Mathematics - 2016-04-19 (page 1 of 3)

12:04 AM

@PedroTamaroff Strange this is how it appears for me m.imgur.com/sqzUjVH , is this a bug? I'm viewing it using Chrome on Android

PVAL

@Bysshed It should be fixed now

It didn't render for me either due to the unnecessary enters after the $$

Bysshed

@PVAL Thanks, it now renders for me too.

PVAL

There's some other weirdness though with the quotient groups after the "on the other hand"

user19405892

12:37 AM

hi

why does $(A \cap B') \cap (B \cap A') = A \cap A' \cap B \cap B'$

Samuel Yusim

you can derive this from the following facts: $X \cap (Y \cap Z) = (X \cap Y) \cap Z$ and $X \cap Y = Y \cap X$.

(first I might use the second fact to swap $B$ and $A'$, and then use the first fact to say $(A \cap B') \cap (A' \cap B) = ((A \cap B') \cap A') \cap B$)

user19405892

12:57 AM

Would it be okay to say the domain of $f(x) = \dfrac{1}{x}$ is $\mathbb{R}$?

Krijn

What would be your value for $0$?

What is your codomain?

user19405892

in my book it says the image is a subset of the range

Krijn

What does your book say is the range?

user19405892

$f(A) = \{f(a): a \in A\} \subset B$

Krijn

So, let us reprhase your starting question

Your function $f$ seems to be a map $f: A \to B$

user19405892

1:01 AM

yes that's right

Krijn

What is $A$ and what is $B$?

user19405892

it is a mapping for the function $f$ such that $(a,b) \in f$

if for every element $a \in A$ there exists an element $b \in B$

ah so $f(A)$ is a subset

of the range

Krijn

You just said that $f(A)$ was the range?

user19405892

i am confused if they are calling $f(A)$ the range or $B$

is $B$ the codomain?

Krijn

$B$ is the codomain (or sometimes called target set) and confusingly, is sometimes also called the range. Confusing, because $f(A)$ is called the image of $f$ and the image is sometimes also called the range

user19405892

1:05 AM

my book seems to suggest that the image is always a subset of the codomain

Krijn

That is true, yes

user19405892

why can't it be equal?

Krijn

It can

user19405892

oh sorry

I was confusing with proper subset

Krijn

Ah indeed

If it is equal the function is called surjective (or onto)

user19405892

1:06 AM

why is that

Krijn

Erh, it's the definition.

I guess they call it onto because the function maps onto the whole codomain

user19405892

so it is the opposite of a function?

Krijn

No?

I don't get what you mean by that

A function $f: A \to B$ is called surjective if $f(A) = B$

user19405892

a function maps the domain onto the codomain

an onto function maps the codomain onto the domain?

Krijn

No, an onto function maps the domain onto the codomain, in such a way, that every element of the codomain is reached

So for every $b \in B$ there is some $a \in A$ such that $f(a) = b$

But try to figure out for yourself why this is the same as $f(A) = B$

user19405892

1:14 AM

what do you mean by maps onto?

Krijn

Like the image on the right here

Codomain

In mathematics, the codomain or target set of a function is the set Y into which all of the output of the function is constrained to fall. It is the set Y in the notation f: X → Y. The codomain is also sometimes referred to as the range but that term is ambiguous as it may also refer to the image. The codomain is part of a function f if it is defined as described in 1954 by Nicolas Bourbaki, namely a triple (X, Y, F), with F a functional subset of the Cartesian product X × Y and X is the set of first components of the pairs in F (the domain). The set F is called the graph of the function. The set...

On the left in the preview, apparently

user19405892

so you are saying an onto function maps the domain onto the codomain? Doesn't it do the opposite? It maps each element in the codomain to an element in the domain.

Krijn

No

Look at the image

user19405892

so in the image the domain X is mapped onto the codomain Y?

Krijn

Okay the term onto may be confusing, let's use surjective instead

A surjective function $f: A \to B$ is a function such that $f(A) = B$

user19405892

1:19 AM

yes

i agree

Krijn

The functionin the image is not surjective, as there is some part of the codomain that is not reached by $f$

There are no elements of $A$ mapped to those elements by $f$

user19405892

yes

i was just wondering what you meant by onto

Krijn

So now back to your original question

user19405892

ok

i just dont get what you mean by maps onto and i actually have never heard that phrase before

Krijn

I usually don't use it either and prefer surjective

Jessy Cat

1:27 AM

I've got a 3rd question tonight

0

Q: Use contraction mapping theorem to show that the integral equation has a unique continuous solution on $t \in [0,3]$

I have to use the contraction mapping theorem to prove that the integral equation with continuous functions $K(t,s)$ and $f(t)$, $\begin{equation*} x(t) = \lambda \int_{0}^{3} K(t,s) x(s)ds + f(t), \, 0 \leq t \leq 3 \end{equation*}$ has a unique continuous soltuion on $t \in [0,3]$ for suffi...

user19405892

is the function $f(x) = x$ onto?

Krijn

Depends on the domain and the codomain of $f$

$f: \mathbb{Q} \to \mathbb{R}$ with $f(x) = x$ is not surjective

$f: \mathbb{R} \to \mathbb{R}$ with $f(x) = x$ is surjective

user19405892

or if $f: \mathbb{R} \to \mathbb{C}$ with $f(x) = x$ is not surjective

Krijn

Yes!

user19405892

but really changing the range of the function doesn't change the function at all right?

sorry codomain

Krijn

1:38 AM

It does change the function, but I get why you think it doesn't

The thing is, all of the three functions above are different functions

It's just semantics really

You are probably used to saying that $f(x) = x$ is a function

But if you want to do things proper, a function is a thing with a domain, a codomain and an $f$, so to speak

user19405892

so something must be a function in order for it to be a possible to be a surjection?

Krijn

Yes, a surjection is a special kind of function

user19405892

What is the difference between $f(x) = x: \mathbb{R} \to \mathbb{R}$ and $f(x) = x: \mathbb{R} \to \mathbb{C}$?

Krijn

The codomain

user19405892

yes but the functions map the same points

Krijn

1:43 AM

Jup

user19405892

so there is no difference

Krijn

That's why I say that it's just semantics

These small details are what make mathematics rigorous

Jessy Cat

If anybody can answer any of my three recent questions I've posted, please, please, please do. Nobody ever even looks at them.

Krijn

@JessyCat I've had difficulty getting answers to functional analysis too

user19405892

I don't see the difference between them. They seem like equivalent defintions

Krijn

1:46 AM

@user19405892 They are really really really really alike, except the latter has a bunch of points with no things mapped to them

However, surjectivity is an important concept in mathematics, so the codomain is very important to specify

user19405892

You are just graphing $f(x) = x$ in the complex plane versus the real plane so I don't see the difference

sorry

Krijn

So what do you propose?

When we define functions

user19405892

so basically subjectivity ensures no missed y coordinates

Krijn

Surjectivity is more than that, as there are much more functions than just functions to the line

user19405892

why wouldn't it be possible for a something to not be a function and be surjective?

Krijn

1:52 AM

?

That's like being not an apple but being a red apple

A red apple is an apple that is also red

A surjective function is a function that is also surjective

user19405892

oh so surjective is a word that describes functions only not other types of mappings?

Krijn

Oh, like that

Well, this definition describes functions only

You could describe surjectivity for other types of mappings yes

Jessy Cat

2:09 AM

@Krijn, I got someone to look at it, and now they won't answer follow-up questions I have about their answer!!

Axoren

$\mathbb R / 2\pi\mathbb R$. Is this the proper notation for the set of all unique angles on a circle?

Or am I using quotient notation wrong?

anon

@Axoren $2\pi \Bbb R=\Bbb R$ so what you've written is $\Bbb R/\Bbb R$, which has only one element, namely $0$

you're thinking of $\Bbb R/2\pi \Bbb Z$

Axoren

@anon I forgot, for any number $x \in \mathbb R$, $\frac x {2\pi} \in \mathbb R$, lol.

The Substitute

Hello. Off topic algebra question. There's a well-known theorem that says a polynomial $f$ is separable iff $(f,f')=1$. Does the statement $(f,f')=1$ hold in the original field or only in the splitting field?

Axoren

As for whether or not I mean the second one, does $\mathbb R / 2\pi \mathbb Z$ contain all legal radians?

I don't immediately see that the relation is defined there, and thus how the quotient leads to being the correct set.

@anon This would make all integer multiples of $2\pi$ equivalent to 0, but it won't make any non integer multiples of $2\pi$ equivalent to each other, or am I missing something?

anon

2:23 AM

it will make x+2pi=x for all x

do you know what A/B means in this context?

Axoren

It means set quotient, but I'm very shaky on the formal definition of it. I understand it in terms of equivalence relations.

anon

when A has an addition operation, B is a subset closed under the addition operation, then A/B means impose the equivalence relation a~a' whenever a-a' in B and then collect the equivalence classes and call the result A/B.

Axoren

Alright, that makes more sense than how I understood it before. Thanks.

Ted Shifrin

@TheSubstitute: Doesn't matter. Same gcd.

rehi @anon @Axoren

Axoren

@TedShifrin Yo.

anon

2:38 AM

you don't get to say yo

I say yo

Ted Shifrin

I was wondering if that would yo your chain.

But since you didn't return my hello, @anon, I think Axoren was within his rights.

The Substitute

@TedShifrin In the proof that I studied, the auther (Conrad) uses the fact that if $f$ is separable then $f,f'$ have no common linear factor, but what if there is a higher degree common factor that doesn't split in the original field?

Ted Shifrin

Can't happen, @TheSubstitute. Imagine $f(x),g(x)\in\Bbb R[x]$ that have no common factor. Now suppose that they have a common factor $x-c\in\Bbb C[x]$. What does that mean?

This is one of my favorite exercises that I assigned when I taught algebra, BTW.

anon

@TheSubstitute f, f' cannot have a common root (in the algebraic closure) is the stronger implication

Ted Shifrin

But still, understanding that gcd stays the same when you go to a field extension is an important idea.

Same thing with $\Bbb Z$ and $\Bbb Z[i]$, etc.

Do you see it in the $\Bbb C/\Bbb R$ case, @TheSubstitute?

Jessy Cat

2:45 AM

@arctictern, are you also Joel?

anon

no

Ted Shifrin

anon has probably 100 identities, but he's not everyone :)

Jessy Cat

Eep, he's @anon as well.

I'm scared now.

Ted Shifrin

Too many anons.

Jessy Cat

And they all wear those stupid V for Vendetta masks.

The Substitute

2:47 AM

@TedShifrin I don't see it yet since I'm not completely convinced that the gcd doesn't change. Still thinking about it.

Ted Shifrin

Well, how can $x-c$ be a factor of a real polynomial, @TheSubstitute, assuming $c$ is not real?

Let's be concrete. Suppose $x-i$ divides both $f(x)$ and $g(x)$ when we work in $\Bbb C[x]$ but both $f(x),g(x)\in\Bbb R[x]$.

The Substitute

$x+i$ is a factor of $x^2+1$

Ted Shifrin

So can you argue that $x^2+1$ must in fact divide both $f(x)$ and $g(x)$ to start with?

The Substitute

Yes, since it's the minimal polynomial for $i$.

Ted Shifrin

OK ... Now generalize.

You can do an ideal-theoretic proof of my original statement, or you can do it other ways.

Jessy Cat

2:55 AM

The Sobolev-space question I posted is seriously giving me stress...

Axoren

"In the case of the manifold $S^1$, the circle, if we think of the tangent space to 1 as being the imaginary axis (y-axis) in the complex plane, then..." [http://mathworld.wolfram.com/ExponentialMap.html](http://mathworld.wolfram.com/ExponentialMap.html)

I was reading up on the exponential map for an answer to a question I had the other day, but I ended up running into this odd comparison between the exponential function in complex analysis and lie groups. Taking a vector in the tangent space $xi$ back to the circle by the exponential map is only possible if the circle is in the complex pl…

The Substitute

@TedShifrin @anon Thanks for the help (and for pointing out the fact that the gcd doesn't change when you extend the field!)

Ted Shifrin

For matrix groups, the Lie group exponential is actually matrix exponential, @Axoren.

Axoren

Can't edit the post, the link broke: mathworld.wolfram.com/ExponentialMap.html

Ted Shifrin

Sure, @TheSubstitute.

Axoren

2:59 AM

@TedShifrin Right, but we don't have a matrix group here. It's the worst.

Matrix groups make everything so easy, I love them.

Ted Shifrin

I haven't looked at your link. I was responding to what you put visibly here.

Sure, but your $\Bbb R$ is a Lie algebra here.

Axoren

Right, but the lie group is $S^1$

Ted Shifrin

It's isomorphic to $\Bbb R$; it isn't $\Bbb R$.

You're doing $\mathfrak{so}(2)\to SO(2)$, of course.

Axoren

I see. Rather than working on it in terms of $\mathbb R$ and $\mathbb R^2$ I can take it to the matrix group instead.

Ted Shifrin

nods

anon

3:02 AM

@Axoren in the circle in R^2, what are you calling the identity element, and how are you defining multiplication, assuming you aren't just doing exactly the same thing as in C but with the stuff relabelled?

Axoren

@anon The identity element would be the point at $(1, 0)$ and multiplication would be traveling along the circle in arc-length distance from $(1, 0)$ (radians, I guess)

In a sense, I guess it would be just a relabeled complex plane.

anon

then your tangent vectors would be elements of $\{0\}\times\Bbb R$

oh, didn't read the end of your first post

Axoren

But I was hoping to generalize the manner in which the exponential map applied to the complex plane.

anon

bleh

yeah, ted answered: for matrix groups, just use matrix exponential

Ted Shifrin

I have no idea what that means.

Axoren

3:06 AM

What specifically, @TedShifrin ?

Ted Shifrin

(What Axoren just said, not what anon just said.)

"generalize the manner in which ..."

Axoren

So, in a sense, taking the imaginary axis to be the tangent space, we can take any point on the imaginary axis and map it by the exponential onto the complex unit circle.

It just so happens, that $e^{ix}$ is already happily defined for this.

Ted Shifrin

Yes. And it generalizes by taking $e^{x+iy} = e^x e^{iy}$.

Axoren

But what if I wanted to use a different unit vector? Such as $(0, 1)$?

And work in $\mathbb R^2$ instead? Without mapping to $\mathbb C$ or using matrix groups.

Ted Shifrin

But then $e^0 = 1$ gets difficult unless you just change coordinates completely.

user19405892

3:09 AM

hi

how are equivalence relations related to equality?

Ted Shifrin

You can use the exponential at any point (as in Riemannian geometry, following geodesics). But then it will be a shifted matrix exponential. Nothing interesting going on here.

Equality is a special case @user19405892.

anon

@user19405892 equality satisfies properties, those properties are turned into the definition of equivalence relations

Ted Shifrin

@anon: I assume you're feeling better?

anon

yeah

Ted Shifrin

Good :)

anon

3:11 AM

I was only bad for a couple of days

Ted Shifrin

I meant to ask earlier.

I think you left off a final "y" on your name, @wellynaught

well y naught

lmao, that would certainly have been better!

Ted Shifrin

Just making a constructive suggestion :P

Axoren

@TedShifrin Let's consider now that instead of something neat like a circle, I had some nasty curve in some $n$-dimensional space where $n$ isn't $2, 3, 7, 15, 31$ (I won't be able to use anything like a quaternion or complex number). The group operation is addition of distance along the curve with some identity element on that curve $0$ (essentially, the number line twisted and turned about in a higher dimensional space). I can define the tangent space, but it's full of vectors.

What do I even try to do here?

A continuous curve*

anon

what do you think?

(also, might want differentiable so you can talk about length at all)

Axoren

3:23 AM

@anon Right, that too.
I want to be able to define the exponential map. However, in this space, the tangent space is full of vectors. The exponential map as defined on matrix groups is not applicable here. The exponential map on complex numbers, quaternions, and octonions, etc, is not applicable here.

anon

If v is the unit tangent vector, then exp(tv) is the point on the curve a distance of t away from the identity element traced out in the direction of v.

Mike Miller

morning

anon

what are you, in China?

Axoren

He likely just woke up.

anon

<-- jests

Axoren

3:26 AM

@anon Right, but then we have $\exp \{t\vec v\}$, how should I define $\exp : \mathbb R^n \to \mathbb R^n$? I don't know of a definition of the exponential on vectors.

Mike Miller

anything fun tonight

anon

@Axoren you don't define $\exp:\Bbb R^n\to\Bbb R^n$, because the tangent space of the curve you defined is not $\Bbb R^n$ (and certainly your curve itself is not $\Bbb R^n$)

the exponential function takes a tangent vector at the identity and spits out an element of the group

@MikeMiller nope

Mike Miller

darn

Axoren

@anon So even though the curve has points in $\mathbb R^n$ before I apply the group structure to it, once I do, I'm working with a different?

anon

@Axoren huh?

go back to, say, the unit circle in R^2

with the exp function, you don't plug in just any vectors, you plug in vectors tangent to the identity element

Axoren

3:30 AM

Okay, I have some curve in $\mathbb R ^2$.

The identity element is some point on the curve which I've dubbed $0$.

The space tangent to $0$ is some line $ax_1 + bx_2 = c$

anon

waiting for question...

Axoren

Let's make it simpler. Let's say my identity element lies on the $0$ of $\mathbb R^2$

user19405892

if equality is a special case of an equivalence relation, then why call it an equivalence relation?

Axoren

So, elements of the lie algebra are $(x_1, x_2)$, correct?

anon

@user19405892 because it is a binary relation that behaves like equivalence

Axoren

3:35 AM

They're some solution to the equation of the line tangent to $0$.

anon

@Axoren elements of the lie algebra are tangent to your curve

not arbitrary elements of R^2

Axoren

Well, they aren't arbitrary, but they're still elements of $\mathbb R^2$

anon

yes

Axoren

Aren't they?

anon

do you have a question coming?

Axoren

3:36 AM

So then how is the exponential map defined in this case?

anon

depends on your group operation

is it still the distance thing?

Axoren

In this case, it's addition of distances along the curve. Let's call the curve $c(t)$. If we have elements $c(a) * c(b) = c(a + b)$

anon

in that case then

13 mins ago, by anon

If v is the unit tangent vector, then exp(tv) is the point on the curve a distance of t away from the identity element traced out in the direction of v.

Axoren

Oh, I misunderstood that the first time you said it.

That was a definition, not a claim about exp(tv)

That's where my confusion was. Sorry about that.

anon

well, the definition of exp:(tangent space)->group is that (i) the curve exp(tv) parametrized by t has velocity vector v at t=0, and (ii) exp(rv+sv)=exp(rv)exp(sv) for any scalars r,s and tangent vectors v. these conditions uniquely specify the function exp because of existence/uniqueness of solutions to PDEs and stuff. then it's a matter of verifying that the exp(tv) I gave satisfies the defining conditions.

zed111

3:41 AM

Hi, can someone tell me in context of differential geometry.. How does the transformation y_i = x_i for i = {1,2,...n-1} and y_n = x_n - gamma(x_1,...x_{n-1}) flattens/straightens out the boundary? Here gamma is function used for the boundary of U

Axoren

@anon You answered my follow up question with your edit. I'm fairly certain it does.

I guess finding the closed-form for such an exponential map is on me to solve some nutso equations

Mike Miller

@Axoren if you pick some nonzero $v$ in the tangent space, your exponential map is $tv \mapsto c(ta)$, for some appropriate scalar $a$

Forever Mozart

hola

zed111

@MikeMiller can you answer my question

user19405892

hi

Mike Miller

3:57 AM

I don't understand your question, so probably not.

Joshua Lamusga

Can somebody tell me why the polar vectors (26, 45$^{\circ}$) and (12, 30$^{\circ}$) have a resultant direction of 23.3$^{\circ}$? I get 29.5 and 40.3 on different tries. I used calculators that also gave 40.3.

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$Mathematics$ Mathematics