Yes, but how to get to the dual basis 'naturally'.

Now, can I teach any of this? Probably not at this level.

I think the parallogram approach is good.

But it could be a great basis for a lecture or two to interested high schoolers.

I think you can teach it. Some Gary Larson 'and them a miracle occurs' gaps in between, but the general idea is there.

I certainly would not take the time to do this in my (much more sophisticated) multivariable math class. Linear algebra I barely had time to do determinants, other than two days or so, because I wanted to do significant applications.

What parallelogram approach?

Showing that the area does not change as long as the base and vertical height remain the same.

I guess I am looking more at the intuition than a proof.

Oh, shearing, sure. But comparing the column parallelogram to the row parallelogram seems futile. That's where Mike started.

No, Cavalieri's principle is something the students should see/understand here.

And you don't even need to rant about Fubini's Theorem.

Something new for me :-).

Yes, this isn't a question of whether the students are capable of this argument, but whether or not there's time... there's not.

@MikeMiller that is very sensible imo

The issue is the geometric relationship between a basis and the dual basis.

Right. I always wanted to get to Spectral Theorem, quadrics, and some systems of ODEs. Most colleagues got to none of that.

I'm skeptical I will.

I am so glad I learned some of this stuff as an engineer first.

There are so many places one can get bogged down in.

Yup, it helps (for me) to have some 'physical' intuition.

@copper: So when you talk about the geometry, you're identifying the dual space with the original space via dot product?

Yes.

I guess geometry for me is coordinate.

OK, that's compatible with "Ted's favorite formula" $Ax\cdot y = x\cdot A^\top y$.

yes, I think it is ok to learn that as a rule first.

ok = good, i mean.

Well, I certainly teach (taught) that early, and never at this level discussed the dual map.

But I did say that that formula is why the transpose is important/significant.

I think that is a good approach. Like Von Neumann supposedly said, you never really understand, you get used to it :-). So, the sooner you get exposed (without necessarily) understanding immediately) the better.

Exactly my pedagogical view on differential forms.

You have to actually use/apply them for a month and then you start to get it.

Certainly works for me.

One topic that exposure did not work for me was an optimization technique called conjugate gradients. It was an algebraic grind until someone showed me an approach using Krylov subspaces and suddenly it was clear what was happening.

So I'm actually following your suggestion and trying to compute some dual bases explicitly in this context.

Oh, duh, you just rotate $\pi/2$, with perhaps a sign change because of orientation.

It sort of straddles the geometry and algebraic approaches, I think.

Gorgeous morning here, just got back from a nice 2 hour cycle around the Oakland docks :-).

So we should be able to see that if $v$ is an eigenvector of $A$ ($2\times 2$), then $Rv$ is an eigenvector of $A^\top$, with $R$ the rotation.

Pretty here, too. I will go for a walk soon.

Hi folks. Can convergence of $\int_1^{\infty} \frac{\cos(x)}{x} dx$ be determined from Limit or Direct comparison tests?

@MikeM: I don't know if you care any more, but that's the way to do $2\times 2$ for a naive audience. Show that the $\pi/2$ rotations of the eigenvectors are the eigenvectors of the transpose.

@Jeff I doubt it.

I like it!

You need its oscillating nature, so it's more sophisticated.

@MikeM: I wish I had to do another edition of my linear algebra book. This would go in as an exercise :)

@TedShifrin Because the DCT requires the functions to be nonnegative and Limit doesn't work, comparing to $1/x^2$, right?

You have to integrate by parts and then you can think about comparison (because then it will converge absolutely).

Yet my boss, the prof. said to use direct comparison.

Can't compare when it's conditionally convergent. Just cannot.

You are misunderstanding your boss.

@TedShifrin Oh?

Can you elaborate?

Did you read what I wrote?

I prefer to say that you are misunderstanding rather than to say that your boss is wrong.

@copper You agree?

@TedShifrin No. I missed that. We've actually already integrated by parts. We have $-\frac{\cos x}{x} - \int_1^{\infty} \frac{\cos(x)}{x^2} dx$ and she's said to finish the second integral using direct comparison.

Oh good grief. That's exactly what I told you to do ages ago.

Your first term is incorrect, of course.

First, the function is wrong. Second, you need to evaluate at endpoints.

And your integral is also wrong. Maybe your original question had $\sin x$ in it.

annoyed

@TedShifrin I stepped away, give me a moment, I presume you are referring to the $2 \times 2$ case?

Yes, the original is $\sin(x)/x$. If you're annoyed,then nevermind.

hi chat

Hi @Astyx

hiya

Yeah, @copper, I was.

Let someone elseanswer.

good evening

Hi, Simone.

hey Ted.

@TedShifrin ...still struggling with the computation for a $2 \times 2$ upper triangular. Can't blame wine for slowness...

Self-hiding knot Reidemeister moves

Dunno 100% how much this breaks things

Oh it breaks things a lot I'm now realizing

would be impressive to see anything that's not broken

I'm realizing now that it makes everything equal

You can turn a trefoil into an unknot pretty easily

I mean, maybe it doesn't break tangles

(tangles being little "pieces" of a knot - each of those pictures above are tangles)

yea

idk how it doesnt

So obviously I can't change the number of endpoints of a tangle through these rules

@Jeff As Ted says it is conditionally convergent. That means that $\int_1^\infty\left|\,\frac{\cos(x)}x\,\right|\mathrm{d}x$ diverges. This means that limit and direct comparison will fail.

If I start with the first figure on row *3, I'll always have a tangle with four ends

But can I get rid of that little segment in the middle through these rules, turning it into a normal crossing?

1st figure on row 2? wait you count invisible things?

Oh no sorry row 3

oh ok

Typo

np

it seems even MO is suffering these days

though the two rules on row 2 are pretty broken because they mean I can always pull an unknot out of anywhere

I mean, I can always pull any knot out of anywhere, from "behind" a string

@Thorgott definitely looks like homework, to me.

Riemannian geometry homework lol

That's not Riemannian geometry

Oh I see that's the tag he used

"Riemann" is used and "interval" is somewhat geometric...

@AkivaWeinberger: perhaps they know that we won't do their homework, and they are trying out MO.

They need to try HWO

or Yahoo! Answers

(can i burn a luigi board?)

or try thinking for themselves

students hate this trick

@AkivaWeinberger be sure not to release a daemon

@Thorgott Riemannian geometry. Lovely.

random thought

What most discussions of the trolley problem lack, I think, is a time limit

Trolley problem?

no way

Trolley approaching a switch in the track, straight ahead there are five people tied up, on the other track there's one person tied up, do you pull the switch (and kill one person but save five) or do nothing

It will arrive in ten seconds. Decide

So it can't stop, huh?

Time's up. Five people died

So why is it so hard to say $-1$ is better than $-5$?

ethics

Would you murder a person to save more people?

Maybe it's $-\infty$ vs $-\infty$?

Doing nothing has a 5 times worse casualty.

Murder 5 versus murder 1 ?

Doing nothing does not make you innocent.

I don't get it.

Although I realize that the world is full of versions of this.

@TedShifrin this is the whole question, kinda

What if, instead, there's no switch, only five people tied up, but there's a very fat man standing near the tracks

If you push him onto the tracks, the trolley will stop and the five people will be saved

Or you could let the fat man live and the five people die

or instead one has $\aleph_0$ people and the other has $2^{\aleph_0}$ people

OK, maybe that's a slightly better (worse) variant.

I mean: We are used to being complicit in all sorts of murders, and we just shrug.

@copper: Did you check out my rotation claim?

@TedShifrin Sorry, really slow at the moment. If $A v_i = \lambda_i v_i$ then $v_2^T A v_1 = \lambda_1 v_2^T v_1 = \lambda_2 v_2^T v_1$ so, if $\lambda_1 \neq \lambda_2$ then $v_1, v_2$ are rotations of each other.

This has become a bit of a meme

For example

DogAteMy: My problem with that picture is that I would hesitate because I wasn't sure which position of the switch was which.

oh no, Reimann again

Saving 1/12 of a life!

@copper.hat No, that's not right! It's only right when your matrix is symmetric to start with.

Reimann lol

Reimann and the damn -1/12

what if the 5 people on the one track are higher topos theorists

Kill them with high multiplicity.

@TedShifrin Aargh. Like I said, really slow.

OK last one:

I think DogAteMy has got the room out of control now.

"for"?

I love the loop one so much

I will not defend the grammar mistakes of memes

Let's just say the people who make memes are not exactly up to your expectations in general, Ted.

Well, @anakhro, I'm not up to my expectations. What's your point? :D

Ted and his impossible standards. :((

@Thor: The five get a scenic death as they're dragged along.

@TedShifrin What I meant to write was $Av_1 = \lambda_1 v_1$ and $u_2^T A = \lambda_2 u_2^T$ then if $\lambda_1 \neq \lambda_2$ then $u_2^Tv_1 = 0$.

Ah, so you're starting with eigenvectors of each. Fair enough. I was amused to see that the actual rotated basis vectors made the dual basis (up to a sign).

This is independent of linear maps.

@TedShifrin It is sad to say that I needed to go to Octave to arrive at the algebraic conclusion.

Going to bed early... goodnight all!

My pen & paper calculations made every conceivable mistake.

Buona notte!

grazie :D

@Astyx How are you creating these drawings so quickly?

I am not

They're not mine

I would just choose the left track

Every time

Minimizing curvature?

I read an article about similar experiments to compare various situations. It seems that people generally favor dogs over cats when making these decisions.

Because outdoor cats are the worst.

I think there are more dog people than cat people. I'm one of the latter, although I have nothing against dogs.

When I go to a restaurant, I quickly pick one selection and then spend my time improving the choice. Same with most decisions I make.

In other words, the quick pick leads to many recriminations.

:-), not really, I don't announce my decision until the waiter comes over.

I like dogs, but I think it is because they address some emotional deficit that I have (ask my kids).

Most of the cats I had were more dog-like in personality/behavior.

Cats just dig their claws into my crotch when they get comfortable.

Maybe something to do with me, maybe not.

Well, the cats know your vulnerabilities.

@TedShifrin This is a bottle of wine discussion :-). Many things I would like to write that would fall into the do not star category.

I wonder if Elephants can sense when tsunamis are approaching

LOL, back to math.

Apparently they have excellent memories.

I was chased by an elephant once.

And one spent its time trying to compress me between overhead branches and its back.

This was on safari or in your childhood?

Thankfully I was in a car when I was being chased.

Better being chased than chaste.

A personal safari with my wife in a rental in Kruger National Park.

Drove around a corner, encountered an angry elephant.

A few panicked moments trying to shove the car into reverse.

And realising how difficult it is to reverse into the cloud of dust you created getting there.

Strange computations go through your head, cost or losing the deposit vs death by trampling.

LOL ... This would make an excellent short subject.

That was when I actually had a life.

But no income at the time. Choice, choices.

You mean ... before children and responsibilities.

Yes indeed!

But I also had infinitely more energy.

does this hold more generally?

@robjohn TY

I'm referring to first part with measurability

Does what hold more generally?

And why do they say "measurable spaces"?

The 8th letter of the alphabet is 8ch

measurability, for weaker conditions

Doesnt need to be bounded

You also don't need $\mu$ to be finite in general.

@Astyx if they are going to use the zeta regularization to sum 1+2+3+4+... =-1/12, then they need to do the same thing with 1+1+1+1+...=-1/2, actually saving 5/12 lives.

If the function is integrable w.r.t. the product measure this holds. But there might be some slightly funky stuff since then this integral is only defined almost surely

@robjohn How thoughtless of them!

all I know is that my function is positive, integrable in y

@TedShifrin I hate most arguments using regularization to give finite values to divergent series.

@Jakobian if it is positive you dont need integrability

Thats tonellis theorem

I have never understood any of this, @robjohn.

@Jakobian I've usually seen that written as $\mathrm{d}\mu(y)$

Yeah, that source is unusual. Also referring to measurable spaces.

I thought Tonelli is for sigma-finite. That part generalizes to arbitrary measures?

You mean any measure?

@Jakobian are you worrying about the sum of an uncountable number of terms?

Okay no clue I dont think one can generaloze this arbitrarily

@robjohn wdym

Ask @AlessandroCodenotti he did a seminar on some weird friedman theorem

@Jakobian sums of an uncountable number of terms do not make sense unless only countably many are non-zero.

what about it

many things will fail there. Not real interesting except for counter-examples.

as far as I know.

hmm. I see. Well this is a book about Markov processes I'm reading but author neglected some measureability conditions. I guess I'll switch to another one

say you had a bunch of magmas given by presentations. you could have them be free if you really wanted, so just some generators. could you take the free product of these? like that is the coproduct for magmas, right?

just the magma generated by all the generators

(assuming it's free)

(if not, then also subject to all the relations)

@Jakobian can you show an example?

@Jakobian which book is this?

Dynkin's Markov Processes

the category of magmas is an algebraic category, whence complete and cocomplete

so it has a coproduct, but is it the one I said/

yes

it's just the free product

it's gonna be some bs involving non-associative words, but yeah

the spirit hardly changes

cool cool. say we took a bunch of free magmas on one generator, countably many of them, and took the free product of all of them- you get the free magma on countably many generators. so for each $n$ I guess I can see how it would be one big tree with each n-ary tree hanging around, but how do I visualize the countably generated one? maybe by hooking up all the other ones at one point where it is not a "locally finite" graph (idk the term, it might be that)

@Jakobian Why should a probabilist work with anything but finite (or sigma-finite + density) measure?

@Thorgott yeah so Catalan number many parenthetizations of a given word

Idk, it just confuses me if it's not written down

I mean yeah lebesgue, but may be that the author stated somewhere at the start that all measures are supposed to be finite if there's nothing else stated there (like sigma-finite)

wait if the operation is binary, the trees will always be binary. the arity of the operation gives the "arity" of the tree

idk how you're viewing magmas as trees

label nodes with elements of the magma

you continue to move away from the root the bigger the word gets

it was translated by multiple people, so perhaps some assumptions were lost

@Jakobian do you have a page reference

Dynkin mentions explicitly when he's considering a finite measure however. Perhaps he assumed sigma-finiteness in appendix but that's in volume 2, not avaible online

start of chapter 2

I mystically got hold of a copy

volume 1 though

yeah I also have volume 1

there he defines the transition functions

but he explicitly mentions that the transition kernels individually add up to at most one

dunno if that's what you're referring to, though

heck of a lot of semigroup theory there

my problem is his definition of probability function from probability density

@user2103480 your mysticism is better than mine, it appears.

@Jakobian page?

next page

at the beggining of chapter 2

THe discussion has been quite colourful today I see.......from discussing how many people to murder to convo about cats and dogs........

Hello, everyone. I have a question. Consider an integer k greater than or equal to 2, such that k is not a power of 10. Suppose S is a non-empty sequence of base-10 digits which do not start with 0. For example, S could be the sequence "2304" or "13". Must there be a power of k such that that power of k starts with S in its decimal representation.

48 to be explicit

So for example, there is a power of 3 that starts with "257" , there is a power of 20 that starts with "49" etc.

@Jakobian number?

48

for instance at 2.1.gamma he says that the measure defined by integrating against the density is a probability measure

but I get what you mean

I don't know whether I can apply fubini there since he just says "measure"

and it's also an arbitrary state space, not a borel measure or anything

yeah. This is basically my concern. I guess Dynkin could of just forget the necessary assumptions

what's certain is that they're not there

There's nothing wrong in assuming those I think. I'll just do so

@MikeMiller are you at all familiar with the basic theory of $\wp$?

um. Actually I need Tonelli theorem anyway...

Ah, nevermind I figured it out!!!!!

@user107952 yes

@LeakyNun What is the proof?

@Jakobian the question is whether you actually ever use the measure $\mu$