in some forum which proposed to discuss the mathematics olympiad.

@FrankScience i never took it

maybe anon did. did you @anon ?

@Eugene And then a user pointed out that it's trivial when we try to apply the trick $\sqrt a-b=(a-b^2)/(\sqrt a+b)$ again and again.

@FrankScience i think that to do with continued fractions

@Eugene How?

anyway i don't get what you want to ask

I think it hard for me to guess the result $3$, and the trick (although it is useful in $\sqrt{f(x)}+\sqrt{g(x)}$, but in such complicated problem, I have no idea).

@FrankScience and?

@Eugene Is it because I'm too silly?

@FrankScience i don't think so. i wouldn't have known this a few years ago either

@Eugene This trick is ordinary. For example, prove that $\lim_{x\to\infty}(\sqrt{x^2+1}-x)=0$.

@Eugene But why I can't get it when I deal with $\sqrt{1+2\sqrt{1+3\sqrt\cdots}}$.

@FrankScience did you try the user's trick to see if it works?

Oh, no. I was too lazy yesterday. Now I'll take some scratch papers.

@FrankScience on a closer look it looks like whoever said that it's trivial when applying the trick might not have actually tried it

Not having calculated it, I felt too embrassed.

It's justified.

@FrankScience does it work?

@Eugene I have missed something. The original problem is $a_k=\sqrt{1+ka_{k+1}}$ but stops at some $a_{n+1}=1$.

@Eugene The image on wikipedia doesn't suppose that: upload.wikimedia.org/wikipedia/en/math/5/8/c/…

@Eugene and the original problem is to calculate $a_2$ as $n\to\infty$.

@Eugene Let $b_k=a_k-(k+1)$.

@FrankScience do you mean a_n?

@Eugene $a_2$, because the $a$ is exactly constructed from $a_{n+1}$ then $a_n$ and so on, eventually $a_2$.

@FrankScience oh i see.

well again look at continued fractions

$a_k=\sqrt{1+k a_{k+1}}$ does not appear to define $\sqrt{1+2\sqrt{1+3\sqrt{\cdots}}}$; they are very different.

@anon Eh?

@anon Yesterday's problem was $\lim_{n\to\infty}\sqrt{1+2\sqrt{1+3\sqrt{\cdots+n\sqrt{1+n}}}}$, and you've mentioned that it's Ramanujan's.

@Eugene Sorry, $a_{n+1}=\sqrt{1+n}$

@Eugene thus $\displaystyle b_k=\frac k{k+1+\sqrt{1+ka_{k+1}}}b_{k+1}$.

@Eugene So $\displaystyle|b_k|\le\frac k{k+1}|b_{k+1}|$.

sorry it's like 2.40am here and i'm not really thinking to hard about this problem

@Eugene Continued fraction?

@Eugene Don Knuth uses continuant $K(a_1,a_2,\ldots,a_n)$ to denote the continued fraction.

@FrankScience i think this is a typical continued fractions problem

let it go guy, there are better things in math

where $K(a_1,\ldots,a_n)=K(a_2,\ldots,a_n)+a_1K(a_3,\ldots,a_n)$.

didn't you look at my link?

Let what go?

never mind, I meant the problem you are solving

@Eugene You mean that infinite continued fraction?

@FrankScience yup

Oh, infinity looks like adj.

@Eugene Only definition, also mentioned by Don Knuth in CMath.

@Eugene He talked about the relationship between Stern-Brocot tree, continuants and continued fractions.

@Eugene His observation about continuants seems naughty. It's about Morse code, which is used in SOS.

i've never read knuth

For example

@FrankScience read this mathworld.wolfram.com/NestedRadical.html

@Eugene I think it's only ramanujans' that is nontrivial.

@Eugene In that page.

well you can read that page. it might help

And the one of $e$ is interesting.

2AM? Do you need a sleep?

yes. i'm reading about algebraic geometry though

It might be harmful for your health to stay up.

yeah. sorry i'm not really thinking too hard about this problem

But it's better for tomorrow.

well i'll be off then. good night

@Frank : Do you know physics

I have just noticed that my Chinese surname is the same as Terry Tao's. (陶).

@JonasTeuwen True dat.

@MattN Meow?

There is a serious lack of crazy cat ladies in this very chat room.

A very basic question: $f(x,y)=\sqrt{x^2+y^2}$.

Is $f$ in $\mathcal C^\infty$?

What space?

$\Bbb R^2$

$\mathcal C^\infty$ means infinite differentiable.

Oh, no

It is even not differentiable

8-).

Yes, how about $(0, 0)$?

it is on R^2 minus the origin

@anon The space is $\mathbf R^2$!

Exclamation!

hey guys, how can I show $|g(x)-g(y)| < |x-y|$ for the function $g(x)=\sqrt{1+x^2}$ on the interval $x \geq 0$? I've made some examples like $x=0$ and $y=2$, but this isn't ideal I guess

You guess?

lol

Do you know the mean value theorem?

I mean, I can show $g(x) > x$ for every $x$, but I guess this doesn't help me either

sec, let me translate that

ok yeah, its the Mittelwertsatz der Differentialrechnung, I (should) know it

oh I think I just need to derivate g and do some stuff, sec.. thanks for the tip!

$$\left|\sqrt{y^2+1}-\sqrt{x^2+1}\right|=\frac{|y-x|(y+x)}{\sqrt{y^2+1}+\sqrt{x^2+1}} $$

now, $\sqrt{y^2+1}>y$ and same for $x$, so $y+x$ is less than the denominator..

anon wants to do it the crazy way :-(.

@anon The calculus book claim that, if $f(x,y)$ is $\mathcal C^n$ in bounded open set $\mathcal M$, and $\mathcal L$ is the edge of $\mathcal M$, and $f(x,y)$ can be $\mathcal C^n$-extended into some neighborhood of $(x_0,y_0)$, wherever $(x_0,y_0)$ is a point on $\mathcal L$, then $f(x,y)$ can be $\mathcal C^n$-extended into $\Bbb R^2$.

The proof seems so complicated for me to understand.

It's only a lemma for a general theorem.

Hmm, makes sense right?

What they say is that stuff makes sense in the (some) neighborhood of $(0, 0)$. This means NO BLOWUP. Yay.

So basically they claim you have something like $x/x$. Goes wrong for $x = 0$, but is ok nevertheless.

@JonasTeuwen There never were more than 0 since all of the cat ladies are sane. : ) (And of course, also the cat Sires, if you count robjohn as cat sir.)

@FrankScience So basically, if all the limits of the function to $(0, 0)$ are like what you would "expect" them to be then you just redefine the function in that one point and kickass!

@MattN But they will always be crazy to me :-).

@FrankScience So lots of sense it does make!

@MattN And you will always be the Crazy Cat Bro as well.

@JonasTeuwen what?