Mathematics

Jordan

12:00 AM

Center 1, 0

I think there is some weird algebra trick at play here

and I dont like it

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

user19161

@Jordan What is center 1, 0? That does not make sense.

Jordan

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

@JasperLoy Center of the circle

user19161

You need to be careful in reading and communicating.

user19161

@Jordan That just follows from the trigonometic identity.

leo

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

Jordan

12:05 AM

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?

Peter Tamaroff

@Jordan How are you doing, pal?

Jordan

@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

Peter Tamaroff

@Jordan Welcome to the math world!

Jordan

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

Peter Tamaroff

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

Jordan

12:24 AM

Welcome for what?

Peter Tamaroff

@Jordan Just kidding, for the help.

Jordan

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

Peter Tamaroff

@Jordan Leaving? Leaving what/where?

Jordan

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

Peter Tamaroff

@Jordan Hahaha, how did you get that?

Jordan

12:33 AM

Maybe I just read too much into things

Jordan

12:49 AM

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

Peter Tamaroff

@Jordan Always think about the implications of your statements.

Jordan

I have no idea how to even approach this next problem

I dont think I know of any methods to do it

$y = 1 + 3x$

Peter Tamaroff

@Jordan You want to write that in polar coords?

Jordan

yes

Peter Tamaroff

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

can you solve for $r$?

Jordan

12:58 AM

what is that from?

Peter Tamaroff

$$y = r\sin \theta $$

$$x = r\cos \theta $$

Jordan

oh

I see

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

Peter Tamaroff

@Jordan It is.

Jordan

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

Peter Tamaroff

@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?

Jordan

1:01 AM

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

Peter Tamaroff

@Jordan Just a sec!

$$\eqalign{
& r\sin \theta = 3r\cos \theta + 1 \cr
& r\sin \theta - 3r\cos \theta = 1 \cr
& r\left( {\sin \theta - 3\cos \theta } \right) = 1 \cr
& r = \frac{1}{{\sin \theta - 3\cos \theta }} \cr} $$

Remember you need to solve for $r$!

Jordan

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

Peter Tamaroff

@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.

Jordan

1:29 AM

This is why I fail every math class the first time

leo

1:40 AM

ouch

Jordan

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

Peter Tamaroff

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

Jordan

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

leo

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

Peter Tamaroff

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

Jordan

1:49 AM

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

leo

@PeterTamaroff That must be the record!

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

Jordan

@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

leo

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

Jordan

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

Peter Tamaroff

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

leo

1:59 AM

@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

anon

and who doesn't love big parties? amirite?

leo

And that's true. You are doing well!

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

Eugene

@robjohn doesn't look all that different actually.

anon

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

Eugene

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

anon

2:02 AM

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

Eugene

@anon oh i see it now. =p

Peter Tamaroff

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

leo

well have a good night. See you soon

Peter Tamaroff

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

@PeterTamaroff nerd

2:03 AM

I don't even like LAN parties

Eugene

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

Peter Tamaroff

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)

Rajesh D

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

Peter Tamaroff

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

Rajesh D

2:19 AM

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

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

Peter Tamaroff

@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)?

Rajesh D

yes ofcourse

they vary with $k$ right?

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

Peter Tamaroff

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

Rajesh D

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

Peter Tamaroff

@RajeshD Could you rephrase that¿

Rajesh D

2:29 AM

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

render

@PeterTamaroff

Peter Tamaroff

@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?

Rajesh D

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

Eugene

magma is super-cool.

leo

2:48 AM

compact operator!

Rajesh D

2:59 AM

Whoa I didn't realize today is sunday!

Jordan

3:14 AM

That is because today is Saturday

Rajesh D

@jordan : Its Sunday morning 8:46 am here

Peter Tamaroff

3:38 AM

@anon Could you give me a hand?

anon

okay

Peter Tamaroff

@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 $-s \notin \alpha$."

anon

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

Peter Tamaroff

@anon Any $r>0$

anon

some $r>0$?

Peter Tamaroff

3:46 AM

@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.

anon

... glad to help.

Peter Tamaroff

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 $\alpha$.

anon

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

(in $\Bbb Q$)

want me to mspaint it? :P

Peter Tamaroff

@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!

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$Mathematics$ Mathematics