Mathematics - 2018-01-20 (page 1 of 2)

So we can embed $N$ in $\mathbb{R}^3$ which is nice, and therefore we have an embedding of $S$ and this embedding of $S$ in $\mathbb{R}^3$ will bound a ball by Alexander's Theorem. If that ball is in $N$, then we are done. So if it doesn't... then now we are up to the task of building some homotopy

Secret

Ooooo, you guys are talking about spherinders...

Leaky Nun

1:01 AM

hi @Secret

Secret

hi

I wonder what happens if we do $S\times (1-,1)$ instead

That is still an uncapped spherinder, but now the boundary is open in the (forgot) topology

Leaky Nun

the boundary is the same

Forever Mozart

sigh. I'm finally tired of teaching these entitled brats. Time for a new job.

Jacksoja

@ForeverMozart lets talk it out, am all ears

Prototank

@ForeverMozart what kind of entitled brats?

Leaky Nun

1:13 AM

@ForeverMozart I tagged you just to continue the chain

Secret

Actually going back to cylinders, what is the difference in topology between S1 X [-1,1] vs S1 X (-1,1)?

Forever Mozart

freshman calculus students

so much whining and impatience

Leaky Nun

@Secret one is closed, the other open

Jacksoja

@ForeverMozart well do you remember when you were a freshman ?

and how your teacher felt the same thing as you feel now

this is part of the job, teaching and learning are the beauty of being human, also the most annoying but that is a mistery

If am not making sense ignore me because am high

Secret

@LeakyNun Then the case for spherinders should be similar, since they can be built from a stack of cylinders along the 4th dimension?

Leaky Nun

1:18 AM

you don't need the fourth dimension

Prototank

All of these things fit in $\mathbb{R}^3$

Leaky Nun

$S^2 \times [-1,1]$ perfectly fits in $\Bbb R^3$

Prototank

woops sorry didn't mean to talk over you

Leaky Nun

@Prototank nah, you should have said "SNIPEDDDD"

as is the tradition of this room

Prototank

I think there may be multiple applications of Alexander's theorem

since the boundary of $N$ consists of two more spheres

but I am very new to these arguments and I'm not taking a class or anything

Secret

1:20 AM

hmm... so $S \times [-1,1]$ is a spherical shell with some thickness?

yep

yes

SNIPED

:P

user was banned for this post

Secret

1:21 AM

Hmm, so both the spherinder boundary and the spherical shell are defined by the same Cartesian product... I need to check...

Spherinder

In four-dimensional geometry, the spherinder, or spherical cylinder or spherical prism, is a geometric object, defined as the Cartesian product of a 3-ball (or solid 2-sphere), radius r1 and a line segment of radius r2: D = { ( x , y , z , w ) | x 2 + y 2 + z 2 ...

Leaky Nun

oh I didn't know what a spherinder was

it's a different thing then

it even has 4 dimensions

right, spherinder = $B^3 \times [-1,1]$

Prototank

that looks right

Leaky Nun

its boundary (product rule) = $S^2 \times [-1,1] \cup B^3 \times \{-1,1\}$

with some gluing between the two objects

specifically, gluing $S^2 \times \{-1,1\}$ with $S^2 \times \{-1,1\}$

Prototank

makes sense

Leaky Nun

it looks like $S^3$ to me

Prototank

1:27 AM

the boundary?

Leaky Nun

yes

Secret

Meanwhile, I have not found the set notation for spherical shell yet, but Prototank is talking about this shape:

Spherical shell

In geometry, a spherical shell is a generalization of an annulus to three dimensions. It is the region between two concentric spheres of differing radii. == Volume == The volume of a spherical shell is the difference between the enclosed volume of the outer sphere and the enclosed volume of the inner sphere: V = 4 3 π R 3 − 4 3 π r ...

Prototank

I think you are right

Leaky Nun

yes, we were talking about this object

dem

Prototank

one ball glued on the inside

the other glued on the outside

BAM $S^3$

Leaky Nun

1:29 AM

exactly

Prototank

sweet

Leaky Nun

very amazing

Prototank

So yeah, the theorem I'm trying to prove (or lemma, or whatever) is that if a spherical shell $N$ has a sphere inside of it, then that sphere either 1) bounds a ball inside of $N$ OR 2) the sphere can be wiggled around into being one of the boundary components of $N$.

Leaky Nun

let's just say $N = \{ \vec v \in \Bbb R^3 ~:~ 1 \le \|v\| \le 2 \}$

Prototank

fair enough

Leaky Nun

1:32 AM

and what do you mean by a sphere?

Prototank

an embedding (category $C^\infty$) $S^2\hookrightarrow N$

Leaky Nun

ok...

I imagine things can get quite complicated if it isn't a real sphere

Prototank

we can control a lot of that with the categorical restriction

but yeah could still be pretty bizarre

Leaky Nun

I mean, there are a lot of endomorphisms of $S^2$ that aren't null-homotopic

Prototank

you could imagine the embedding elongating the sphere into a tube with caps and then tying that up

every endomorphism of $S^2$ is homotopic to the identity or the antipodal map

Leaky Nun

1:35 AM

I think you meant automorphism

because afterall $\pi_2(S^2) = \Bbb Z$

Prototank

that ain't right

Leaky Nun

sorry, fixed the subscript

Prototank

Alexander's theorem does a lot of the leg work here, I think

Jacksoja

whats a field extention ?

Prototank

A field extension is when you have a pair of fields $K\subset F$ so that they share zero and one elements

for example, $\mathbb{R}\subset\mathbb{C}$

Leaky Nun

1:48 AM

We don't know if $\Bbb Q \subset \Bbb Q(\pi+e)$ is a field extension

Is... is that true?

nobody knows

my goodness

We don't even know if $\pi +e$ is irrational

orbit-stabilizer

2:05 AM

I don't even know if i am irrational

Secret

::thinking of a group made of irrationals such that some of these are orbit stabilisers...::

3

A: Irrationals: A Group?

To speak to the spirit of your question a bit: the rational numbers are a so-called normal subgroup of the reals (since the reals form an abelian group, and all subgroups of an abelian group are normal), so we can talk about the quotient group of the reals by the rationals, $\mathbb{R}\ /\ \mathb...

orbit-stabilizer

2:25 AM

hah

DarkRunner

3:40 AM

Hi all;

I'm solving AMC 10A 2016 P#10, and I had a question

I solved the question and got 3; I did by doing the following:

First, I represented each of the areas via 2l+2, 2l+10, and 2l+18 (from least to greatest area)

Then, we know the areas form an arithmetic series, and the average of an arithmetic series is it's last term,

So, I did (6l+30)/2=(2l+18)

This gave me an answer of 3; however, that is not one of the answer choices

If anyone could help that would be great

Faust

its 4

DarkRunner

what did I do wrong? sorry for the naive question

Faust

te

let me read what u wrote.

DarkRunner

ok

Faust

well the total area should be 5(4+l)

DarkRunner

3:52 AM

? Let me check

Faust

the area of the smallest rectangle should be 5l

?

less i am too brain fried to think correctly

DarkRunner

No, it's 2l+2w, which is 2l+2

I'm pretty sure

Faust

whats l

and whats w

DarkRunner

l for length, and w for width; Those are variables I use

Faust

well the total height of the rectangle is 5 and the total length is 4+l where l is the inside rectangle

so there product should yield the total area

then make a progression that makes sense no?

DarkRunner

3:56 AM

That is true; I should consider this

Thanks

I got it!

Faust

i dont think you have any idea what the total length is

but you do know the height

kk

DarkRunner

thanks for helping

Faust

np

Meow Mix

4:20 AM

hi

Faust

4:36 AM

meaow\

Meow Mix

does anyone want to check my rigorous-ish proof of an integral

(no fundamental theorem of calc)

WDUK

is there ever a situation in maths where imaginary numbers is a valid x-intercept or do all mathematics consider that to mean there are no x-intercepts if the solution is imaginary

Meow Mix

if thy dot mind

@WDUK not in the sense of "x-intercept"

but more generally as "roots"

WDUK

i see

Meow Mix

any $n$-th degree polynomial is guaranteed $n$ complex roots (real numbers are complex too!)

by the fundamental theorem of algebra

so if you look at something like $x^6 - 1 = 0$

Adeek

4:40 AM

@MatheinBoulomenos here ?

WDUK

is there a difference between roots and intercepts here?

Meow Mix

you instantly thing $-1$ and $1$

but there are 4 more roots

in fact $x^n - 1 = 0$ have roots that form a circle in the complex plane

that's called the roots of unity

well intercept is usually taken to mean intersection with an axis

Adeek

@Narcissusjewel maybe you can discuss this with me ?

Meow Mix

since you can't plot complex numbers on the real plane it doesnt make sense to consider those roots as "intercepts"

WDUK

ah yeah i briefly learnt roots of unity - kinda sucked at it, need to re-revise it at some point

ive seen imaginary numbers on a plane for number lines

guess thats bit different really

Meow Mix

4:42 AM

that's because it was on the complex plane

if you graphed a real function on the complex plane it would just go back and forth on the real-axis

WDUK

ok so if it doesn't specify its on a complex plane, i should assume intercepts are only valid on real numbers

Meow Mix

as far as i know, yes, complex roots aren't considered $x$-intercepts

WDUK

ok thanks for clarifying :)

Meow Mix

no problem

Meow Mix

5:04 AM

lol was looking at spivaks online and teds book appeared

Abcd

7:26 AM

Is it okay to write $\dfrac{dQ}{dT}= \dfrac{dU}{dT}+\dfrac{dW}{dT}$ in physics?

because $dQ= dU+ dW$

BeginningMath

hello, i am trying to fix a question i've asked and i am wondering: how would you correctly say in english that a linear mapping(transformation) follow some rules?

abide maybe?

blat

8:22 AM

hi

what topological space is referred to as $T^n$ ?

Leaky Nun

@blat torus

blat

i thought so, but it doesn't make sense in my script

Leaky Nun

could you show us?

@BeginningMath satisfy

anon

@BeginningMath "satisfies," "such that," maybe some other phrases depending on what it is you're trying to say

blat

my bad, it's indeed a torus. thanks :)

Silent

11:52 AM

Can a function have exactly one right inverse and no left inverse?

anon

12:04 PM

exists right inverse iff map onto
exists left inverse iff map one-to-one

if there is no left inverse, the map is not one to one, so there exist two things in the domain that map to the same element in the range, which implies there are at least two right inverses (i.e. sections)

Silent

Thanks!

Do we have to use Axiom of Choice in "onto implies right inverse"? @anon

Leaky Nun

@Silent yes

anon

existence of right inverse

yes

Silent

and we get "one to one implies left inverse" without using AoC, right?

skullpatrol

@LeakyNun you should change your username to the "Leaky Sniper." :P

Leaky Nun

12:10 PM

yes

anon

yes

Silent

thank you!

skullpatrol

got'em again

:-)

anon

completely unrelated, but as room owner when moving messages one gets literal crosshairs to select which messages to remove from the room

Leaky Nun

lol

skullpatrol

12:12 PM

The man with the golden gun?

Alessandro Codenotti

@Silent it's equivalent to AC over ZF actually

Silent

oh!

Vrouvrou

12:35 PM

hello, i have $\ker g \subset \ker h$ why this means that there exists $\lambda$ such that $h=\lambda g$ ?

user21820

@AlessandroCodenotti @Silent: That's not correct. If S is non-empty and f injects S into T then the inverse relation to f is simply { (x,y) : x∈T ∧ ∃y ( f(y)=x ) }, which does not at all require AC. This inverse relation is already a left inverse of f.

Silent

@user21820 He is talking about onto, right invrse.

user21820

@Silent: Oh okay. @AlessandroCodenotti: Sorry I thought you were replying to Silent's last message.

Leaky Nun

@user21820 follow arrows :P

@Vrouvrou context?

Vrouvrou

X is a banach space and $h,g\in X'$ @LeakyNun

user21820

12:41 PM

@Silent: By the way, you accepted this answer but it is talking nonsense, because the author does not actually understand Godel's theorem. See also here.

Silent

@user21820 I am so sorry, I have unaccepted it. I fell in love with foundations and started reading somethings about it, but since i study on my own, there was a big stock of unanswered but (to me) very important questions, that haunted me every time i tried to move forward. I was mainly using Tourlakis' 2 volumes "Lectures in Logic and Set Theory", but soon it became unbearable.

Studentmath

@Balarka think I have the proof. Requires playing with Hall's marriage theorem

user21820

@Silent You are welcome to ask whatever you want about logic in the Logic room, where both I and LeakyNun will probably be there to discuss with you.

Silent

I am not free till June, but if I want to continue with Mathematical logic, which book is good to start?@user21820

@user21820 Wow!

user21820

@Silent No hurry. I'll give you some suggestions over in the other room now.

Silent

12:51 PM

Thank you! :)

skullpatrol

or you could just talk here :-)

Alessandro Codenotti

1:12 PM

Speaking of AC I asked a question that I discussed yesterday in chat about how much AC is needed for functional analysis, if someone knows an answer the question is here

Balarka Sen

1:47 PM

@Studentmath I don't know what that is.

Jacksoja

Hi chat

Semiclassical

This looks relevant: gilkalai.wordpress.com/2012/11/25/happy-birthday-ron-aharoni

jublikon

I have to prove that f is continuous in $x_0 = 0$
$$f(x)=\frac{2x^3+x^2+x\cdot sin(x)}{(e^x-1)^2}$$

Well, if I check the limit for f it is like $\frac{0}{0} $(I have tried L'hopital but failed doing that).
Is it a good way to check if $\lim_{x \to x_0^+} = \lim_{x \to x_0^+} = L$ or is it best to use a different method here?

Semiclassical

namely because it includes a statement of "Hall’s marriage theorem for hypergraphs"

Balarka Sen

I'll have to read it later

I don't know any serious combinatorics

Semiclassical

1:52 PM

i have no idea about the marriage theorem myself either

if it's not easy to do with generating functions, I usually don't know it :P

Abhas Kumar Sinha

hii

anyone of here knows lens formula? (in physics)

1/v - 1/u = 1/f

anyone here?

i'm in trouble with that

Semiclassical

Ask, don’t ask to ask.

Abhas Kumar Sinha

if 1/v - 1/u = 1/f so 1/u - 1/v should be equal to 1/f?

but that doesn't happens to be

Balarka Sen

Interchanging the position of the object and the image has the effect upon the lens that it changes from being convex to concave (or the other way around)

Semiclassical

Then either it doesn’t hold or you have the wrong values of u,v,f

Balarka Sen

2:00 PM

So, no, 1/u - 1/v would be 1/(-f)

Which is consistent with the formula

Abhas Kumar Sinha

okay, let 1/v - 1/u = 1/f, so squaring both sides = (1/v -1/u)^2 = (1/f)^2, which is equal to (1/u - 1/v)^2 = (1/f)^2 cancelling 2 exponent both sides then its 1/u - 1/v = 1/f

Semiclassical

One point re: the formula is that you never need to use it. You can always make the computations by appealing to ray-tracing diagrams and similarlity of right triangles

Abhas Kumar Sinha

as (a - b)^2 = (b - a)^2

but that doesn't happens

Semiclassical

(-2)^2 = (2)^2 hardly implies 2=-2...

Abhas Kumar Sinha

I've only studied about this formula, other formula is not in my book @Semiclassical

yep! But, its true as my teacher said that square roots can't be negative

for iit preparation

square roots are always positive

Balarka Sen

2:04 PM

x^2 = y^2 does not imply x = y

Semiclassical

I doubt that. Ray-tracing diagrams are what you use to obtain the formula

Balarka Sen

That is 8th grade algebra error

Abhas Kumar Sinha

its given in my book, what's error?

Semiclassical

x^2=y^2 implies |x|=|y|

Balarka Sen

Burn your book.

2

Abhas Kumar Sinha

2:06 PM

(a - b)^2 = a^2 + b^2 + 2.a.b which is equal to (b - a)^2

@Semiclassical how?

(b - a)^2 also is equal to the previous one as a^2 + b^2 - 2.a.b

woops forgot - in 1st equation

Semiclassical

If x^2=(-1)^2, what can you conclude about x?

Abhas Kumar Sinha

minus symbol

skullpatrol

@BalarkaSen Fahrenheit 451?

Abhas Kumar Sinha

1 or -1 @Semiclassical

both

Semiclassical

Right. x=1 or x=-1

Abhas Kumar Sinha

2:09 PM

@Semiclassical so, you mean even exponents can't be cancelled?

Semiclassical

Which is equivalent to |x|=1=|-1|

Abhas Kumar Sinha

what is ||?

Semiclassical

Absolute value

Abhas Kumar Sinha

okay, omission of sign, you mean mods?

how you can use mods in practical gemoetery?

William Oliver

Can anyone give me an intuitive explanation of the Ultra Limit? I am failing to see how it relates to the standard topological/metric definition

Abhas Kumar Sinha

2:13 PM

@Semiclassical is there any proof for Mods? I've not studied much about that.

anyone here knows how can I do IIT without coaching (specially mathematics)?

@Semiclassical I got that document to get sine and cosine values though calculation. That's a great idea, I've never thought of doing that, its great!, taking a triangle in 36, 36, 72 deg ratio and dividing them by bisecting... its a great idea!

Alessandro Codenotti

2:52 PM

Sanity check: Let $(X,||\cdot||)$ be a Banach space, then any $A\subset X$ which is compact or weakly compact is bounded and has empty interior (but for different reasons in the weakly compact case), right?

Daminark

3:16 PM

@Alessandro in the case of strong topology, then yeah, because if it had non-empty interior, it would contain a closed ball

That's a closed subset of a compact set, which would make it compact. But it isn't. Rip

In the case of the weak topology

I think that's just false

A Banach space is reflexive iff its closed unit ball is weakly compact (and weakly sequentially compact)

(That's a hard theorem in general but like, take the dual space, its closed unit ball is weak* compact by Banach-Alaoglu, so if that space is reflexive, it means weak* compact = weakly compact)

Alessandro Codenotti

@Daminark but the closed unit ball has empty interior in the weak topology

weakly open sets are unbounded

Daminark

... I'm an idiot

Okay maybe that's right, let's see

Alessandro Codenotti

You're not, I realized this morning that weakly open sets are unbounded even though I should have noticed months ago so...

Daminark

Yeah you're just right

Alessandro Codenotti

Let's work with nbhds of $0$ because you can just traslate stuff otherwise. $0$ has a nbhd basis of the for $\bigcap\limits_{i=1}^k f_i^{-1}(-\varepsilon,\varepsilon)$ as $k,f,\varepsilon$ vary in $\Bbb N$, $X^\star$ and $\Bbb R$ respectively

Daminark

3:23 PM

So in particular you're gonna contain the intersection of the kernels of those $f_i$

And each one has codimension 1, so in total if your space is infinite dimensional then yeah intersection of finitely many kernels is a subspace, so in particular it's unbounded

Alessandro Codenotti

Indeed

Daminark

Okay that's dank

Alessandro Codenotti

So not only open nbhds of $0$ are unbounded, but they actually contain closed, infinite dimensional subspaces

Alessandro Codenotti

3:39 PM

(This works because weakly compact set are bounded wrt the norm on $X$, which is easy to prove)

Vrouvrou

4:04 PM

X is a banach space and $h,g\in X'$ i have $\ker g \subset \ker h$ why this means that there exists $\lambda$ such that $h=\lambda g$ ?