Mathematics - 2018-11-15 (page 2 of 2)

In a Hilbert space if $Y$ is a closed subspace, is the map $x\mapsto x-Px$ always surjective on the orthogonal complement of $Y$ (where $P$ is the orthogonal projection onto Y)?

Liad

i know that even $a = (f_1,...,f_r)$ @TobiasKildetoft

m is of this form also

Tobias Kildetoft

1:25 PM

@Liad That says only that it is finitely generated. Maximal ideals have a much nicer form

Liad

they are principal ?

Tobias Kildetoft

no, very much not. They are generated by precisely as many elements as the number of variables

Liad

hmm. didn't know that.

Tobias Kildetoft

remember that maximal ideals will have to provide singletons in the affine space by the various properties

Liad

doesnt it singletons or empty sets ?

by the factorization of $V(m)$ to irreducible components

Tobias Kildetoft

1:30 PM

well, a priori sure, but definitely singletons

Have you covered Nullstellensatz?

Liad

not quite but i heard about it many times in the past few days ^^

so i know what it says

(and i think i can use it.)

Tobias Kildetoft

Well, one thing it basically says is that the closed set corresponding to a maximal ideal consists of precisely one point

Though really, what you need for this is knowing what maximal ideals look like, which is a fairly easy exercise once you have Zariski's Lemma

Liad

it says that $I(V(a)) = \sqrt{a}$

if a maximal then $\sqrt{a} = a$

why this means that $V(a)$ is a point?

Tobias Kildetoft

what would the statement say if it was empty?

Liad

if $V(a) = \emptyset$ then $I(\emptyset) = k[x_0,..,x_n]$

Tobias Kildetoft

1:35 PM

so the radical of $a$ is...

Liad

everything

so a is everything

Tobias Kildetoft

right

Liad

and this is a contradiction because m couldn't be the whole ring?

Tobias Kildetoft

yes, that was the original assumption, right?

Liad

(im not sure if the whole ring is maximal ideal or not)

Tobias Kildetoft

1:37 PM

maximal ideals are proper ideals

(the $0$-ring is not a field)

Liad

the original assumption was $a \subset m$ where $m$ maximal

so we got that $m$ is the whole ring

ok i believe it. thanks! @TobiasKildetoft

Tobias Kildetoft

I had forgotten how much harder one has to work when doing these things in the classical way

Liad

yea and it doesn't really feel like we got a contradiction at the end ^^ although i understand we did.

Tobias Kildetoft

The thing that "really" happens is that the maximal ideal has the form $(x_1-a_1,\dots, x_n-a_n)$, and the point $(a_1,\dots, a_n)$ is the one at which any polynomial in $a$ will vanish

Liad

i can see why its true for the case $n=1$ from the fact that each f splits over k

for general n im not sure how to get it.

Tobias Kildetoft

1:44 PM

yeah, it is not quite trivial. The main ingredient is Zariski's Lemma stating that any field which is finitely generated as a $k$-algebra is in fact finite over $k$

Rithaniel

A continuous function doesn't need to be bijective, right?

Tobias Kildetoft

@Rithaniel right

user131753

@Rithaniel Consider a real valued constant function of a real variable. Is it continuous?

user131753

@TobiasKildetoft Oops!

Rithaniel

Okay, that's what I thought. In that case, I believe I need to restrict my domain to make the function bijective

Math geek

2:23 PM

I understood. How the open set in $I \times I$ look?

@TedShifrin

For any neighbourhood of $p\in I \times I$ I can find a connected neighbourhood marked in blue. $p$ was arbitrary, so $I\times I$. locally connected.

Won't the marked blue neighbourhood be pathwise connected?

Still I don't get a suitable point to show that given $I\times I$ under dictionary order is not locally pathwise connected

CaptainAmerica16

2:55 PM

@AlessandroCodenotti No, that sounds interesting though.

Anjani Gupta

Let G, a Lie group act continuously and transitively on manifold M. Consider a point x in M and it's stabilizer H. H is closed Lie subgroup and hence G/H is a smooth manifold. Prove that the orbit map f:G/H->M is a homeomorphism. I know how it's a continuous bijection but can't show that the inverse is continuous.

Rithaniel

3:25 PM

Hmmm, what would a non-path-connected, but connected space look like?

Math geek

$[0,1]\times [0,1]$ under dictionary order.

but I am not able to prove given space is not locally pathconnected.

:47630123

Rithaniel

Okay, I get your meaning, though. Something where values are sufficiently scrambled such that a path cannot be constructed?

Lucas Henrique

Hey there.

So I'm studying the basics of topology and I came across this problem:

The author wants me to show that $\mathbb{d}$ is a metric

My question is not about the solution at all: I want to understand how that Cartesian product is well-defined.

I could choose $\Lambda = \Bbb R$, and have $\lambda \in \Lambda$ in any arbitrary order. So the Cartesian product would depend on the choice, I guess.

(Well, if the class of all the metric spaces has cardinality of at least $\aleph_1$)

Oops, $\frak{c}$

Buddhini Angelika

3:55 PM

Hi, I came up with a definition about the closure of a graph: the closure of a graph of order n is defined as the graph obtained by joining non adjacent vertices whose degree sum is at least n until no such pair remains. Does that mean if there are no vertices whose degree sum is at least n , there is no closure for that graph?

As an example if I have a graph with 75 vertices which is 4-regular the sum of degrees of any 2 non adjacent vertices is 8, so there are no vertices whose degree sum is 75. Then can we say there exist no closure for that graph?

Semiclassical

@BuddhiniAngelika given that the definition includes that "until no such pair remains" step, I could see it argued that this graph would be its own closure, rather than the closure not existing

Lucas Henrique

Yeah.

Semiclassical

in particular, that'd mean that cl(cl(G))=cl(G)

i.e. that the closure of a graph is its own closure

(also, the word 'closure' seems to have a few different usages in graph theory. what you're referencing is specifically known as the 'Hamiltonian closure' of a graph)

Alessandro Codenotti

4:33 PM

@Rithaniel topologist's sine curve

Liad

why does $nil( (K[x_1,x_2,...)/(x_1,x_2^2,x_3^3,...)) ^n) \ne 0$ for all n?

Leaky Nun

5:14 PM

because $x_{n+1}$ is there

Semiclassical

5:25 PM

there's an extra ) in there btw

should be only one between the dots and ^n

hmm. just noticed a question on MSE which doesn't appear to have been answered directly on this site, but has been answered on MO in the past

Would that count as a duplicate?

(the question isn't research-level by any means)

Ted Shifrin

5:46 PM

@Mathgeek But what if your point $p$ is $(x,0)$ or $(x,1)$ for some $x$? Now is there a path-connected neighborhood?

Érico Melo Silva

yoyoyo

Ted Shifrin

heya yoyo

Secret

abs.gov.au/websitedbs/a3121120.nsf/home/…

Érico Melo Silva

just got back from gre

Ted Shifrin

What excitement brims ...

Secret

5:47 PM

gotta be glad that time only flows forward, otherwise we will not be able to distinguish A cause B vs A just happened to correlate in time with B or vise versa

Érico Melo Silva

the uber driver made an unsavory joke about latinos and i just sat there stone faced thinking “how can u not tell”

was p bad

Ted Shifrin

These are the racistly emboldened times in which we live ...

Érico Melo Silva

v untactful

im hibyeing i gotta meet w sid in an hr or two

Ted Shifrin

Both the pre-college interviews I've done so far have been with racial minorities. I tried to stay unpolitical, but I couldn't help a snide remark or two.

Oh cool. Have a great meeting with Sid. Say hi. :)

Rithaniel

So, cofinite topology on $\mathbb{R}$ is connected, as you can't have disjoint open sets, but not path connected, because you can't construct a continuous function from $[0,1]$ in the standard metric topology to $\mathbb{R}$ in the cofinite topology (due to one being hausdorff and the other not being hausdorff). Is this correct?

(Also, googling topologists sine curve)

Érico Melo Silva

5:51 PM

i think i’ve been unabashedly hard left openly in front of profs and it seems to have not hurt me yet

but it might soon lol idk

ok bubye

Ted Shifrin

Hmm, certainly there are continuous functions, since any open set in the cofinite topology is open in the usual topology.

Rithaniel

Okay, fair enough. What about cofinite on the naturals?

Semiclassical

@ÉricoMeloSilva yuck

Ted Shifrin

What about it, @Rithaniel? :)

Buddhini Angelika

Thanks @Semiclassical

Semiclassical

5:54 PM

np

Rithaniel

Can you still construct continuous functions?

Ted Shifrin

From what to what?

Buddhini Angelika

But in that graph anyway the degree sum is not going to be at most n though

Rithaniel

from $[0,1]$ under standard topology to $\mathbb{N}$ under cofinite.

Ted Shifrin

Oh, I see.

Alessandro Codenotti

5:56 PM

Rithaniel it doesn't matter what $X$ and $Y$ are, there will always be some continuous functions $X\to Y$. You tell me which functions I'm thinking about though

Ted Shifrin

He's trying to violate path connectivity.

and hi, demonic Alessandro.

Buddhini Angelika

Can someone please guide me to understand how to know whether a graph has a projective arrangement, and draw it if it exists? Any good reference texts to follow in this subject?

Ted Shifrin

@Rithaniel: Aren't the only continuous functions in that case constant maps? Can you prove that?

Rithaniel

I think so. I'll have to think about it a little bit, though.

Also, yeah, constant functions are always continuous, that was oversight on my part.

Abcd

hello @TedShifrin

Rithaniel

6:16 PM

Well, I have a proof in the other direction, that a map from cofinite $\mathbb{N}$ to standard $\mathbb{R}$ cannot be continuous. However, that doesn't imply that the inverse can't be continuous.

Aleksa

6:30 PM

Hey could someone help me out?
https://math.stackexchange.com/questions/3000065/irrational-equation-sqrt9-4x-p-2x

Rithaniel

6:47 PM

Hmmm, is it true that a continuous, nonconstant function from $\mathbb{R}$ must be to an uncountable space?

user330477

How do I show that closure of the set of matrices of rank $k$ is actually the set of matrices of rank at most $k$ for Zariski topology on the set of $m$ by $n$ matrices? I have shown that matrices of rank at most $k$ are closed.

The following result is also to be used: Let $S \subset A^n$ be a subset, and let $F : A^1 \rightarrow A^n$ be a polynomial map. Suppose that $F(t) \in S$ for infinitely many $t \in A^1$. Then, $F(t) \in S$ for all $t \in A^1$.

Abdullah UYU

7:10 PM

Hi all, @Semi I had the analysis exam today.

Daminark

Seems I have stumbled upon the Nerdematics chat

How's it going folks?

MatheinBoulomenos

Hi @Daminark

Rithaniel

Well enough, Daminark.

MatheinBoulomenos

Pretty well, thanks. Attended a really cool colloquium talk

Daminark

Nice

Rithaniel

7:17 PM

Currently working on proving that the natural numbers under the cofinite topology is not path connected.

MatheinBoulomenos

@Daminark And yourself?

Daminark

I'm actually heading on Friday to a seminar at a nearby school that someone I met told me about, on algebraic combinatorics

MatheinBoulomenos

I know nothing about algebraic combinatorics, I know that representation theory of $S_n$ is related to combinatorics, but not sure if it's about that

Semiclassical

I know a bit about analytic combinatorics

Daminark

I also don't know anything about it, though it's something I'd like to look more at

MatheinBoulomenos

7:21 PM

yeah, sounds good

Semiclassical

It's not too complicated. Express you're counting as coefficients of a generating function, then express those coefficients as complex contour integrals over that generating function using Cauchy's integral formula

Bam, you've turned your counting problem into a complex-analysis problem

Daminark

I wonder if it's ever useful to go the other way

Semiclassical

What it's mostly used for is finding asymptotics of said coefficients

which in turn usually means the saddle-point approximation

(Or Laplace's method, I forget what the standard name is)

lush

Hi guys!

Alessandro Codenotti

@Rithaniel do you need a hint? There is some nontrivial facts needed (Also this is referred to a message you wrote earlier, $\Bbb R$ with the cofinite topology is path connected)

MatheinBoulomenos

7:33 PM

hi @lush

Rithaniel

I actually thought it was. It'd be useful to see what I had missed there, but I'm now working on $\mathbb{N}$ with the cofinite topology not being path connected (or is that one also path connected?)

Alessandro Codenotti

This one isn't

And as you might suspect by now there is some cardinality argument involved

Rithaniel

Excellent! That's what my argument is working along.

Shobhit

If X is a normed linear space and $M$ is a linear subspace of $X$, then show that $\overline{M} = \cap \{kerf | f \in X^{} \text{ and } M \subseteq kerf\}$, where $X^{}$ denotes the dual space of $X$.

Alessandro Codenotti

To show that $\Bbb R$ with the cofinite topology is path connected: pick any injection $[0,1]\to\Bbb R$ and show that it is continuous

Rithaniel

7:37 PM

Basically, since ($\mathbb{N}$, cofinite) is $T_1$, points are closed, so the inverse of any continuous function partitions $[0,1]$ into countably many closed sets. (From there, I still need to consider what to say, but I think I have what I need)

Alessandro Codenotti

@Rithaniel that's a good start. Is such a partition possible?

Shobhit

If $X$ is a normed linear space and $M$ is a linear subspace of $X$, then show that $\overline{M} = \cap \{kerf | f \in X^{*} \text{ and } M \subseteq kerf\},whereX^{*}denotesthedualspaceofX$.

Alessandro Codenotti

Use \ast if the asterisk is messing with the chat's formatting

Rithaniel

Well, such a partition is possible, but I was thinking that would imply the function is a constant function, which is not allowed in a function being used to show that a space is path connected (am I mistaken?)

Alessandro Codenotti

Why is it possible?

Shobhit

7:41 PM

If $X$ is a normed linear space and $M$ is a linear subspace of $X$, then show that $\overline{M} = \cap \{kerf | f \in X^{\ast} \text{ and } M \subseteq kerf \}$, where $X^{\ast}$ denotes the dual space of $X$

How to prove this

I took a member in closure in M

say x

Rithaniel

Ah, hmmmm, I think I see. There's always going to be some overlap between closed sets in the standard metric topology, if they cover all of $[0,1]$, which is going to violate it being a function.

Shobhit

then there exists a sequence in M such that it converges to x

now what to do?

how to show that this x is in R.H.S

Jake Rose

Here’s a pretty basic differential equations question

why canwhen solving a differential equation why couldn’t you have $A\sin(x-1) +Bsin(x)$ as your gener solution?

Akiva Weinberger

Hey guys

Rithaniel

Greetings Akiva

Akiva Weinberger

7:47 PM

@JakeRose That's the same as $A\sin x+B\cos x$, probably

The latter is more of a standard form

(Not for the same $A$ and $B$. I mean the set of functions as $A$ and $B$ vary.)

If you know any linear algebra: $\{\sin(x-1),\sin(x)\}$ and $\{\sin(x),\cos(x)\}$ are both a choice of basis for the same space

Jake Rose

Ahh I see. It’s more so just a case of simplicity

Akiva Weinberger

@Rithaniel Have you learned about quotient topologies yet?

Rithaniel

Lightly, but not in much depth.

Akiva Weinberger

A space quotiented by an equivalence relation

Rithaniel

I know what they are, in other words.

Akiva Weinberger

7:53 PM

I just want to point out two things that I think are interesting

If you take the space $[0,1]$ and quotient $(0,1]$ to a point (that is, $x\sim y$ if either $x=y$ or $x,y\in(0,1]$), then the result is the Sierpiński space

a non-T1 space with two elements

And secondly

if you take $\Bbb Q$ and quotient by the equivalence relation $x\sim y$ iff $x$ and $y$, when reduced, have the same denominator

then I'm pretty sure the result is the same as $\Bbb N$ with the cofinite topology

Rithaniel

Well,that's the starting topology in the $\mathbb{Q}$ case? Subspace under standard metric?

Akiva Weinberger

Yeah

Rithaniel

because that's very interesting, if so.

I'm having a little trouble envisioning all cases for that quotient.

Akiva Weinberger

It's weird 'cause the result isn't T2

Rithaniel

How do you get the set of all natural numbers except, say, 2520, for example?

Akiva Weinberger

8:00 PM

Take the set of all rationals other than the ones with denominator 2520

The map here (from $\Bbb Q/{\sim}$ to $\Bbb N$) is, each equivalence class maps to the denominator

The set of rationals that do have denominator 2520 is a discrete set of points

Rithaniel

Alright, yeah, and the interval $(0,2)$ is mapped to all of $\mathbb{N}$

Akiva Weinberger

(Let's say the denominator of $0$ and all the integers is $1$, to make this defined on those)

@Rithaniel Yeah

(Like $\frac01$ and such)

I need to go

Rithaniel

while $(\frac{1}{2},1)$ is mapped to $\mathbb{N}$ minus $\{1,2\}$

Cya

Akiva Weinberger

Yeah

Rithaniel

Thanks for the interesting math fact.

Akiva Weinberger

8:04 PM

And $(0,\frac1n)$ is $\Bbb N$ minus the first $n$ numbers

I guess $0$ isn't in $\Bbb N$ for this. Oh well

Rithaniel

Well, yeah, but that's the standard for the American sense of $\mathbb{N}$.

Akiva Weinberger

Oh, also note that, while $\Bbb R$ with the cofinite topology is path-connected, $\Bbb N$ with the cofinite topology isn't

(I think)

Rithaniel

Yeah, still working on showing that the second statement is true. It's proving tricky, now that I'm at the point of saying that it's impossible to write $[0,1]$ as the union of countably many, disjoint closed sets. I might just invoke a theorem, though I want to prove it, if I can.

Akiva Weinberger

8:56 PM

So you have a countable collection of closed sets $[a_i,b_i]$, and you want to construct a point that's not in any of those

Hm

Alessandro Codenotti

@Rithaniel Extensive hint: Suppose that such a countable collection of closed sets $C_i$ covering $[0,1]$ exists. For every $i$ you can pick a closed interval $F_i$ in $X\setminus C_i$ (why?). Can you arrange it such that $i<j\implies F_i\supset F_j$?

Akiva Weinberger

I guess you need to take care to make sure that no $F_i$ is completely contained in any $C_j$

Alessandro Codenotti

@AkivaWeinberger I don't think this is needed

Akiva Weinberger

If $F_2$ is contained in $C_4$ by accident, then $F_4$ will have to simultaneously be a subset of $F_2$ and disjoint from $C_4$ somehow

@AlessandroCodenotti

($F_4$ will never be contained in $C_2$, because $F_4$ is a subset of $F_2$ which is disjoint from $C_2$. So this is only a concern for $F_i$ and $C_j$ with $i<j$)

Alessandro Codenotti

Ah, I see, that's annoying indeed

It should be possible to ensure it doesn't happen though

Akiva Weinberger

9:11 PM

It would be unavoidable if $C_0\cup C_1\cup\dotsb\cup C_j=[0,1]$ for some finite $j$

but that can never happen

and I think that's the only way for it to be unavoidable

I mean, basically, once you're choosing $F_i$, it's gotta be in $X\setminus(C_1\cup\dotsb\cup C_i)$, yeah?

So choose one of the connected components of that

That connected component surely is the union of more than one $C$ thing

Alessandro Codenotti

@AkivaWeinberger No wait, don't you need to avoid $C_j$ for $j>i$?

Akiva Weinberger

so make $F_i$ be an interval starting in one $C$ thing and ending in another

@AlessandroCodenotti It can't be a subset of those, it doesn't need to be disjoint

Alessandro Codenotti

Oh, ok, I see what you meant

Akiva Weinberger

I should point out that, unless "countable" means "countably infinite", there's a stupid counterexample

which is, write $[0,1]$ as the union of one closed set

(itself)

But clearly we're talking about at least two closed sets

and it's easy to show that it can't be finite ($\ge2$)

Alessandro Codenotti

Agreed

Akiva Weinberger

9:18 PM

So I'm now convinced that the theorem is true

Unrelatedly, I like this proof:

Without using unique factorization or brute force, show that $7$ is not a factor of $5\times93$

and the "trick" is, notice that $15$ is a multiple of $5$ and also one more than a multiple of $7$

so, if $7$ were a factor of $5\times93$, it'd also be a factor of $3\times(5\times93)=(15)\times93$

${}=(14+1)\times93=14\times93+93$

and then $14\times93=7\times(2\times93)$ is clearly a multiple of $7$, so subtracting it off gives that $7$ would be a factor of $93$

which is false, so $7|5\times93$ is false

and I realize that this is incredibly basic

but I was trying to show that $\Bbb Z[i]$ has unique factorization, and I kept on forgetting the details

and I eventually realized that the key to proving that is the $p|ab\implies p|a\lor p|b$ thing, but I kept on forgetting how to prove that one also

so eventually I just took a step back with a concrete example in $\Bbb Z$

and now I get everything I think

and why Euclidean domains have unique factorization, which wasn't really clear to me until now

And unique factorization fails in $\Bbb Z[i\sqrt5]$, for example (eg $6=(2)(3)=(1+i\sqrt5)(1-i\sqrt5)$) because no multiple of $1+i\sqrt5$ is one more than a multiple of $2$, despite having no common divisor other than $1$ and $-1$

You can look at the shape of $\langle d\rangle$ in the complex plane for any $d\in\Bbb Z[i\sqrt5]$ and you can look at the shape of $\langle 2,1+i\sqrt5\rangle$ and you can see how they're different shapes

(I mean, they're both lattices, but the former looks like $\Bbb Z[i\sqrt5]$ up to scaling and rotation and the latter doesn't)

Trey

how do I take the derivative of something like $\frac{u}{du}$ ? cause it's not the same as ${u}{du}$ right?

Akiva Weinberger

9:36 PM

What's $u$ a function of?

What are you differentiating with respect to?

Trey

actually, I meant $\frac{x}{dx}$

trying to derive with respect to x

Akiva Weinberger

I think you would treat $dx$ as a constant

Semiclassical

Where are you coming across this?

u/du is notation which in most contexts would be either a typo or nonsense

Trey

well, I was trying to take the derivative of $\frac{arctan(x)}{1+x^2}$, I applied $u = arctan(x)$, $du = 1+x^2$ but then I ended up with $\frac{u}{du}$ rather than ${u}{du}$ and I got stuck

Akiva Weinberger

That would be $\dfrac{du}{dx}=1+x^2$

or $du=(1+x^2)dx$

In any case, I don't think you want to do substitution like this for derivatives

It works more for integrals

If you do go this way, you'd end up with $\dfrac u{u'}$, whose derivative is $\dfrac{(u')^2-(u)(u')}{(u')^2}$

Oh, not even. What's the derivative of the arctangent?

$\dfrac1{1+x^2}$

So this is all wrong

$\dfrac{du}{dx}$ (or, another way to write it, $u'$)${}=\dfrac1{1+x^2}$

and so your function is $u\dfrac{du}{dx}$ (or $uu'$)

and the derivative can be found using the product rule

and then you can substitute in $u=\arctan(x)$, $u'=\dfrac1{1+x^2}$, and $u''=\dfrac{-2x}{(1+x^2)^2}$

but I dunno if this is all any simpler than just using the quotient rule on your original thing directly @Trey

You don't use substitutions like this for derivatives, generally. It's a thing for integrals

Semiclassical

10:04 PM

I mean, you can do this:
$$\frac{d}{dx}\frac{\arctan x}{1+x^2}=\frac{d}{du}\left(\frac{u}{1+\tan^2 u}\right)\frac{d}{dx}\arctan x$$

and then $u/(1+\tan^2 u)=u/(\sec^2 u)=u\cos^2 u$, so the first derivative is $\cos^2 u+2 u\sin u \cos u$

the trouble is that you need to re-express that in terms of $x$

I mean, you do have d/dx(arctan x) = 1/(1+x^2)

and cos(u) = cos(arctan x) = 1/sqrt(1+x^2), sin(u)=x/sqrt(1+x^2)

So you end up with $$\left(\frac{1}{1+x^2}+2 \arctan(x)\frac{x}{1+x^2}\right)\frac{1}{1+x^2}=\frac{1+2x \arctan(x)}{(1+x^2)^2}$$

which isn't bad by any means. but the real question is indeed whether this is easier than the quotient rule

and I'm dubious of that

Arrow

Hello. Let $R$ be a polynomial ring over a reduced ring (no nilpotents). Let $R_d$ be the ideal of homogeneous polynomials of degree $d$ and let $R_+$ be their sum. Isn't $R_+$ a radical ideal? It seems it contains the generators of the polynomial ring, so dividing by it should give the ring of coefficients, which is reduced by assumption.

Ted Shifrin

10:34 PM

@Trey This is why I get upset when people write nonsense like $du=1+x^2$. First of all, the derivative of $\arctan x$ is $\dfrac1{1+x^2}$. But, more important, when you take $df$ you MUST write $f'(x)\,dx$. You cannot leave off the $dx$. Are you trying to differentiate or integrate that, anyhow? The integral $\int \frac{\arctan x}{1+x^2}\,dx$ does turn into $\int u\,du$ if you use the correct derivative.

Semiclassical

if you've got df on one side and not dx on the other, you've either done something wrong or you're abusing notation

Ted Shifrin

10:49 PM

I think you and DogAteMy totally went down the rabbit hole on this one, Semiclassic.

mick

1

Q: Why is $\sup f_- (n) \inf f_+ (m) = \frac{5}{4} $?

Let $f_- (n) = \Pi_{i=0}^n ( \sin(i) - \frac{5}{4}) $ And let $ f_+(m) = \Pi_{i=0}^m ( \sin(i) + \frac{5}{4} ) $ It appears that $$\sup f_- (n) \inf f_+ (m) = \frac{5}{4} $$ Why is that so ? Notice $$\int_0^{2 \pi} \ln(\sin(x) + \frac{5}{4}) dx = \int_0^{2 \pi} \ln (\sin(x) - \frac{5}{4})...

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Q: Why is $\inf g \sup g = \frac{9}{16} $?

Consider this question here : Why is $\sup f_- (n) \inf f_+ (m) = \frac{5}{4} $? Call that conjecture about $\frac{5}{4} $ conjecture $1$. Let $g(n) = \prod_{i=0}^n (\sin^2(n) + \frac{9}{16}) ) $ Conjecture $3$ : ——- Conjecture $2$ is : $$ \sup g(n) \space \inf g(n) = \frac{9}{16} $$ And ...

calculus geometry fractions limsup-and-liminf products

Any ideas ?

user330477

How do I show that closure of the set of matrices of rank $k$ is actually the set of matrices of rank at most $k$ for Zariski topology on the set of $m$ by $n$ matrices? I have shown that matrices of rank at most $k$ are closed.

The following result is also to be used: Let $S \subset A^n$ be a subset, and let $F : A^1 \rightarrow A^n$ be a polynomial map. Suppose that $F(t) \in S$ for infinitely many $t \in A^1$. Then, $F(t) \in S$ for all $t \in A^1$.

Semiclassical

@TedShifrin lol

probably

Konformist Liberal

11:07 PM

I always looked down on these kind of questions but somehow I really want to hear some thoughts on this one: What are the reasons to compute the homotopy groups of spheres (I can imagine that some of them (especially low dimensional ones or (n,n) pairs) might be "useful" but what about 1007th homotopy group of 4502-dim sphere)?

Semiclassical

Easy to pose, hard to answer?

Konformist Liberal

That's a good one in my opinion, yes :)

Secret

And sometimes, special cases are so unique that it may open doors to other questions

Semiclassical

I'm not a topologist tho

so take that as a guess

Secret

There are areas of mathematics where almost every special cases are their own field, such as in integration, changing even the integrand slightly you get a completely different function

and for geometry related stuff, there are only 3 regular polytopes for dimensions > 4

amd 4 has the most number of them

but we will not know about these if we don't study them

Konformist Liberal

11:15 PM

When I look up some applications of algebraic topology online (in mse or wikipedia for example) I can't find any regarding theoretical physics for example. An relatively easy example on that regard would be nice to hear. Also, would this be a rude question to ask my (algebraic) topology professor?

Semiclassical

I don't think so. You might motivate it as: What got people interested historically in computing the homotopy groups of spheres?

there's a little bit of the history on Wikipedia, btw: en.wikipedia.org/wiki/Homotopy_groups_of_spheres#History

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