Mathematics - 2017-12-08 (page 2 of 2)

Christopher 2EZ 4RTZ

6:38 PM

hello

I need some help with mathjax

I basically have a complicated equation for a question but I can't type it in mathjax

is there any way one of you nice people would be willing to take my equation and put it in mathjax?

@EricSilva could you help?

halp

David Reed

ok i just went through this yesterday

Christopher 2EZ 4RTZ

David Reed

top right of the screen click the link where it says latex in chat

Christopher 2EZ 4RTZ

I tried

David Reed

what happened

Christopher 2EZ 4RTZ

6:47 PM

ohh lol different link

i can view mathjax but I can't write it well

David Reed

in terms of you knowing the language or your computer not rendering it?

knowing the language

oh ok

2nd time i needed it

that I can't help you with other than to give you a link

Ted Shifrin

6:48 PM

@Christopher2EZ4RTZ: You should recognize the first integral as $-b/a$ times the area of a semicircle of radius $a$. The second term integrates to $2ab$, of course.

Christopher 2EZ 4RTZ

@TedShifrin I just need the latex to use in my question :P

David Reed

ok gotcha

Ted Shifrin

But I answered the question :) You want \int_{-a}^a \big(-\frac{b}{a}\sqrt{a^2-x^2}+b\big)dx ... That's the hard part.

David Reed

hold up

Christopher 2EZ 4RTZ

@DavidReed i got it (martin gave it)

Ted Shifrin

6:56 PM

Howdy, Demonark.

Daminark

How's it going?

Ted Shifrin

Going fine, and you?

Done with finals yet?

Antonios-Alexandros Robotis

hi @TedShifrin, @Daminark.

Ted Shifrin

hi @Antonios.

Christopher 2EZ 4RTZ

Hey would math.meta.stackexchange.com/questions/4666/… be ok to post?

or should I edit it?

Daminark

7:00 PM

Just did both my finals this morning

Ted Shifrin

You should have some parentheses inside the integral for clarity, as I put them. Do you want an answer to your question? I already gave you most of the answer.

Christopher 2EZ 4RTZ

nice

Daminark

Hey Antonios!

Ted Shifrin

Wait, two finals already? I guess they're not 3-hour finals.

Tug' Tegin

Hi everyone! What abot link text?

Daminark

7:01 PM

2 hour finals

Ted Shifrin

Aha.

I was used to 3-hour, both as a student and as a prof.

Christopher 2EZ 4RTZ

@TedShifrin I need to post it to main (added some edits btw) so I can show my old math teacher the proof

Ted Shifrin

Hmm, OK.

Daminark

I see. I think I prefer 2 hours since a day like today would've been draining if I had to test for 6 hours almost straight

Christopher 2EZ 4RTZ

@TedShifrin does the newest version look fine?

Ted Shifrin

7:03 PM

I prefer 3 hours to give people time to think and for the exam to be somewhat comprehensive.

Daminark

Yeah that's fair

Ted Shifrin

You don't need the big parentheses, unless you remove the fraction and just write 1/2 times the whole thing.

Christopher 2EZ 4RTZ

kk

Daminark

One of my professors gave us about 4 and a half hours for the test.

Christopher 2EZ 4RTZ

anything else?

Daminark

7:05 PM

Last year she gave them unlimited time and a harder test but someone took 8-9 hours, which she felt was cruel to the TA

Christopher 2EZ 4RTZ

XD

Ted Shifrin

Well, Christopher, it's good enough.

Christopher 2EZ 4RTZ

@TedShifrin what should i title it?

Ted Shifrin

I don't know. It's basically "What is an explanation for this equality?" You do not want to do the integral by some horrendous means. As I said, you want to observe how it's related to the area of a semicircle. And then it's immediate.

Antonios-Alexandros Robotis

One of my friends has extra time for his exams (well – did in undergrad) and they were 6 hours long.

Christopher 2EZ 4RTZ

7:09 PM

0

Q: Proof request for finding the area of a rectangle using an ellipse and an integral.

$$\frac{\int_{-a}^a (-\frac ba \sqrt{a^2-x^2} + b\;)\mathrm{d}x + \frac{\pi ab}2}2=ab$$ The above equation finds the area of a rectangle given A and B (both > 0). I wrote this myself as a challenge. Using desmos I verified that it was true (the desmos). I was wondering if someone could write a pr...

Lucas Henrique

hey @Ted!

Ted Shifrin

hi @Lucas.

How's life with linear algebra? :)

Daminark

@Antonios-AlexandrosRobotis oh wow that's a lot

But yeah these two were tests that I kind of enjoyed

How's it going with you Ted and Antonios?

Antonios-Alexandros Robotis

7:44 PM

not bad. Finally beginning to understand what this algebraic NT stuff is about.

Balarka Sen

7:54 PM

ramification = branch points of covering maps

algebraic NT = geometry done over bad fields

algebra is just geometry made complicated

Alessandro Codenotti

@BalarkaSen algebraic geometry = geometry over the best fields?

Balarka Sen

(I am joking; don't let the pretentiousness get you)

Daminark

Ugh geometry not again

Balarka Sen

@Alessandro complex geometry = geometry over the best fields

the best fields are obviously 1) C 2) C 3) C

Alessandro Codenotti

Of course

Well you need an algebraically closed field and C is the only one, right? :P

Daminark

7:57 PM

Gotta write a geometry diss track

Balarka Sen

@Alessandro Correct

Daminark

"It's char p bro with the finish field flow"

Balarka Sen

Dabs at @Daminark's general direction

Dragneel

Hmm does this question make sense to you guys? Apparently the answer is b.
https://i.gyazo.com/60c38343bddc64b20675c8a01021196e.png

Vrouvrou

Hello, someone know the Lagrange differetial equation ?

PVAL-inactive

8:11 PM

@Dragneel nope

it doesn't make sense

Balarka Sen

I think n%16 = 1 means n = 1 mod 16

PVAL-inactive

oh

Lucas Henrique

@TedShifrin I took a break to study for MathCamp's quiz.

Balarka Sen

I have never seen that notation used anywhere except PARI/GP

Vrouvrou

hello @AkivaWeinberger

Narcissus jewel

8:14 PM

@Dragneel Well, look at the prime factors. How many in the list 0,1,\dots,1000 are divisible by 2?

PVAL-inactive

so you just need to find the proportion of order dividing 4 elements in $\Bbb Z/16$

Akiva Weinberger

sleepily Hi

PVAL-inactive

I think that should be plenty accurate.

Akiva Weinberger

I guess I finally understand the deformation retraction of Bing's house

Balarka Sen

push holes into a pancake

PVAL-inactive

8:16 PM

It seems all of the odd elements have order dividing 4

Akiva Weinberger

No, not the deformation retraction from the *ball into the house

The deformation retraction from the house to a point

The function from ${\rm BH}\times I$ to ${\rm BH}$

The main idea is this image

Balarka Sen

oic

@AkivaWeinberger you mean a ball btw

Akiva Weinberger

I do

PVAL-inactive

I don't think spheres def retract to points.

Akiva Weinberger

They don't

Balls do

Dragneel

8:21 PM

@PVAL-inactive Ah that's interesting. And that's why b is the correct answer, since 50% of the numbers are odd.

Akiva Weinberger

@PVAL-inactive That's equivalent to the fact that there's no retraction from $D^3$ to $\partial D^3=S^2$, isn't it

You can't deformation retract a ball onto its boundary

PVAL-inactive

@Akiva to me that's a different statement.

Vrouvrou

who know how we can solve this ODE $$y(t)=t f(y'(t))+g(y'(t))$$

Balarka Sen

I don't see why they are the same either

PVAL-inactive

They can both be proven using the same machinery.

Akiva Weinberger

8:23 PM

@PVAL-inactive The connection is that $D^3$ is the cone on $S^2$

You know how you can take $S^2\subseteq D^3$ and shrink it uniformly to the center?

Suppose there were a retraction $f$ onto its boundary, and put that shrinking through $f$

You'd get a way to shrink $S^2$ to a point inside $S^2$

i.e. a deformation retract of a sphere to a point

PVAL-inactive

sure I guess that works.

Balarka Sen

Oh, the homotopy S^2 x I --> S^2 to the constant map gives you a map D^3 --> S^2

which is id on the boundary

I see

Akiva Weinberger

@BalarkaSen 'Cause you can factor it through S^2 x I --> D^3 --> S^2

sinxe S^2 x {1} should map to a point

Balarka Sen

Sure, universal property of quotient maps

PVAL-inactive

@Dragneel I think that $\Bbb Z/n$ has non-cyclic units group unless n is prime which should imply every element has order dividing 4 immediately without checking element by element.

Akiva Weinberger

8:26 PM

If that's what it's called, so be it @BalarkaSen

Ah, universal property

I see

Balarka Sen

I fucked up the name

Dragneel

@PVAL-inactive Here's how I thought about it. For any integer $n$, $n*16$ seems to result in an integer than ends with either $0,2,4,6,8$, that is, even digits. And since we're looking for $n\equiv 1 (mod 16)$, only odd $n$ integers will satisfy us because of a remainder of $1$.

Tobias Kildetoft

@PVAL-inactive No, $\mathbb{Z}/n\mathbb{Z}$ can have cyclic unit group without $p$ being a prime. It holds precisely for $n = p^r$ or $n = 2p^r$ for odd prime $p$ (well, plus $n=4$).

PVAL-inactive

I knew it was something like that.

@Dragneel All odd integers though don't have remainder 1 when divided by 16, but it turns out the 4th powers do.

Dragneel

True, good observation.

Anyway, I have to go write an exam, see you later, and thanks for the help guys :)

Balarka Sen

9:15 PM

A cool generalization to all the "exponentio-polynomial" divisibility problems: if $m$ divides $a + d$, $(b - 1)c$ and $ab - a + c$ then $m$ divides $ab^n + cn + d$ for all natural numbers $n$.

There's a short induction proof. Anyone knows a more natural way to think about this?

Induction is basically bashing all the algebra

Akiva Weinberger

3Blue1Brown has a new video out on the following problem: Choose four points on the sphere uniformly at random. What are the odds that the tetrahedron determined by those points contains the center?

Also: Bye, see you all in ~25 hours

brot

Hi

I have a question. The points of $RP^2$ are the lines of $R^3$ which go through $0$. What are the lines of $RP^2$ then? According to my resources it's the planes containing $0$ of $R^3$ but how can I see that? How is the term "line" even defined for an arbitrary space?

Balarka Sen

@brot Cool question. So for projective spaces, lines inside $\Bbb{RP}^n$ basically means 1-dimensional projective subspaces. So, the "linearly embedded $\Bbb{RP}^1$'s".

Since $\Bbb{RP}^1$ is the space of 1-dimensional vector subspaces of $\Bbb R^2$, the lines in $\Bbb{RP}^2$ are precisely the subspace of 1-dimensional vector subspaces of a 2-dimensional subspace of $\Bbb R^3$

That's a bit of a world-clutter there, but I hope that makes sense

PVAL-inactive

The term "line" cannot really be defined for an arbitrary space.

Balarka Sen

The point is if you take a plane $S$ in $\Bbb R^3$ going through the origin, and think of lines lying on $S$ that passes through the origin, those give rise to points on $\Bbb{RP}^3$ which lie on a line.

@PVAL Well we can tell him a bit about geodesics.

PVAL-inactive

9:26 PM

If you have the structure of a Riemannian metric (i,e, lengths of paths) you can talk about geodesics.

In this case there is a natural Riemannian structure on S^2

and the covering map to $\Bbb RP^2$ gives a "spherical" Riemannian structure on $\Bbb RP^2$.

Geodesics are locally length minimizing paths.

Balarka Sen

The point being that $\Bbb{RP}^n$ has a natural Riemannian metric on it, and under that the geodesics are exactly the $\Bbb{RP}^1$'s coming from the way I described, or rather, the "projective lines", as PVAL says

PVAL-inactive

The geodesics of S^2 are just great circles and the geodesics in RP^2 are just the images of great circles under this projection.

Balarka Sen

@BalarkaSen I meant $\Bbb{RP}^2$ here, by the way. Unfortunate typo.

brot

I think I get the idea

But what is a projective subspace? Does $\mathbb R P^n$ have vector space structure?

Balarka Sen

No, it has a projective space structure :)

It's really nothing complicated. A projective subspace of $\Bbb{RP}^n$ is a subspace that comes from a vector subspace in $\Bbb R^{n+1} \setminus \{0\}$ under the quotient map

You can take that to be the definition

brot

9:43 PM

Ok, I see :) and then I can directly observe that in $\mathbb R P^2$ projective lines always intersect at one projective point? that's cool

Balarka Sen

Quite.

There's another picture I like to think about. Do you know homogeneous coordinates?

brot

You mean like $(x_1:x_2:x_3)$?

Balarka Sen

yeah

So that's how you denote a point in $\Bbb{RP}^2$, right?

brot

yup

Balarka Sen

So now consider the set of all points $(x_1 : x_2 : x_3)$ in $\Bbb{RP}^2$ such that $x_3 \neq 0$.

You can scale the coordinates to get $(x_1/x_3 : x_2/x_3 : 1)$

So such points all look like $(x : y : 1)$ in homogeneous coordinates

But there's a bijection between the set of such points and $\Bbb R^2$, right? $(x : y : 1) \mapsto (x, y)$.

On the other hand, consider the complement of the set, which are set of all points $(x_1 : x_2 : x_3)$ such that $x_3 = 0$. Or, well, set of points of the form $(x : y : 0)$. We can ignore the last zero coordinate and just say that this set is bijective to $\Bbb{RP}^1$, bijection being $(x : y : 0) \mapsto (x : y)$

So we have decomposed our projective space as follows: $\Bbb{RP}^2 = \Bbb R^2 \cup \Bbb{RP}^1$

Is this okay?

vanaghka

9:55 PM

hi there, can someone help me with getting the gcd of two polynomials over the field of rationals?

what is the gcd of $x^3 + 2x^2 + 2x + 1$ and $x^2 + x + 2$

Leaky Nun

@vanaghka euclidean algorithm

brot

Yes I can follow

vanaghka

over $\mathbb{Q}[x]$

Leaky Nun

@vanaghka doesn't change

vanaghka

am i right with 2?

Balarka Sen

9:57 PM

Nice. So the way you can think about it is, $\Bbb R^2$ is not compact. But you add a "line at infinity" to it to get a compact object $\Bbb {RP}^2 = \Bbb R^2 \cup \Bbb{RP}^1$. So it's a compactification in a sense, like one point compactification, but with adding something more than a point at infinity @brot

If you think like this, then the lines $\ell$ of $\Bbb R^2$ (not necessarily passing through the origin; just a random straight line) become precisely the lines $\ell'$ of $\Bbb{RP}^2$ after the compactification

So it's just your usual concept of a line, souped up somewhat

Leaky Nun

@vanaghka $\gcd(x^3+2x^2+2x+1,x^2+x+2) = \gcd(x^2+1,x^2+x+2) = \gcd(x^2+1,x+1) = \gcd (2,x+1) = ???$

is $\Bbb Q[x]$ a euclidean domain?

brot

How do you know that $\mathbb R P^2$ is compact?

Balarka Sen

Good question.

Do you know how $\Bbb{RP}^2$ can be defined as a quotient of $S^2$?

vanaghka

@LeakyNun yes. can i show you what ive worked out so far?

brot

Yes, relating opposing points

Leaky Nun

10:03 PM

@vanaghka sure

brot

Ok, and then it's compact. I see :)

right

@brot Precisely

it feels kind of weird

because 2 doesn't exactly divide $x+1$

brot

10:11 PM

I don't quite understand the lines part though. So if I have a line in $R^2$ and then compactify it, what happens with the line on $\mathbb RP^1$?

Balarka Sen

The $\Bbb{RP}^1_\infty$ (line at infinity) is the extra bit you add to compactify $\Bbb R^2$. If $\ell$ was the line in $\Bbb R^2$, in the compactification $\Bbb R^2 \cup \Bbb{RP}^1_\infty$ $\ell$ intersects $\Bbb{RP}^1_\infty$ at one point.

So the line $\ell$ itself gets "compactified" to an $\Bbb{RP}^1$

It's a confusing picture but gets clearer if you think about it

Eric Silva

the line at infinity has a point at infinity

double infinity

but not

Balarka Sen

it's turtles all the way down

worse is if you have RPinfty lmao

Eric Silva

i guess $\mathbf{R}P^{\infty}$ is turtles all the way up

Balarka Sen

these are the kind of bullshit cell decompositions you get when your objects are algebraic though so

Eric Silva

10:17 PM

i think the cell decomp of the projective spaces is p good

it's like a 7/10 for general constructions, but like a 9/10 cell complex

Balarka Sen

eh, it's kinda bad if you want to think about the projective space as an algebraic geometric object

like projective varieties get these kind of structures

Eric Silva

i wouldnt know about that

Balarka Sen

where you have a massive zariski open set as your cell

top cell

and you throw it out

you get a variety again

and take a massive zariski open in it

rinse lather repeat

Eric Silva

I should learn some algebraic geo

Balarka Sen

i agree

Eric Silva

10:20 PM

maybe this winter break

Balarka Sen

i guess you'd immediately go to the complex geometric story

given your background

Eric Silva

yeah

I have a copy of griffiths book on complex algebraic curves

Balarka Sen

coolio

Eric Silva

so i was gonna bring that when i go home for break and give it a read

Balarka Sen

i never progressed much on the Forster thing i had

vanaghka

10:22 PM

@LeakyNun so it doesn't. 2 is the remainder yes. whats the next step after what i've done (if im right??)

Eric Silva

rip

I tried to read this book at some point but didnt get very far cause I was doing way too much analysis

Balarka Sen

i want to restart it again but nobody wants to read it with me

Eric Silva

and now im kind of burnt out on analysis

brot

If I think about $\mathbb R P^1$ as the line at infinity, it's intuitive that $\ell$ intersects it at one point. But why can I do so?

Balarka Sen

well, it's the thing you add to $\Bbb R^2$ to make it compact

like you add a point to $\Bbb R^2$ to make $S^2$

and call that point the "point at infinity"

in this case you add a full projective line, so you call it the line at infinity

brot

10:26 PM

And then parallel lines in $\mathbb R^2$ would intersect $\mathbb R P^1$ at the same point?

Balarka Sen

Yep

That's why we say the parallel lines "intersect each other at infinity"

the point of intersection lies on the line at infinity

brot

How could I show that?

LegionMammal978

Phew, the shortest substitution I could write up from $(ab^{-1})(cd^{-1})$ to $(ac)(bd)^{-1}$ with field axioms uses a lot of associative transformations:

$$\begin{split}
&(ab^{-1})(cd^{-1})\\
=&((ab^{-1})c)d^{-1}\\
=&(c(ab^{-1}))d^{-1}\\
=&((ca)b^{-1})d^{-1}\\
=&((ac)b^{-1})d^{-1}\\
=&(ac)(b^{-1}d^{-1})\\
=&((ac)1)(b^{-1}d^{-1})\\
=&((ac)((bd)(bd)^{-1}))(b^{-1}d^{-1})\\
=&((ac)((db)(bd)^{-1}))(b^{-1}d^{-1})\\
=&((ac)((bd)^{-1}(db)))(b^{-1}d^{-1})\\
=&(((ac)(bd)^{-1})(db))(b^{-1}d^{-1})\\
=&((ac)(bd)^{-1})((db)(b^{-1}d^{-1}))\\
=&((ac)(bd)^{-1})(d(b(b^{-1}d^{-1})))\\
=&((ac)(bd)^{-1})(d((bb^{-1})d^{-1}))\\
=&((ac)(bd)^{-1})(d(1d^{-1}))\\
=&((ac)(bd)^{-1})(d(d^{-1}1))\\

Leaky Nun

@vanaghka oh wait, we're in $\Bbb Q[X]$

$\gcd(2,x+1) = \gcd(2,x+1-2(\frac12x)) = \gcd(2,1) = 1$

vanaghka

11:00 PM

@LeakyNun okay how was that step allowed

user8663905

Question: given $y = e^{x + y} - e^{-x-y}$, can you get all the y's to one side?

LegionMammal978

11:21 PM

@user8663905 I don't believe so, no

Actually, WA gives the real solution $x=\ln\left(\frac12\left(\sqrt{y^2+4}+y\right)\right)-y$

user8663905

what'd you enter into Wolfram to get that? I'm always struggling to get WA to do what I want it to do.

LegionMammal978

"solve y=e^(x+y)-e^(-x-y) in reals"

user8663905

Alright, I wonder what the algebra is to get there. Danke for the info

LegionMammal978

Also, a question for anyone up to it: Given any equation between two expressions consisting of constants, the six basic trigonometric functions of $x$ ($\sin x,\cos x,\tan x,\cot x,\sec x,\csc x$), and elementary operations (addition, subtraction, multiplication, division), what would be a minimal set of trigonometric identities (e.g., $\sin^2x+\cos^2x=1$) required to verify via substitution a general equation in this form?

$x$ is assumed here to be a real number such that all expressions are defined

Simply Beautiful Art

11:46 PM

2

Q: Growth rate of the nth natural number not constructable with n steps of addition and multiplication

While messing around with the idea of ordinal collapsing functions, I stumbled upon an interesting simple function: $$C(0)=\{0,1\}\\C(n+1)=C(n)\cup\{\gamma+\delta:\gamma,\delta\in C(n)\}\\\psi(n)=\min\{k\notin C(n),k>0\}$$ The explanation is simple. We start with $\{0,1\}$ and repeatedly add it...

discrete-mathematics asymptotics

^ Seems to be a fun question. I myself am having fun over it lol

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$Mathematics$ Mathematics