Mathematics - 2020-11-17 (page 1 of 2)

epic_math

03:27

I don't know who is upvoting all of my answers and questions. Please don't upvote all of them blindly. Yes, if one of them is good, it can be upvoted, but not all of my questions and answers are good.

Please tell who is doing it.

Recently 60 rep was deducted from my current reputation because of that.

user 6232128

03:48

2

Q: Why is computer science hard?

Data pretty regularly shows that computer science programs have among the highest failure and dropout rates of any college program. A number of sources all echo the finding that roughly one-third of incoming CS majors do not progress to a second year, higher than most other majors. Computer scie...

Nicholas Roberts

04:05

Would it be possible to find some values $\theta$ and $\phi$ such that $\cos(\theta + \phi) = \cos(\theta)+\cos(\phi)$? In other words, do there exist values in which any of the trig functions are additive?

Rithaniel

Well, the best thing is to try out some values

Nicholas Roberts

I'm trying. Can't seem to find any

Rithaniel

Fist, fix $\theta=\pi$ and then see if you can find $\phi$ such that $\cos(\pi+\phi)\leq \cos(\pi)+\cos(\phi)$. Then see if you can find a different $\phi$, this time such that $\cos(\pi+\phi)\geq \cos(\pi)+\cos(\phi)$.

Nicholas Roberts

Hm, well $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$

Rithaniel

Since $\cos$ is continuous, you would have to have a $\phi$ such that they are equal between those two other values you found

Nicholas Roberts

04:11

Ahh, intermediate value theorem.

Rithaniel

Yep

Nicholas Roberts

I was wondering if there was a way to explicitly solve for these values. Or find like a closed form of them as opposed to proving existence of such

Rithaniel

Hmmmm, probably

Nicholas Roberts

Yeah, maybe by messing with the sum/difference identites. Anyways, thanks!

Rithaniel

Yeah, you'd have to fool around with it and attack it with algebra, but I'm fairly certain you'll find a few points

epic_math

08:23

Hella

The MSE question I gave here has been moved to MO. Here it is:

0

Q: A generalization of partition function to the sums of squares

The well known partition function $p(n)$ is defined as the number of ways to represent $n$ as the sum of natural numbers. An asymptotic formula for $p(n)$ is $$p(n)\sim\frac{1}{4n\sqrt{3}}\exp\left(\pi\sqrt{\frac{2n}{3}}\right)$$ which was obtained by Ramanujan. Recently an interesting idea came ...

nt.number-theory analytic-number-theory partitions sums-of-squares

IamKnull

09:45

I need help in understanding the meaning of this question :

If $Q(v)=\left(D^{2} f(a) v, v\right)$ is indefinite form, then prove that $f$ has a saddle point at $a$.

What does $D^{2} f(a) v$ represent? and is the bracket for inner product

aderchox

11:02

Could someone please explain this to me? What difference would it make if 2^n is prime?

Alessandro Codenotti

The regular operations would make this set isomorphic to Z/2^nZ, which is not a field since it has nonzero elements whose product is 0 (for example 2 and 2^(n-1))

love_sodam

14:00

What is the difference between simplicial complex and delta complex?

Definitions are all same as wiki

Mike Miller

14:17

@love_sodam No they're not

Wiki doesn't even have a page defining delta complex

love_sodam

I mean my definition of those two are same as two definitions in wiki

Mike Miller

The crucial combinatorial criterion which fails for delta-complexes but is demanded for simplicial complexes is that the vertices determine a face/edge

love_sodam

Not two definitions are the same

Mike Miller

Think of the usual delta-complex of the torus

You start with a decomposition of the square with 4 0-simplices, 5 1-simplices, and 2 2-simplices

When you do the edge-identifications you end up with a single 0-simplex (!), 2 1-simpllices (!), and the same 2 2-simplices

Now it's no longer true that the vertices of a given edge or simplex are all distinct (like is the case in a simplicial complex), nor that the edge is determined by its set of vertices

love_sodam

Isn't it 3 1-simplices?

Mike Miller

14:26

Yeah my bad

But the point is that they all have the same set of vertices

Any triangulation of the torus requires many more faces, perhaps 8 is the minimum

love_sodam

So the example you gave is delta complex not simplicial complex

and simplicial complex of torus has more simplicies

When we compute the homology group of a torus, we give delta complex structure and compute its simplicial homology group right?

Mike Miller

That's one way to do it yeah

love_sodam

14:50

My confusion is that if we want to compute the simplicial homology, don't we need to give simplical complex structure on that space?

Mike Miller

no

the reason Hatcher talks about delta-complexes is because this rule that simplices need to be determined by their vertices is an important condition in the combinatorial study of simplicial complexes

But it's completely irrelevant to calculating homology

Much better to work with this complex with (v, e, f) = (1, 3, 2), then with the smallest triangulation of the torus, where (v, e, f) = (7, 21, 14) :)

love_sodam

15:27

One more question, like the case of torus, we usually gives orientation for 1-simplex by just following the identification (following the arrow of the identification). Is this necessary?

In 2-simplex case, does the homology group depend on the choice of orientation?

Thorgott

16:20

is the zero set of a degree 2 homogeneous polynomial discrete in P^1

working over C

feynhat

why would you think that

(I actually don't know if its true btw)

Mike Miller

@love_sodam I am not certain I understand the question. I think there are ways to interpret this where I could say "no" and ways to interpret this where I could say "yes".

@Thorgott Is the zero set of a degree 2 polynomial discrete working in A^1?

Thorgott

you mean A^2?

Mike Miller

No, I mean A^1.

love_sodam

@MikeMiller Never mind. I think I can handle this. But thanks anyway

Mike Miller

16:29

P^1 = A^1 cup infty

if p(z,w) is a homogeneous polynomial, its non-infinite zeroes are in bijection with the zeroes of the one-variable polynomial p(z/w, 1)

Thorgott

ah, thanks

Rajdeep Sindhu

16:43

Hi. So, I was wondering why we need to extend the definition of trigonometric ratios to all angles? A friend of mine told me that it has some useful applications in physics but AFAIK, the unit circle definition came a long time ago, back when physics was not so developed.
So, why exactly did we need to define trigonometric functions for all angles? What motive did the mathematician(s) who did this have in mind? Thanks!

Thorgott

"extend" from what exactly?

Rajdeep Sindhu

The definitions in terms of sides of a right triangle. Like $\sin\theta = \dfrac{\mathrm{perpendicular}}{\mathrm{hypotenuse}}$

Thorgott

but that definition already works for all angles

Rajdeep Sindhu

Not for obtuse angles, does it? It only works for $\theta\in(0^\circ,90^\circ)$...

Astyx

A right triangle doesn't have obtuse angles

Rajdeep Sindhu

16:51

Right, that's why the definition doesn't work for them...

Astyx

The coordinates of the points of a unit circle centered at the origin are given by $(\cos\theta, \sin\theta)$

geocalc33

17:15

Does this metric imply zero curvature? $ds^2=\frac{dxdy}{xy}$

Ted Shifrin

17:59

@love_sodam: You'll be amused to know that I emailed the author of your diff geo text, and he didn't even recognize/remember that the exercise was in his book :D Turns out I know his Ph.D. adviser and we have a lot of friends in common. :D

Alessandro Codenotti

18:26

@user2103480 I seek intuition and I summon thee, master of probability

Thorgott

using intuition and probability in the same sentence, daring

Alessandro Codenotti

I prefer optimistic

Sal

Another one of my dumb questions.

Is the expression $\lim_{x\to+\infty} \dfrac{1}{x} = 0$ just a postulate, or is there a proof for it?

user2103480

Alessandro Codenotti

lmao

user2103480

18:32

if you find that master of probability you're talking about, call me

what do you seek intuition for?

Alessandro Codenotti

So in a seminar today I was introduced to the concept of Brownian motion

Ted Shifrin

@Thorgott LOL

Alessandro Codenotti

But I'm not sure what some parts of the definition are telling me intuitively

user2103480

Which exactly? There are 4, and one of those is that it starts at 0, and the other one is that the paths are continuous, so I guess these aren't the trouble makers

Ted Shifrin

@Sal it's a consequence of the Archimedean principle .... it is the statement that the positive integers are not bounded above. This is a consequences for , I suppose, of these least upper bound property.

Alessandro Codenotti

18:36

Right, there's an independence condition and a "jumps are Gaussians condition

user2103480

normal distribution of increments should be ok as well, in principle. You can imagine it as diffusing outwards with paths on average landing in a certain distance away

Alessandro Codenotti

The latter makes sense, it's telling me the distribution of "where will the point be after time $s$ starting from time $t$" and I guess you can play around with weird stuff here but Gaussian is the easiest thing

user2103480

@AlessandroCodenotti The independence condition means that using events that are measurable with respect to the preceding path, we cannot say anything about the future behaviour of the brownian motion

Note that the independence condition is "normalized"

I.e. From the path up to time s

Alessandro Codenotti

So it's like where I'm going to jump depends from where I am, not how I got there?

user2103480

we cannot say how W_t-W_s will behave

Mike Miller

18:38

@TedShifrin Erico and I are confused about something

Ted Shifrin

BTW, howdy, demonic @Alessandro et al.

Oh oh. OK, @MikeM.

Alessandro Codenotti

Hi Ted

user2103480

@AlessandroCodenotti Exactly, it's a kind of markov condition (and, not surprisingly, the brownian motion satisfies several markov properties)

Mike Miller

Let M be a Riemannian manifold. Pick a function f which vanishes on the boundary and pick normal coordinate charts on the boundary.

Alessandro Codenotti

Ok that makes sense

Thanks

Ted Shifrin

18:39

Oh, I thought this was going to be something from my book. Silly me.

I've never thought about normal coordinates on the boundary. Hmm.

user2103480

You can imagine it as saying: If I paste two brownian motions at their ends, these are together one brownian motion

and the decomposition works the other way around as well

Alessandro Codenotti

Uh here all Brownian motions were assumed to be defined for all positive times

user2103480

uhh at one start and one end. you know what I mean. Although their ends would be fine too if I reflect one of them properly

Mike Miller

Erico claims that if 0 < i < n, f_ii = - lambda_i f_n, where lambda_i is principal curvature.

In particular the second derivative of f ALONG the boundary can be nonzero 0 when the boundary is not totally geodesic.

user2103480

@AlessandroCodenotti Then crunch together one brownian motion to a finite one (you can do that) and paste the other one after that, hah!

Alessandro Codenotti

18:42

Ok fair enough

Mike Miller

He has derived the formula via moving frames and also it's in Yau's papers.

Thorgott

can you interpret this gluing sheaf-theoretically

Alessandro Codenotti

Can you not do that

Ted Shifrin

OK, so I'm going to just assume it's an arbitrary hypersurface you're asking about, because I don't know about normal coordinates on the boundary. All I know is that $f$ vanishes on the hypersurface.

user2103480

@Thorgott ???

Mike Miller

18:43

But it's also absolutely preposterous. f_ii can be computed internal to the boundary (iterated partials with one coordinate can be computed along lines).

Yes, seems fine.

Oh

So the problem is normal coordinates aren't adapted to the hypersurface.

user2103480

@AlessandroCodenotti Yes I will not do that

Ted Shifrin

Well, you're choosing adapted frames here to make the statement, but, yeah, if the hypersurface isn't totally geodesic, you don't stay adapted.

Mike Miller

Thanks. Your rephrasing solved the problem.

Ted Shifrin

But you're asking about the derivatives just at the point.

Alessandro Codenotti

I guess nothing stops you from considering sheaves with values in Brownian motions, but why

Ted Shifrin

18:44

LOL, it did? I'm still figuring out the question.

user2103480

@AlessandroCodenotti The why stops me

Ted Shifrin

Right ... the second derivative isn't really along the boundary.

I see why you were confuzled.

Thorgott

lol im just playing

Ted Shifrin

Hi @Thor

Thorgott

hi Ted

Mike Miller

18:46

Yes, @TedShifrin, when someone takes a coordinate chart, if they care about a submanifold, I tend to assume their coordinates are adapted.

The boundary thew me off ... whereas with a hypersurface you realize normal coordinates do what they want to do

Ted Shifrin

Gotcha. Yeah, normal coordinates won't be very adapted.

Glad I could help doing nothing. :) Let me know if anything fun happens with the kidlet. :)

user2103480

is there anything I have to consider when working with sums in hillbert spaces. I always just shuffle everything around freely when I know that what I start with is well-defined

Ted Shifrin

@Sha !!

Sha Vuklia

:000 TED

:D

Akiva Weinberger

(joke)

(source: Michael Kinyon)

Sha Vuklia

18:53

Are there any number theorists here, actually?

Ted Shifrin

Nope.

user2103480

@AkivaWeinberger POTUTT

Ted Shifrin

We have only logician, algebraists, probabilists, and topologists (and me).

user2103480

@ShaVuklia @EdwardEvans

Sha Vuklia

Hmm, I'm always afraid I'm missing out because I don't have an inherent interest in number thy, but I guess here is not the place to be convinced otherwise.

Thorgott

18:54

so he says, but Lukas self-identifies as number theorist in a starred message

Ted Shifrin

@Sha: I never did, either. I survived. But I did come to believe that elementary number theory is important as a tool in teaching abstract algebra concretely.

Lukas is a moving target, and he's been MIA for a long time. So perhaps so, @Thor.

He didn't used to be ;P

Astyx

I have a NT course for what it's worth

Sha Vuklia

@TedShifrin Yea I can imagine that

user2103480

@TedShifrin but that scares off students that don't like number theory

Ted Shifrin

Anyhow, I was mostly joking. No insults intended.

Sha Vuklia

18:55

I guess I'll just take the course algebraic number theory at some point, and that will (most likely) be it then for me.

Ted Shifrin

@user2103480: Not in my experience.

user2103480

Sadly, algebraic topology and diff geo are harder to introduce quickly

Alessandro Codenotti

I wouldn't call myself a logician tbh

Thorgott

I'm likely gonna do ANT next semester

starting the descent

Ted Shifrin

Diff geo is fine to introduce quickly ... People feel like they have to do abstract crap instead.

Astyx

18:56

is ANT algebraic or analytic ?

Ted Shifrin

After all this, @Alessandro? What do we call you, then?

user2103480

@TedShifrin Okay, fair, you are probably right. I was extrapolating from myself who couldn't have cared less for any number theory examples

@Astyx lmao fair question

Sha Vuklia

@TedShifrin Ah, that explains your somewhat dislike(?) of Lee's book.

user2103480

@TedShifrin Oh I mean the algebra applications in diff geo

Ted Shifrin

@user2103480: I never really did, either. But I developed an appreciation writing my algebra book. Algebra books tend to be too much formality and symbol-pushing. Of course, I also did a bunch of geometric stuff, too :D

Thorgott

18:57

@Astyx do I look like a mutant to you

Sha Vuklia

I must say, after a year of reading Lee, I've fully embraced it:D

Ted Shifrin

What algebra applications are you referring to?

user2103480

Cohomology, algebraic invariants

Ted Shifrin

People misuse the term differential geometry, @Sha. I'm NOT talking about a standard differentiable manifolds course.

Astyx

@Thorgott I'm not sure which way that's meant to point towards, but I'm guessing you meant algebraic :p

Alessandro Codenotti

18:58

I guess a (descriptive) set theorist based on the subjects I'm thinking about for my PhD

Thorgott

indeed

Sha Vuklia

Oh:0 Then my confusion remains about your statement on Lee's book some time ago (which is fine).

Ted Shifrin

I don't see differential geometry in what you said, @user2103480.

Edward Evans

@Sha yeah do ANT

Ted Shifrin

Lee and I are longtime friends, but his pedantic style doesn't mesh with me.

Astyx

18:59

@EdwardEvans check discord

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$Mathematics$ Mathematics