which letters one uses to represent things is not very robust and hence not a reliable means of interpretation always

so (derivative plus 3t times the function)y=(t^2 x)

I don't disagree with that, but I know the interpretation that I'm stating is what my teacher had in mind

how do you know that?

she told us to solve it like this:

First we choose x1 and x2 as two arbitrary functions

and a and b are real constants

Then replacing a x1 in the equation we get

$y_1' + 3ty_1 = t^2 (a x_1)$

(We're actually using ax_1 as our function and not just x_1)

Similarly we get

$y_2' + 3ty_2 = t^2 (b x_2)$

Now if we put

$x_3 = ax_1 + bx_2$

We get

$y_3' + 3ty_3 = t^2 x_3 = t^2 (ax_1 + bx_2)$

And basically then you have to show $y_3 = ay_1 + by_2$

looks like we're doing that essentially

define $H(x,y):=y'+3ty-t^2x$, you're checking it's linear as applied to $(x,y)$

If you can do that, the you have proved the system is linear

Hmmm

I was interpreting it as simply $ H(x) = y$

our teacher is not the very best at mathematics, I think I should mention that

(meaning that she can be wrong)

It wouldn't make sense for the DE to describe $H(x) = y$, right?

correct, that wouldn't make sense

Because more than one $y$ can satisfy that DE for each function $x$ leaving the system ambiguous

if we had something like $y - x = t^2$

Then it would make sense

Because then our system is simply $[H(x)](t) = x(t) + t^2$

And we can easily talk about whether this is linear or not

The system here is clearly defined

how do you know it wouldn't be $H(y)=x+t^2$ or $H(x,y)=y-x$?

there aren't any derivatives

How is $H(y) = x + t^2$ different from what I wrote?

Are you thinking of $t$ as a number or some function?

x and y are variables of t

@Alraxite sorry I meant $H(y)=y-t^2$

okay

Because

our teacher has always been using $x$ as the input function and $y$ as the output function

and as for H being a system of two variables

I mean two functions

we haven't talked or studied anything about that in class. so far we have only been dealing with single function systems

If we do interpret that DE as describing a two function system

Then why would one write $y' + 3ty = t^2 x$

I mean, how do you get $H(y, x) = y' + 3ty = t^2 x$

from that equation

oops

$H(y, x) = y' + 3ty - t^2 x$

I meant this^

if we're interpreting H as applied to x and y, then collect x and y terms on one side and everything else on the other

which would give $y' + 3ty - t^2 x = 0$

why would anyone write that to describe $H(x, y) = y' + 3ty - t^2$?

doesn't that equation describes a relationship between x and y

describe*

I think this problem itself as our teacher has given us is wrong

take x'+2x=1 for instance. we can interpret this equation as Hx=1, where Hx:=x'+2x.

so when we see y'+3ty-t^2x=0, we can interpret this as H(x,y)=0, where H(x,y)=y'+3ty-t^2x

So it defines a system and also gives us a particular value of the system for some x and y?

Is this a standard way to describe systems?

or are you just trying to guess what our teacher could possibly have meant?

it's a standard way to write equations using linear operators

okay

put input functions on left side, other functions on the right side, name the left side some linear operator, and now the equation may be rewritten as operator(input)=otherstuff

and usually problems of this sort would proceed to ask us what input can be?

for operator(input) to equal otherstuff?

because then the equation is not only describing a system it also is giving us a particular value in the range of the system

(like 0 in the case of H(x,y) = 0)

well yeah, equations are not functions, equations restrict what kind of values the things in the equation can be

yeah

like x+y=5, you can't have x and y be anything independently

or x^2+x=5. we can treat that as f(x)=5 where f(x)=x^2+x.

I think the problem as our teacher has given us is incorrect

it asks us to simply tell whether the system is linear or non-linear

if the the equation is meant to be understood as describing a system and also giving a particular value for the output then giving the second piece of information wouldn't make sense if you only want to find whether the system is linear

it's like asking whether the function on the set of real numbers described by x^2 + x = 5 is linear or not (here linear having a different but analogous definition for functions on the set of real numbers)

@arctictern okay thank you for your time

I understand better what the problem with the question as stated is

@0celo7 Yep, sorry.

I am around now.

@0celo7 OK. Why should I care?

@0celo7 Yes, considering many, many people (who are, as far as I am concerned, human) actually do point-set topology as research.

@0celo7 Homological orientation is awesome cool.

There is no extra intuition to be gained, no, as they are equivalent for smooth manifolds. But homological orientation is much more powerful, as it works in a more general setting. It's fine if you don't care for it though, nobody's forcing you on that.

The problem with defining orientation for smooth manifolds is that it's a parametric definition. You set an orientation on each fiber in the tangent bundle and ask it to be smooth. That makes it sort of coordinate-dependent.

Whereas, homological orientation has this simple version: an orientation on a manifold is choice of a generator in the top homology, whenever it's isomorphic to $\Bbb Z$. If it's not isomorphic to $\Bbb Z$, the manifold is nonorientable.

However, this does have an analogue in smooth topology: an orientation is the same as choice of a top dimensional form on the manifold.

@BalarkaSen You mentioned the other day that you wanted to prove it's equivalent to the standard one.

@BalarkaSen I don't see how people doing it makes it worth their while. Best not respond to that.

@BalarkaSen The book only considers smooth manifolds, so it's strange they use the homological definition.

@BalarkaSen The definition using nonvanishing top-degree forms isn't coordinate dependent...

@0celo7 I did; that's not equivalent to saying I wanted to read it up from a book.

@0celo7 Read again; I did mention that.

The point is, the definition using forms isn't so different from the definition using singular homology.

@BalarkaSen Ah

It's in fact exactly the same definition with "singular cohomology" replaced by "de Rham cohomology".

@BalarkaSen I figured

Although the homological definition uses the relative homology

I don't know how relative de Rham works

It's in Bott & Tu, but tonight I want to see if I can understand Picard-Lindelöf

@0celo7 You mean the $H_n(U, U - p)$ thing (also known as local homology)? That's the analogue of the coordinate dependent definition.

@BalarkaSen I'm saying I don't know what $H^n_{dR}(U,U-p)$ means.

I don't if that can be translated to forms: probably you'd want to ask for compactly supported forms.

Yes, I think it'd be the collection of all forms compactly supported around $p$. Or something.

Maybe you'd want to play with that idea. I think it should essentially be an analogue of germs of functions at $p$. You have to do the same thing with forms so that you can make the homology groups appropriately.

$$f_n(x)=\max (n-n^2|x-1/n|,0)$$

The worst function

It's a peak at $1/n$.

Well, of height $n$.

It has really bad behavior as $n\to\infty$

part goes to zero, part diverges

Yes, $f_n$ converges to the 0 function pointwise but not uniformly.

$\int f_n$ are all the same, namely, $1$.

Hi @AlexWertheim

Hello @Balarka. How goes it?

So-so. What about you?

Not bad. Recently resumed mathematical activity after a long break, so that's been nice.

Nice. What have you been doing, mathwise?

Just shoring up general knowledge of algebra. UCLA has an algebra qual, which I've already passed. But the old exams have lot of good problems, so I've been writing solutions to them.

In time, hopefully I'll add some problems from Atiyah-Macdonald to that. I should also be learning smooth topology, but we'll see where that goes. :)

Ah, nice. I haven't done algebra in months.

You?

Studying smooth topology and complex analysis, mostly.

Things are going a bit slow.

Good stuff. Need to know parts of both of those, but I don't anticipate knowing much for a while. Hopefully in 6 months, things will be better.'

Slow is good. :) What are you doing, specifically?

Specifically, as in?

Well, smooth topology and complex analysis are very broad, is all.

Maybe what you've been doing recently. :)

Oh, OK. Just the very basics, from G&P and Stein&Shakarchi.

Onto oriented intersection theory in the former, and going through bunch of exercises from the latter from the first 3 or so chapters. Plan to study Fourier analysis from chapter 4, see what it's about.

Mostly using complex analysis as a gateway to actual analysis: people suggested that path would be more motivating than doing real analysis first.

It helps that Stein&Shakarchi has a Fourier analytic theme with it.

Nice, sounds like good stuff. Complex analysis is beautiful, and S&S is supposed to be very good, though I don't know anything about it personally.

Makes sense, since Stein is a famous harmonic analyst.

Right

Hello?

Ah, never mind

Hello now?

Question: is there a way to transform the minimal polynomial of 2 cos(2 pi/11) to the Chebyshev polynomial T_5(x), and if so how?

@ParclyTaxel Unless my interpretation is wrong, I thought T_5(x) is fixed?

Something like a Tschirnhaus transform, which I've tried and found not to work

Context for this problem is math.stackexchange.com/questions/1876590/…

So the question is

wait, I've said it above!

@ForeverMozart Interesting idea, but there are still many additions you have not defined, such as $0 + 0'$ and $0' + 0$ (I have no idea if it can be completed into an example).

Question.

What is x^10 - 11x^8 + 44x^6 - 77x^4 + 55x^2 - 11 factored over Q[sqrt(11)]?

Can I use Wolfram Alpha for this?

so our Functional Analysis prof recommended Rudin for the course but the first chapter he says will be taught is right at the end of the book

Can anyone up now answer my question.

@ParclyTaxel why would adding the squareroot of $11$ allow you to factor it?

Apparently

the polynomial is the minimal polyn. of 2 sin pi/11

Therefore it is cyclic and can be factored over Q[sqrt(11)]

I don't see any easy way to do it by hand

So writing it as a polynomial in $x^2$ you get something irreducible of degree $5$ over the rationals

The basis for my question was actually asked above

The post that says "something like a Tschirnhaus transform..."

I'd also like to know the exact method Gauss used in DA to solve for trigonometric angles in terms of radicals

Well, I'm a bit disappointed. Spent all this time getting some Double-Shift QR algorithm to work so I can solve for eigenvalues of a matrix. Turns out it's quite bad. It will get only some of them. I thought that the example procedure in the PDF they ran was "intermediate output" or output that represented what it might look like when running, but it turns out that I get the same thing when I run the same matrix (It's the end result). And to get only 75% of the eigenvalues is terrible. Sigh...

Need convergent nets be bounded? (topology)

@EsX_Raptor being bounded need not even make sense in general.

@TobiasKildetoft Good point. Suppose a convergent net goes from a directed set to the reals with the standard topology. Then need it be bounded?

@EsX_Raptor Yes, then it certainly has to be bounded (same proof as for sequences)

In fact, as far as I recall, nets behave basically like sequences in every way when the space is nice enough

@TobiasKildetoft Then I'm confused. Define a net $(0,1)\to\mathbb R$ by $x\mapsto1/x$. Then it converges but is unbounded.

I keep getting the feeling I might be missing something.

@EsX_Raptor ahh, right, it can be unbounded in the other end of course

been too long since I thought about nets

I think the deal is as long as the domain of the net is a directed set, then I can come up with examples like these.

Well, by definition the domain of a net is a directed set, lol. I need some sleep.

@TobiasKildetoft Thanks, Tobias.

@EsX_Raptor But I think you can always cut off some part of the domain and get a bounded net with the same convergence behavior

though I am not certain

@robjohn hey! How is it going? You're rare again these days here.

@0celo7 how can you know about homology when you don't know relative homology? it is totally fundamental. you might wish to study homology systematically

@user1618033 you are into that Ramanujan series stuff, aren't you?

@CRAZYGAYSHERIFF lack of intelligence

ok i can tell you about

for "nice" topological pairs (X,A) the relative homology H(X,A) is the reduced homology of the quotient X/A

this is how i think of it intuitively

@CRAZYGAYSHERIFF I wanted to be but lately I realized I'm not that good. Therefore these days I play more with pokemon math.

formally it is the homology of the chain complex S(X)/S(A)

btw i always wanted to read of the proof of that series for 1/pi

@user1618033 can you recommend me a proof using complex analysis?

reasonably self contained?

well he gave many series

@CRAZYGAYSHERIFF no

i guess i should specify which one, just the most famous one

why not?

i thought you are an expert in this stuff

@CRAZYGAYSHERIFF No, I'm not.

@user1618033 series, integrals, isn't that your speciality? surely you must have come across a proof of that most famous series?

@CRAZYGAYSHERIFF I never think of speciality, I just do in math what I like.

@CRAZYGAYSHERIFF Then I trained my mind particularly for real analysis approaches (and I wanted to stay away on purpose from complex analysis - for a while).

but why no complex analysis? don't you think that it is very useful for real analyis?

it can make proofs much shorter, and i mainly want short proofs

when i am reading about that strange series, it appears that it is very deep actuall

y

it seems to be basically an evaluation of an equality of modular functions

@BalarkaSen Your kind of book!

Oh, I hate it when they do that and the proof isn't obvious.

"@Balarka: The request to avoid words like "trivial," "obvious," and "clear" in mathematics is a valid and good one. :) - jul 29 at 20:06 by Ted Shifrin " i disagree

Luckily, the proof is obvious.

@Krijn This one pissed me off:

What book are you reading? @0celo7

@Krijn First one is from an analysis book that I'm skimming, second was from a diff geo book I read a while ago

Technically I'm reading Bott & Tu, who instead of saying things are obvious, make claims that take pages of work to verify

Or maybe they would be obvious if I knew Husemoller from back to front

There is this video on YouTube from Serre where he discusses this and calls it "cheating" in mathematics.

You should watch it once if you have the time

Link?

I've seen a bit of that video, it's pretty great

How are you supposed to write truly easy proofs

@user1618033 busy, but I'm here quite a bit.

@robjohn OK

@user1618033 what's up with you these days?

@robjohn Working as usual on some of my stuff, but slower than before, I feel pretty tired lately. Thanks.

Hi everyone

someone know an example about a groupoid that is not a group?

@user1618033 yeah. We got new garage doors this last weekend. There is a lot of preparation and cleanup that needed doing.

@robjohn hehe, you'll have to do a lot of stuff, an opportunity to take a break from math (which I miss). ;)

@HirotoTakahashi off the top of my head, take the set of all bijections from any open subset of $\mathbb{R}$ to any other, with the operation being composition. This works since you can only compose functions sometimes, but everything still has a left and right inverse, and composition is associative.

for a more trivial example, you can just take two groups at once, where the multiplication exists only when you take a pair of elements both from the same group, in which case it's just the old multiplication

@SamuelYusim Thanks!

one other popular one (if you're familiar with topology) is the fundamental groupoid, which is like the fundamental group at a point, but you just take all paths from any point to any other, with the multiplication just being composition, which only exists when one path starts where the previous one ends

(of course, this is a group if your space is a point but whatever)

also, no problemo

I have a sequence of points (x_i, y_i), i=1,...,N and I want to find a part (x_i, y_i), i=N_1, ..., N_2 which is well approximated by an affine line. Is there any theory/algorithm about finding this part of the data set? Part of the problem is that I want the "longest" part, which still gives a "good" fit; but I don't know how to make these terms precise. I could do a linear regression on the part N_1, ..., N_2 of the data, but then I would have to try all possible "parts", which could be a lot.

BBL

@SamuelYusim that's a crappy space tbh

well true but it's a space nonetheless

not everything can be hausdorff

Hausdorff? I like mine second countable and locally compact, too

And compact, too

fundamental groupoid is nice

from a formal point of view at the very least

@Krijn I like the cool topology picture he draws there.

@CRAZYGAYSHERIFF I don't.

what?

Follow the small grey arrow in the message.

ah that

well these words are traditional in maths

but i guess instead of saying "it is clear that P " it is better to say "P"

for a proposition P, just no argument at all

I mean, "it is clear that" is equivalent to the trivial argument

logically yes

but saying "it is clear" is actually of no value

it is like saying "let x be ANY element of R" when you can just say "let x be an alement of R"

it doesn't really add any content

sure I agree, but if you want to add some words to string your propositions together, why not use "it follows that P" or "we have P"

true dat

at the same time this whole discussion is anyway kind of trivial, isn't it :D?

at least that way if I'm too stupid to get the point I won't feel bad about it

true

personally i never felt bad when they write "it is trivial" and i didn't understand it

that could be for several reasons: a) just skimmed thru the previous parts, b) didn't think carefully about definitions

getting offended by "trivial, clear" is mainly for people that overestimate themselves

@0celo7 I think of homology as a generalization of homotopy in the following way. $M$ be a manifold, then elements of $H_n(M)$ are equivalent classes of closed $n$-dimensional submanifolds of $M$ with equivalence the following: $n$-submanifolds $X, Y$ of $M$ are equiv if there is an embedded $(n+1)$-manifold $Z \subset M$ bounding $X \sqcup Y$ (pro-tip: I'm lying a bit; this definition doesn't actually work).

Then elements of $H_n(M, A)$ are equivalence classes of not-necessarily-closed $n$-submanifolds of $M$ with boundary in $A$. The correct equivalence relation is harder to understand.

This is not a simple way to think about it or anything, but I find all of this pictorially exciting.

(irrelevant, but intersection theory gets a more formal context in this "representation": if I have a closed $n$-manifold $X$ and a closed $m$-manifold $Y$, I can take intersection to get a closed $k$ manifold $X \pitchfork Y$. This indeed gives a well-defined map $H_n(M) \times H_m(M) \to H_k(M)$. If you dualize by Poincare duality, you'd get the cup product aka analogue of wedge product in de Rham cohomology - you can compute $k$ explicitly by transversality)

@BalarkaSen: do you know what's a standard reference for the fact that every cco surface of genus >= 2 admits a pants decomposition?

@Huy why so much topology lately

what's a pants decomposition? writing it as union of a pair of pants and disks?

why doesn't just drawing the decomposition suffice, then?

@BalarkaSen: just pants

hm sure one can generalize

@0celo7: idk I find it much more interesting and surprising than other stuff.

@Huy aka a surface which bounds a circle on one side, and two disjoint circles on the other?

@BalarkaSen: btw I think I managed to somewhat rigorously do the proof that a chain of an even number of scc is always nonseperating

the standard terminology for that is "pair of pants".

pair of pants = sphere with 3 removed disjoint open disks

I don't wear pants

@Huy Yes. So what's a pant?

it's the same

that's what I meant by it

I'm not an English major

OK, I guess I was confused by your message "just pants"

oh, you mean, not disks, just pair of pants. OK.

@BalarkaSen I think that's the view that Bott & Tu take

We'll see when I get to homology in that book...

I doubt.

It use a forms-point of view. Actually making sense of what I said required a lot more work.

@Huy Ah?

They say that homology is a "higher-dimensional analogue" of homotopy in the preface

That's an intuitive notion; I don't think they base their stuff on it.

Because it's nontrivial to do that, is my point.

(e.g., look at Steenrod realization problem)

Maybe Bott & Tu is trivial if you read Steenrod.

They have random statements about vector bundles that are taking a long time to verify

And it doesn't help that I spent all of yesterday reading analysis, not topology

That seems completely irrelevant to what we were talking about...

@BalarkaSen if you want I can write it down tomorrow, one needs to be a bit more careful than I thought

to rigorously argue that the boundary of the tubular nbhd is always a circle

OK. I still believe you can induct that out.

yes, it's via induction still

maybe you had the right thing in mind and I misunderstood you

I didn't write down the proof though :P

yeah :d

You shouldn't believe everything I say.

I don't believe everything I say either.

you're not a belieber

:(

that last typo is dangerously close to being "bieber"

;)

I love Bieber

I don't expect anything different...

@BalarkaSen Hmm?

Never mind.

I gotta show that the inversion map $z \mapsto 1/z$ takes a circle to a line/circle. Not entirely sure where to begin. Any hints?

Perhaps write out the equation for a circle and do the transformation, @Kari?

The standard $|z-z_0|=r$?

I recently ran into this form as well. Would you recommend one over the other, @Danu?

I don't know

How many hours of math do you do daily (sure, on average)? It's a question addressed to all.

@Huy

@Krijn

Does it seem to me or the channel is really weirdly silent today? OK, I got that! Everybody is working hard on math.

Good then.

No, I'm not.

No more questions.

I was just watching a series.

@Krijn Which one?

Hannibal

Never watched it.

When I'm at uni, like 6-8 hours a day, I'd say.

Yeah, 8 hours seems pretty fine (especially if working hard).

Although a lot of the time, I'm working with other students so it's not mathematics all the time

I see.

How do the students there seem to you in the age of Pokemon Go? Are they interested in reaching (top) performance?

When I'm working on a project I do work hard and make a lot of hours in spurs

Most do not know enough mathematics to understand it, but they pass the tests and that's all they care for

Some are exceptionally good and perform really well, but care too much for grades

Only very few are very good, don't care for grades and go for understanding

I see.

I don't blame them though

It might have sounded weirdly if I said instead "reaching (top) performance" just "having a lot of fun". The former form may annoy some people, but by that I simply understand having much fun with mathematics.

Even among top students there are more ways to have a lot of fun than just mathematics, though

That's fair.