@Semiclassical but they're just differential equations rewritten with integrals, right?

That's the very simplest kind. There are harder versions.

ah ok

wanted to make sure I wasn't abusing terminology.

Here's some stuff on a specific kind of integral equations: en.wikipedia.org/wiki/Fredholm_theory

@Semiclassical turns out that my conjecture regarding my operator was never talking about differential equations. In reality this whole time i have been intending to talk about the associated integral equation of a differential equation. That's because integrals can in theory produce things that antiderivatives cannot. For instance, integrals do not need to be differentiable.

that doesn't sound right. the last sentence, I mean.

$\int_0^x \lfloor t \rfloor dt$ does not have to be a differentiable function.

think carefully about it

but short of making floor undefined at integer points, there's not really a true antiderivative that holds for all points on floor(t)

That doesn't seem correct.

i mean...if you really think about it there is no antiderivative at sharp corners.

...sure there is. The area changes smoothly.

@Semiclassical take the derivative of the antiderivative of floor(t) at 5

@Semiclassical i meant that one cannot have an antiderivative containing sharp corners without implying that the original function had nonexistence somewhere.

I thought you meant that the antiderivative of a function with a sharp corner wouldn't exist at that point.

@Semiclassical oh no not at all.

Not, an antiderivative can't have sharp corners.

Which is true so long as you're not doing Dirac-delta stuff.

@Semiclassical precisely, but an integral is just an area function if we mean a definite integral

writing that as $F(x)=\int_0^x \lfloor t \rfloor \,dt$

exactly

and that does exist in a logical sense

Hmm, what's the most lazy way in which I can do that integral.

I guess it'd be $F(x)=0$ for $0\leq x\leq 1$.

subtract away the step function best approximating the jump discontinuities from x*floor(x)?

oh and add 0

No c. This is the definite integral.

XD

mine earlier had a "c"

?

Then $F(x) = x-1$ for $1\leq x\leq 2$.

i posted that same integral earlier with c

Ahh. $F(x)=1+2(x-2)$ for $2\leq x \leq 3$.

Okay, fine now.

It's just x*floor(x) - floor(x)^2/2 - floor(x)/2

Eh, that's presumably equivalent.

the latter two terms come from the n-series btw

it is

presuming my memory of the n-series is correct

I'd be lying if I said I was patient enough to actually work it out, though.

ive used that particular formula in quite a few places

Sign of the kind of weird person I am: I'm tempted to take the Laplace transform of that.

the latter two terms essentially are the step function matching the jump discontinuities of the first term's graph

@Semiclassical actually that is the standard method

hah, good to see I still haven't forgotten things.

@Semiclassical my conjecture is that one doesn't need to use the laplace transform. You just use a different kind of integral/antiderivative and find the continuous solutions.

presumably... it is easier. XD

(most of the time, but the few that elude me would be increasing too fast for laplace anyhow)

Well, I wasn't thinking Laplace transform at first. I was literally just thinking "how much does the area increase as I increase $x$.)

ah

e.g. $f(x)=0$ for $0\leq x<1$, so $\int_0^x f(t)\,dt = \int_0^x 0\,dt =0.$

And past that you've got $(\int_0^1+\int_1^2+\cdots+\int_{\lfloor x \rfloor}^x) f(t)\,dt$

My thought behind my method is basically that floor(x) is essentially constant so x*floor(x) provides the "slope" and a step function subtracted or added to it brings in the necessary "continuity" required for an integral. I hope that makes sense.

@Typhon I would agree with you.

Oh finally that worked.

@LeakyNun yes well the conjecture is that the heart of that mentality extends to integral equations in general.

@Typhon for example?

or rather.. the integral equations corresponding to differential, equations.

And then $\int_k^{k+1} f(t)\,dt = k$, so from all but the last term one gets

@LeakyNun I cannot write the integral equations corresponding to differential equations so I will provide a differential equation.

@Typhon go ahead

$y'' + 3floor(x)y' + 2floor(x)y = 0$

$0+1+\cdots+(\lfloor x\rfloor -1)=\frac12 \lfloor x \rfloor (\lfloor x \rfloor-1)$

by the conjecture, one can solve it via auxiliary equations... sort of

$y'' + 3 \lfloor x \rfloor y' + 2 \lfloor x \rfloor y = 0$

this gives $y = c_1(x)e^{\lfloor x \rfloor} + c_2(x)e^{2\lfloor x \rfloor}$

And then the last term is just $\int_{\lfloor x \rfloor}^x f(t)\,dt = (x-\lfloor x\rfloor)\lfloor x\rfloor$

So that presumably combines correctly, modulo any silly mistake.

the c's are step functions and the conjecture proposes that the continuous solutions of that thing are the solutions to the integral equation

@Semiclassical not sure what you're doing

In $\Bbb R[[x]]$, $~x$ can be written as an infinite polynomial in terms of $x^2+x$.

^^this contains a description of the operator I'm referring to

well, what I just got would be $F(x) = (x-\lfloor x\rfloor)\lfloor x\rfloor+\frac12 \lfloor x\rfloor(\lfloor x\rfloor-1)$

the one by which I'm 'solving' the differential equations.

(Cont'd) I think that follows from the implicit function theorem or some such (or induction also works probably)

which hopefully agrees with what you had?

(I'm not certain I didn't make a silly error)

turn off the syntax and I'll check it. XD

pff

I don't know how to quote

(x-floor(x))*floor(x) +1/2*floor(x)*(floor(x)-1)

Hey, if a ring $R$ contains $\Bbb R[[x]]$, need $(a_0+a_1x+a_2x^2+\dotsb)y= a_0y+a_1xy+a_2x^2y+\dotsb$ hold?

Need that equation even make sense?

which I guess would just be floor(x)*(x-1/2*floor(x)-1/2) if I combine that a little.

I'll just name floor y.... (x-y)y + y/2(y-1) = xy - y^2 + y^2/2 - y/2 = xy - y^2/2 -y/2

you did it correctly

weeee

good job

So yeah, it can be done just by direct integration if one is careful.

(what a shock /s)

interestingly enough finding the continuous solutions to that diferential equation I mentioned is the same way one finds them for this one. It's just a matter of looking at the individual solutions separately and ensuring they are each continuous.

Semi, any idea on my ring question?

nooope

so the final solution is just...

hi @BalarkaSen

Hey

physics exam went fun

Had a diff. forms question which I wondered if you'd know the answer to. (It's a terminology question more than anything)

@BalarkaSen Oh, nice.

I have to run to school in a few minutes but ask on

Sure.

Suppose I have a 1-form $\omega$. If I wedge that with a p-form, I get a (p+1)-form.

So that gives a mapping from the space of 0-forms to the space of 1-forms and so forth.

$y = c_1e^{\lfloor x \rfloor - \frac{\lfloor x \rfloor^2}{2} - \frac {\lfloor x \rfloor}{2}} + c_2e^{2\lfloor x \rfloor - \lfloor x \rfloor^2 - \lfloor x \rfloor}$ is the solution to $y'' + 3 \lfloor x \rfloor y' + 2 \lfloor x \rfloor y = 0$ The trick is auxiliary equations mixed with the stuff I mentioned regarding the continuity of the previous solution. @Semiclassical

@Semiclassical Turns out there was a question on deriving the capacitance of parallel conductors. I mentioned the appropriate boundary value problem w/ Laplace equation in it, just to sound smart.

But if I wedge twice with that, I'll just get zero b/c antisymmetry

@BalarkaSen lol

$V=ax+b$, $V(0)=0$, $V(d)=\Delta V$.

@Semiclassical Sure. A map $\Omega^n \to \Omega^{n+1}$.

sorry... I lied

$y'' + 3 \lfloor x \rfloor y' + 2 \lfloor x \rfloor y = 0$ has no solutions

XD

Is there a name for that sequence of mappings?

I'm referring to the corresponding integral equation

anyways

@Semiclassical that's what I got bored of which prompted me to mess with continuous matrices otherwise known as "integral transforms"

and also number theory

@Semiclassical That's an interesting question. I don't think there's a nice terminology.

Hmm.

make one up?

It seems a lot like how the exterior derivative itself acts.

Yeah.

something something integrability... hmm...

It's not the same, I imagine, but in both cases the image of one map is the kernel of the next

Frobenius or something?

@Semiclassical I got a formula for that thing from earlier. Let me just think out how to exactly write it. It's quite simplistic.

nvmd Edit: nvmd the nvmd

@BalarkaSen The other thing is that I could have instead used a k-form instead of a 1-form

$\sum_{n=0}^{\lfloor \log_y(x)\rfloor} (x \mod y^n) == 0$

In which case I'd be getting the sequence $\Omega^n \to \Omega^{n+k}\to\cdots$

I sorta want to say the word 'filtration' fits in here but

I really don't know

"==" is the function that returns 1 if equal and 0 otherwise

it's a little tricky to write so I omitted it

I have definitely seen $d\phi = \omega \wedge \phi$ appear somewhere, but I can't remember where.

Connection forms?

hmmmmm

Yeah, that really does seem familiar.

In any case, the issue is there's no reason to believe $d\phi = \omega \wedge \phi$ for a universal $\omega$ for all $\phi$. I think if there is such a universal $\omega$ that corresponds to integrability of an appropriate distribution (kernel of $\phi$?) or something.

phew

got that edit in right in time

@BalarkaSen go to school already :P

Sure, I agree.

@BalarkaSen quit ditching class

I'm more just curious what kind of structure we're getting.

Seems like one can construct a cohomology complex just based on that?

e.g. look for p-forms which wedge with $\omega$ to 0 but which aren't of the form $\omega \wedge \phi$

hmmmm

@BalarkaSen but yes, go to school :P

This is interesting.

meh school can wait for me

@Semiclassical @LeakyNun the proof I posted is not nonsensically

It seems to me trivial actually

Then you're wrong.

@Zee what is your proof?

I posted it above, what is the issue with it?

@Semiclassical Let $f(x,y) = \sum_{n=0}^{\lfloor \log_y(x)\rfloor} (x \mod y^n) == 0$. Then the collatz conjecture function can be written as $C(x) = \frac {x}{2^{f(x,2)}} + 1$.

@Zee this one?

Yes

n=6 is even but 3/2*n = 9 =3*3 is not of the required form.

This is my response

^

That's the same issue I'd bring up.

6 becomes 3 , 3 is odd, so that becomes 4, which even, so it becomes 2

6 -> 9

@Zee no, 6 doesn't become 3

This isn't Collatz.

I thought the rule was $x\mapsto3x/2$ for even $x$

It is.

3 also doesn't become 4

That what dodsy posted, idk even know whose collatz

(In this not-Collatz thing we're doing)

....

morning, chat

That's not what Dodsy posted.

Hi @Soham

then learn to read better

3 becomes 2 sorry

@Semiclassical I am more interested in what one can say about the distribution given by kernel of $\phi$. I am pretty sure it's going to be integrable.

6->9->5->3->2

Yes that's it

i.e., it generates a foliation

@BalarkaSen that does sound like what you'd be interested in :P

@Zee take n=28. $28 \mapsto 42 \mapsto 63$. I don't see how you can say that $p$ goes to 2, or that eventually you would reach 6.

14->21->11->6->9->5->3->2

It eventually does in that case, but you've given no argument for why it must do so in general.

if you need that much step, then your claim is basically equivalent to the conjecture itself

The Dodsy Conjecture: Take any integer, if the integer is even, multiply it by 3 and divide by 2. If the integer is odd, add one and divide by 2. The number will always eventually reach 2.

Without that, it's not a proof of any kind.

Oh wow it is equivalent

Compare:

Dodsy: 14->21->11->6->9->5->3->2

Collatz: 13->20->10->5->8->4->2->1

(Going two steps at a time for odd things in Collatz)

Dodsy's version is just shifted up one number.

hmm...

So Dodsy's is equivalent to Collatz.

well damn

nice observation.

But it makes sense, come to think of it. the way dodsy got it was by flipping the even/odd steps and doing some more stuff.

@Dodsy ^^^^^^

@Balarka do you know what RKMVU is like nowadays?

3->2
4->6->9->5->3
5->3
6->9->5
7->4
8->12->18->27->14->21->11->6
9->5
10->15->8
11->6
12->18->27->14->21->11
13->7
14->21->11
15->8
16->24->36->54->81->41->21->11
17->9
18->27->14
19->10
20->30->45->23->12
21->11
22->33->17
23->12
24->36->54->81->41->21
25->13
26->39->20
27->14
28->42->63->32->48->72->108->162->243->122->183->92->138->207->104->156->234->351->176->264->396->594->891->446->669->335->168->252->378->567->284->426->639->320->480->720->1080->1620->2430->3645->1823->912->1368->2052->3078->4617->2309->1155->578->867->434->651->326->489->245->123->62->93->47->24

Typhon's Conjecture: Take any integer. If it is 1 mod 3 then multiply by 12 and subtract 5. If it it is 2 mod 3 then add 2. If it is 0 mod 3, then divide by 3 and subtract 7. Show that successive iterations upon any integer eventually reaches a value below 12.

@SohamChowdhury My impression is it's not great. Most of the faculty is gone.

@LeakyNun pls

@Typhon ...

so shifting by one would in effect do the same thing of swapping evens with odds

Hm. Will they be able to connect me to people elsewhere, then?

Is this page outdated?

hmm, @PVAL-inactive. Would you have any insight on a question we had above?

You should keep a contact with the faculty. Have you talked to SB in a while?

lemme see

yeah the university pages are outdated

No. I think he doesn't want to talk to me any more :P

Hey @Balarka, @Leaky, and @PVAL

Lemme restate it a little.

Here's Collatz starting at 27:

Well, it says someone's joining in July, so it can't be that outdated

Revised Typhon's Conjecture: Take any integer. If it is 1 mod 3 then multiply by 12 and subtract 7. If it it is 2 mod 3 then add 2. If it is 0 mod 3, then divide by 3 and subtract 7. Show that successive iterations upon any integer eventually reaches a value below 12.

@Daminark hi

Given a 1-form $\omega$, the mapping $\phi\mapsto \phi\wedge \omega$ gives a (long exact?) sequence $\Omega^0\to \Omega^1\to\Omega^2\to\cdots$

Well, I don't know, @Soham. MM is not associated to it anymore, at the very least, and the chunk of the faculty I know are elsewhere.

They're not actually letting me post it 'cause it's too long

Hm, fair.

Why would you think he doesn't want to talk to you? :P

My question was, is there a name for that sequence? It seems like an obvious construction.

I don't think its really a sequence

It's just the wedge.

I sent him an email after returning from Mathcamp, but a long time after. He didn't reply.

The name of the operation already has a name.

It's wedging with the same form each time, though.

How's it going @Leaky?

@Balarka holy crap Peter May is aggressive

Which is pointless after the first time

@Daminark Oh, yes, I saw that in the trailer.

@Daminark how so?

What are y'all talking about?

google "Fifty Shades of May"

Not really. Sure, everything in the image of the first map is in the kernel of the latter.

but daminark was making a point about his pedagogy i guess lol

Has Peter May finally realised his calling as a Golden Age Hollywood leading man?

@Typhon Take $n=3k+1$. After two iterations, you end up with $36k+7 = 3(12k+2)+1$. Rinse and repeat until you go without bound.

Yeah so you're talking about the wedge of a 1-form and how it acts on various k-forms

12p - 7 is 5 which is 2 mod 3. Then you get 12p - 9 which is 0 mod 3. Then you divide by 3 and subtract 7 which gives 4p-10 which is 2 mod 3...

@SohamChowdhury Try again. Stuff gets buried down.

Yeah.

That's just called the wedge product.

So there's the main atop REU talk (which is right now more difftop, but there will be some framed cobordism so that's new) and then there's a side one he's started as well, which is more his style of atop

hrm.

Today he was just on fire

@PVAL What if I take the cohomology of the complex?

It's a chain complex isn't it?

You get something boring

*co

I guess there's nothing which makes any particular form $\omega$ all that interesting.

Do I?

huh

@Balarka anyway, if I can have access to any of the two number theorists on that page, I'm happy. Or, indeed, anyone from that page.

Yeah the ker=img

Oh right bluh

@BalarkaSen I guess I should now

Hmm.

Started by defining categories, functors, natural transformations, then he defined (but did not yet construct) Eilenberg-MacLane spaces, and defined cohomology groups out of them. And did some stuff I didn't fully follow with loop spaces and suspensions

@SohamChowdhury If you want, I can get you connected with MM. :)

A very well known guy here has complained about May's classes.

Does $\omega\wedge \alpha=0\implies \alpha = \omega \wedge \beta$? @PVAL-inactive

MM; Mike Miller?

That would be great. If you do, please mention that I'm studying at the Belur college.

Mark Mammon?

@LeakyNun 12p - 7 is 5 which is 2 mod 3. Then you get 12p - 9 which is 0 mod 3. Then you divide by 3 and subtract 7 which gives 4p-10 which is 2 mod 3. Then you do more stuff. Umm.... you messed up.

Akiva: neither.

Monmouth?

Moon Moon?

@Daminark Cool stuff! I'll talk to you about them when I get back from school

"Everyone attack! Retreat!"

If so then I'll agree that it's all trivial. (And I can almost believe that it's true.)

Mars Missions

@SohamChowdhury Sure thing. I'll shoot him an email after school.

@LeakyNun "Take $n=3k+1$. After two iterations, you end up with $36k+7 = 3(12k+2)+1$. Rinse and repeat until you go without bound." you're adding 7 instead up subtracting 7.

@SohamChowdhury What are you doing, Lee

That'd be nice because I've got no clue what's up with loops and suspensions

cya

I think you should get involved with this crew.

That's the only thing the battle of Monmouth brings to mind for me, @Akiva.

But yeah see everyone

Get back on your feet!

for now

why is everyone posting so fast?

@PVAL lmao

Because we are many

@Typhon (12p-7) + 2 = 12p-5 not 12p-9

this feels like a youtube chat

We are legion

Peter May is fun but he's tough to understand sometimes because he just kinda defines things in rapid succession

Any relation to Theresa May?

@Semiclassical It should as long as the dimensions aren't too big.

hmm.

Revised Typhon's Conjecture 2: Take any integer. If it is 1 mod 3 then multiply by 12 and subtract 7. If it it is 2 mod 3 then add 1. If it is 0 mod 3, then divide by 3 and subtract 7. Show that successive iterations upon any integer eventually reaches a value below 12.

He said that he's kinda experimenting the approach of defining cohomology first using the spaces and then defining homology

yeah, I think I agree.

@Daminark I think it's better to take that kind of thing as a "recall that ..." instead of a "y'all will not know this, it's pretty cool"

@LeakyNun fixed that annoying bug.

I guess that's not quite right

actually... I fumigated.

There's things like dx_1 wedge dx_2 and dx_1 wedge dx_3

@BalarkaSen Great, thanks!

As opposed to the approach he dislikes in textbooks of defining homology with chains or something

@Soham "that kind of thing"?

though it should be true for 1-forms I think.

If you think about it in coordinates.

Hmm.

@Daminark Rapid-fire definitions.

@Typhon looks good to me; let me test it.

Yeah its true for 1-forms

Even in that 2-form case, it follows the spirit of the objection in that they share some 1-forms.

Windows have this problem with bugs. They tend to congregate there because they like what is inside. If anyone tries to explore with a net, they get even more swarmed because bugs hate explorers with nets. The solution to the bugs is to just hire either a shiny metal car plate for your window or get a flaming canine to sit and eat them.

I wonder though what happens if you look at that sequence for other forms

Lol I mean that's his general style, a good number of us don't know algebraic topology at all

@Typhon it's no good; the 3k+1 still grow very large

e.g. $\omega$ a k>1 form?

@LeakyNun crap

13->149->150->43->509->510->163->1949->1950->643->7709->7710->2563->30749->30750‌​->10243->122909->122910->40963->491549->491550->163843->1966109->1966110->655363-‌​>7864349->7864350->2621443->31457309->31457310->10485763->125829149->125829150->4‌​1943043->503316509->503316510->167772163->2013265949->2013265950->671088643->8053‌​063709->8053063710->2684354563->32212254749->32212254750->10737418243->1288490189‌​09->128849018910->42949672963->515396075549->515396075550->...

So it's tricky to recall it :P

but once it hits case two it bounces between cases 2 and 3 right?

forever?

Yeah for like 2-forms on R^4 for instance

(He is luckily not assuming you already know the stuff, so quick as it may be, it's self-contained)

Yeah.

So could be something interesting.

@Typhon I don't think so

@LeakyNun to be fair though, it is ironic that the Typhon Conjecture get's very large.

There was a question which @balarka had as well but which was past me, lemme find it

@LeakyNun darn. What if we replace 12 with 4?

You should get something like dx_1^dx_2 kills like everything besides the dx_3 ^ dx_4 component.

@Typhon why is it ironic?

@LeakyNun Typhon is literally a giant living hurricane giant.

@Typhon 28->105->28 disproves your conjecture

@Typhon then how is it ironic?

because it keeps getting bigger

like the numbers