Mathematics - 2017-04-14 (page 3 of 5)

Akiva Weinberger

15:00

Oh, found it

Bolzano-Weierstrass Theorem

BAYMAX

Anyway,Pigeonhole principle gives us many nice results that sometimes is not intuitive,strange .

DHMO

Nice

SBM

could you please state it in simple terms?

Semiclassical

What's not immediately obvious to me is the region of integration.

Secret

Today's exercise taught me one thing: $\Bbb{Z}[\sqrt{q}]$ and $\Bbb{Q}[\sqrt{q}]$ are both dense

In the past back in my linear algebra class, I thought they look like lattices squished onto a line

Akiva Weinberger

15:02

@SBM Explain what in simple terms?

Secret

and thus have "gaps" in it

Akiva Weinberger

The Bolzano-Weierstrass Theorem?

DHMO

@Secret $a+br$ is dense for any irrational $r$

Jack Lam

Yes, I get that, then I can substitute (1+xy)² = 1+2ux + x²

but how does 1-xy vanish?

SBM

Yes

Akiva Weinberger

15:03

Do you know what limit points are?

Semiclassical

I don't see $1-xy$.

Secret

@DHMO I suppose that holds for rationals as well, since rationals are dense in the reals

Semiclassical

What I see is $\dfrac{dx \,dy}{1+x y}=\dfrac{dx\,du}{(1+x y)^2}$.

Daminark

Hey everyone!

Meow Mix

Hi Astyx

And... Daminark?

Daminark

15:04

How's it going?

DHMO

@Secret $a+b\dfrac12$ is not dense.

Meow Mix

@Daminark tired

Semiclassical

And if $(1+xy)^2=1+2xu+x^2$ as you say, I don't know what else there is to say. @JackLam

Daminark

Wait wouldn't you have woken up very recently?

Secret

@DHMO oops

SBM

15:05

What does it mean to be dense?

Astyx

And hi chat

DHMO

@SBM it means that every real number can be approached by elements in the set arbitrarily close

Akiva Weinberger

@SBM A set is dense if every real number can be written as the limit of a sequence of things in your set

Astyx

@Ted How can that be true ? Isn't $f:x\mapsto -x^2$ a counterexample ?

Akiva Weinberger

For example, the rationals are dense

Daminark

15:05

Closure is the whole space, intersects any open set

DHMO

@Astyx yes it is.

He assumed (wrongly) that the function is positive.

Akiva Weinberger

You can approximate $\pi$ (for example) by the following sequence of rationals:$$(3,3.1,3.14,3.141,\dots)$$

SBM

Oh, thank you,

DHMO

@Astyx And chapeau for using $\mapsto$

SBM

I didn't know the terminology for that earlier

Astyx

15:06

Haha thanks

DHMO

@Secret Use linear algebra to find the minimal polynomial of $\sqrt 2 + \sqrt 3$

Akiva Weinberger

@SBM There's a related notion of an open interval, which is written like "$(a,b)$" and means "the set of all numbers between $a$ and $b$ exclusive"

@SBM So another way of defining dense is: A set is dense if every open interval contains something in your set.

Astyx

I must have spent a few hours trying to find why $f (\pi) $ being negative was a contradiction :p

DHMO

@Astyx I am sorry for that.

Akiva Weinberger

(And, in fact, it'll have to contain infinitely many things in your set, exercise.)

(So every open interval contains infinitely many rationals.)

SBM

15:08

OK

Astyx

Don't be, I should have been more careful

Anyway back to work

Secret

Is $\{a+b\sqrt{2},a,b\in \Bbb{Z}, a,b > 0\}$ not dense, since there are no elements in (0,1)? thus there is no way to approach 0 hence making every point dense?

Akiva Weinberger

Yeah. In fact, it's discrete.

DHMO

Interesting.

SBM

Discrete as in discontinuous maybe

Secret

15:10

so the density property is pretty much because of 0 becoming a limit point?

DHMO

@SBM Discrete as in, given an element in the set, you can always construct an open interval containing only that element but not other elements

@Secret it's because it is an additive group.

and 0 being a limit point.

SBM

Ok

Secret

ah yes

Semiclassical

Discrete as 'isolated', more figuratively.

Akiva Weinberger

Discrete can also be characterized as "no limit points"

DHMO

15:11

@Semiclassical as 'every element is isolated'

Semiclassical

Right.

Secret

I suspect $\{a+b\sqrt{2},a,b\in \Bbb{Z}, b > 0\}$ and $\{a+b\sqrt{2},a,b\in \Bbb{Z}, a > 0\}$ are both dense since you can always find one element in (0,1) and once that happens, powers will ensure you to converge a sequence of them to 0

DHMO

How would we prove that it is discrete?

@Secret My intuition tells me that they are both dense

Akiva Weinberger

@DHMO Finitely many of them less than any given $N$

DHMO

@AkivaWeinberger thanks, I'm stupid

Akiva Weinberger

15:14

My guess is that $\{a+br:a\in\Bbb Z,b\in\Bbb N_{>0}\}$ will always be dense

for irrational $r$

(Same for switching $a$ and $b$)

DHMO

Yes

Semiclassical

Well, sure.

Secret

It is also interesting to note that for $n > 0$, $x^n$ has 0 as an attractor and a basin of attraction of $(-1,1)$

SBM

attraction ... ?

DHMO

@Secret what is attractor? and are you thinking about my question?

Jack Lam

15:15

@Semiclassical Oh

Akiva Weinberger

If you repeatedly apply $x^n$, anything close to $0$ goes to $0$

Jack Lam

I am very, very, very blind

thanks

Secret

@DHMO I used that property to wrote my proof, in fact, also what Akiva said

Jack Lam

I get it now

Akiva Weinberger

and "close to $0$" means "in $(-1,1)$" in this case I think

Semiclassical

15:16

np. I should stress, though, that this only answers the algebraic aspect.

Jack Lam

indeed

DHMO

@Secret are you thinking about my question?

Semiclassical

I don't understand why the region of integration is so simple.

(I can verify that it is by doing a parametric plot in mathematica, but I don't really understand why it works.)

Jack Lam

No, that went out the window as soon as I saw the form of P+Q

Secret

Ok, now to figure out why $r$ is rational it will be discrete...

Semiclassical

15:17

@DHMO Not really.

Akiva Weinberger

OH! Duh.

Jack Lam

I thought they combined it together into a square

but they just substituted to leave it linear

DHMO

@Semiclassical I pinged the wrong guy

Semiclassical

oh.

Jack Lam

what other aspects do you think are also important?

Akiva Weinberger

15:17

Wait. A little less duh. Hold on

Jack Lam

I know there's that weird geometric proof using hyperbolas

Akiva Weinberger

Oh, OK, I see it

DHMO

@Secret $a+b\dfrac pq = \dfrac{aq+bp}q$

Akiva Weinberger

It follows trivially from $\{a+br:a,b\in\Bbb Z\}$ being dense.

DHMO

@Secret $aq+bp \in \Bbb Z$, $q$ is fixed. $\{\frac nq:n \in \Bbb Z\}$ is discrete. I'm sorry for spitting out the totality of the proof.

Semiclassical

15:18

Well, what I mean is that I can explain why the new integrand is what it is (like I said above)

Jack Lam

yeah no, everything works

I just thought the P+Q was 2/(1-x²y²)

Semiclassical

Ahh.

It's not obvious to me, though, why the limits of integration in $u$ are just $-1$ to $1$.

I mean, I can verify in Mathematica that it works.

Jack Lam

I'll investigate that and see what I can come up with

Akiva Weinberger

We need shorter names. Let's call these sets $S_,$ (the dense one), $S_{,>}$ and $S_{>,}$ (the ones we want to prove are dense), and $S_{>,>}$ (the discrete one)

DHMO

12 mins ago, by DHMO

@Secret Use linear algebra to find the minimal polynomial of $\sqrt 2 + \sqrt 3$

Semiclassical

15:19

Mmkay.

DHMO

@Secret ^

Jack Lam

you, in the meanwhile can focus on DHMO's question

Thorgott

Can I freely change the order of integration of an iterated integral when the limits of both integrals are equal and independent of the variables and when the integrand is continuous in the whole domain?

DHMO

@JackLam the question wasn't addressed to @Semiclassical . i pinged the wrong guy

Semiclassical

@JackLam One thing that I think may help is to treat the limits of integration in $x$ as $(-1,1)$ not $(0,1).$

Jack Lam

15:20

oh ok

DHMO

@Thorgott that's a theorem.

Secret

@DHMO Ah I see. btw I was a bit caught up with the dense thing, I am still think about the minimal polynomial question

Akiva Weinberger

A sequence approaching $0$ in $S_,$ can be turned into a sequence approaching $0$ in $S_{,>}$ by multiplying the right elements by $-1$.

Semiclassical

The limits still transfer in that case, as a Mathematica plot verifies.

Thorgott

@DHMO how is it called

DHMO

15:21

Fubini's theorem

In mathematical analysis Fubini's theorem, introduced by Guido Fubini (1907), is a result that gives conditions under which it is possible to compute a double integral using iterated integrals. One may switch the order of integration if the double integral yields a finite answer when the integrand is replaced by its absolute value. ∫ X ( ∫ Y f ( x , y ) ...

Akiva Weinberger

So it contains $0$ as a limit point, and it's either approached from the right or from the left (or both, most likely). Assume wlog it's approached from the right.

This immediately implies that all positive numbers are limit points, since it's closed under addition.

Thorgott

Thank you very much

Akiva Weinberger

Since this set is closed under subtracting integers, all negative numbers are limit points as well, and the set is dense, QED.

Semiclassical

ffs WolframAlpha stop being stupid.

Faust7

Why do you only need to find a one side limit in C?

SBM

15:23

WolframAlpha what's that?

DHMO

@Faust7 what the hell is C?

@SBM wolframalpha.com

Semiclassical

useful, that's what it is.

Akiva Weinberger

@SBM wolframalpha.com

Faust7

z=x+iy

DHMO

@Faust7 and what are you referring to?

SBM

15:23

Complex numbers maybe

Faust7

the complex plane

Secret

@DHMO Uh, I don't see how I can build a matrix using that expression of numbers and thus constructing a minimal polynomial from it?

Faust7

something my prof said in passing that i didnt understand

DHMO

@Secret of course you can't. You don't even have a basis.

Faust7

functions analytic except at a point you only need to show the limit aproaching one side of the point exists

SBM

15:24

analytic?

DHMO

alright I don't know.

@SBM able to be expressed as a power series

For example, $e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots$

Faust7

in the complex plane everything is analytic ( C^{\infty} $ or its C^{0}

SBM

Oh

Faust7

well i guess or less than C^0 but thats not very intresting

Semiclassical

Secret

15:27

Btw, how does one use rational numbers as a basis, cause in order to approach anything not rational, you will need to add up countably many elements but a basis set is always finite linear combination of elements?

DHMO

@Secret when did i ask you to use rational numbers as a basis?

Is that related to my question?

Semiclassical

@JackLam For reference, this is the parametric plot I have in mind (namely, parametric plot of $(x,u)=(x,y+(x/2)(y^2-1))$ for $(x,y)\in[-1,1]^2$.)

Faust7

@DHMO thats the defintion in the reals "expressable by a taylors eries as your approach that point something like xe^{-x^2} is something that is not analytic on the reals but is a C^{\infty} that doesnt happen in the complex numbers

Secret

@DHMO yes, I am guessing you either want me to use the rationals, or $\Bbb{Z}[\sqrt{q}]$ as a basis. I don't see how one can get a standard basis here

Jack Lam

Somehow, treating x as constant, justifies the entire thing

DHMO

15:29

@Secret My basis is finite.

And my field is $\Bbb Q$

Semiclassical

@JackLam For $0<x<1$, at least.

@JackLam Maybe this plot helps:

This is $u=y+(x/2)(y^2-1)$ plotted as a function of $y\in(-1,1)$ for various values of $x\in(0,1)$.

Jack Lam

it does seem to uniformly cover the entire square (half square?) for all values of x.....

and the jacobian gives us the correct differential measure to integrate with respect to

I sort of understand why the region is like that

Semiclassical

Had to step away for a bit.

I think the point is that each of those curves gives a one-to-one mapping from [-1,1] to itself.

That's easy enough to prove: $\partial^2 u/\partial y^2 = 1>0$, so $u$ is strictly increasing in $y$. on the other hand, $u(1,x)=1$ and $u(-1,x)=-1$ for all $x$.

Jack Lam

15:49

yeah

that makes sense

Ben Millwood

I guess this is really more of a LaTeX question than a maths one, but can anyone remind me how to align these "equations" in a way that doesn't look terrible: math.stackexchange.com/a/2234095/29966

I think I want the alignment I have there but without the weird extra space, but I'm willing to consider other alignments as well

Zophikel

Does anyone have a pdf of the handbook of Integration on hand

The site for the book is here: mathtable.com/hoi

DHMO

@BenMillwood put an ampersand before and after every operator

Jack Lam

I have a pdf

But I think it's highly questionable to be distributing such information

is unaware of the rules

Zophikel

@Jack Lam how come it's highly questionable

Jack Lam

16:06

well

isn't it always?

unless there's some ignorance rule I'm unaware of

DHMO

@Secret are you still here?

1 hour ago, by DHMO

@Secret Use linear algebra to find the minimal polynomial of $\sqrt 2 + \sqrt 3$

is anyone else interested in my challenge?

Jack Lam

Zophikel, what is your email?

I will send the PDF to you

Semiclassical

What does minimal polynomial mean in this context?

Integer polynomial with lowest degree such that it has that as a root?

DHMO

yes

Zophikel

@Jack Lam [email protected] thanks for sending the book owe you one mate :)-

Jack Lam

16:13

it may take a while to send, the file is quite large for a pdf

Semiclassical

No idea about linear algebra, but not really seeing why you'd need to make it more complicated than just the product of the four terms $x\pm \sqrt{2}\pm\sqrt{3}$.

In which case you get a fairly simple quartic.

DHMO

well, it's really a practice of linear algebra

Semiclassical

Hmm.

Jack Lam

using √2 and √3 as basis vectors

right?

DHMO

@JackLam not enough

Semiclassical

16:16

I guess the linear algebra question I'd be more interested in is to find the smallest matrix (in dimension) which has $\sqrt{2}+\sqrt{3}$ as an eigenvalue.

Jack Lam

@Zophikel I wasn't even aware I already had the book, I just did a quick search of my old maths pdfs folder

DHMO

@Semiclassical that's interesting. One can use the charateristic polynomial?

Zophikel

@Jack lam lol

Jack Lam

that file is like

last year

or something

I don't know

it's hard to keep track of what I have anymore

I'm getting old

is actually 18

Zophikel

@Jack Lam math is an old man's game

Secret

16:21

@Semiclassical pick any 2x2 diagonal matrix with one of its entry being $\sqrt{2}+\sqrt{3}$

Ben Millwood

@DHMO looks better, thanks

DHMO

@Secret not interesting.

I'm thinking of constructing a 4x4 matrix with characteristic polynomial $x^4 - 10 x^2 + 1$, the minimal polynomial of $\sqrt2+\sqrt3$

Secret

@DHMO Is $\{1,\sqrt{2},\sqrt{3}\}$ the basis you have in mind for that problem?

DHMO

An algorithm would be given here

@Secret almost

Ben Millwood

general vague memories of Galois theory make me think you also want $\sqrt 6$ in there

DHMO

16:25

correct

Secret

ah yes, you cannot make $\sqrt{6}$ from linear combination of $\sqrt{2}$ and $\sqrt{3}$

DHMO

$\begin{pmatrix}
0 & 0 & 0 & -1 \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 10 \\
0 & 0 & 1 & 0
\end{pmatrix}$

Ben Millwood

you can represent multiplication-by-$\sqrt 2$ (likewise $3$) itself as a matrix wrt that basis, and thus you can represent multiplication-by-$\sqrt 2 + \sqrt 3$ fairly easily

DHMO

$\begin{pmatrix}0&0&0&-1\\1&0&0&0\\0&1&0&10\\0&0&1&0\end{pmatrix}
\begin{pmatrix}-10r+r^3\\r^2-10\\r\\1\end{pmatrix}
=
r\begin{pmatrix}-10r+r^3\\r^2-10\\r\\1\end{pmatrix}$

Semiclassical

Neat.

DHMO

16:31

where $r=\sqrt2+\sqrt3$

@Semiclassical how do you call a field wherein $x^4-10x^2+1=0$?

Semiclassical

oof.

I should know that.

field extension?

DHMO

I guess $\Bbb Q/[X^4-10X^2+1]$?

Semiclassical

$\mathbb{Q}[X]/(X^4-10X^2+1)$, I think.

DHMO

thanks

Daminark

How's it going everybody?

Secret

16:41

Just realise my minimal polynomial memory have pretty much gone down the drain after reading wikipedia, and also minding with my quantum chemsitry calculations

Thorgott

I was trying to evaluate an iterated integral and got a value of $\pi /2$, yet the solution should be $-\pi /2$. I thought my change of order of integration might have been unjustified, but I don't possess any knowledge about measure spaces and such to check whether Fubini's Theorem (which was linked to me earlier) can be applied here. Can anyone correct my mistake?
$\int_0^{\infty}-2b\int_0^{\infty}e^{-b^2(1+y^2)}dydb=\int_0^{\infty}\int_0^{\infty}-2be^{-b^2(1+y^2)}dbdy=\int_0^{\infty}[\frac{e^{-b^2(1+y^2)}}{1+y^2}]_0^{\infty}dy=\int_0^{\infty}\frac{dy}{1+y^2}=\pi /2$

Ben Millwood

@Thorgott: just eyeballing it I think the second-last $=$ is wrong

Meow Mix

@Daminark Just working on Ted-cercises

DHMO

T/F: A strictly increasing function must be injective.

(The definition of strictly increasing function is here)

Ben Millwood

evaluate at $b = \infty$ you get $0$, evaluate at $b = 0$ you get 1, then subtract the latter from the former, and not the other way around :)

Daminark

16:46

@Meow Today I had measure theory, we did Radon measures and Lebesgue measure, next week we're gonna do integration

Ben Millwood

@DHMO sorry, I haven't been here long and have only been paying half attention, do you intend that just anyone answers?

DHMO

yes

Meow Mix

@DHMO T. Suppose it mapped distinct $x,y$ to the same element (it was non-injective). We must have that either $x < y$ or $x > y$. In either case, we have neither $f(x) < f(y)$ nor $f(x) > f(y)$, a contradiction

Ben Millwood

oh, well, $< \implies \not=$, so

DHMO

@MeowMix not really. the msitake is in the second line of your proof (excluding T)

Ben Millwood

16:48

"injective" can be defined as "preserves inequality"

Thorgott

@BenMillwood Thanks, I completely forgot the subtraction

Meow Mix

Why can't we have that $x<y$ or $x>y$??

DHMO

@MeowMix who said that our domain is totally ordered?

Ben Millwood

@DHMO you quoted a definition that referred specifically to real numbers

Meow Mix

^

Daminark

16:50

tfw the domain wasn't actually $\mathbb{R}$

Hey @Mike!

DHMO

> In mathematics, a monotonic function (or monotone function) is a function between ordered sets that preserves or reverses the given order.

between ordered sets

This is the first line of the website I quoted

Ben Millwood

@DHMO but that's not a definition of "strictly increasing", which isn't defined until the section about calculus / analysis

DHMO

ok, you win.

Meow Mix

@Daminark What's lebesgue measure all about? :P

debrief me.

DHMO

@MeowMix extending the usual concept of volume/area/length

Meow Mix

16:51

I know that

I just don't know how it's done...

Daminark

Oh that grittiness of the process is somewhat long

Basically, you need to create an algebra of sets and additive set function which forms a relative outer measure

Meow Mix

So, all open sets have non-zero lebesgue measure, right?

Ben Millwood

@DHMO I guess the cardinality of a set is an example of a function that is "strictly increasing" (in a certain sense) but non-injective

Daminark

Which you can, using pain that our prof didn't feel like going into, show is satisfied by taking finite unions of intervals

DHMO

@BenMillwood in which sense?

Daminark

16:53

(Or boxes in $\mathbb{R}^n$)

@Meow Aside from the empty set, yeah

Ben Millwood

@DHMO: in the sense of "$x < y \implies f(x) < f(y)$

"

Daminark

But basically, once you take your finite unions of boxes and define the measure the way you expect, you use something called the Caratheodory extension theorem

Meow Mix

@Daminark Oh, pfft

DHMO

@BenMillwood in which sense of $<$?

Ben Millwood

@DHMO $\le$ but not equal

DHMO

16:54

in which sense of $\le$?

Ben Millwood

where $\le$ on sets is $\subseteq$ and on cardinalities it's... y'know

DHMO

alright

Daminark

Which extends what we defined to a much larger class of sets

DHMO

then cardinality isn't strictly increasing

Ben Millwood

@DHMO restrict to finite sets

DHMO

16:55

oh, alright

Ben Millwood

for that you deserve: T/F: every injective function has a left inverse

DHMO

seen that before

Ben Millwood

damn it :P

Transcript for

$Mathematics$ Mathematics