Mathematics - 2020-08-11 (page 2 of 2)

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5:00 PM

So, it makes me feel that higher mathematics isn’t not about manipulating the expressions to get what we want.

@Knight Depends on what you mean by higher mathematics

Alessandro Codenotti

@abhas_RewCie depends on whether you have a question that I find interesting :P

@abhas_RewCie It means the field in which current researches of mathematics is going on

@Knight It also depends on the branch of math. In general, algebraic topics are less computational (not completely non-computational, just less so) than analytic topics. If you read some analysis answers, you will see a lot of manipulation.

@Knight Yes, in jr school mathematics is all what we know what we are doing, in higher mathematics we have to find what we are doing.

5:05 PM

@Knight analysis is not without theoretical arguments and algebra is not without computation, but each tends to the other side of things.

@AlessandroCodenotti Does Matrix and Determinants even make sense?

@robjohn Ok... But how do we see that? Do we have touse inequalities?

@robjohn My findings are contrary to it, I find Analysis less manipulative and more like a subject which requires thinking (what epsilon to choose to falsify the claim). High School Algebra doesn’t require more thinking than practice.

@Alessandro a 1-dimensional domain should work, no?

cause then you always have the chain $(0)\subsetneq\mathfrak{p}$ for any non-zero prime ideal $\mathfrak{p}$, which can't be extended, so $\mathfrak{p}$ is already maximal

hmm, trying to think of a 1-dimensional non-domain where some prime isn't maximal

Alessandro Codenotti

@abhas_RewCie why shouldn't they?

@Thorgott yes

5:12 PM

@AlessandroCodenotti What is the use of Matrix, why are they multiplied in a very strange way and why matrix and determinants have different rules.

?

Hi, demonic!

Hi, @Thor.

What benefit I get by using them?

@TedShifrin Hi Ted Shiffin

@MaryStar For $a\lt0$, it is obvious. For $0\lt a\lt1$ and then for $a\gt1$, we can use Bernoulli

Hi, abhas.

Hi Ted

5:13 PM

Heya, @robjohn

I don't know enough examples of 1-dimensional non-domains

@TedShifrin coffee or tea?

What do you mean matrices and determinants have different rules?

There's a very good reason for why matrices are multiplied the way they are.

It makes the multiplication of matrices correspond to the composition of linear maps.

Alessandro Codenotti

@abhas_RewCie the thing is that a matrix is a way to represent a linear map, and multiplication is defined in such a way that the map represented by $AB$ is the composition of those defined by $A$ and $B$. Not sure what you mean about the rest

5:15 PM

@TedShifrin for example when you take common a number, they are taken from either a row or column but in case of matrix, it's taken away from all elements.

Yup. I explained that in detail in one of my early YouTube lectures.

@Knight The fact that high school algebra and college algebra are both called "algebra" is a disservice. High school algebra is about solving equations and is actually more applicable to analysis. College algebra deals with groups and such.

@TedShifrin link please?

The link to all the lectures is in my profile.

linear maps?

5:16 PM

You'll have to figure out which one(s) to watch.

okay, let me check it then!

okay, I'll check all videos

I'm free tonight! :-)

The problem is that most people define Ax in terms of dot products with rows, but it's essential to take linear combinations of columns (for many reasons, including the definition of matrix multiplication).

@robjohn Did Analysis get completed in just two years of University course?

@abhas_RewCie matrices and linear maps are started in high school algebra and then abstracted in college linear algebra

They don't really do anything at all with linear maps in high school, @robjohn. Unless you're talking about high schools that teach multivariable calc and linear algebra — which drives me nuts.

5:18 PM

@TedShifrin I've 2 books, in one, the matrix multiplication is ith row times ith column and in other one it's ith row to nth column

But, they all give different answers

@Knight Anyalysis is a big topic, as is algebra. Analysis contains, but is not limited to, real, complex, and fourier analysis.

I don't know what you're talking about, @abhas.

And, more generally, various sorts of differential equations @robjohn

@TedShifrin "not limited to" ;-)

Well, that was a biggie.

@TedShifrin for example, Higher algebra by hall and knight has different rules for determinant multiplication and school book has a different

5:19 PM

@abhas: I don't believe you're correct.

@robjohn Will these kinds of manipulations still gonna linger in Real Analysis?

Okay... Let me check again....

@Knight some manipulations will always linger in any branch of math.

:)

Not to mention calculations.

5:21 PM

@robjohn Is Real Analysis a matured field now? I mean further development in Real Analysis is unlikely.

@TedShifrin some types of differential equation studies more resemble algebra than analysis.

Well, lots of $C^*$ algebras and functional analysis is certainly like that. The dynamical systems course I studied one year of grad school was definitely manifolds + serious analysis, and all the PDE I know doesn't resemble algebra.

@Knight That probably depends on which expert in the field you ask.

the natural context in which to study differential equation is clearly by working on smooth topoi

smacks Thor

5:25 PM

@robjohn For me, your view matters quite a lot. So, what you think about it’s maturity?

@TedShifrin are there multiple ways of multiplying determinants?

Alessandro Codenotti

A determinant is a real number, how do you multiply numbers?

@AlessandroCodenotti by adding n times m

for n x m

@Thor: I will make one final remark re the sphere bundle stuff we were discussing ad nauseam. What I argued was clear was that the restriction of the $\omega_1^j$ to the fiber gave a coframing of the fiber. Perhaps the equation $s^*\tilde\omega_1^j = \omega_1^j$ has some base $1$-forms in it, too, i.e., it holds mod $\omega_1,\dots,\omega_n$. But since we're interested in the volume form, we're going to wedge with $\omega_1\wedge\dots\wedge\omega_n$ and any non-fiber contribution goes away.

What are you talking about, @abhas?

@TedShifrin to multiply 2 numbers $m$ and $n$, we add $m$ times $n$? right? (I'm answering Alessandro)

5:29 PM

ah wait, $\mathbb{Z}\times\mathbb{Z}$ is 1-dimensional and $(0,1)$ is prime, but not maximal

That's not what he's asking you, @abhas. And what you just said makes sense only for positive integers.

for some reason I thought this wouldn't work earlier, but it does

Hmmm.... okay....

@Knight If you are simply asking about real analysis as taught in the first and second courses in college, then that is pretty mature, unless you start talking about non-standard analysis, which still needs a lot of base work to make it palatable.

Okay, I'll come back tomorrow with required materials...

I'll check your channel too :-)

5:30 PM

OK

hi :)

My lectures start very low-level, but they get very sophisticated, particularly when it gets into more advanced linear algebra and multivariable calculus/analysis.

Hi, @Suisse. Comment ça-va?

Okay... If they cover basics, then it shouldn't be a problem with me :)

If $\alpha<0$ then $$\lim_{n\rightarrow +\infty}\left ( (n+1)^{\alpha}-n^{\alpha}\right )=0-0=0$$

If $\alpha> 1$ then $\left (1+\frac{1}{n}\right )^{\alpha}\geq 1+\frac{a}{n}$ and so $$ n^{\alpha}\left ( \left (1+\frac{1}{n}\right )^{\alpha}-1\right )\geq \frac{a}{n^{1-\alpha}} $$
Taking the limit $n\rightarrow +\infty$ we get $$\lim_{n\rightarrow +\infty}n^{\alpha}\left ( \left (1+\frac{1}{n}\right )^{\alpha}-1\right )\geq \lim_{n\rightarrow +\infty}\frac{a}{n^{1-\alpha}} \Rightarrow \lim_{n\rightarrow +\infty}n^{\alpha}\left ( \left (1+\frac{1}{n}\right )^{\alpha}-1\right )\geq \infty$$ …

@TedShifrin hey, merci, je vais bien, e toi? je ne parle pas francais hehe .. allemand ^^

Hey I have a very dumb question..
how do I read this formula?

user image

5:34 PM

Aber auf deutsch sagt man nicht "Suisse" :P

@MaryStar $\left(1+\frac1n\right)^a-1\ge0$ if $a\ge0$

You're asking about the formula for $\tau^*$?

@TedShifrin Du sprichst auch alle Sprachen? :D

Aber ja :P

yes that Tau .. how exactly do you "read" it?

tau* = smalles argument of Tau?

and what does that | mean?

I am very much lost. I should start from zero

5:36 PM

they're taking the absolute value of the two conditional probabilities ... and asking for the smallest possible absolute value you can get as $\tau$ varies.

@TedShifrin Welche Sprachen sonst noch?

Ich hab' auch ein Bißchen russisch studiert.

I know |x| => means absolute value of x... but there is sometimes just one |

p(c=C1|x1=T)

how do you read this?

probability of c which is the same as C1 OR x1 which is Tau?

oO

It's conditional probability. $P(A|B)$ is the probability that $A$ occurs given that $B$ has occurred.

The probability that $c=c_1$ given that $x_1=\tau$.

@robjohn Ah yes! And with that we are done, right?

5:40 PM

aha thanks

@MaryStar I think so.

@Suisse Bitte.

hehe :D

@robjohn Great!! Thanks a lot!! :-)

@Ted I'm taking a break from thinking about the sphere bundle to prepare for my algebra exam next week. I'll come back to the topic after, perhaps it will be clearer to me then (I also posted it as a question on main earlier, so maybe it resolves that way). In any case, thanks for all the putting up with me so far.

5:50 PM

how does it feel to be considered as the most smart humans?

@Thor: No problem. I just figured I'd throw in that last comment for completeness.

ρ → |T |

what does this → means?

in a sentence:
"The experiment in Figure 3.9 has used a large value of ρ (ρ → |T |,
little randomness, large tree correlation) to make sure that each tree
decision boundary fell within the gap."

Hi everyone. Can anybody answer this :If f(x) =$x^2+ ax +b$ be an intergral function of variable x , then can we comment on the nature of a and b?

@Suisse: I assume it means "approaching"

@Bhavay: You mean integral. What does that word mean in your sentence?

@TedShifrin oops, yes .

5:56 PM

@TedShifrin aha like in limes? y -> x means y tries to approach x but will never touch it?

Limits, yes.

nice

thx

I don't know why $|T|$ is considered large, but you have some context.

@Bhavay: So explain what an integral function is.

@TedShifrin In which y belongs to Z ?

For what sort of x?

5:59 PM

@TedShifrin For every x i.e. the domain of x.

What is the domain?

@TedShifrin All the real axis.

That is impossible.

Polynomials are continuous, so the only way they can take on only integer values is if they are constant functions.

But isnt domain of a polynomial function R?

Not for your question.

That's why you must understand the context of the question when you ask it.

6:02 PM

For f(x) to be a constant function shouldn't the coefficient of $x^2$ must be 0?

I'm telling you you need to find the correct definition for your question to make sense.

@TedShifrin Can f(x) be ever a constant function ? I think i need a little more hint.

user434058

Hi @TedShifrin I am stuck at solving a fjnctional equation, could you please help:

@Bhavay: I've told you that you do not understand your own question. You need to find out what the correct definition is.

@FakeMod: I'm not particularly good at that kind of stuff.

user434058

> If $f$ is a continuous function satisfying $f(f(x))=1+x \:\:\forall x\in \mathbb R$, then find $f'(1)$.

user434058

6:07 PM

@TedShifrin Oh, no worries. I'll just leave the question here, so if someone's interested, they could answer it.

You do NOT want to solve the functional equation. You only want to answer that one calculus question. What does that suggest?

user434058

@TedShifrin There's probably some symmetry that I ought to use?

I was thinking creative use of chain rule. They said $f$ is continuous but asked for $f'(1)$, so it seems they need to say $f$ is differentiable.

user434058

@TedShifrin Let's for a moment assume $f$'s differentiability, what next?

I don't know. Play around with chain rule and see what you can get.

What is $f(f(0))$?

6:12 PM

@Balarka yooo it looks like Kreck will host a topology seminar next semester where we are going to work through Milnor's construction of exotic 7-spheres

I'm very excited

Oh, all sorts of classic geometric topology, Kirby-Siebenmann, etc., I guess.

user434058

@TedShifrin Applying chain rule to the expression as it is, gives me: $$f'(f(x))f'(x)=1$ But this expression doesn't really give a hint of what to do next.

Yeah, well, what $x$ is relevant?

user434058

@TedShifrin I don't know. The question is all I have.

I know that.

Some thinking is needed.

user434058

6:15 PM

Hmmm...

Anonymous

6:28 PM

Could someone point me to a proof of the theorem that there is a unique order-preserving bijection from an initial segment of a well-ordered set $A$ to an initial segment of any well-ordered set $B$?

@S.D. induct on $A$

Anonymous

@LeakyNun I think I somewhat get the idea, but I also need to show that the bijection is maximal on at least one side. I'm not sure how to show that

user434058

I got this far: $f(f(f(x)))=f(1+x)$, but $f(f(f(x)))=1+f(x)$. Thus, $$f(x)+1=f(x+1)\iff f(x)+n=f(x+n)$$ But now I don't know what to do next.

@FAkeMod: I'm busy with something else, but will get back to you in a bit.

user434058

@TedShifrin No worries :-)

6:42 PM

@FakeMod: It's more instructive to try to end up with $f'(1)$. How can you do that? In particular, what $x$ value should you start with to finally end up with $1$?

From my fiddling around, you're going to need to find $f'(0)$ somehow.

yeah, a quick manipulation shows that $f^{\prime}(1)=f^{\prime}(0)$ if they exist

write down the difference quotient and apply your previous observation to it

yields to @Thor's cleverness

user434058

@Thorgott Heh, it was just a black out moment. I get what you say :-)

user434058

@TedShifrin Doesn't Thor yield his Mjolnir? :P

My observation was that $f(f(0))=1$, so we want $f'(f(f(0)))$.

user434058

6:51 PM

@TedShifrin But doing it, is just giving me either equations with unknowns, or redundant equations.

there must be more to the original equation $f(f(x))=1+x$, just using $f(x+1)=f(x)+1$ will not suffice

because the latter is also satisfied by the floor function, which fails differentiability at integer points

user434058

@Thorgott Yeah, true.

user434058

I have now fiddled to get a few redundant (already true) equations. I am not able to extract any more information.

You have to use chain rule a bunch.

Hellou.

6:55 PM

observation: $f$ is bijective

user434058

@Thorgott Now, that isn't obvious to me. How so?

Note that $g(x)=1+x$ is bijective.

user434058

@TedShifrin which implies $f\circ f$ is bijective.

So $f$ is both injective and surjective.

I thought I needed to use this, but I didn't yet. I think Thor sees more than I do.

I'm afraid I'm not seeing any further than this so far

observation: if $f$ is differentiable, it is a diffeomorphism

7:01 PM

I don't know what $f(0)$ or $f'(0)$ is, unfortunately.

This is the type of problem I never like. Oh well.

it's really bothering me whether the hypothesis imply $f$ differentiable, this isn't clear to me at all

user434058

@TedShifrin yeah.

do we know any solution of this equation?

user434058

@Thorgott I suppose we can assume $f$ to be differentiable.

user434058

@TedShifrin heh. Same 'ere.

user434058

7:03 PM

@Thorgott I do have one, but I don't really follow its line of reasoning and the method seems too unintuitive to me. Here it is:

No, if they just gave continuous, we're supposed to deduce it somehow, I would think. Or maybe it has to be differentiable at $1$ specifically.

@Thorgott Yes, sure.

There's an obvious affine solution.

of course, $x\mapsto x+1/2$

Yup. But this helps me how?

not really, I just wanted an example to stare at intently

meta-logic dictates that therefore $f'(1) = 1$, or the problem would have been posed differently

7:08 PM

true

user434058

user image

LOL ... so $f(1)=f^3(0)$ and $f'(1) = (f^3)'(0)$ (where I'm writing $f^3$ for $f\circ f\circ f$).

user434058

@MikeMiller The answer is 1, btw.

Oh, so you have a solution.

You prove that the obvious solution is the only solution?

Looks awful.

user434058

7:09 PM

@TedShifrin But I don't like it :\

So probably correct.

ah, that makes sense

It's funny that they introduce x = a for some reason.

user434058

@Thorgott Does the solution make sense? How? How does equation/observation (1) help in funding the function.

Is it difficult to show that $\Gamma(\frac{1}{3})$ cannot be expressed by elementary functions?

7:11 PM

funding the function?

@Huy Who knows.

sure I member

@TedShifrin You probably do!

Hell no.

Is it still unknown?

7:12 PM

It's a corollary of what I said that I have no idea.

I'm working on some cool stuff guided by an article to "compute" values like that

user434058

Alright, so is the solution correct, or is it not correct?

I'm done with it.

user434058

@TedShifrin Alright, I understand :-)

Did they ever prove that $f$ had to be differentiable?

user434058

7:14 PM

@TedShifrin Nope. I suppose it was implicit.

yea we all member

No, they never differentiated $f$. They just showed that continuity implied that $f$ is the obvious function.

So the approach I was taking was totally hopeless.

Not surprising.

There's a reason I never do math like this.

user434058

Like, I am just a high schooler, our teachers don't really take the pain to clarify every minute detail, and it's often the case that continuity and differentiability go hand in hand unless mentioned.

user434058

@TedShifrin Is f(x) being the obvious function neccessary?

user434058

(It's okay if you don't respond) ^

7:17 PM

@FakeMod The proof they're giving is that f(x) = x+1/2, by showing that f(x) - x - 1/2 is zero. The proof that the difference is zero is gimmicky and irritating. These kinds of questions are suitable for olympiad math, and not much more.

tee hee

user434058

@MikeMiller But why did they conclude that g(x) needs to be zero. I don't see any reason whatsoever.

It seems India is infinitely more into competition math than anywhere else.

Their justification seems to be the line "Considering..."

Putnam problems, although sometimes annoying, often are a bit more substantial.

7:18 PM

I don't find this question very interesting, sorry.

I hate the Putnam.

I agree it's still better math.

user434058

@MikeMiller Neither do I ;-)

Yeah, their 'considering' is hardly a complete argument.

Then why did you come in here and ask it, @FakeMod?

user434058

Putnam is nice as long as I am able to solve the questions involved :P

user434058

@TedShifrin I am preparing for a competitive examination (see JEE Advanced). It's not my choice to leave out and not study the questions which I don't like. I can't really go unprepared just cause I didn't like that certain thing :)

Well, you should start by turning their "solution" into a mathematically valid one. I don't see how they've justified the last step. It may be obvious, but it's up to them to make it clear how.

7:22 PM

on second look, the argument doesn't seem all that clear to me either

actually, it seems incomplete

@TedShifrin The nicest question I've seen recently I was sent from MSE. The question was: if $x \in H^2(M;\Bbb Q)$ satisfies $x^{k+1} = 0$, is there some multiple $nx$ and a map $f: M \to \Bbb{CP}^k$ so that $f^*(H) = nx$, where $H$ is the hyperplane class?

In other words, we know that $\Bbb{CP}^\infty$ classifies degree-2 classes. If the classes are truncated in this way, can you also truncate the $\Bbb{CP}^\infty$, maybe at the cost of scaling your class?

We're asking if $x$ has to be ample (as a divisor).

They assumed $M$ is a closed even-dimensional manifold.

Not even a complex manifold, huh?

user434058

So are we sure that the answer is definitely the obvious one? If even this hypothesis is incorrect, I don't see it worth the effort to spend more time behind this question.

7:24 PM

I can prove this when $k = 1$ or $k = (\dim M)/2 - 1$, without scaling the class. I have no idea about what's in between.

So the first case where I don't know what to do is I suppose cube-zero classes on an 8-manifold.

I only know how to approach such things with complex line bundles on complex manifolds.

Yeah, I think they're just asking smooth here. Note that $n$ can be negative.

user434058

Alright, now I gotta sleep. Thanks a lot for helping me out through this, have a nice day! :-)

@FakeMod: I would suggest you try to figure out their last step. The rest looks straightforward, but I didn't check it.

user434058

@TedShifrin Yeah, it's the last step which looks as an unjustified leap to me

7:28 PM

The proof when $k=1$ uses that the dual submanifold has trivial normal bundle, so you can use that normal bundle to construct a map to $\Bbb{CP}^1$ whose fiber above 0 is the dual submanifold. The proof when $k = (\dim M)/2 - 1$ is that you can construct a map to $\Bbb{CP}^{\dim M/2}$ classifying the class, but it has degree 0, so by a theorem of Hopf you can homotope it to miss a point; at which point you can crush it down to the next-dimension-down complex projective space.

Interesting, @MikeM. You're using techniques alien to me.

I figured. Thought you might be interested anyway.

No idea how to do the general problem.

user434058

7:52 PM

Can any >10k rep user please share a screenshot of this deleted question?

user434058

The above question, in it's exact form has been asked several times on Math.SE: math.stackexchange.com/questions/1711383/…, math.stackexchange.com/questions/3370554/…, math.stackexchange.com/questions/1808941/…, math.stackexchange.com/questions/2657883/… (deleted)

9:31 PM

user image

if $y_1$ and $y_2$ are integral curves flowing "upward" then let $y_3=y_1y_2$

I am confused, can you even describe $y_3$ as a vector field?

sorry, the integral curves are not flowing

should have drawn arrows

9:50 PM

I was driving once when I thought of a certain "impossible egg theorem".

Namely: it is impossible (I think) to draw a closed curve whose curvature is defined and positive everywhere, and which has a point of minimum curvature and a point of maximum curvature, and whose curvature changes monotonically in between those points.

This is probably a well-known fact (or possibly a well-known falsehood). Anyone know offhand what the theorem (or counterexample) is called?

10:04 PM

isn't an egg (well, a the boundary of a slice thereof) such a curve?

2 hours later…

11:40 PM

@TerranSwett Do you mean "one point of maximum curvature" and "one point of minimum curvature"?

otherwise, an ellipse fits the bill.

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