LOL

?

just plotted that

Anyone?

What are you allowed to use?

No la hopital

And thats it i think

How can i know what else i dont know

Series expansions?

No

if only you butts hadn't scared away that integral person

I bet he'd know

have you seen some standard limits?

As in, say, $\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1$?

Yeah

Ummm

Ln(1+x)/x ?

Yes

x->0

(you should use mathjax - see the chat description)

Im on mobile

But i can go with that if u have a elementary proof of it

@user379685: I don't know what you know, but I need to use $\ln(1+u) \approx u$ for $u$ near $0$.

is it generally frowned upon to self-learn from books that are not "popular"?

Often that limit is standard

It's the defn of the derivative.

Hiya chat!

Heya @Fargle

and if you know that limit, you can use that $\lim_{x \to -\infty} \frac{\ln(1+a^x)}{a^x} = 1$

i.e. learning from a free book

How goes it @Ted?

Pretty well, and you?

Quite alright. Trying to keep reading, haha.

Meaning your limit becomes $\lim_{x \to -\infty} \left(\frac{2}{3}\right)^x$

(2/3)^x ?

yup

What're you reading these days?

@Axoren I disagree with one of your post in Meta. I hope don't disturb you, and I am saying then I am sorry. My opinion is that such OP tried help (do a comparison between the activity of such OP versus a moderator that also answered the question) to a younger student. Notice that this site democratizes the access to mathematics to all students. I believe that this message from me to you isn't 100% right, thus I am saying I am sorry, but I wanted said it. You are welcome.

I dont get it how from ln(3^x +1)/ln(2^x+1) get to (2/3)^x

whoa, it's the other way around, no? Numerator has $2^x$ and denominator $3^x$?

@Ted Baby Rudin right now, just trying to close some gaps.

Ln((3^x)+1) is in he numerator is that a clear writing?

I mean denominator

No, the point is your question before had $2$ in the numerator and $3$ in denominator

ah, okay

Numerator can be rewritten as 2^x * ln(2^x+1)/2^x; and the denominator by 3^x * ln(3^x +1)/^3^x

now the second factor in both of those is just the limit I mentioned before, which is $1$

Oh thank you

Whoa did I miss the magic?

hi chat

@Semiclassical QM test is so bad the prof is holding optional problem sessions

lol

What sort of problems showed up, anyhow?

it sounded like it was pretty broad

Yeah I'll tell you when I get it back. I'm looking at a 60 based on the solutions

With a curve that might be an A, I hope

@Fargle: Remember that just reading isn't so good. You must write up exercises and get someone to criticize them.

could be

@0celo7 mmkay

@TedShifrin If you'd be willing to do that, I'd be interested. I've been doing exercises as well--I find the only way I absorb knowledge is to use it.

Reminds me, I need to do the exercises I set for myself

I don't mind doing it occasionally, @Fargle, if you LaTeX them up and email me the .pdf.

I'll be traveling a week from now for 2 weeks, though, but ...

That's the trouble with rewrites, @Semiclassic.

Yeah.

@TedShifrin I'd be more than willing to do that. Far better than scanning it--my handwriting is atrocious at best, haha.

I asked them to turn the originals in as well, so I should be able to bump up their scores based on that

@Semiclassical well I can tell you what was on there, but I don't need another person telling me it was trivial and making me feel terrible

I'd rather know what I got first

Ah, fair enough

One of my former colleagues allows his students to do test corrections, so he has to do two rounds of grading. Since I post solutions (unless it's senior-level courses), I absolve myself of such options.

I'm more just curious what the problems were

Only the third one was hard, and I got it completely right.

But @0celo is right. We live to demean him ... our main purpose for living.

@0celo7 nice

The first one was purely linear algebra, and I got every part wrong.

Linear algebra truly is super important ... all over math, physics, engineering, economics ...

@TedShifrin Not our main purpose, but that's how we procrastinate.

um, completely right or completely wrong

I was actually being sarcastic, @Balarka, unlike you :)

@TedShifrin Medicine, too, since there's stuff like tomography

Well, sure, I include that in engineering.

Fair.

It's discrete implementation of integral geometry analysis, actually.

@TedShifrin I don't understand it

I don't know what to do

Have you done a down-to-earth course on it, @0celo?

Yep

Part of the problem with 'linear algebra' is that it can mean different things depending on how abstract you're talking

So what do you not understand, specifically?

Taught by a famous topologist who didn't give a fuck about us and a well meaning TA who just apologized for the prof.

That's why I said down-to-earth, @Semiclassic :P

@0celo7 eugh

@TedShifrin Too bad.

The prof preferred to rant about liberals than teach.

Or communists.

Well, @0celo, OK ... pick topics you know you didn't learn and sit down with a decent book (I won't even insist on mine) and do a few problems. You can also look at one or two of my YouTube lectures if you want to get some intuition or see examples.

Linear algebra is essential, but I don't apply much of it often. Need a refresher.

We're not going to politics, 0celo.

@Danu: Have we agreed on multiindices or are we going to have a fist fight?

Check the room ;D

I'm sorry I got more insistent, but you were truly messing up that poor guy.

I didn't know what the word "multiindex" means.

Google is your friend?

@TedShifrin

pfft i have no friends

I was on my phone during all of that conversation :\ Texing was hard enough :P

LOL, ok, @user3502615, if you insist.

Ok, time to math.

Honestly, though, @Danu, what you wrote and your examples are nonsense.

the professor replied to my e-mail :D

I'm sure it is

When you write $\sum dx_{j_1}\wedge dx_{j_2}$, I write down all possible terms. If $n=2$, I get $dx_1\wedge dx_1+dx_1\wedge dx_2+dx_2\wedge dx_1+dx_2\wedge dx_2$... which is a longwinded way of writing $0$.

If you indicate $\sum_{j_1<j_2}$, then I write down just $dx_1\wedge dx_2$.

But, in general, there are $\binom nk$ terms when I use increasing $j_\ell$'s.

What email, @user3502615?

Just check out the dedicated chat room---you can find the link on the page. I'm not doing anything different from what you meant, I just don't know at all what the terminology is, as far as I can tell.

Am I not yet done, @Danu?

i'd say something snarky like "just do $dx_{(i}\wedge dx_{j)}$ but I cant actually remember whether that's symmetrization or antisymmetrization

i sent an e-mail to a professor at a college near me, asking if he could assist me if i had questions

No, when you started writing $dx_{2_2}=dx_2$ ... all that stuff was nonsense.

@TedShifrin I wrote out a reply there a while back already

...plus, what I just wrote looks pretty horrible

Yeah, but physicists like to do that, @Semiclassic.

We do.

I like Ted's notation by indexing over $k$-subsets of $\{1, 2, \cdots, n\}$. Much more neat.

@TedShifrin I don't understand any linear algebra it seems

I think one would be more likely to encounter it like so: $\omega=\omega_{ij}dx_i\wedge dx_j$ with $\omega_{ij}=-\omega_{ji}$

i didnt like linear algebra

Than $dx_{i_1} \wedge dx_{i_2} \wedge \cdots \wedge dx_{i_k}$, I mean, which is ugh.

I couldn't find eigenvalues of a 2x2 matrix

I prefer to sum only over increasing multi-indices, @Semiclassic.

I can see the advantage, looking at that

Working with basis vectors is preferable, @Semiclassic, in general.

@0celo: You couldn't meaning you don't know how ... or you just made a silly mistake?

@TedShifrin By basis vectors do you mean $dx_i\wedge dx_j$?

I mean all those for $i<j$.

@TedShifrin mistake

You include $dx_i\wedge dx_i=0$ and $dx_j\wedge dx_i = -dx_i\wedge dx_j$.

right, and about half of that is redundant

OK, 0celo, so you should practice a few. You can write down the characteristic polynomial for a $2\times 2$ in a millisecond.

And, as I repeatedly told my students, if you can't find a non-trivial eigenvector, you didn't really have an eigenvalue after all.

Without that ordering assumption I think one introduces a factor of $1/k!$ in general which becomes a mess too.

@BalarkaSen Can't argue with that

On occasions, it's convenient to sum over all multiindices, @Balarka. Note that even if $I$ and $J$ are increasing, when I concatenate, $IJ$ is very unlikely to be. So even my definition of wedge came along before the increasing convention.

That's a fair point.

That's another reason why notation like $A_{(i} B_{j)}$ for (anti?)symmetrization is annouying: It's not clear whether one includes a division by $k!$ in it

what you'll get is (anti)symmetric either way, of course, but it doesn't necessarily map an antisymmetric thing back to itself. it depends on how it'bs being used, and that's tiresome

I have never seen that notation. Looks horrible.

here, for instance: physics.stackexchange.com/questions/95133/…

Urgh.

it's convenient in some cases, I guess

but I don't do GR, so i've never had to deal with messes of tensor indices

That multi-index notation might've come in handy for me in some many-electron calculations I've seen, though

@TedShifrin what does that mean?

I had the right eigenvectors, I could write them by inspection

But I had the wrong eigenvalues

If you knew the eigenvectors, @0celo, by direct substitution you could find the eigenvalues. Just multiply.

I know. I wrote them down wrong.

But I had the wrong matrix anyway

What I was saying is this: If you thought 3 was an eigenvalue, but then you solved for the nullspace of $A-3I$ and got only the trivial solution, that would tell you $3$ cannot in fact be an eigenvalue.

Yeah, too many things wrong to sort out and save yourself.

h

From mismultiplying two vectors

But, seriously, you might want to just watch a couple of my lectures and see if I'm just a smidgeon better than your professor who didn't care was :P

If there really are a few topics you are sure you don't understand but should.

@MikeMiller Hi.

@Balarka: Are you playing with parallel translation? :)

I guess I shouldn't say anything.

Hello.

I was thinking about something that frustrated my mind.

What's the derivative of $x=0$ ?

Doing that, yes, @Ted.

I mean if we try to write it as a function we get into trouble.

@MikeMiller For a change, correct.

That isn't a function @Mahmoud.

You're fixing an $x$ value.

$y=\frac x0$

What???

Which isn't defined

You can only talk about functions that are functions.

Ok.

@TedShifrin Rude.

It makes sense if we don't count it as a function.

I just read that the boundary of the Mandelbrot set has Hausdorff dimension 2, my brain isn't happy to accept this fact

Thanks @TedShifrin

I am entitled to be rude.

@Alessandro Check out Osgood curve.

@TedShifrin I saw one of your lectures, you're a great teacher (Professor).

All though I could only keep track with the first Introduction to Integrals :P

@BalarkaSen On the other hand, the Mandelbrot said has "visibly 1-dimensional" boundary.

Go to the beginning of the first course, @Mahmoud, and you can start with vectors.

The linear algebra at the beginning should be fun for you.

@TedShifrin Okay I'll do it.

True, I am seeing what you mean.

@Balarka that's also very counterintuitive but the Mandelbrot looks "more" 1-dimensional intuitively

@TedShifrin But think about it,

@Balarka: What did I mean about which?

@TedShifrin If we try to compute the derivative from the graph

It will be infinite

You don't have a graph when the function makes no sense, @Mahmoud.

Unbounded*

Sorry, I was replying to Mike's message, not yours, @Ted.

No ... there is nothing.

Ohhhhhh. See how confuzling it gets in here? :)

Also, naively, one would think the Hausdorff dimension of a Jordan curve is always 1, even if the lebesgue measure is positive.

IIRC the Lebesgue measure of the boundary of Mandelbrot is zero but I don't remember.

:/

No, @Mahmoud, there's no such thing.

$\infty$ is not a number.

@MikeMiller Wikipedia says that's unknown.

It's wrong I know

Dang.

It's a concept

Weird stuff.

@MikeMiller I'm not sure that combination is possible. I'm pondering.

No, @Mahmoud, it's a symbol. Functions need to have numerical values.

@Mahmoud just in case you were wondering you need \infty instead of \infinity for the $\infty$ symbol

@TedShifrin I don't know this stuff.

$\infty$ Weird :/

I directed an MA thesis called "Demented Dimension" :P

Infty ??? Why ?

because it's shorter

Did the developer make a spelling mistakes ? :D

@TedShifrin I probably won't do that.

He probably didn't.

Sadly, I don't think I have an electronic copy ...

Again I'm wondering @TedShifrin The derivative is linear operator right ?

Oh, yippee, I did find an electronic copy.

Do you even know what that means, @Mahmoud?

It satisfies Additivity

I spelled that wrong

And

Scaling

I think.

OK, true, if you know what it means.

@TedShifrin Fractals are nice.

I watched a video, did anyone ever talk about writing the derivative as a Matrix @TedShifrin ?

So here's an interesting conundrum to mull over, @Balarka. Invariance of domain tells us that a space-filling curve cannot be an injective mapping. How badly non-injective must it be?

meh

LOL, of course, @Mahmoud, but not the way you're thinking. The space of differentiable functions is very, very huge ...

I think it has to be non-injective on a cantor set or something

dunno

@TedShifrin I mean is it a common thing ?

Can you make it just 2:1 at bad points or must it get worse? ... Not that I know the answers ... But this came up in that MA thesis work.

It blew my mind at the first while.

@Mahmoud: Yes, if you work on the vector space of polynomials, or if you define derivative in multivariable calculus.

how's a space filling curve defined? A continuous surjective map from an interval into the plane?

unit square

Yup, @Alessandro.

Well, unit square ... or Hilbert cube :P

@TedShifrin Oh, that kind of thing. I am pretty sure at least one preimage must have infinite cardinality.

I don't think that's true, @Balarka. I'll have to go back and check Andy's thesis.

Interesting; I'd be surprised otherwise.

There is a great great video here @Alessandro YouTube

@TedShifrin One more thing, what EXACTLY is an Axiom ?

It's something you accept as fundamental, without proof.

Does this have several meanings ?

no

For example en.wikipedia.org/wiki/Axiom_of_extensionality

This is like a game rule, a formal definition of a fundamental concept.

Is it an axiom because we accepted it without proof ?

@Balarka: In one of the constructions of a space-filling curve, there are only countably many points hit more than once and none gets hit more than 4 times.

Wow.

He did this in terms of base 4 decimals (analogous to using base 3 for Cantor).

Ah.

What's going on ?

I think it's a construction due to Hilbert (Peano's was different).

according to this question whose answer I don't really understand that happens with the Peano curve too

G'Bye. And thanks again @TedShifrin

@Alessandro That seems to only ask that it's not 1-1.

the answer shows that no point is hit more than $4$ times according to the poster

oh, in the answer. interesting.

Bubye, @Mahmoud.

Yes, Andy in his thesis commented that you can repeatedly divide the unit square into $k$ squares (Hilbert is $k=4$, Peano is $k=9$) and for any $k$ there are a countable number of points with more than one preimage and always at most 4.

neat

This was 12 years ago, so I've forgotten, but I did understand it at the time :)

Anyhow, back to parallel translation !

does anyone here know about martingales? i have an extremely elementary question

@BalarkaSen are there squirrels in India?

lots

A bound is a number, right?

@Monad I am suspicious of the question.

Hey Ho

@AntonioVargas why? P:

because you provide no context

a bound could be an element of a partially ordered set, or an asymptotic for a function, or maybe something I can't think of off the top of my head

What is the discussion about ?

Physics.

Thanks, Ocelo

@Ocelo7 Why is your blog called "Einstein and the Evidence" ? Evidence for what ?

@user243301 I'm sorry you disagree with a question of mine.

Oh jeez, my fart comment is starred even more than before.

hey there @Brody

aloha @BalarkaSen

how are you?

@MikeMiller I'm not sure how a group action preserves the underlying structure of a set in the case of an action on a Linear Space.

@Axoren That's how group action on a vector space is defined.

$G$-action on $V$ preserves the linear structure if $V \to V$, given by $x \mapsto gx$ is linear.

Hi there axo

Hey, @Brody

@Brody fantastic. how about you?

Thanks for posting a certain earlier question in chat. I'm now acquainted with polynomial interpolation @Axoren

(a useful theorem thereof, that is)

@BalarkaSen Great to hear. I'm doing well

@BalarkaSen So this is a specific definition for Linear Spaces, but if we don't utilize the linearity of the space and equip it with a group action, does that group action necessarily have the same property as the one defined with the linearity of the set in mind?

I don't know if I asked that question very well.

@Brody Glad I could help I guess, lol. I think that might actually have been the result of a tangent off of a question I asked.

Let me think of rewording that question, @BalarkaSen

@Axoren nah. a group action on the underlying set of the vector space need not preserve the linear structure.

but that's not interesting because you're forgetting you have a linear structure

@BalarkaSen Okay, so if that's the case, then a group action on the linear space must be chosen with the linearity in mind for it to be linear in $x$.

I wasn't sure if the transitivity of the group action is what magically gave it that property.

sure.

Thanks.

whenever we say group action on a set equipped with [blah] structure, we assume it to preserve that [blah] structure

eg, group action on a topological space must act continuously (so multiplication by $g$ is continuous for all $g \in G$).

Yeah, I'd read that today when doing more investigation into this.

brb

or, group action on a group must act by homomorphisms (multiplication by $g$ is a homomorphism)

etc etc

I'm not 100% on selecting which mapping preserves algebraic structure for arbitrary groups and sets, but after the other night, I'm fairly confident with linear spaces and now I understand the necessary pieces for it.

Like $S_n$ acting on $\mathbb R^n$.

The most natural action on $\Bbb R^n$ is action of $\text{GL}_n(\Bbb R)$, by multiplying by matrices.

Right. I'm not clear on when something is the "natural action".

I presume your symmetric group action is by switching around the 1-dimensional factors?

Is it just whatever operation is normally associated with arguments in that set?

Yeah

Like are single-argument functions the natural action on real numbers?

@Axoren It's natural in the sense that it's the "largest possible" group that can act on $\Bbb R^n$. If $G$ acts on $\Bbb R^n$, then multiplying by $g \in G$ gives a linear map. A linear map is literally a matrix, so this realizes $G$ as a subgroup (I lie here a bit) $\text{GL}_n(\Bbb R)$.

It actually gives a map $G \to \text{GL}_n(\Bbb R)$, but whatever.

@BalarkaSen I assume it's missing some isomorphism there, right?

Okay.

What do you mean by that?

$G$ isn't necessarily a subgroup of $GL_n(\mathbb R)$ but it is isomorphic to one, right? Is that what you meant by "I lie here a bit"?

Nah, the point is you can go as far as representing every element of $G$ as a nxn matrix, but you might have a group action for which two different elements are represented by the same matrix.

Aka, the map $G \to \text{GL}_n(\Bbb R)$ might not be injective

Oh, that's interesting. I hadn't thought about if $G$ were a much bigger set than $GL_n(\mathbb R)$

Yeah.

That makes sense.

Like you could have $\mathrm{SO}(3)$ acting on $\Bbb R^2$ in the natural way.

Yeah, I had just thought of that exact example, lol

Rotating a circle in $n$ dimensions

Even though it's just spinning flat on its plane.

Yup

lol

In fact your $S_n$-action is not injective either, @Axoren :)

@BalarkaSen How so? Isn't there a one-to-one correspondence between permutations and matrices that perform those permutations?

Do you mean into $GL_n$?

$S_n$ is a much larger group. It has order $n!$, whereas $\text{GL}_n(\Bbb R)$ has order ...?. Permutation matrices do NOT correspond 1-1 with $S_n$.

Yes, action on R^n.

Um.

$\mathrm{GL}_n$ is uncountable.

It's an $n^2$ dimensional manifold.

do there exist integers a,b,n such that $(\sqrt[3]{2}-1)^n=a+b\sqrt[3]{4}$?

Oh, for some reason I was working over a finite field. Nevermind that.

You're scaring me, Balarka.

I thought I knew a thing and you made that disappear instantly.

I am being silly, forget what I said.

Hey, @TedShifrin

Go to sleep, @Balarka!

Hi, Axoren.

@TedShifrin Planning to.

Hello, Ted.

Don't put it off until tomorrow.

It's already tomorrow.

Hi @Brody. Long time.