Mathematics - 2016-03-31 (page 1 of 2)

Akiva Weinberger

00:20

@GPhys Well, it is, but it doesn't look like it

GPhys

@AkivaWeinberger Well, no. It's true for all classically defined sets.

For example, there doesn't exist a set that contains only the standard natural numbers

Akiva Weinberger

Yeah, but the axiom of specification only talked about classically defined properties in the first place. @GPhys

Technically, it was an axiom schema (a collection of axioms, one for each property). So, IST just uses the exact same collection of axioms. @GPhys

GPhys

@AkivaWeinberger Right, or it would be a contradiction

but my point is this fact is subtle enough to muck up somebody unfamiliar with having to think about whether or not a subset even exists

Forever Mozart

@BalarkaSen

Suppose you have a bounded space in $\mathbb R ^3$, above the $x$-axis.

Akiva Weinberger

@GPhys Hence the "it doesn't look like it" bit

GPhys

00:32

@AkivaWeinberger Sure, then

Forever Mozart

you take a plane somewhere above the space, and project perpendicular from that plane, onto the $x,y-plane$

is that uniformly continuous?

so I am projecting a space in $\mathbb R ^3$ onto a plane

@MikeMiller

@PVAL

PVAL

check from the definition.

Forever Mozart

ok so I have to figure out the equation

it will depend on the plane I project from

but intuitively you think it is?

Akiva Weinberger

I'm not sure I get what you're doing, but I think you're doing a continuous map from a compact domain so yes

Or, at least, it can be extended to a compact domain

Forever Mozart

kindof

the original space might not be compact

but it is bounded

Akiva Weinberger

00:39

So it's a subset of a compact thing

Forever Mozart

the projection should be extendable to its closure

so really it is a restriction of a map on a compact space

which should be uniformly continuous, right?

Akiva Weinberger

Isn't it extendable to all of $\Bbb R^3$? (Not compact, but still)

Forever Mozart

yes

but I think the restriction of a uniformly continuous map is uniformly continuous

Akiva Weinberger

So than it's extendable to its closure, its convex hull, its whatever.

@ForeverMozart Yeah, pretty sure it is

Forever Mozart

ok thanks

PVAL

00:42

@Akiva No

@Forever Check it from the definition.

Forever Mozart

but the definition could be strange

of the projection

because the plane could be weird

PVAL

planes aren't weird

Forever Mozart

so what is the general definition of a plane in $\mathbb R ^3$?

PVAL

That seems like something you'd want to know if you were going to ask/answer this question.

solution set to some equation ax+by+cz=d with one of a,b,c nonzero.

Forever Mozart

but tell me why this doesn't work...

If $A\subseteq\mathbb R ^3$ is bounded and above, then a projection from a plane is extendable to its closure, which is compact. Therefore the extended projection is uniformly continuous. Then the restriction to our original space is also uniformly continuous...

Akiva Weinberger

00:48

How do you project a 3D object from a plane to a plane?

PVAL

$f(x)=x^2$ extends to the closure on any bounded set and extends over $\Bbb R$.

It is not uniformly continuous

Akiva Weinberger

@PVAL But it's uniformly continuous over every bounded set

Forever Mozart

yes

Akiva Weinberger

We're defining this as a function $f:{\rm bounded~thing}\to\rm plane$

not $\Bbb R^3\to\rm plane$

so the former can be uniformly continuous while the latter isn't

Right?

Forever Mozart

@AkivaWeinberger furthermore, if the projection of the space is one-to-one, it should be a homeomorphism

right?

PVAL

00:59

Well the bounded hypothesis was not linked to the message I saw.

Such a function is unif. cont. on all of $\Bbb R^3$, and you should be able to prove that.

Akiva Weinberger

Wait, if you're mapping $\Bbb R^3$ to a plane, it can't be one-to-one and continuous.

Forever Mozart

but a subset of $\mathbb R ^3$.

PVAL

Hopefully both of you can at least write it out in coordinates.

Forever Mozart

not the entire thing

Akiva Weinberger

Still no, I guess — consider $\{(\cos\theta,\sin\theta,\theta):0\le\theta<2\pi\}$ onto the $xy$-plane

Forever Mozart

01:02

a projection from some subsets of $\mathbb R ^3$ could be one-to-one

Akiva Weinberger

It's one-to-one, but it's homeomorphic to $[0,2\pi)$ and it maps to a circle.

Forever Mozart

huh

I cant see the picture

Akiva Weinberger

It's a piece of a spiral

r9m

@RandomVariable I was thinking of the $\cosh$ sum and product formulas while simplifying the first expression .. so thought I'd do the same for the other one too :) (hence that $\pm$ in the write up .. ) :-)

Akiva Weinberger

I'm projecting a bit of a spiral down so that it becomes a circle @ForeverMozart

Forever Mozart

01:06

but the projection is not one-to-one

Akiva Weinberger

Yeah it is. It's just a piece of the spiral

Forever Mozart

oh

ok so that is something to consider

luckily my bounded space should not have that problem

If I solve the problem I will mention you in my paper :)

r9m

@RandomVariable what do you think of the $4$-th power of $\arccos$ one? :-) (I mean if it looks to ugly .. )

Random Variable

@r9m Is there a similar one on MSE?

Akiva Weinberger

01:35

@GPhys It is consistent with ZF that $\cal P(\Bbb R)$ has no total order.

ZFC, however, proves that not only does every set have a total order, every set has a well order.

Does IST prove the statement (weaker than choice) that every set has a total order?

I'm not sure

You know what, I'm 99% sure that IST proves this.

(IST without choice, that is.)

Yeah, say we want to order $X$.

Let $E$ be the finite, nonstandard set that contains every standard element of $X$. (This exists, right?)

Let $\le$ be the total ordering on $E$ (which exists 'cause it's finite).

Then $^S\!{\le}={}^S\!\{x\in X\times X\mid x\in{\le}\}$ should be a total order on $X$.

That is, for standard elements $x$ and $y$ of $X$, $x$ precedes $y$ in this total order iff $x$ precedes $y$ in $E$'s total order.

r9m

@RandomVariable yes there is :-) the post might be a complete spoiler (I couldn't fetch the apple too far from from tree) .. you want it?

Akiva Weinberger

And use standardization to define it for nonstandard $x$ and/or $y$.

Random Variable

@r9m Is it with $\arccos(x^2)$ raised to the second power?

r9m

01:51

@RandomVariable it was with $\arcsin^2 x^2$ .. this one here

Akiva Weinberger

@GPhys Hm… apparently, Tychonoff's theorem is equivalent to choice. Maybe try seeing if IST can prove it?

(Product of compact spaces is compact under product topology)

r9m

@AkivaWeinberger Tychnoff , compactness thm and choice are equivalent :-) (as far as my understanding goes)

Random Variable

@r9m Why is your answer not the accepted one?

Akiva Weinberger

@r9m Compactness theorem in logic? I thought that's only equivalent to Tychonoff for Hausdorff spaces (which is weaker than choice). I might be wrong, though.

r9m

@RandomVariable I have no idea ,.. OP must have seen something interesting in Jack's expression (I got that too at first but couldn't make headway from there .. )

GPhys

01:57

@AkivaWeinberger I was eating, I'll look at this in a sec

r9m

@AkivaWeinberger you are right ..

Akiva Weinberger

@GPhys Honestly, my guess is that, without choice, IST can only prove the Boolean prime ideal theorem, which is weaker than choice (and equivalent to Tychonoff for Hausdorff spaces and the fact that every set has a total order).

Actually, not sure if it's equivalent to every set having a total order. Ignore that bit.

Yeah, no, internet says that the ordering principle (every set has a total order) is strictly weaker than those.

GPhys

@AkivaWeinberger for any set $E$, there exists a finite $A\subseteq E$ such that the standard elements of $A$ are the standard elements of $E$. It's not necessarily nonstandard.

This is a consequence of Idealization

Akiva Weinberger

@GPhys I think you meant $\supseteq$ there

GPhys

@AkivaWeinberger I did not

Akiva Weinberger

02:04

Right, yeah. Sorry

You're right

GPhys

For example, take the set $\mathbb{N}$, there exists a finite subset of $\mathbb{N}$ that contains all the standard elements of $\mathbb{N}$

In this case it's easy to explicitly construct; $[1,n]$ where $n$ is unlimited

but such a set exists for any set

Akiva Weinberger

Right. So the $E$ in my proof exists?

GPhys

and, moreover, it's not necessarily nonstandard

but yeah, such a thing exists

Akiva Weinberger

I should have said "possibly nonstandard"

Well, it would have to be nonstandard if $A$ is infinite, no?

GPhys

@AkivaWeinberger Can you elaborate what you are doing here

@AkivaWeinberger Yes, this is true

Akiva Weinberger

02:06

In essence I'm saying that I can totally order any finite subset of $X$

GPhys

I ask because, standardization only applies to a property of a standard set

although this property doesn't need to be classical

Akiva Weinberger

So there's a (possibly nonstandard) total order on all of $X$

@GPhys $X\times X$ is standard here so it's OK

GPhys

Oh I see what you're doing now

make sure to specify you are proving for a standard $X$ at the start (although a lot of authors get lazy about this)

Akiva Weinberger

Right, yeah

and then use transfer at the very end to make it work for all sets

GPhys

usually you only even cared about a standard $X$, so it gets left off even if it's vital to the proof

Akiva Weinberger

02:10

So, my guess is that IST can't prove full choice, but it can prove BPI (linked above), which is slightly stronger than what we just proved.

en.m.wikipedia.org/wiki/Boolean_prime_ideal_theorem

GPhys

@AkivaWeinberger You need an extra step where you apply transfer if you're being extra explicit, but yeah it looks fine

since you only have a total order for the standard guys

Random Variable

@r9m The other day I was messing around and discovered that $$\int_{0}^{\infty} \sin \left(\frac{1}{a^{2}+x^{2}} \right) \, dx = \frac{\pi}{2a} \left[\cos \left(\frac{1}{2a^{2}} \right) J_{0} \left(\frac{1}{2a^{2}} \right) + \sin \left(\frac{1}{2a^{2}} \right) J_{1} \left(\frac{1}{2a^{2}} \right) \right] \ , \ a>0 $$ But I used two hypergeometric function identities.

GPhys

@AkivaWeinberger This is my statement of Idealization: i.imgur.com/J3jtkWE.png

r9m

@RandomVariable interesting .. (I am not very familiar with hypergeometric identities .. )

Akiva Weinberger

@GPhys Pretty sure you don't need $E$ there

GPhys

02:18

@AkivaWeinberger almost nobody puts it

Akiva Weinberger

My version is:

GPhys

I don't list it in the other definition, but I list it there to be explicit for how I use it in that section of the intro

To use it to prove the finite part lemma, let $E$ be any set and $R(A,y)$ the relation $A\subseteq E$, $A$ is finite and $y\in A$

Akiva Weinberger

$$\forall^{\rm stfin}x'\exists y\forall x\in x'~A\iff\exists y\forall^{\rm st}x~A$$

where $A$ is an internal formula

Putting $E$ in there is unnecessary, and it's not obvious that it's equivalent

(Your $\sf s$ is my $\rm st$)

GPhys

@AkivaWeinberger it's not, it's more restricted

Akiva Weinberger

So then what you have is weaker than IST.

Mike Miller

02:22

@EricStucky: I see you in the email about people flying in. Funny.

GPhys

nah, I cite the broader axiom as the basis @AkivaWeinberger, but I'll probably change it more anyway

(this version is easy to prove from the broader one)

I need sleep, but

Akiva Weinberger

By the way, @GPhys, there's a way to interpret IST where there are no infinitesimal objects at all. We just introduce two new symbols, $\forall^{\rm st}$ and $\exists^{\rm st}$, which have a purely syntactical, rather than semantical, use. But that's not the best way to look at it, I don't think

GPhys

finish this proof first

Mike Miller

@AkivaWeinberger: Set theoretic stuff in general is just not to my taste, but all the power to people who like it.

Akiva Weinberger

I meant to take GPhys, sorry

Mike Miller

02:25

We have someone here in a couple weeks talking about finite set theory, which is to say ZFC where instead of the axiom of infinity you take its negation. That'll be fun.

No worries.

GPhys

@AkivaWeinberger Anyway, we need for every standard and finite subset $F$ of $E$ to produce an $A$ such that for every $y\in F$, $A$ is a finite part of $E$ containing $y$

we just choose $A=F$

so by idealization there exists a finite part $A$ of $E$ containing all standard $y\in E$

and that's where that comes from

Forever Mozart

no infinite set?

that would kill me

of course there is an infinite set

duh

negation of axiom of infinite would be "ever set is finite"

Akiva Weinberger

Getting rid of the axiom of infinity is basically the same as working in Peano Arithmetic, if I recall correctly. Like, any statement in one can be translated into a statement in the other, and they can prove the same things

Forever Mozart

that makes sense

Akiva Weinberger

@ForeverMozart Sorry, you're right, I misread.

My bad!

Wagh.

Mike Miller

02:33

Yeah, but we're doing more than getting rid of it, right?

Akiva Weinberger

Negating it

Mike Miller

Right, but I mean, Peano doesn't include a negation, does it?

Akiva Weinberger

? Pretty sure it does

It has all logical symbols, no?

GPhys

@AkivaWeinberger I hope that makes sense, but that's an extremely common lemma of idealization to use

Akiva Weinberger

Oh.

Mike Miller

02:35

Hm, that doesn't parse.

Akiva Weinberger

Peano doesn't have infinite sets because it doesn't have sets.

Mike Miller

Right. But this finite set theory still has sets, and probably still has nonstandard sets.

As in, sets that aren't in bijection with {0,...,n}.

I hope the talk isn't just about Peano arithmetic or it would be quite dull :)

Akiva Weinberger

Yeah… every logical theory has models of every infinite cardinality, so there are nonstandard models with weird sets. Or something.

GPhys

@AkivaWeinberger Using idealization with $R(x,y)$ given by $x\neq y$ and an infinite set

Akiva Weinberger

@GPhys Looks right

GPhys

02:37

you get an $x$ such that $x\neq y$ for every standard $x$, that is, every infinite set contains a nonstandard element

Akiva Weinberger

In any case, I might be wrong, but I think the map between hereditarily finite sets and integers goes like this:

Mike Miller

hmmm

16

A: What are the consequences if Axiom of Infinity is negated?

ZFCfin, ZFC with infinity replaced by its negation is biinterpretable with PA, Peano Arithmetic. This result goes back to Ackermann, W. Ackermann. Die Widerspruchsfreiheit der al lgemeinen Mengenlehre, Math. Ann. 114 (1937), 305–315. You may want to read about the early history of this theo...

GPhys

as a contrapositive, if every element of a set is standard, that set is finite

Akiva Weinberger

$\{a,b,c,\dots\}\to2^{\bar a}+2^{\bar b}+2^{\bar c}+\dotsb$

where $\bar\cdot$ is the map

GPhys

In fact the following statements are equivalent: 1) every element of the set $E$ is standard 2) $E$ is standard and finite

you also need transfer to prove it this strongly, though

Akiva Weinberger

02:39

@GPhys Yup

'Cause the first thing is $\forall x\in E\exists^{\sf s}y,x=y$

GPhys

ehhh you're right @AkivaWeinberger, I should really rewrite that axiom

Akiva Weinberger

Which becomes $\lnot\forall x\in E\exists^{\sf s}y,x\ne y$

which becomes… etc

@AkivaWeinberger I forgot to mention that $\emptyset$ maps to $0$, or else the inductive definition has no base

GPhys

@AkivaWeinberger @ infinitesimals -> I just define infinitesimals as a particular type of element furnished for reals by idealization

I have: $x\in\mathbb{R}$ is infinitesimal $\iff$ $\forall^\mathsf{s}y>0,\lvert x\rvert <y$.

Akiva Weinberger

Right, of course

GPhys

I like this definition because it gives a good platform to teach an easy application of the transfer axiom

Fix $x$ to be standard, then you can apply transfer to $\forall^\mathsf{s}y>0,\lvert x\rvert <y$

furnishing $\forall y>0,\lvert x\rvert <y$

Akiva Weinberger

02:46

Right. No standard number is infinitesimal except $0$

GPhys

giving, classically, that $x=0$. In other words, $0$ is the only standard infinitesimal

which is I think a good first example of applying transfer

pedagogically, at least

Akiva Weinberger

What I'd like is a nonstandard proof that for any differentiable complex function $f$ on a domain $D$ and a closed contour $\gamma$ that doesn't wind around any point outside of $D$, then $\int_\gamma f=0$.

(Cauchy's theorem)

I mean, it's probably still pretty hard to prove, and I doubt you can find a fundamentally different proof than the standard ones

But it might be interesting, I dunno

GPhys

@AkivaWeinberger $\mathbb{Z}/\nu\mathbb{Z}$ for unlimited $\nu$ interested me for a while

Akiva Weinberger

Anything specific about it?

GPhys

I was just hoping to prove anything in number theory in a non boring way with NSA

Akiva Weinberger

03:00

I suppose certain details depend heavily on $\nu$ (ex: is it even? Prime?)

GPhys

I didn't succeed, but I proved a lot of things about unlimited prime fields

for an unlimited prime field, there's always an unlimited primitive root

There's also an unlimited prime field with $only$ unlimited primitive roots

and unlimited prime fields with finite primitive roots

(by unlimited prime fields, I mean $\mathbb{Z}/\nu\mathbb{Z}$ for $\nu$ an unlimited prime)

Akiva Weinberger

That makes sense, I guess, though I'm not 100% sure how to prove it

GPhys

some of those were not straightforward at all

Akiva Weinberger

Remind me how to prove that every prime field has a primitive root?

Forgetting about IST

GPhys

sec phone

Akiva Weinberger

03:07

@GPhys When you say an element of $\Bbb Z/\nu\Bbb Z$ is unlimited, you mean that it's not congruent to any limited integer, right?

GPhys

yes

Akiva Weinberger

'Cause every limited integer is congruent to an unlimited integer ($n\equiv\nu+n$).

GPhys

yes

Akiva Weinberger

I'mma see if I can translate all of those into equivalent non-IST statements.

@GPhys There is a $G$ such that for any $g>G$, there is an $N$ such that for all $\nu>N$, $g$ is a primitive root of $\Bbb Z/\nu\Bbb Z$

(not sure if that's equivalent)

GPhys

not all nonclassical statements have classical equivalents

Akiva Weinberger

03:15

@GPhys For any $G$, there exists a $\nu$ such that all primitive roots of $\Bbb Z/\nu\Bbb Z$ are greater than $G$

GPhys

you can prove things with IST that you can't proof without IST

Akiva Weinberger

@GPhys Don't think so

GPhys

(that is: statements explicitly about IST itself)

Akiva Weinberger

IST is a conservative extension of ZFC, also. Anything provable with IST that can be written in the language of ZFC is provable on ZFC

GPhys

that's what I'm saying: there are things provable with IST that can't be written in the language of ZFC

normally we're not trying to prove those things because nobody cares

Akiva Weinberger

03:17

@GPhys There is a $g$ such that for any $N$ there are $\nu>N$ such that $g$ is a primitive root of $\Bbb Z/\nu\Bbb Z$

Not 100% sure if any of them are equivalent

GPhys

I think I proved there is always an unlimited primitive root by counting the number of primitive roots

if i recall correctly

Akiva Weinberger

But everything in IST is equivalent to something in ZFC, I'm pretty sure. (That statement only makes sense for things independent of IST, because things that IST proves are equivalent to $\rm True$)

Not 100% sure though

Lo me'a achuz

GPhys

that's exactly what I mean: statements that aren't independent of IST

yeah, I proved for any unlimited prime $\nu$, the field of integers modulo $\nu$ contains an unlimited number of primitive roots

and containing an unlimited number of primitive roots implies at least one of them must be unlimited

Akiva Weinberger

Everything that's not independent of IST is probably equivalent to either $\rm True$ or $\rm False$ and thus trivially has classical equivalents.

Nelson gives a pretty algorithmic method of turning IST statements into ZFC statements, though

It's pretty neat

*provably, not probably

GPhys

@AkivaWeinberger maybe I am too tired to understand the point here (I haven't slept in some 30 hours)

or maybe I do not know enough logic

I don't disagree; I just don't understand the application to what I was trying to say

but I'm sure your knowledge on the matter exceeds mine

at any rate, back to the proof

the number of primitive roots is finite, take the max

if there did not exist an unlimited primitive root, the max would be limited

if the max was limited, then the number of primitive roots would exist in a limited interval from 1..max, contradicting the number of primitive roots being unlimited

therefore the maximum primitive root must be unlimited

hence there exists at least one unlimited primitive root

^ this is an interesting proof, though

you get information about one of the elements of a set only from the size of said set

@AkivaWeinberger I gave the equivalent IST statement for a limit here: math.stackexchange.com/questions/1595793/… (but you've probably seen this)

I gave a very explicit proof of it, though

@AkivaWeinberger oh haha, I see you commented in it

That was quite a long multi-question discussion where people were arguing about...I'm not really sure what.

Anyway, I'm extremely sleep deprived at this point so I'm going to bed @AkivaWeinberger

highlight me if you have any thoughts about IST...I'm always looking to try something new

I suspect maybe you are right about IST not being able to prove the full choice, but I will try the problem again tomorrow

Mike Miller

03:44

@Daniel Are Fredholm operators dense in bounded operators? (In unbounded densely defined operators?) If not, do you have an explicit example of an operator without a nearby Fredholm operator?

SoumyoB

04:07

is anyone online rn?

Dharmesh Sujeeun

04:48

Yee yee!

If I type something in latex in this message, how would I display it, in like mathematical format?

SoumyoB

05:01

it automatically gets displayed in mathematical format to those who've loaded the 'Start Chatjax' bookmark on this page

arctic tern

see "$\LaTeX$ in chat" in room description or starboard ------>

Adeek

06:51

hi

@MikeMiller can I ask you a quick question

I just need something for my talk

or @PVAL if your around

what does it mean this thing ? : If f : S^m —> S^m has degree m ?

nvm I get it

Balarka Sen

07:18

@Adeek it's the m-times map in homology.

user17629

07:56

Hello

I have a short question

can a subset proper be equivalent to the original set?

I meant. Let V a vector space and let S a proper subset from V, then is possible that $V \equiv S$?

$V \approx S$

Tobias Kildetoft

08:19

@user17629 Sure, infinite dimensional vector spaces can (will) have proper subspaces which are isomorphic to the original space

Eric Stucky

In finite dimension this is impossible, but in general it is possible.

beat to the punch :P

skill patrol

Snooze ya lose :P

Eric Stucky

It's a microcosm of research mathematics: nobody even has any idea the problem exists for 20 years (minutes) and then suddenly two people produce independent solutions simultaneously :P

user17629

I understand

thanks

Balarka Sen

Hi @EricStucky.

Eric Stucky

08:24

hellu

Balarka Sen

How's things?

Eric Stucky

hehe well

I'm currently experiencing that dissonance

where I have some things to do but

it feels like I have nothing to do

Generally this happens because there is something I have to do but don't want to, and which has an ambiguous due date; this is indeed the case

hbu?

Balarka Sen

Yeah that happens often with me.

Try finding a reason for why you want to do whatever you need to do. That might motivate.

Eric Stucky

I have a hard time with creating motivation from nowhere

Balarka Sen

@EricStucky I am studying some algebraic geometry.

Eric Stucky

08:29

Generally the way that these things resolve for me is

I just do other work I want to do more, and then eventually I really must do the thing and then it happens

Balarka Sen

Nice, then that doesn't sound so bad.

Eric Stucky

Which is it's own kind of magic non-scalable solution, I suppose

anywho :)

Hartshorne?

Balarka Sen

Haha no. Shafarevich.

@EricStucky A couple months ago you (and Alex) were working with tropical geometry, not?

Eric Stucky

hehe yes

plus a few others

Balarka Sen

Right.

Eric Stucky

08:57

Do you think there is any quick way to explain to a freshman non-major what it means to "work in characteristic 2", without going through the whole spiel about clock arithmetic?

Tobias Kildetoft

@EricStucky just tell them you replace all numbers by their parity (i.e. "even" or "odd")

But $2$ is such a bad characteristic to work in. It always requires all sorts of special treatment

Eric Stucky

it's better for explaining homology though because ughhhhh orientations in dim 2+ are too much trouble

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

10:17

Akiva Weinberger

11:04

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ In the "protip" bit, it's the trig one, right?

skill patrol

11:24

@AkivaWeinberger yes

Evinda

Hi!!!

skill patrol

hello

Balarka Sen

@AkivaWeinberger Hi.

Akiva Weinberger

Hi

Balarka Sen

What's up?

Evinda

11:29

If we want to check if the limit $\lim_{(x,y) \to (0,0)} \frac{\sin{(2x)}-2x+y}{x^3+y}$ exists, we can consider the limit along the line y=0 and the limit along x=0 and we will see that they are not equal. Is this the only way to see that the limit does not exist?

Hey @robjohn :)
Do you maybe have an idea?

Mambo

11:45

This isn't the only way

this is enough though

Balarka Sen

It's probably the simplest.

Evinda

@Mambo @BalarkaSen Could you tell me an other way?

Balarka Sen

You can probably find out other curves approaching along which you have different limits by yourself. I am not going to try and find one.

Evinda

@BalarkaSen So you mean that we could pick y to have an other form?

Balarka Sen

Yes. E.g., approach along $y = (2x)^3/3!$. Then the numerator becomes a term of order $x^5$ (go Taylor series) and the denominator becomes something of order $x^3$. Things cancel and the limit becomes $0$. Whereas approaching along $x = 0$ gives you $1$.

robjohn

11:51

@Evinda why would you need another way? Can you not think of another way?

Balarka Sen

Hi @AlexClark.

Evinda

I meant if we can just see that the limit does not exist, setting y as some function of x that is equal to 0 for x=0. @BalarkaSen @robjohn

user147690

Hey @BalarkaSen, I'm just doing more lie theory atm

robjohn

@Evinda $$\frac{\frac y{x^3}-\frac43}{\frac y{x^3}+1}$$

Balarka Sen

@AlexClark Such Lies.

user147690

11:55

@BalarkaSen :P

user147690

I've actually learned heaps during this research project, the sad thing is that it probably doesn't look like it to Mcnamara

Balarka Sen

Mcnamara?

Evinda

How did you find this? @robjohn
Dividing by $x^3$ at both the numerator and the denominator we get $\frac{\frac{\sin{2x}}{x^3}-\frac{2x}{x^3}+\frac{y}{x^3}}{1+ \frac{y}{x^3}}$...

user147690

@BalarkaSen My adviser for the project

user147690

smp.uq.edu.au/people/PeterMcNamara

Balarka Sen

11:59

Ah. Why do you think he feels like you haven't learnt a lot?

user147690

His current task for me is pretty much, understand these two paragraphs, and it is taking me aggggeees

user147690

6 conditions for a basis of canonical type, and 6 conditions for a basis of dual canonical type

skill patrol

where is his CV?

Balarka Sen

@AlexClark Maybe the stuff you want to understand is hard and requires more background, which is taking time to pick up. Does he give the impression that it's not hard?

user147690

No idea, but he is from Stanford and MIT fairly sure

robjohn

12:03

@Evinda what is $\frac{\sin(2x)-2x}{x^3}$ for small $x$? approximately

user147690

@BalarkaSen Yeah it seemed there was much learning background stuff to get to understanding some of these conditions, I am almost done though, I'll email him tonight hopefully

user147690

@BalarkaSen I don't know what impression he gives honestly in terms of difficulty, he seems like a nice guy though

Akiva Weinberger

OMG @BALARKA

in.mobile.reuters.com/article/idINKCN0WX0S2?irpc=932

Balarka Sen

@AkivaWeinberger Yeah

robjohn

@Evinda How can you be asking questions about Fourier Transforms and differential equations and not understand this? What courses are you taking?

Evinda

12:06

@robjohn I am taking PDES: Theory of weak solutions, algebra and coding theory....

Balarka Sen

Actually the death count is much more than 14. There were hundreds down the collapsed bridge.

robjohn

@Evinda you really should learn limits well before attempting courses like those.

Balarka Sen

@AlexClark If he doesn't give any impression, seems like a nice guy, then why do you think he doesn't feel like you haven't understood much/took a long time? Seems to me you're the one who thinks that ;)

robjohn

4 mins ago, by robjohn

@Evinda what is $\frac{\sin(2x)-2x}{x^3}$ for small $x$? approximately

@Evinda That is, what is $\lim\limits_{x\to0}\frac{\sin(2x)-2x}{x^3}$?

Evinda

@robjohn 0/0

robjohn

12:15

@Evinda Is that really what you would answer on a test?

Evinda

Aaa that's what you asked me? It is -4/3. @robjohn

robjohn

@Evinda yes. Now do you see where I got $\frac{\frac y{x^3}-\frac43}{\frac y{x^3}+1}$

Akiva Weinberger

@GPhys Hey. Let $N$ be an unlimited natural number. Let $A$ be the set of numbers in $(0,1)$ such that their $N$th binary digit is $1$. Is $^S\!A={}^S\!\{x\in(0,1):x\in A\}$ a measurable set?

As in, Lebesgue measure

'Cause I think that IST without choice proves the existence of nonmeasurable sets, and I think this is one of them

Evinda

$\lim_{(x,y) \to (0,0)} \frac{\sin{(2x)}-2x+y}{x^3+y}=\lim_{(x,y) \to (0,0)} \frac{\frac{\sin{(2x)}-2x}{x^3}+\frac{y}{x^3}}{\frac{x^3}{x^3}+\frac{y}{x^3}}= \frac{\lim_{(x,y) \to (0,0)}\frac{\sin{(2x)}-2x}{x^3}+\lim_{(x,y) \to (0,0)}\frac{y}{x^3}}{\lim_{(x,y) \to (0,0)}\frac{x^3}{x^3}+\lim_{(x,y) \to (0,0)}\frac{y}{x^3}}=\frac{-\frac{4}{3}+\lim_{(x,y) \to (0,0)}\frac{y}{x^3}}{1+\lim_{(x,y) \to (0,0)}\frac{y}{x^3}}$
@robjohn

Kari

12:33

Is there any convenient/concise way to write the set of all linear combinations of a set of vectors?

Akiva Weinberger

@GPhys I don't actually have a reason for thinking it's nonmeasurable, other than "it's so crazy it might just work".

@Kari ${\rm span}(S)$, I think

Kari

Thanks, @Akiva!

Transcript for

$Mathematics$ Mathematics