Mathematics - 2024-09-14

Mad

08:42

Hello, why is there a bigger equal sign instead of equality here? $A \subset U$
$U=\bigcup_i U_i , A=\bigcup_i A_i$
equation in question: $\mu(U-A) \leq \mu(\bigcup_i (U_i -A_i))$

i would thought you have :
$\mu(U-A) = \mu((\bigcup_i U_i -\bigcup_i A_i)) = \mu(\bigcup_i (U_i -A_i)) $

where mu is the lesbegue meassure.

Ryder Rude

09:02

hi

i want to understand some sections of this answer physics.stackexchange.com/a/14944

the sections : "what finitary means", "ordinal religion", "ordinal analysis" and "other interpretations"

> Given a sequence of ordinals which approaches the Church-Kleene ordinal, the theories corresponding to this ordinal will prove every theorem of Arithmetic, including the consistency of arbitrarily strong consistent theories.

later on :

>To go further requires an advance in our systems of ordinal notation, but there is no limitation of principle to establishing the consistency of set theories as strong as ZF by computable ordinals which can be comprehended.

> Doing so would complete Hilbert's program--- it would removes any need for an ontology of infinite sets in doing mathematics. You can disbelieve in the set of all real numbers, and still accept the consistency of ZF, or of inaccessible cardinals (using a bigger ordinal), and so on up the chain of theories.

what does this stuff mean? I know that the consistency of a theory can be proved by a bigger theory, but then the consistency of the bigger theory is in question

is this stuff expressing the same idea

The Empty String Photographer

09:33

Be careful when squaring numbers

$n = n+1$

$n^2 = n^2 + 2n + 1$

$0 = 2n + 1$

$n = -0.5$

Ryder Rude

09:57

@TheEmptyStringPhotographer this is really interesting. If n=n+1 were an axiom then n=-0.5 would be a theorem

but if n=-0.5 were an axiom then n=n+1 wouldn't be a theorem

this is because the former axiom is inconsistent, so everything is a theorem

ILikeMathematics

11:16

Btw @BenSteffan will I hear back from Bonn Studierendenwerk again or only in the last 3 weeks before moving in to my apartment there? How can I be sure that I will get an apartment there if it's such a short period of time after being notified

Since I already put up this request one year in prior, maybe I will be prioritized?

Ben Steffan

@ILikeMathematics I would hope they will notify you sooner

@ILikeMathematics As far as I understand it, the system works off of something like a wait list

so the earlier you put your name on that list the better

Ben Steffan

11:52

@Mad Let $U_0 = U_1 = [0, 1]$, $A_0 = \emptyset$, $A_1 = [0, 1]$, and compute both sides :)

Bml

Hi everyone. I ask the question again because I have not received an answer: is there a method to prove that $\frac{2}{3}-\frac{2}{3} \sin\left(\frac{pi}{6}- \frac{\arccos \left(\frac{54 abc}{(a+b+c)^3} - 1\right)}{3}\right) > (\sqrt{a} - \sqrt {b})^2 + c$ holds $\forall a,b,c > 0$?

Sine of the Time

the expression is not defined for all $a,b,c>0$

Bml

@SineoftheTime Why?

Sine of the Time

you tell me

Bml

12:14

@SineoftheTime I don't know

Sine of the Time

what can be problematic in your expression?

Bml

@SineoftheTime The argument of $\arccos$, I guess.

Sine of the Time

yeah

Bml

@SineoftheTime So we have to verify also $-1 < frac{1}{3} \frac{54 abc}{(a+b+c)^3} -1 < 1$, right?

Sine of the Time

no, you don't have to "verify" it. This is the domain

Bml

12:30

@SineoftheTime Yes, the domain is $\frac{abc}{(a+b+c)^3} < 1/27$, which is consistent to Semiclassical's answer to my question on Math SE.

@SineoftheTime So I rephrase my question: is there a method to prove that $\frac{2}{3}-\frac{2}{3} \sin\left(\frac{pi}{6}- \frac{\arccos \left(\frac{54 abc}{(a+b+c)^3} - 1\right)}{3}\right) > (\sqrt{a} - \sqrt {b})^2 + c$ holds $\forall a,b,c > 0$ such that $\frac{abc}{(a+b+c)^3} < 1/27$?

Sine of the Time

ok

Bml

12:56

@SineoftheTime Do you have any idea?

Sine of the Time

13:11

no, but you can try to apply Semiclassical's method

Bml

@SineoftheTime How, in what ways?

Ben Steffan

...how about you think about this on your own for a bit? :)

nickbros123

does the indefinite integral operator, say from polynomial space to (??) map into some kind of quotient space?

perhaps $\int:P(\mathbb{R}) \to P(\mathbb{R})/ \text{span}(1)$

Bml

@BenSteffan I have been thinking about this for a bit, and as I mentioned yesterday, I cannot define a new function as Semiclassical did because in this case I have three parameters, not one. Also, this is not an inequality to solve in one variable, but to verify that the inequality holds for the given values of $a,b,c$.

nickbros123

@nickbros123 Ok so we can still identify the codomain with $P(R)$ since conveniently span(1) happens to be the kernel of the differentiation operator on $P(R)$

I basically want to make a counter example for the claim (where $S,T$ are linear operators) that $ST=I$ implies $S$ and $T$ are invertible

Ben Steffan

13:25

@nickbros123 You can make it map into any quotient space by post-composing with any quotient projection you like :)

nickbros123

@BenSteffan in this case i wanna do the reverse of that :)

Ben Steffan

what's the reverse of composing with a quotient projection?

nickbros123

i dont know? in this case though I know that $P(R)/span(1) \cong P(R)$ so I can use that isomorphism (call it g) so that $\int \circ g$ becomes a linear operator again

or the other way round

Ben Steffan

@Bml You have 3 parameters, the question answered by Semiclassical has 4, essentially. One way to verify that a given inequality holds for a range of parameter values is to solve it.

nickbros123

tbh never mind, all this can be avoided if I just defined the map $\int:P(R) \to P(R)$ as $x^j \mapsto \frac{x^{j+1}}{j+1}$

Ben Steffan

13:30

As for "I have been thinking about this," well, why don't you tell us your thoughts? You come in here with a big inequality and essentially ask how one solves it, and offer us nothing else.

This is not the main site, but at this point the motivation of people to just solve problems like this here for you is rather slim, I'd reckon.

Jakobian

@RyderRude I believe they mean relative consistency, yes

@RyderRude finitary means (of relation or operation) with finitely many inputs

Bml

@BenSteffan My idea was to express the RHS in a form similar to $\frac{abc}{(a+b+c)^3}$ at the LHS, so I could set it equal to a parameter and write a function that has the latter as an independent variable. The problem is that I cannot transform the RHS into a usefully advantageous form.

Jakobian

Ordinal analysis is some kind of study of ordinal functions I'm pretty sure.

The rest I don't know, not being a set theorist or logician

@nickbros123 the quotient of polynomials by relation p ~ q iff p, q differ by a constant

@nickbros123 yeah sure taking $Sp = p'$ and $Tp(x) = \int_0^x p(t)dt$ works. Note this is not an indefinite integral.

@nickbros123 sure you don't even need to mention integration

@BenSteffan depends on what you find interesting I'd say. For example, I don't find inequalities to be interesting

But I agree with the conclusion

@nickbros123 I'd say easiest way to see this is by vector space of real sequences. You have two shifts operators, shift to the left and to the right.

nickbros123

13:56

@Jakobian ah, neat

Bml

@BenSteffan I did not ask to solve the inequality, but if you had any ideas to enable me to handle the methods that would allow me to solve it. In the previous message I said what I had in mind to do, without success.

Jakobian

14:46

@Bml about similar form, the RHS of that inequality is not symmetric so I don't know what you mean by that

Ryder Rude

@Jakobian apparently, they meant a somewhat different idea

but it is similar to relative consistency

it is this idea en.m.wikipedia.org/wiki/Gentzen%27s_consistency_proof

Bml

@Jakobian Actually I was wrong, the RHS is $1-\frac{2\sqrt{ab}}{a+b+c}$. So the inequality is $\frac{2}{3}-\frac{2}{3} \sin\left(\frac{pi}{6}- \frac{\arccos \left(\frac{54 abc}{(a+b+c)^3} - 1\right)}{3}\right) > 1-\frac{2\sqrt{ab}}{a+b+c}$ holds $\forall a,b,c > 0$ such that $\frac{abc}{(a+b+c)^3} < 1/27$

Derso

15:18

0

Q: Is the angle between tangent vectors of a parametrized ruled surface constant along rulings despite singular points?

Definition. Let $M$ be a complete riemannian 3-manifold and $I\subset \mathbb{R}$ an open interval. Consider a smooth curve $p:I\to M$ and a unitary vector field $v:I\to TM$ along $p$. We define the ruled surface with base curve $p$ and directional curve $v$ as the image of map $\mathbf{X}:I\tim...

Soumik Mukherjee

15:59

Hi all

Ben Steffan

@SoumikMukherjee hi :)

Soumik Mukherjee

If an ideal is contained in union of three ideals then it is not necessary that the ideal is contained in any one of them. The usual example of this is of the Z2×Z2 one. Can someone give some other examples?

Thorgott

on the level of underlying groups, any example is a pullback of this one

actually not sure if that's true, your set-up is slightly more general than what I was thinking of

Mad

16:40

@BenSteffan thank you

Joe

Is there an intuitive reason for why if $A$ is a (commutative) ring and $S\subseteq A$ is multiplicatively closed, then the prime ideals of $S^{-1}A$ are in one-to-one correspondence with the prime ideals of $A$ that are disjoint from $S$? I suppose one idea is that if $x\in S$, then after localizing $x$ will become a unit, and hence cannot belong to any prime ideal.

Jakobian

Containing an element of $S$ will make it trivial after localizing

as in the whole ring

yeah you said that

If $f:A\to B$ is a surjective homomorphism, and $P$ is a prime ideal with $\ker(f)\subseteq P$, then $f(P)$ is a prime ideal

Now take $B = S^{-1}A$. What is $\ker(f)$? It's precisely the set of $a\in A$ such that there is $s\in S$ with $sa = 0$.

Ben Steffan

17:01

The map $A \to S^{-1} A$ is generally not surjective.

Thorgott

1. every ideal of $S^{-1}A$ is generated by (the image of) some ideal of $A$ (idea: division is invertible)
2. every ideal of $A$ meeting $S$ generates the unit ideal of $S^{-1}A$ (idea: you said it already)
3. every prime ideal of $A$ not meeting $S$ generates a prime ideal of $S^{-1}A$ (idea: the image of $S$ in $R/\mathfrak{p}$ does not contain $0$, so localizing that still yields an integral domain)
4. no two distinct prime ideals of $A$ not meeintg $S$ generate the same ideal of $S^{-1}A$ (idea: a fraction with nominator in $\mathfrak{p}$ and denominator in $S$ disjoint from $\mathfrak…

Jakobian

yeah. I just realized that

Joe

Thanks to all for your help. I need to go now, but I'll look back at these answers soon.

Jakobian

No, don't read what I said it was wrong. Read what Thorgott said

Thorgott

@BenSteffan it is, however, epimorphic :P

Ben Steffan

17:10

Here's another intuitive take for the direction "$\mathfrak{p} \cap S = 0$ implies extension of $\mathfrak{p}$ is prime": What does it mean to be a prime ideal? If $ab \in \mathfrak{p}$, then at least one of $a$ and $b$ is in $\mathfrak{p}$. But when you're localizing, all you're doing is making the elements of $S$ invertible, and if one of $a, b$ is a unit, say $a$, then $ab \in \mathfrak{p}$ iff $b \in \mathfrak{p}$, so there's no new elements in $S^{-1} A$ that could violate the rule.

maybe this is too handwavy

@Thorgott oh, that's a fun fact :D

Sine of the Time

algebraists taking over the chat

that's not good news :(

Thorgott

another fun fact is that the map is always surjective if $A$ is zero-dimensional and has finite spectrum

@BenSteffan ah I like this explanation

Ben Steffan

ty

Pizza

Calculate the triple integral on the domain $V$ , $\iiint_V z dxdydz$, $V$ is the set inside the tetrahedron limited by the planes $x=y=z=0, x+y+z = 3-\sqrt{3}$ and outside the sphere of center (1,1,1) radius

1

Sine of the Time

@Pizza spain without s

Pizza

17:16

I tried to do the integral on the tetrahedron, $\int_0^{3-\sqrt{3}} dz \int_0^{3-\sqrt{3}} dx \int_0^{3- \sqrt{3} -x} z dy$

It already seems wrong to me

@SineoftheTime :'(

Ben Steffan

@Joe One final tangential thought about this localization business: If you replace $S$ by its saturation, i.e. the maximal multiplicative subset $S \subseteq S' \subseteq A$ generating the same localization as $S$, then $A \setminus S'$ is a union of prime ideals.

There's other ways to characterize this $S'$ as well; one would be (I believe) that if $f\colon A \to S^{-1} A = S'^{-1} A$ is the canonical map then $f^{-1}((S^{-1} A)^\times) = S'$. So you really get pretty clean picture if you choose to work with saturated multiplicatively closed sets.

Sine of the Time

@Pizza this is only inside the tetrahedron?

Pizza

@SineoftheTime Yes but on geogebra the sphere seems to be outside

Sine of the Time

I'm having another conversation rn so I may be slow answering your questions

@Pizza you sure?

Jakobian

17:32

@Thorgott do we have a theorem similar to the one I described for epimorphisms?

Pizza

Isn't it external?

I insert: y = 0, z = 0

(x-1)² + (y-1)² + (z-1)² = 1

x + y + z = 3 - √3

I didn't write x = 0 to better show that in my opinion the sphere is external

Sine of the Time

yeah

the distance of $(1,1,1)$ from the plane is $1$

so the plane is tangent

Pizza

17:50

now I don't know what to do

Since the sphere is external

Sine of the Time

you can ignore the sphere

and integrate inside the tetrahedron

Thorgott

@Jakobian no, it generally fails for localizations

the image of a non-zero prime ideal of $\mathbb{Z}$ under the inclusion $\mathbb{Z}\rightarrow\mathbb{Q}$ is not a prime ideal

(and they trivially contain the kernel cause that map is injective)

Pizza

@SineoftheTime Oh ok, the integral is written above, is it correct?

I did it by layers

Sine of the Time

so you fix $z$ right?

Pizza

Yes

Sine of the Time

18:07

there's something wrong with the order of integration

Pizza

Where ?

Sine of the Time

the formula I'm familiar is: if $V=\{a\le z\le b, (x,y) \in D_z\}$, then $\int_V fdxdydz=\int_a^b(\int_{D_z} fdxdy)dz$

Jakobian

@Thorgott a pity

Pizza

@SineoftheTime Same

Sine of the Time

what is $D_z$ in your case?

Pizza

18:11

wait

Sine of the Time

I would have use the other formula

Pizza

I was wrong ! :D

I'm going to correct it

Sine of the Time

you can look at pag. 232 of Marcellini Sbordone, esercizi di analisi 2

Pizza

Ah ok found it!

Thanks !

@SineoftheTime a pag 232. é fatto per fili, per strati si può fare ?

Maybe I missed -z in the equation y = 3 - √3 - x

Jakobian

@Thorgott what's your reasoning behind re-opening this question? It clearly is a PSQ. If you think the question is interesting, appropriate thing to do is not to re-open it, but potentially ask question yourself.

Ben Steffan

18:26

Is there a reason you're bringing this against Thorgott but not Joe?

Sine of the Time

@Pizza sì, però è più complicato

Joe

You can me ask me as well :)

Sine of the Time

devi intersecare un piano perpendicolare all'asse z e calcolare l'area del triangolo in funzione di z

Ben Steffan

Also, is the whole lecturing-on-what-you-should-do thing really necessary

Fwiw I'd also be interested to know why y'all reopened this

Jakobian

@BenSteffan I just didn't see that Joe did that too

Ben Steffan

18:28

ok

Jakobian

@Joe same question to you

Sine of the Time

9 upvotes and 7 downvotes is crazy

Pizza

Thanks

@SineoftheTime ah ok

Sine of the Time

apparently, a lot of people are interested

Jakobian

@BenSteffan yes? It's important to educate people how this site works

Ben Steffan

18:29

All parties implied in this have >10k reputation

Jakobian

Then that means they are either ignoring the issue, purposefully, or aren't aware of how the site works

Ben Steffan

You're sure quick to jump to accusations here.

Jakobian

What did I accuse someone of and who did I accuse

This is not something personal

You are seemingly trying to make it personal. Why? Do you have anything to gain? You are playing favourites?

Sine of the Time

no need to jump the guns :(

Joe

Basically, my view is that if a question is clear, well-motivated, and useful to potential future visitors of the site, then I am supportive of it. To me that counts as reasonable context. I think what "context" is is actually much less clear than people make out. For example, some people think giving an "attempt" counts as context, but I've seen others argue that this is actually the lowest form of context, or even detracts from context.

Jakobian

18:33

@Joe where is the motivation

A problem followed with a bunch of questions is not motivation of any kind

Joe

I think it is well-motivated in the sense that people involved in the area care about it, and find the question natural and interesting. You can see evidence of this in the comments. I don't see why explaining why a question is interesting is going to help anyone. The people who find it interesting don't need an explanation, and people who don't find it interesting are unlikely to be convinced.

Jakobian

Again, that doesn't matter. If you find a badly written question interesting the appropriate thing to do is repost that question.

Motivation in the context of closing a question is supposed to be written inside the body of the question.

It doesn't matter how well-motivated the question can potentially be, the issue is that this motivation is not written anywhere in the body of the question

And you can't write it for the author

Ben Steffan

Agree. I've reclosed it for the time being.

The whole post is a fine mess.

Joe

Well, I didn't find it badly written, except for a few grammatical errors which I myself edited out. Obviously people in the community are going to disagree about what constitutes a good question, which is why they can vote either way. It's far from a perfect system, but I think it works well on the whole

Jakobian

It's badly written in the sense that it's very clearly not conforming to the math.se standards

Ben Steffan

18:43

I do think your take on this diverges pretty heavily from site policy and how we usually handle things, Joe.

My impression is also that the category theory people feel... different about context in questions.

Jakobian

I mean it's understandable - it's probably a new user. I am wondering if the question will be edited at all with any additional context. I think no. And so reposting is probably better to do

Ben Steffan

They don't seem to have interacted with the post at all since posting. I'd wait.

Jakobian

The annoying thing is that no one tried to educate the new user about how this site works, yet they jump right to complaining

This is backwards

Ben Steffan

yeah

as I said, it's a fine mess all around

@Joe One thing to consider here maybe is fairness: This question is interesting in and of itself, but it is structurally not much different from, say, somebody posting an integral and asking for a solution. But the latter question will almost certainly be closed for being without context, while this one is to remain open, even though really all the questions differ by is topic and field of study.

One point of having rules like we do is to address issues like this. What's interesting or not is largely a question of audience, not of question quality.

Jakobian

Yeah its just favouritism. I get it, it's a way more interesting (to some people at least) question.

But it can't influence your decision about whatever or not the closure is appropriate

Ben Steffan

18:56

Also, perhaps one should point out that making an attempt at finding counter/examples to this question is not exactly witchcraft. The examples in the existing answer (or similar) are pretty easily attainable; you just need to sit down for 5min and think about the definitions.

12 mins ago, by Ben Steffan

My impression is also that the category theory people feel... different about context in questions.

that's what I was referring to here: for some reason, category theoretic questions just get let off the hook for this (sometimes, by more people than should be the case)

Thorgott

@Jakobian I thought it was fine

there's a better question I've been meaning to ask for a long time and which I never got around to, but that's besides the point

Jakobian

I see. Well, that's fine, but keep in mind that the reason those questions are closed is so that the author can add necessary details. So if a question doesn't contain those details, like here, the appropriate thing to do is to educate the author.

So that they can eventually get to it

Joe

19:12

@BenSteffan I see your point about fairness. If somebody asked a question about an integral that I thought was interesting to the community, and it was not a duplicate, then I would be supportive of it. Anecdotally, I have found that a clear majority of the new questions here that are at the high-school level or lower undergraduate level are duplicates, and so I most often vote to close them.

Ben Steffan

Duplicates are a separate thing. There's no reason not to close a duplicate, as far as I know.

Joe

Yes, I always vote to close a question if I know that it is a duplicate. My point was just trying to elaborate on how I handle lower level questions that are of a similar "structure" to the question about category theory.

Jakobian

Well then your voting is not in accordance with the policies of this site, is all.

Also about duplicates, the difference is that those questions which get closed as duplicates stay here. So it should be somewhat reserved for the questions that are actually interesting and received good feedback. Or something like that. @XanderHenderson said this once

Not what I am saying but I don't remember exactly

I.e. if a question is horrible and a duplicate then it probably shouldn't be closed as a duplicate

Mad

Hello Friends

All of my life, i have used $ lim_n $ instead of $ lim_{n\rightarrow \infty}$
When it is clear that the limit is infinty. This notation is varied from standard notation. I am writing a book, and i am using this notation (after i declared it very clearly)

Do you think it is arrogant or immature to do this? I did this mostly because it spared me writing symbols, i am fairly lazy.

Jakobian

Its easy to engage in this tribalism and see it as "they want to close our interesting questions" or something. But that's not the issue

Ben Steffan

19:21

@Mad Arrogant? No. Immature? Maybe a little. A good idea? Probably not.

Thorgott

@Jakobian with all due respect, you told me just last week that rules are merely suggestions, so I don't feel like getting lectured by you on them now

Ben Steffan

Having standardized notation is an asset

Thorgott

@Mad I sometimes even omit the index

Jakobian

@Thorgott doesn't matter

You are doing a fallacy when you disregard everything a person said because of one thing a person said

But feel free to be privileged

Joe

Perhaps my views are not in accordance with the policies of the site, then, but given that 5 people voted to reopen the question, and there are currently 3 reopen votes, I hardly think that my viewpoint is uncommon. On the whole, the community moderates itself (which is part of the Stack Exchange model), and in this sense I certainly feel like I follow the "policies" of the community.

Jakobian

19:24

They're just ignorant is all

Or maybe not ignorant, well, could be a lot of things. But I think some people voting are just ignorant

Thorgott

@Jakobian well, I hardly disregard everything you say

Jakobian

I mentioned that thing that you are saying before I knew that you could get suspended for it

and if you don't feel like getting lectured then why don't you show me that you have proper knowledge of the policies of this site

Joe

This conversation is not really going anywhere...

Jakobian

by actions instead of by speaking to me. I don't want to be lectured because you are this and that. It's just excuses

Sine of the Time

19:39

4 reopen votes

Thorgott

@Jakobian you are in no position to make demands of me

Jakobian

Okay, this is immature

Thorgott

I'm glad we agree

Sine of the Time

the fact that Kevin Carlson's comment is the most upvoted says a lot

Jakobian

It only says a lot about how humans evolved from tribes

Thorgott

19:43

it doesn't say that much, in reality Kevin wants to answer a question more interesting than the one that was posted

Jakobian

I don't know how you came to the conclusion that upvotes somehow imply anything about what Kevin wants to do

or somehow matter for that

it's clear why Kevin was upvoted
people want to see themselves as beloning to a group
they imagine, in their head, that there's two groups, those who want to close the question and those who want to open it

do you really think they even read most of the contents? Of course not

it's tribalism

only dichotomies exist, it's us versus them, and so on

Thorgott

I don't know how you came to the conclusion that upvotes somehow imply anything about what people see themselves as

Jakobian

And I didn't?

Sine of the Time

"people want to see themselves as beloning to a group
they imagine, in their head, that there's two groups, those who want to close the question and those who want to open it"

I meant this

and when someone disagrees with site policies, I've notice it's almost always the most upvoted comments

Jakobian

Yes, people want to see themselves on the side of justice

no one really want themselves to be on the side of taking the criticism by those people who feel like they are on that side

I'd say this is why

but this is too simplistic to me

taking sides that is

Sine of the Time

20:00

reopened

lol

Jakobian

It was bound to be done eventually

Jakobian

20:14

wrong thing

Policies, such as they are, are enforced by the community. There is far too much traffic here for the moderators to vet every post. Hence policies are enforced selectively by the people who (a) see the post, (b) know the policy, and (c) choose to enforce the policy. There are plenty of places where things fall through the cracks. — Xander Henderson ♦ Aug 5, 2022 at 18:46

Xander gives a good summary here

In theory this should be how it works. It doesn't

Or I suppose it does? For closing questions at least

For opening questions people seem to go off of their whim

Mad

@Thorgott interesting so i am not the only one

Sal Rahman

21:23

Is it generally true that if two $m$-degree polynomials intersect at $m^2$ specific points, then a third $m$-degree polynomial that passes through $m^2-1$ points, then it must pass through the $m$th point as well?

Joe

@SalRahman: For the last clause in your question, you mean "pass through the $m^2$th point", no?

Sal Rahman

@Joe yes! Thanks for pointing that out.

Rephrasing what I said: is it generally true that if two $m$-degree polynomials intersect at $m^2$ specific points, then a third $m$-degree polynomial that passes through $m^2-1$ points, then it must pass through the $m^2$th point as well?

Ben Steffan

...but two polynomials cannot intersect at more than $m$ points (unless they're equal) (?)

(of degree $m$)

am I misunderstanding the question?

Sal Rahman

@BenSteffan I always thought that by Bézout's theorem, a degree $m$ polynomial will intersect with a degree $n$ polynomial through at most $mn$ points.

Joe

I think the question is: suppose $p,q$ are polynomials of degree $m$, and $p$ and $q$ intersect at $m^2$ distinct points. If $r$ is a polynomial which passes through at least $m^2-1$ of those points, then must it pass through all of those points? Is that right?

Sal Rahman

21:31

@BenSteffan The Cayley-Bacharach theorem wouldn't work had two cubics not been capable of intersecting at 9 points.

Ben Steffan

disregard what I said, I made a silly mistake :)

Sal Rahman

@Joe yeah, that sounds right

Ben Steffan

I was thinking about number of values $x$ s.t. $f(x) = g(x)$

...I'm making no sense today.

rather, I was thinking of polynomials in one variable over $\mathbb{R}$

ILikeMathematics

@BenSteffan Alright, thanks

Joe

Bezout's theorem is about polynomials in two variables $x$ and $y$. Are you talking about polynomials in $x$ and $y$?

(At least, I think the classical version of Bezout's theorem only concerns itself with 2 variables.)

Sal Rahman

21:40

@Joe Yeah, $x$ and $y$.

Joe

I'm trying to think of an example/counterexample but I haven't found anything yet. I'll let you know if I come up with something.

Jakobian

@BenSteffan huh... I think you're making perfect sense

anyone would assume we are talking about standard polynomials and not ones in two variables

Ben Steffan

thank you

Jakobian

22:04

@SalRahman what exactly do you mean here by intersect? If $p = q$ then $p, q$ always intersect each other. Are you sure that you aren't misinterpreting those theorems you cite?

both of these results seem to be about algebraic curves, which are sets of zeros of polynomals

Sal Rahman

23:16

@Jakobian I don't think I am understanding your question. Probably because I always that thought it's not necessary that $p = q$ for all points, but there are points where $p = q$, and the total number of points where they are equal is not exceeding the product of their degrees, by Bézout's theorem.

And if $p = q$ on all points, then that means that the polynomials are exactly equal.

Ryder is not Rude.

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Ben Steffan

@RyderisnotRude. ?

Ryder is not Rude.

Ben Steffan

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Ryder is not Rude.

For those interested in CS.

Derso

23:38

I was reading your idea! Where is it? noooo

Ben Steffan

I realized mid writing that I probably misinterpreted something

the thing is, "degree 2 polynomial" probably means that all terms have order 2

Derso

You said there was no reason to think about two algebraic curves intersecting on finitely many points or something

Ben Steffan

so $a xy + bx^2 + cy^2$

yeah

Jakobian

@SalRahman I see, so you seem to be misunderstanding Bezout's theorem then. I myself don't quite understand it, but clearly even if two polynomials are equal, that means they are equal on infinite amount of points. Even if we don't take equal polynomials, they can still be equal on infinite amount of points

Derso

@Jakobian True.

Jakobian

23:41

$x+y$ and $y$ for instance

Ben Steffan

I was saying there's no reason to suspect that if $p$, $q$ are polynomials of degree 2 in 2 variables, then there's no reason to suspect their graphs intersect in only finitely many points, even if $p \neq q$. But my counterexample relied on a lower order term :(

Derso

What we have is a common component

Jakobian

@BenSteffan $x^2+x+y$ and $x^2+y$ then, I suppose

Derso

For example $(x^2+y^2+1)x$ and $x$ has all the points in common, but they're not equal

Ben Steffan

@Jakobian sweet

But I think what you're meant to consider is the solution sets $p(x, y) = 0$

that's what the term "curve" means

Derso

23:42

*algebraic, yeah

Jakobian

Yes, but still, this is a misunderstanding I think. Because polynomials aren't their sets of zeros.

Ben Steffan

yeah

that's what I'm trying to say I think

Derso

Oh, maybe he's talking about coincidences(? I believe that's the term), i.e. $(x,y)$ such that $p(x,y) = q(x,y)$?

Jakobian

There is also a possibility that Sal Rahman is just using wrong language to explain this, instead of misunderstanding those theorems from theory of algebraic curves

for example if there is some language barrier/they learnt concepts in another language

but their English is quite good so I don't know

Derso

Well, when he says "And if p=q on all points, then that means that the polynomials are exactly equal", that's indeed true if by "all points" he means "every $(x,y)\in \mathbb{R}^2$".

In fact, coinciding on an open set would already suffice for that.

Jakobian

23:51

A mathematician wouldn't say $p = q$ on all points, I'd think. Instead, $p(x, y) = q(x, y)$ on all points, or something like that

Ben Steffan

this seems close to the Cayley-Bacharach theorem mentioned above, so probably the setting is similar en.wikipedia.org/wiki/Cayley%E2%80%93Bacharach_theorem

Jakobian

Which makes me think that maybe they come from different background

I can imagine a situation when a programmer is trying to apply algebraic geometry to find solutions to some number theory problem, and asks such questions

but I can't be sure

If you click on their profile they seem to have the most points on stack overflow which is the programmer stack exchange

Derso

In that case, he can try applying the resultant method to find intersections

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