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16240472 No.16240472 [Reply] [Original]

[math]/\mathfrak{mg}/[/math]

Platonic universe edition
Talk maths, formerly >>16187402

>> No.16240517
File: 1.12 MB, 1140x4777, official mg curriculum.png [View same] [iqdb] [saucenao] [google]
16240517

>> No.16240527

first for number theory

>> No.16240545

>cardinal arithmetic
>ordinal arithmetic
>transfinite induction and recursion
>NBG set theory
>type theory
>surreal numbers

Are these concepts useful/do these concepts commonly arise in the context of (other) modern mathematics research (say, on algebraic topology or ergodic theory)? Or is it all just metamathematical ramble/computational nerd shit

>> No.16240588

I'm gonna graduate at 30 am I cooked

>> No.16240589

>>16240545
>is it all just metamathematical ramble/computational nerd shit
yes

>> No.16240593

>>16240545
Cardinals and ordinals sometimes appear as examples or counterexamples in topology. Type theory had its revival by being combined with topology to homotopy type theory ten years ago.

>> No.16241140

>>16240588
i got my first job at 31 lol

>> No.16241233
File: 131 KB, 658x610, hatcher_simplicial.png [View same] [iqdb] [saucenao] [google]
16241233

I'm trying to understand why we get the signs in the boundary homomorhpishm in simplicial homology:
[math]
\partial([v_0, ..., v_n]) = \sum_k (-1)^k [v_0, ..., \hat{v_k}, ..., v_n]
[/math]
hatcher says that these signs count the induced orientation on the kth face of the standard n simplex and that this is just obvious.
why is it obvious that the face we get when we remove the kth vertex inherits the orientation corresponding to [math](-1)^k[/math]?

What is rigourously even meant by orientation here?
I've been trying to make it more precise by thinking of a simplice as a convex subset of a certain affine subspace of [math]R^m[/math],
which through the ordering of the vertices get an ordering of the vectors spanning it but this approach turned messy.
I also had the faint hope that the boundary homomorphism would be uniquely defined by the property that
[math]\partial_n \circ \partial_{n-1} = 0[/math] and [math]\partial_1([v_0, v_1]) = [v_1] - [v_0][/math]
as I understand why we want these properties when we're trying to algebraically count holes of various dimensions (the motivation behind homology?).
but this fails to uniquely define a homomorphism...

>> No.16241238

>>16240545
surreal numbers are useful in combinatorial game theory, which is probably one of the furthest fields from metamathematical ramble. the other stuff shows up on occasion but only at a low level. last week i was doing some homological algebra and needed some ordinal arithmetic to construct one of the counterexamples. in general if you don't enjoy them don't bother learning though.

>> No.16241271

>>16240472
What actually compelled Kepler to believe this was the case? He was a smart guy but this was pretty batshit

>> No.16241294

>>16241271
i thought this was like the ancient greeks, not kepler. the ancient greeks had some great ideas but a lot of it is batshit insane. idk how they came up with it i think they just had no clue so they made shit up.

>> No.16241297

>>16241294
Part of the reason that it took so long to move away from the idea of epicycles was because it was inconceivable that heavenly bodies could move in anything other than a perfect circle.
These ideas lasted until very recently, all things considered

>> No.16241305

>>16241297
yeah i mean it took thousands of years until people thought "maybe aristotle was full of shit and just made all this up" so i wouldn't doubt it

>> No.16241363

Can i get some measure theory books recommendations. Not just the wich but the why.

>> No.16241381

>>16241363
measure theory is pseudoscience

>> No.16241399

>>16241297
>>16241305
It had much more to do with the invention of telescopes that enabled more precise measurements to be made.

>> No.16241410

>>16241271
>What actually compelled Kepler to believe this was the case? He was a smart guy but this was pretty batshit
Not at all, actually. There were only 6 known planets at the time, so it made sense to correlate them with the platonic solids. His model also closely adhered to the known planetary distance measurements. Finally, it wasn't understood that objects could follow orbits by only a central force; it was assumed that there must be an attractive force and an opposing force (just how there is for electron orbits, hint hint) and the natural shape that two opposing forces creates is a platonic solid. Kepler was a smart dude.

>> No.16241414

>>16241363
folland is great. it also just so happens to be what my school uses

>> No.16241421 [DELETED] 

>>16241410
>6 planets
>6 platonic solids
>woah… these le numbers are, like, the same number??? Bruv I have a le theory!!!

>> No.16241427

>>16241421
There's only five, but no worries... not everyone can be Kepler caliber.

>> No.16241430

>>16241421
this is literally what the ancient greeks did. plato mocks it (or maybe he's just that stupid) in the republic [587b - 588a].
>>16241427
kek

>> No.16241433

>>16241410
>6 planets
>5 platonic solids
>woah… these le numbers are, like, kinda similar??? I have a theory!!!!

>> No.16241437

>>16241433
If we weren't on this Mongolian grasshopper board I'd think you must be a woman

>> No.16241547

>>16240472
How old were you when you realized that the Microsoft Equation Editor is good enough for 99% of the times you would othewise tinkertranny with a hodge podge of various LaTeX modes?

Not only does Word let you enter Latex syntax, you can use the faster Unicode form.

>> No.16241709

>>16240545
>useful
if you come into maths with that outlook for existence then you have the wrong disposition

>> No.16241719

>>16240527
lol applied computer science fail

>> No.16241745 [DELETED] 

>>16241233
>What is rigourously even meant by orientation here?
You know how in computer graphics if you specify three vertices in order, a triangle appears, but only from one side? And if you swap two vertices in the order, the triangle switches the way it's facing?
That exactly corresponds to what's meant by orientation here. The induced orientation is the order in which to specify the vertices so all the faces of the simplex are visible from the outside.
The way you can do that is by rotation. Define a_k = [v_0,..,v_k-1, v_k+1,...., v_n], and suppose you take v_0 as one of the faces as a convention. You want to obtain v_(k+1) to be the same orientation as v_(k) (starting with v_0), by rotating the vectors v_0,...., v_n. Clearly you do that by a linear map that moves swaps v_k with v_k+1 and leaves the other vertices untouched. This matrix has determinant -1, therefore it reverses the orientation, therefore you multiply by -1 to account for it so that the orientation stays the same.

>> No.16241752

>>16241233
>What is rigourously even meant by orientation here?
You know how in computer graphics if you specify three vertices in order, a triangle appears, but only from one side? And if you swap two vertices in the order, the triangle switches the way it's facing?
That exactly corresponds to what's meant by orientation here. The induced orientation is the order in which to specify the vertices so all the faces of the simplex are visible from the outside.
The way you can do that is by rotation. Define a_k = [v_0,..,v_k-1, v_k+1,...., v_n], and suppose you take a_0 as one of the faces as a convention. You want to obtain a_(k+1) from that of a_(k) (starting with v_0), by rotating the vectors v_0,...., v_n. Clearly you do that by a linear map that moves swaps v_k with v_(k+1) and leaves the other vertices untouched. This matrix has determinant -1, therefore it reverses the orientation, therefore you multiply by -1 to account for it so that the orientation stays the same.

>> No.16241765

>>16241233
It is just a consequence of wanting del^2 = 0

You can do things purely symbolically to arrive at the properties without any simplex "story".
Let d be the operator defined by dv = 1 - vd, d(const)=(const)d.
Applying this to a sequence s=(v0)(v1)...(vn) and then pushing d to the right does what you want.
ds = (something1)[s] + (something2)[s]d
dv = 1-vd
ddv = d(1-vd) = d - (1-vd)d = vdd
dds = sdd
Clearly (something1)^2 = 0,
(something1)(something2) + (something2)(something1) = 0,
(something2)^2 = Id.
call (something1) del

d(v0) = 1 - (v0)d
d(v0)(v1) = (v1) - (v0)d(v1) = (v1) - (v0) + (v0)(v1)d
d(v0)(v1)(v2) = (v1)(v2) - (v0)(v2) + (v0)(v1) - (v0)(v1)(v2)d
Etc.
d(v0)(v1)(v2)...(vn) = del[v0,...,vn] - (-1)^n*(v0)(v1)(v2)...(vn)d

d is very similar to the usual differential operator, D, satisfying Df = f' + fD. The minus sign in d gives a kind of alternating product rule which makes the (del)^2 = 0 work.

Probably not the kind of answer you want but it is what it is.
You can use this to generalize del to get del_k satisfying (del_k)^k = 0 by requiring dv = 1 + v*exp(2*pi*i/k)d.

>> No.16241910

I need to create a rectangle of X square feet with an aspect ratio of Y. Help a dumb engineer out.

>> No.16241936

>>16241910
[math]
width := \sqrt{X Y}, height := \sqrt{\frac{X}{Y}}
[/math]

Assuming positive X, Y.

>> No.16241944

>>16241936
Thanks homie, I’d hire you any day.

>> No.16241953

>>16241765
what is d? it acts on what and results in what? what is v? what is a sequence?

the general setting I'm thinking about is a free abelian group [math]C_n[/math] with a basis a set of n simplices, and then we're interested in obtaining a boundary homomorphism
[math]\partial_n : C_n \to C_{n-1}[/math]

I fail to see how any of what you wrote makes any sense in this context

>> No.16241971

What are some nice math communities where people study together?
I want to find study groups.

>> No.16242071

Any anons know of a good youtube series that gets you up to speed on stuff like calculus and matrix/vector multiplication? I'm trying to get into electronics and programming as a hobby and I'm running into shit that I haven't studied since high school.

>> No.16242122

>>16242071
try khan academy?

>> No.16242135

>>16242071
Why does it have to be youtube? There's good books for that.

>> No.16242161

>>16241363
I also back folland.

>> No.16242163

>>16241953
d acts on the terms v0, v1, ... that are "multiplied" together similar to how the differentiation operator acts on functions.
The monomials you get correspond to your [] terms (assuming the v's don't commute)
For the usual derivative operator,
Df = f'+fD
Dfg = (f'+fD)g = f'g + fg' + fgD
This is how multiplying works for differential operators in the usual sense (evaluation of D remains pending).
The simple rule Df=f'+fD reproduces the product rule.
I used d as modified version of D.
It is basically related to Weyl algebra. You have dx+xd = 1 instead of Dx-xD=1 in the weyl case.
It is a bit different since Weyl algebras have {Di} and {xi} and [Di,xj] = kronecker(i,j), [Di,Dj]=0, [xi,xj]=0
I am only using 1 d and many v where d(vi) + (vi)d = 1 and vi do not commute (technically you could allow the vi to anticommute for your case).
Just think of d as acting on the term (v0)...(vn) where the decision branches for each factor that is encountered by d from the left.
For dv = 1 - vd, think of the the 1 as the action of removing v and the -vd as skipping past v and picking up a factor of (-1).

I told you this is purely mechanical and has nothing to do with the simplex "story". My point is this is the only way to get something that behaves like your boundary operator "del" and the alternating signs are just necessary to get del^2 = 0.

>> No.16242208

>>16242163
My whole motivation was just to have some operator give ddv = vdd for each variable v since this would give a del that satisfies del^2 = 0 and operates on things of any dimension.
Since del returns objects of 1 dimension less, d must behave like dv = A+B*vd
This gives ddv = (A+BA)d + BBvdd.
For ddv to equal vdd, B must be -1. This B = -1 explains why removing vk has a factor of (-1)^k.
The value of A doesn't really matter (besides not being 0) since A^k will just keep track of how many times the dimension is reduced which is redundant since the number of terms in the monomials also keep track of this.
You can conclude
d(v0)(v1)...(vn) = A*del[v0,...,vn] + (something)d without even computing anything.

>> No.16242244
File: 522 KB, 720x720, 1000011234.webm [View same] [iqdb] [saucenao] [google]
16242244

>2015: took calc 1
>got sick and almost died
>2015 summer: took calc 1 again
>2015 fall: took calc 2
>got a biz degree in sucking boomer dicks and getting meme'd on
>2024: taking calc 2 again going back for engineering and math minor
>have collected a full library of recreational mathematics by now
>the only thing I know is I need to know linear algebra, data structures & algorithms to become gigaChad
Feels chud, homers enemy

>> No.16242246
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16242246

>tfw I could have avoided a decade of pain if I had listened to my highschool nerd friends instead of my boomer barons

>> No.16242250
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16242250

When did you realize math was for you?
For me it was Calc 2 seeing the Fast Fourier transform and being gatekept from the hyperbolic functions. It was the first time I felt like I invented math AND I stood on the shoulders of giants.

>> No.16242256

>>16242122
I'll give it a shot, thanks.

>>16242135
Because I don't need a book's worth of knowledge, just the cliff notes.

>> No.16242261

>>16241971
My university has quite a few if you can put up with the REDDIT AURA. They have a "Geeks with Beers" Saturday meet up at a bar. No classy people in sight. So much blight. It always ends in a Discord detour. It's hard because I want to be genuinely friendly instead of social credit NPC script. So many of my personal projects have become dusty beyond my own grasp. People come and go. Americans are hard to host and harder to invite and even harder to keep around. I have kept much foreign company. Chinese, Saudi, Russian guests. When you meet someone super good they tend to travel a lot. I wish we had tea time.

>> No.16242264

>>16242246
How?

>> No.16242270
File: 98 KB, 1080x930, 1000011321.jpg [View same] [iqdb] [saucenao] [google]
16242270

>>16242264
My nerd friends did AP Calc in highschool, CAD, and so many AP classes they basically did 95% of their associates degree in highschool. I met homeschool kids who straight after middle school did their associates.
During HS I saw some rich kids "social skill" their way to the big bucks and thought shmoozing with them would give me such an opportunity. It failed. I sandbagged academia for the sake of desperate temporary grifts that were never enough. Those nerds never got gfs, never partied, never rubbed shoulders with big wigs, but they got out of school early and got the best jobs while the world was more upside down than we could have imagined. Those friends were like the big doge. Boyscouts. Based even. In the mean time I shagged my way to the Prodigal Son's shame.

>> No.16242576

>>16242163
>>16242208
it looks like you're trying to say something programming language theory parsing related, but you're unable to make it precise, comprehensible or connect it to the general situation of sequences of free abelian groups and homomorphisms between them such that any composition becomes 0.

>> No.16242605

>>16242250
when i watched the 3blue1brown "the hardest problem on the hardest test" and realized "hey this math shit is cool".

>> No.16242654

>>16242576
>something programming language theory parsing related
Df=f'+fD is just Weyl algebra. It is how big boys multiply differential operators beyond the baby ones in diffeq with constant coefficients.
I constructed d from the free algebra R<v0, v1,...vn,d> modulo the ideal generated by all of the d(vi) + (vi)d -1 = 0.

Each monomial corresponds to your [] cycles in the obvious way if there are no repeats.
You just add multiply, distribute the way you normally would.

>connect it to the general situation of sequences of free abelian groups
I am defining the boudary operator since that is what was required.
There was a bunch of WHY questions about del. I am answering the why question by getting at something more abstract but arguably more illuminating (since you can cook up what del must be just from the specification del^2 = 0 which would be a nightmare if you just stayed in the "simplex" story).
I handed you del and its recipe. You figure out how it can be used.

I've noticed you haven't opined on how the simple rule dv = 1 - vd gives the correct answer for del for all dimensions.
Maybe start there.

>> No.16242846
File: 86 KB, 800x450, disappoint.jpg [View same] [iqdb] [saucenao] [google]
16242846

>>16242654
as I stated in my original post it is in fact wrong that as you said [math]\partial[/math] is uniquely defined by
[math]\partial_n \circ \partial_{n+1} = 0[/math] and [math]\partial_1 \left( [v_0, v_1] \right) = [v_1] - [v_0] [/math].
Since already for n = 2 we can define
[math]\partial_{2} [v_0, v_1, v_2] = -[v_1, v_2] + [v_0, v_2] - [v_1, v_2][/math].
Here we have the signs being the opposite of what they are in the usual definition. and yet [math]\partial_1 \circ \partial_2 = 0[/math], as well [math]\partial_2 \circ \partial_3 = 0[/math], with the usual [math]\partial_3[/math]!

Thus we have choices as to how we actually define [math]\partial_n[/math] for arbitrary n. what I was interested in was whether there is a compelling reason for why the definition of the boundary homomorphism looks EXACTLY like it does for arbitrary n. not why is it an alternating sum, or any other question. such a justification could be along geometric lines in a way that clearly justifies it for general n simplices. or it could be along algebraic lines. but as I just showed we can in fact not provide such an algebraic justification for the two minimal reasonable requirements. since we can specially define [math]\partial_2[/math] with the opposite signs as usual and everything still works out!

from more research I'm settling on the conclusion that the exact form of the boundary homomorphism is defined the way it is simply because it 1. confirms to our geometric conventions in low dimensions, 2. because it's a convenient formula and 3. because it does confirm to the requirement that [math]\partial_{n} \circ \partial_{n+1} = 0[/math].
this is somewhat disappointing.

beside all that, your language and notation isn't well defined.
dv = 1 - vd is not a well defined expression in this context.
>d acts on the terms v0, v1, ... that are "multiplied" together similar to how the differentiation operator acts on functions
this is of course nonsense without more precisely specifying.

>> No.16242938

>>16242250
Calc 3 when I accidentally found a formula for the volume of a pyramid during an exam

>> No.16243061

>>16242846
>Thus we have choices as to how we actually define
Obviously you can just multiply my del by arbitrary constants c_n to get your "choices" of del_n since it is linear.
That isn't interesting. The ratios of the constants multiplying the "faces" remains the same.
>this is of course nonsense without more precisely specifying.
I worked examples.
Start with the monomial (v0)(v1)...(vn) where the vi don't commute.
multiply by d
d(v0)(v1)...v(n)
= (1-(v0)d)(v1)(v2)...(vn)
distribute in the usual sense
=(v1)(v2)...(vn) - (v0)d(v1)(v2)...(vn)
= (v1)(v2)...(vn) - (v0)(v2)(v3)...(vn) + (v0)(v1)d(v2)(v3)...(vn)
...
=del[v0,...,vn] - (-1)^n * (v0)...(vn)d

You might think this is just programming language or parsing but this is just how things are when you modulo things in algebra. You still retain the richness of the ring structure so it isn't just simply string rewriting.
I encourage you to at least look into weyl algebra (since I assume you know basic calculus) to even understand the flavor of what is going on.
Weyl algebra is useful because sometimes you can recast your problem in terms of the mechanics of differential operators then use analysis techniques to approximate the answer when a closed form is not available.
I'm surprised you are learning abstract math yet are so opposed to me taking your problem and abstracting it then recasting it in a different form to get the answer easily.
Better not look at how laplace/fourier transforms are used to turn differential equations into algebraic equations.
God forbid you look into generating functions and see functions that you never plug in a value for the variable but still add and multiply them to do combinatorial operations with the coefficients.

I solved your problem and a whole family of generalizations with some slick abstract algebra which I think is pretty cool.
Please tell me how you would find an operator del with the property del^k = 0 with your simplex "story".

>> No.16243110

>>16243061
>I'm surprised you are learning abstract math yet are so opposed to me taking your problem and abstracting it then recasting it in a different form to get the answer easily.
you fail to do that and you fail to make any sense.
>Please tell me how you would find an operator del with the property del^k = 0 with your simplex "story".
oh ok you were just a literal ranting schizo this whole time.

>> No.16243158

>>16243110
>oh ok you were just a literal ranting schizo this whole time.
The answer is to just replace -1 with e^(2*pi*i*m/k) in the alternating sum in the pic >>16241233
I gave you the answer. Now good luck proving it satisfies del^k = 0

>> No.16243378
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16243378

I'm looking into the fact that a fractional brownian motion is not a semimartingale, but I'm having a hard time understanding a certain point in the proof : it is said that if a process has a quadratic variation of 0, then for it to be a semimartingale, it must have a finite 1-variation, but I found no proof of this, I have a hunch that it's due to the decomposition as a local martingale and a process of bounded variation, but I really dont know what to do with all that, could you fellows give me a hand ?

>> No.16243464

>>16242256
>just the cliff notes.
Look at math books for physicists or engineers. They're pretty good for just distilling the things you'll need.

>> No.16243900

>>16243378
If you have a continuous semimartingale [math]X=X_0+M+A[/math], for a local martingale [math]M[/math] and a BV process [math]A[/math], its (predictable) quadratic variation [math]\langle X\rangle=\langle M\rangle[/math].
Wlog you can take each [math]M^{\tau_n}[/math] bounded, for a localizing sequence [math](\tau_n)[/math], and also [math]\langle M^{\tau_n}\rangle=\langle M\rangle^{\tau_n}=0[/math], implying that [math](M^{\tau_n})^2[/math] is a martingale with [math]\mathbb E ((M^{\tau_n})^2)=0[/math], implying [math]M^{\tau_n}=0[/math] a.s.
This extends to [math]M=0[/math] a.s. because for any [math]t \geq 0[/math], [math]\mathbb P(M_t\neq0)\leq\sum_n \mathbb P(M^{\tau_n}_t\neq0)=0[/math] and modifications of continuous processes are indistinguishable.
So if [math]\langle X\rangle=0[/math] then [math]X=X_0+A[/math] which is BV.

>> No.16244173

>>16243900
Got it, thanks man

>> No.16244512
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16244512

>>16240472

>> No.16244609
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16244609

>>16241294
You should read this book.

>> No.16244611

>>16241363
Axler's book is free on his website.

>> No.16245290

>>16241363
Why would anyone waste their time studying a theory? Study some measure facts

>> No.16245445

>>16242244
Sounds like my dad's college experience

>> No.16245814

what's the most esoteric out of touch branch of mathematics which will never have any practical application?

>> No.16245901

>>16245814
large cardinals

>> No.16245907
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16245907

If you throw a ball from height h meters and with velocity v m/s, what does the angle alpha need to be in terms of h and v to maximize the distance in which the ball lands when it falls to the ground?

Hint: it is not 45 degrees because 45 degrees only applies for h=0.

>> No.16247100

>>16245907
>homework

>> No.16247603

>>16247100
Not homework. I came up with the problem for recreational math

>> No.16247628

>>16245907
Range can be expressed as a function of angle. Differentiate and find the roots