@Vrouvrou ? Did you confuse it with $\setminus$? I'm talking of the quotient space.

@EricStucky Are you at the UMN presently? I had the impression that you were elsewhere for a bit

"a bit"

I went to Kansas this weekend :P

I'm not coming into the office this summer, except for the 2nd and 3rd week in June

But I'm still here.

gotcha

i tend to stick around campus regardless of the term/season

(i am very boring)

jeje naw I respect that.

Honestly, I didn't really spend much time on campus even during the school year, so this isn't much of a departure for me either :P

I like my keyboard at home a lot XD

on an entirely different note: mathematica, I bite my thumb at thee

i've had this computation running overnight and for the last few hours, and when i checked this morning it was plugging along

So Eric

but about an hour ago it appears to have stopped giving any output, despite the fact that mathematica still lists it as running

in what way do you think the floor stuff would be useful?

gnash

*shrug* eh

I mean, there might be something interesting that comes out of the foundations.

Because a lot of people do the same thing you do; they think about the analytic expressions of functions symbolically, rather than function-theoretically.

We expect that the symbolic approach is feasible

because differential galois theory is a thing

But it would be nice to see it done.

I'm pretty blind to the big picture of analysis.

I don't know how much the floor function, in particular, would be insightful.

Ce sont mes pensées.

"The analytic expressions of functions symbolically"

do you mean graphs and properties of graphs?

No

Could you explain a bit more what you mean by that then?

Sorry, that sentence was a bit convoluted: I meant "they think about (the analytic expressions of functions) [symbolically], rather than [function-theoretically]."

() = noun phrase, [] = adverb.

So what do you mean by symbolic?

I mean, you'll write things like "the function contains floor"

Ah

i see

so I describe a branch of functions abstractly

No, not really.

ok

sorry, I am trying to generate an example

For instance, consider this function

Ok

$f(x) = \sum_{n=1}^\infty \frac{1}{2n-1} \sin(\pi(2n-1)x)$

So that is function theoretically I presume?

No, I'm just telling you a function

Ok

Whether you are thinking about it as a function or a bunch of symbols, depends on what you are thinking about.

You just lost me

It doesn't matter

are you talking about the ye old concept in computer science of interface vs implementation?

Hmm maybe

let me look that up.

Kk

Hmm I don't see the dichotomy between these two things.

shrugs

It seems like symbolically vs function-theoretically is an issue of implementation.

im not sure what you are getting at so it was a shot in the dark

oh ok

But I don't see interfaces involved here :P

Fair enough

i assumed it was sin

sin is a function that you just accepted is what it is

Anyway the function I gave you is not the one I was going for

:/

then later on you actually learn the equation defining it

And what is the equation defining it, if you don't mind me asking :P

Taylor series

Ah, okay.

or was it maclaurin

:p

XD

there's a good quote about this somewhere

I can't keep it straight

i just know it is

kind've sounds like the difference between convergent power series and formal power series

1 + x + x^2/2! + ... x^n/n!

wait no

thats e^x!

That's exp :P

Sin is the fen terms

cos is the odd terms

or was it the other way around?

it's the other way around. otherwise, cosine would have to vanish at zero.

Thanks

Wait...

does that mean you can add sin and cos together to get e^x?

O.O

$e^{i x}=\cos x+i \sin x$

so therefore $e^x = \cos(-i x)+i \sin(-i x)$

which becomes a lot less bizarre when you realize that cos/sin at imaginary arguments is pretty much just cosh / sinh at real arguments

and therefore it becomes $e^x = \cosh x+\sinh x$

Interesting

I never would've thought to mix I into a series

no reason not to, really

well, assuming you're not doing real analysis of course :p

semic, do you know why this isn't the square wave? (Besides the infinities, of course) It seems to agree with the formula I keep seeing.

That's the infamous conversion formulae for using polar forms of complex plane positions isn't it?

Euler's formula, yeah

you also see generalizations of it in the context of matrices, which see a lot of use in physics

yeah I know

though i'm not remembering the convention off the top of my head. google pauli matrices and you'll probably find something

an upper class man was asking about a practice problem

do you have a reference for said square-wave formula?

i can't remember what it was

i don't remember it off the top of my head

this one below fig 3. Agrees with Wolfram

mathworld also has it: mathworld.wolfram.com/SquareWave.html

Interesting fact: I actually considered the concept of the square root of negative one before learning about i.

Yeah.

Not sure what Euler's formula means but I /do/ know that square root halves angles. That's pretty easy to consider. After all multiplying -1 mirrors a point on the number line, so half multiplying should half mirror.

oh lord, i see the issue. god WA is dumb sometimes

WA?

Oh

look at the small grey text to the right of the plot command @eric

LOL

GOOD

"durr, that's a pi(n). i bet that's the prime-counting function!"

Where

in Eric's link from a bit earlier

anyways, change it from pi(2n-1) to pi*(2n-1) and its fine

Yeah

Lol

that is so dumb

it really is

Btw eric

do you know indicator functions?

yeah

Believe it or not

this is a nice picture, gibbs phenomenon and all: link

wow, my computer was struggling at N=20 and then it opened your link just fine… the miracle of optimization :)

hah

oh gosh it's already 2:00 I need to copy+past mike's text into a blog post so I feel like I've done something today.

@EricStucky $\lfloor 1 + \frac x {\sqrt {x^2 + 1}} \rfloor = I[x>=0]$

Piece wise functions are a subset of floor

if you want to see the effect of using the dirichlet kernel to smooth out the gibbs phenomonen, here you go

@EricStucky that's another good reason I just say "floor functions" because literally all of them can be built from floor

except abstract algebra

which is beyond me

No, it's really genuinely not true.

The rule about real functions is

There are always real functions that do what you don't want them to do

always

Hmm?

Are you talking to me Eric?

Yeah

I meant all of the discontinuous ones, not all functions, lol.

well barring asymptotes

Yeah, me too :P

other

There are bigger monsters in the real line than asymptotes, Duck :)

That aren't just piecewise?

there be dragons out there

It depends very precisely what you mean by piecewise. But yes, if you are using standard definitions.

I mean, in some sense, every function is a piecewise function, if you just let the pieces each be a single point :D

But this isn't a very useful way to think about things.

I meant piecewise in the sense that we don't use regular expressions

i imagine the following could be true: any function can be produced as the limit of a sequence of piecewise functions

^it is. People have don't that to me in jump series questions in a smart alek fashion

Hmm even nonmeasurable functions, semic?

no clue there

I'd imagine we mean functions that we can measure or get values of

@Semiclassical depends on how many pieces you allow

@EricStucky so when you mean bigger monsters what exactly are you referring to? Any piecewise expression based upon numerical x ranges is expressable with floor

If you only allow finitely many pieces I don't think so (I don't think the indicator function for the rationals will be of that form for example)

one thing to bear in mind: i've never taken a course on measure theory. so my knowledge is limited to the realm of "i know things can get screwy"

Duck: The essential issue is that there are more subsets of $\Bbb R$ than you think.

as Eric was pointing out, the pointwise limit of measurable functions are measurable

Once you know weird subsets, you can make weird functions pretty easily.

For instance, by taking the indicator function

Although, you can do even worse.

i do know that the cantor distribution is an example where things are pretty weird

i.e. en.wikipedia.org/wiki/Cantor_distribution

ya

*headdesk* Mike thanks for reminding me of the painfully obvious :/ :/

I was about to post a question too :P

Anywho I'm not done quoting you yet.

the cantor example isn't necessarily the best one, though, since it's an example of a singular continuous measure

so it does have a measure, weird as it may be

the csntor function is a continuous function, so certainly it's the linit of piecewise continuous functions :D

pfffft

yeah

when are quals, Eric?

a month ago

I passed real, alg, and top. So I am very happy :)

@DanielFischer is my proof mean that $T$ is continuous at 0 ?

Complex will be hard because I am starting from nothing, pretty much :/

I know that there is a thing called the Cauchy-Riemann equations, but I don't know what they are. (I know they are simple, but I still don't know them.)

when is that test? I expect you'll be fine.

Either August or April.

yeah, just spend a little time learning and you'll be good by August :)

@EricStucky when you try to define the derivative $f'(z_0) = \lim_{z \rightarrow z_0} \frac{f(z_0 + dz) - f(z_0)}{dz}$ for a complex function, you see you need to ensure that the value of $f'(z_0)$ is the same no matter what direction $dz$ goes in, just as you do this when definin real-valued derivatives, in ensuring this will hold you'll arrive at the Cauchy-Riemann equations.

Any thoughts on 'intersection theory' in differential topology? Seems like it's like a coordinate-free generalization of the fact that you define a system of non-linear equations by the intersection of surfaces?

I usually think of intersection theory as an algebraic geometry thing; I suppose you can ask your varieties to be smooth.

That's about the extent of my thoughts.

So did you mean things like all whole numbers = 1?

cause that's just I[x mod 1 = 0]

and testing for equality can be made with floor

but I see your point

fractional vs irrational

things like that

but anyway. A good chunk of piecewise is derivable from floor

did you know that multiplying indicator functions is the same as AND ing them?

|a - b| = a OR b

1 - a = NOT a

Interesting, eh?

wait

not OR

XOR

@EricStucky remember how I said that subtracting the jump series fixes the implied integral

behold

$\int^c_a + \int^b_c = \lim_{n -> c^-} F(n) - F(a) + F(b) - \lim_{n -> c^+} F(n) = F(b) - F(a) - the difference of limits

see

it is confirmed

Woah, sorry for the pause

My browser stopped informing me of new messages.

Yeah, algebra of indicator functions is something I'm pretty familiar with :)

I'm not sure how your calculation has anything to do with subtracting the jump series.

I'm guessing that the difference of the limits is a feature of implied integration, because in ordinary integration this difference will be zero for any "function with a jump series" (as I defined in that one question of yours).

@TheGreatDuck

@EricStucky Can I tell you something interesting about complex analysis?

Integrals are really nice.

@MikeMiller I don't know how to do this offhand. For irreducible 3-manifolds this is harder than obstructing surgery on a knot (and thats already too hard for me. Seem like there might be some tricky way to do it if you allow reducibility.

Hey guys! I'm back :) How was your days

You know how, with real functions, you can integrate them over an interval? And, if they have an antiderivative, you can compute it by subtracting the values of the antiderivative at the endpoints?

@PVAL I do allow reducibility. But I could ask Tye when I see him next. He probably knows.

Certainly you can't do anything dumb like $mM \# n\overline M$.

With complex functions, you have to integrate over a curve, since now specifying two endpoints doesn't define how you get from one to the other.

@MikeMiller Ya I'd think reducible examples would be known.

@EricStucky But, if our function has an antiderivative, it turns out that the exact curve doesn't matter; we can still just subtract the values of the antiderivative at the endpoints! In particular, integrating these over a loop gives $0$.

Did you see the Stein message?

What's the question?

The existence of contractible 4-manifolds which aren't Stein is proved in Mark and Tosun's paper.

(Note that $1/z$ doesn't have an antiderivative; $\ln z$ isn't continuously defined over the whole complex plane.)

As it turns out, this is also true of any function that simply has a derivative!, as long as it's defined everywhere inside the loop.

Reason is, you can break the interior of the loop into a bunch of small triangles. And then the integral over the (oriented, by the way) loop is equal to the sum of the integrals around each of the small triangles, since wherever two triangles meet, the orientations cancel out.

There's a program at Banff or something next year that listed that as an important open problem.

What's an example?

Any contractible 4-manifold bounding $Sigma(2,3,13)$.

One is in a paper by Akbulut-Kirby and in a paper by Casson-Harer.

"The function has a derivative" means the function is locally linear. So, on each of those small triangles, the function is approximately linear, and linear functions have antiderivatives! This means that the integral around each triangle is doubly small, because it's a small loop and it almost has an antiderivative.

It turns out that these are small enough that, when you add them all up, we can show that the total thing is smaller than $\epsilon$ (for some $\epsilon$ depending on the way we break the interior of the loop into triangles).

Choosing increasingly smaller triangles lets us show that it's smaller than every $\epsilon>0$, and so the whole thing is zero. So the integral of any differentiable function on a loop, as long as it's defined everywhere on the interior, is $0$.

In this paper for instance math.berkeley.edu/~kirby/papers/…

Thanks

hellloooo

@EricStucky the full function of jump series is just a series of differences of limits being added together. So, it winds up being that the implied integral and jump series is a way of restructuring another method called "splitting the integral".

@EricStucky however, it could still be useful outside the context of integration. Besides, splitting the integral is a way of taking it over continuous periods so as to avoid discontinuity. My method is still right. It's that I can now certifiably prove that it is correct as an integration method.

Can someone briefly explain Hidden Markov Models to me? I think I'm misunderstanding the model.

We have a Markov Chain that is unobservable, but it's image under another function $f$ is observable. Is this correct?

Or is there something missing?

Hi

I am struggling to understand when can i get out with differentiation from the absolute value

in a book i have that $\frac{d}{dt} \nabla X(a,t)= \nabla g(X(a,t)) \nabla X(a,t)$

and then i have $\frac{d}{dt} |\nabla X(a,t)| \leq |\nabla g(X(a,t))| | \nabla X(a,t)|$

And i just dont understand - whats the reason/theory behind such operations?

@Max whenever I see absolute value and derivatives in the same place, I panic and just replace $|a|$ with $\sqrt{a^2}$. Good luck with that.

oh i should've written $\frac{d}{dt} sup(a) |\nabla X(a,t)| \leq sup(a) |\nabla g(X(a,t))| sup(a) | \nabla X(a,t)|$

it is supremum norm

I see what was missing from what I was saying, now. The observations aren't just a function, they're a statistical process dependent on the current state. If it were just a function's image, we could always just redefine the chain in terms of the image space to make a Visible Markov Chain directly.