Mathematics - 2018-02-06 (page 3 of 4)

Jaideep Khare

15:00

@BalarkaSen Is this about bobs and vagens?

0celo7

@JaideepKhare I have trace amounts of Indian heritage

I feel the bob urge

it's real

Balarka Sen

@JaideepKhare Yep, a meme we all love and adore

Being an Indian I give that meme a strong bob/10

Jaideep Khare

Guess I got to see dark side MSE today :P

Balarka Sen

@0celo7 Hm, I don't understand your arbitrarily thin comment. Can you elaborate?

0celo7

go to the physics chat if you want memes

@BalarkaSen I'm currently freaking out about a philosophy assignment I should have done yesterday

I'll catch up with you later

Jaideep Khare

15:02

I don't like PhSE

Balarka Sen

Just do it

0celo7

@JaideepKhare how dare you

@BalarkaSen I am

but doing non-math while stuck on math is fucking hard

Balarka Sen

:keeps making faces like Shia Lebouf having a massive stroke:

0celo7

wot

Jaideep Khare

@0celo7 Every ducking question is closed as Homework.

Even if I show attempt

Balarka Sen

15:05

im just bringing up the JUST DO IT meme from it's grave aren't i

Jaideep Khare

For ex : physics.stackexchange.com/questions/324300/…

Unlike MSE

0celo7

MSE is a cesspit of homework help

the PSE chat is great

@BalarkaSen and I always talk about genocide

somehow

Jaideep Khare

But MSE doesn't have any homework tag, every question lacking attemt of OP is clsed here. And with attept is NEVER closed, unlike PhSE

0celo7

The leaders of PSE think that MSE is a deplorable craphole. Not kidding

They're never changing the homework policy

It's been tried a billion times

there's a post on the homework policy where every word is a link to another post on the homework policy

Balarka Sen

MSE closes down good questions as homework while leaves bad, actual homework questions open

It's a total mess

Jaideep Khare

15:10

@BalarkaSen It's never like that. Never ever.

Balarka Sen

Yes, it is.

0celo7

@JaideepKhare lies

Balarka Sen

If you don't agree with that you haven't seen a good proportion of the questions in MSE. Your sample is biased.

Given that majority of the more active users agrees with my conclusion

Jaideep Khare

Well I would love to see some examples supporting question?

MatheinBoulomenos

0

Q: Prüfer domains are Arithmetical rings

Suppose $R$ be a Prüfer domain. How should I Prove that it is an arithmetical ring?

commutative-algebra

See the comments by Pete Clark

Jaideep Khare

15:12

@BalarkaSen Well, arguably , currently I'm most experienced MSE user here

MatheinBoulomenos

This was closed as homework by people who had no idea what the question is about

this wouldn't have been closed on MO

Balarka Sen

2

Q: finding a differential curve tangent to a distribution

I have the following distribution on $\mathbb{R}^3$ $${\cal{D}}_{(x,y,z)} = \langle\{\partial_x,\partial_y + x\partial_z\}\rangle$$ I want to show that for any $(x,y,z)$ in $\mathbb{R}^3$, there exists a path $\gamma$ from $0$ to $(x,y,z)$ tangent to $\cal{D}$ i.e. $\dot{\gamma}(t)\in{\cal{D}}_...

differential-geometry

This is nowhere near as complicated as @Mathein's example I suspect, but it's sufficiently nontrivial to leave it open

"off-topic" is totally bullshit

There are millions of examples like that

Jaideep Khare

There Is NO ATTEMPT

That's why it's closed

Balarka Sen

There is no attempt because it's a hard question.

MatheinBoulomenos

So what? It's a bullshit policy that you always have to include an attempt

Balarka Sen

15:17

@JaideepKhare You have a high reputation but your reputation distribution accumulates to your top tags, which are calculus/analysis and algebra/precalc. On what basis are you labelling yourself as the most experienced MSE user here? That's completely bollocks dude

Jaideep Khare

There is no policy saying that hard questions are left open, even if there is no attempt

@BalarkaSen Experience $\neq$ Knowledge

Balarka Sen

Yes, we are criticizing the policy here.

It's a good question which MSE sees as a bad one, unjustly

@JaideepKhare I'm saying your experience is limited to those tags that you roam about. Which is fine, I roam around like 2 or 3 tags. I don't label myself as the most experienced :P

Jaideep Khare

You may be more knowledgeble @BalarkaSen about Mathematics , I may be more knowledgeble Mathematics SE

Mike Miller

I voted to reopen

MatheinBoulomenos

I don't think showing an own attempt necessarily makes a question good and not showing an attempt makes a question not neccesarily bad

Jaideep Khare

15:21

But I see the flags and close votes on various questio beyond my area, too.

Daminark

Presumably the point is that if most of your interaction with MSE is on the level of much more elementary questions, you're likely less qualified to judge the quality of higher level questions

MatheinBoulomenos

The highest voted unanswered questions in the tags I frequent don't show an own attempt

Daminark

Perhaps there may be a rule of the sort but the complaint is less about execution of existing rules and more that the rules are bad

0celo7

@BalarkaSen Considering the GP example: take $y\in R\setminus Q$, so $\Sigma=u^{-1}(y)$ has no critical points. Of course $\Sigma$ is just a collection of isolated points. I claim that if $U$ is a neighborhood of $\Sigma$, say $\cup_i (x_i-\epsilon_i,x_i+\epsilon_i)$, then $\liminf\epsilon_i=0$

Mike Miller

I like to search intags:mine score:1 answers:0

Balarka Sen

15:24

@0celo7 Oh I agree but that can't happen in your compact case.

This example is just too specific

If $\Sigma$ is compact your neighborhood can't shrink away infinitely

Jaideep Khare

What would you say about physics.stackexchange.com/questions/324300/…

Why was this closed?

And why is this still closed?

Balarka Sen

Thanks @ all for reopen votes. Maybe I should get off my ass and start fixing my deleted answer of that question

Leaky Nun

@MatheinBoulomenos if a Z-bilinear map $f:A \times B \to C$ is $R$-balanced, where $A$, $B$, $C$ are $R$-modules, does this mean that $f$ is $R$-bilinear?

MatheinBoulomenos

no

Leaky Nun

do you have a counter-example?

0celo7

15:30

@BalarkaSen Hmm, that's what I said originally but changed my mind, hmm...

s.harp

arxiv.org/pdf/1203.6902.pdf

0celo7

Yeah I guess the point is that $\text{dist}(\partial U,\Sigma)$ is bounded below by a positive number.

s.harp

> The number of gods has come out to be an integral number. This rules out any demi-gods or lower devas, as they are known from Greek or Indian mythology. However, the divine cardinal can get negative; with the obvious interpretation of these gods being devils.

MatheinBoulomenos

First, you want $A$ to be a right $R$-module and $B$ a left $R$-module (that stuff is important with noncommutative rings) $C$ can be any abelian group

Leaky Nun

commutative rings

and then how?

MatheinBoulomenos

15:33

For commutative rings, it's more common to work with bilinear maps instead of $R$-balanced maps

0celo7

all rings are commutative

MatheinBoulomenos

@0celo7 matrices don't exist

0celo7

@MatheinBoulomenos not a ring bro

Balarka Sen

nah they do but they just don't form a ring

0celo7

that's a noncommutative ring

just like manifolds with boundary are not manfolds

they are manifolds with boundary

anyway @BalarkaSen I think a compactness argument will work

Leaky Nun

15:34

@MatheinBoulomenos so I'm asking whether there is a map that is $R$-balanced but not bilinear

0celo7

I'll ask my professor (research advisor?, collaborator?) who is a GMT wizard if there's a slick way to do it

Leaky Nun

i.e. $f(x,ry) = f(rx,y)$ for all $r \in R$, $x \in A$, $y \in B$, but not bilinear

0celo7

I might send you a proof if I have to do it "classically"

Leaky Nun

I mean, in French fields are called "corps commutatif", so...

0celo7

I would be nice to avoid Morse theory, tbh

Mike Miller

15:35

Nice find @s.harp

Balarka Sen

> The number of gods in a universe equals the Euler characteristics of
the underlying manifold

:ok_emoji:

0celo7

@BalarkaSen it's 1

Christianity confirmed

Mike Miller

Seems about right

MatheinBoulomenos

@LeakyNun Consider a commutative ring $R$ and a non-commutative ring $S$ such that $R$ is not contained in the center of $S$. For example, you can take $R= \Bbb C$ and $S= \Bbb H$. Then the multiplication $S \times S \to S$ is $R$-balanced, but not $R$-bilinear

Of course I want $R \subset S$ as a subring, I forgot to say that

0celo7

@BalarkaSen wait so what's the issue with scrunching up the GP result and rolling it up to an $S^1$?

MatheinBoulomenos

15:48

@LeakyNun wait that's a strange definition of $R$-balanced. I thought you'd want $f(xr,y)=f(x,ry)$

0celo7

the image would be unbounded, which is an issue, but...?

maybe truncate it and then do that?

Leaky Nun

@MatheinBoulomenos but that's the same

MatheinBoulomenos

No

Leaky Nun

(I'm still assuming commutativity)

MatheinBoulomenos

Even over a commutative ring you can have different left and right module structures on the same abelian group

Okay we take two copies of $\Bbb H$ and consider both as vector spaces over $\Bbb C$. On the first copy $V_1$, we multiply from the right with complex scalars. On the second copy $V_2$, we multiply from the left with complex scalars. Then the multiplication map is $V_1 \times V_2 \to \Bbb H$ where we multiply from left on the target $(x,y) \mapsto xy$ is $\Bbb C$-balanced, due to associativity of multiplication in $\Bbb H$, but it is not $\Bbb C$-bilinear

Balarka Sen

15:55

@0celo7 It's not translation invariant, so you can't scrunch it up in an immediate way. If you do it in $[0, 1]$, say, you'll get a discrete set as preimage as you should.

By doing in [0, 1] I mean enumerate the rationals in [0, 1] by 1/n's for n = 1, 2, 3, ...

And bump

MatheinBoulomenos

Because for example $i(j,k)=ijk=-1$, but $(j,ik)=jik=1$

Balarka Sen

Ok I gotta scurry

gotta skrra

skibi pop pop

Leaky Nun

@MatheinBoulomenos what if we require multiplication to be on the left and the three modules to be the same?

Faust

Morning

MatheinBoulomenos

In the example I gave, you would get a map from multiplication that is neither $\Bbb C$-balanced nor $\Bbb C$-bilinear

@LeakyNun in the example I gave, you can still formally consider the multiplication of $\Bbb H$ with complex scalars from the right as a left $\Bbb C$-vector space, because $\Bbb C$ is commutative

Faust

16:01

Anyway to make latex chat work on my phone?

MatheinBoulomenos

@LeakyNun you can modify the example I gave such that it satisfies these criteria, $\Bbb H$ as a $\Bbb C$ vector space with left multiplication is isomorphic to $\Bbb H$ as a $\Bbb C$ vector space with right multiplication

the map will just look more complicated

Leaky Nun

thanks

Jaideep Khare

Anybody interested to chat about Superheroes Comics and their Cinematic Universe such as Marvel and DC?

Faust

I think you are in the wrong nerd chat

Jaideep Khare

If intereseted, you can come here - chat.stackexchange.com/rooms/72766/marvellous-room

s.harp

16:10

Is the locally constant sheaf allowed have different stalks in differnt connected components?

or is it supposed to be the sheaf of locally constant functions into the module or whatever

Mike Miller

Connected components of the space (as opposed to open set)? Yes

s.harp

I think I hate sheaf cohomology

Mike Miller

Sorry to hear that

MatheinBoulomenos

You can take as your space a disjoint union of a point and another space which is not locally connected, then I think you should get different stalks

@LeakyNun I think a big conceptual difference between $R$-bilinear maps and $R$-balanced maps is that you don't need a $R$-module structure on the target for the latter

s.harp

Maybe its just this book, but they list two axioms for intersection homology and say they are obviously the same if the intersection sheaf is constructible, ie its homology sheaf is locally constant on strata, but I think the definitions are clearly different if you have non-connected strata or just two copies of the same space..

Akiva Weinberger

16:16

@Faust Yes, the same link works

I'm using Chrome, the browser, on my iPhone and it's working

Mike Miller

@s.harp I am also learning intersection homology now!

s.harp

:D

what references are you using?

MatheinBoulomenos

You can take for example any $\Bbb R$-balanced map from two $\Bbb R$ vector spaces $V_1 \times V_2 \to \Bbb R$, then you can compose that, say with any group homomorphism from $\Bbb R \to \Bbb Q$ (for example, take a $\Bbb Q$-basis for $\Bbb R$ and sum the coefficients), then the result will still be a $\Bbb R$-balanced map, but it's impossible to make $\Bbb Q$ into a $\Bbb R$ vector space

Mike Miller

Kirwan-Woolf

s.harp

ah, thats the book I am reading right now

Mike Miller

16:18

I don't see the claim you say

s.harp

with the two apparently the same axiom systems that look to be different to me

on page 109 and page 111 there are two axiom systems

Mike Miller

Ah yes

I don't see the contradiction. Can you be painfully explicit for me?

s.harp

in one of them they say that the codimension of those points for which $H^{-i}(j_x^* E^*)\neq0$ has to be greater than some number, in the other they say that on the stratum of codimension so and so $H^{-i}(j_x^* E^*)$ has to be $0$ if $i$ is smaller than some number

Mike Miller

If you have disconnected strata (of the same dimension) their union has the same codimendion though?

s.harp

Also the sheaf has to have locally constant cohomology on the strata

Mike Miller

16:23

I do think those are equivalent

s.harp

but if my stratum is disconnected, maybe I can imagine a sheaf that has locally constant homology that is zero on one connected component (for hte right $i$) and non-zero on the other connected component

this means the points of homology zero contain a space of the right codimension, but not the entire stratum

Mike Miller

But the I should be the same for both strata

As they have the same codimension

Ah I see

Yeah so you have a stratum on which the local system is allowed to be nonzero but on one component it is zero

s.harp

yes, so that the set where the things are zero has the right codimension

Mike Miller

I don't think that's a problem. That's still a local system

s.harp

and the first axiom is fufilled, also everything is locally constant after homology, but hte second axiom does not allow it, because there are points in the stratum that do not have $i$-th homology zero on the stalk

Mike Miller

16:26

Just not one constant on the whole disconnected space

Just to clarify, this issue isn't a problem with the numbers not matching up, right? But rather possibly with the definition of local statement

System

Actually I am a little confused

s.harp

No, its not a problem with the numbers, but with one statement being "the set on which =0 has to contain a codimension blah blah set" and the other saying "this specific codimension blah blah set has to have =0"

the axioms AX_p[S] are only formulated for constructible sheaves and that has locally constatn homology

Mike Miller

Yeah.

Apologies for being slow. I do believe that these should be equivalent but I took it on faith at the time

s.harp

They probably are^ but I think putting it into an exercise is wrong, because there this subtlety which makes it hard to understand for me

Mike Miller

Alright, laptop out, time to think for a sec

I'm going to ignore the costalk condition since that's just there for duality purposes

s.harp

yes

I think if you just take two copies of a space, on one you have the usual intersectoon chains and on the other you do some totally crazy shit but that has locally constant stalks on strata after homology, you should have the second axiom system covered

Mike Miller

16:39

@s.harp It is also not obvious to me. I think maybe you're right, but that you should conclude that intersection homology is a very powerful invariant?

Like in the second case where you have disconnected strata and the local system is zero on one component but not on others - maybe that indicates a different stratifications

oh but that's obviously nonsense, isn't it? since the cohomology sheaf is an invariant of isomorphism classes of things in the derived category, and there's only one intersection sheaf

so the local system on the top stratum should determine it in lower strata?

Sorry, I think I have not been very helpful

s.harp

No, you have given me courage that my misunderstanding is not because I am retarded^

Also in the paper by goresky and mcphearson they have the second axiom system:

I'm guessing you are not allowed to look at disconnected spaces, as then taking the zero sheaf on one component will satisfy the axioms, and it is probably impossible to change the coefficient system from one component of the stratum to the other without destroying something like the constructability condition, but that really seems like a hidden point

I have to give a talk about this on thursday, I'll see whatr the professor thinks about these ideas, maybe ill have a little bit of a concrete counter-example by then

Mike Miller

16:56

@s.harp If you take the zero sheaf on the other component, it's just the intersection chains with coefficient group 0

I think there's no counterexample. I think you have good ideas but probably this just says it's hard to make sheaves whose homology is local systems on each strata that aren't intersection chains

The next chapter introduces perverse sheaves (as things satisfying those cohomological vanishing conditions) and shows that they can be constructed as extensions of the "skyscraper intersection complexes" associated to a closed stratum $\overline X_k$

So either that theory will show me right or will construct a counterexample for you :p

Xander Henderson

aw, man @s.harp, you got my hopes up. I saw $\dim$ in tiny font, and got all excited, and then it turned into manifolds. :(

mick

0

Q: $a_2^{a_3^{... ^{a_n}}} = n $

$a_2 = 2$ $a_2^{a_3 } = 3$ So $a_3= \ln(3)/\ln(2)$. I wonder about all solutions $a_n$ such that $a_2^{a_3^{... ^{a_n}}} = n $ For all $n.$ How does $a_n$ behave ? What are the best asymptotics ? Ofcourse $a_n$ goes quickly towards values between $exp(1/e)$ and $1$ that is trivial. But I...

asymptotics tetration power-towers

s.harp

I see about the coefficients on the other component, this also destroyed another picture I had (one point union of two manifolds, with intersection chains on one end and zero on the other end)

@Xander what do you mean?

Mike Miller

On the other hand I don't see why the new set of axioms immediately implies the old (perhaps look at the discussion of the "canonical stratification" in 7.3 if you haven't yet)

s.harp

The old imply the new I think, I don't see why the new imply the old (or what do you mean?)

Xander Henderson

17:08

@s.harp I do fractal geometry; question about "dimension" are interesting there. I don't do manifolds, where questions about dimension might be interesting, but are above my pay-grade.

That being said, manifolds are not generally pathological enough to have interesting dimensional properties.

I have similar complaints about the sheaves and stalks and whatnot in algebraic geometry. Not enough pathology. :(

s.harp

pseudomanifolds can get pretty singular (ha-ha), but not in their dimension, thats controlled

Mike Miller

@s.harp that's what I meant, sorry

Actually that's what I said

s.harp

oh jeeze right ^^'

Mike Miller

Nbd pal

Eric Silva

@Semi lol m8 got reeeeekt by E&M

Bennett

17:19

@TedShifrin @Tuki Found the 'proof' I was looking for yesterday that didn't sound believable through a cached version of PlanetMath. Could you guys confirm whether it's wrong or not?

For any finite family $(a_i)_{i\in I}$ of real numbers in the interval $[0,1]$ we have $$\prod_i(1-a_i)\ge 1-\sum_ia_i\;.$$ Proof: Write $$f=\prod_i(1-a_i)+\sum_ia_i\;.$$ For any $k\in I$ and any fixed values of the $a_i$ for $i\ne k,$ $f$ is a polynomial of the first degree in $a_k$ Consequently $f$ is minimal either at $a_k=0$ or $a_k=1$ That brings us down to two cases: all the $a_i$ are zero, or at least o…

Silent

It is given that if $a_n=\frac{n^n}{n!}$ then $\lim_\limits{n\to \infty} \frac{a_{n+1}}{a_n}=e$, how to find $\lim_\limits{n\to\infty} \frac{n}{(n!)^{1/n}}$ from that?

s.harp

@Bennett seems fine to me

Bennett

17:36

@s.harp the bit where we consider $f$ as a polynomial of first degree in $a_k$ keeps going over my head! Why can we do that? Also, what's the purpose of $i \ne k$?

s.harp

Assume you have some tuple $(a_1,..,a_n)$ so that $f(a_1,..,a_n)<1$. This means the map $x\mapsto f(x,a_2,...,a_n)$ must assume values smaller than $1$. But this map is takes it minimal value either at $x=0$ or at $x=1$. If $x=1$ you get a contradiction as then $f ≥1$ at such a point. This means that $f(0,a_2,...,a_n)<1$ if there exists some $a_1$ so that $f(a_1,..,a_n)<1$. Now do the same with the next components to see that $f(0,0,a_3,...,a_n)<1$ must be true until you arrive at $f(0,...,0)<1$

but that istn true, so your assumption at the beginning was false

Bennett

@s.harp Excellent! Thanks.

jublikon

How to check if $\sum_{k=1}^{\infty}{ \left( \frac{7k-2}{8k-3 \sqrt{k}}\right)}^k$ converges?.

$\begin{align}
\sum_{k=1}^{\infty}{ \left( \frac{7k-2}{8k-3 \sqrt{k}}\right)}^k &=
\lim_{n \to \infty} \sum_{k=1}^{n}{ \left( \frac{7k-2}{8k-3 \sqrt{k}}\right)}^k \\
&= \lim_{n \to \infty} \sum_{k=1}^{n}{ \left( \frac{7\frac{k}{k}-2\frac{1}{k}}{8\frac{k}{k}-3 \frac{\sqrt{k}}{k}}\right)}^k \\
&= \lim_{n \to \infty} \sum_{k=1}^{n}{ \left( \frac{7-2\frac{1}{k}}{8-3 \frac{\sqrt{k}}{k}}\right)}^k \\
\end{align}$

Dattier

17:55

Hi,

$f\in C^2([0,1])$ with $f''$ convex and $f(0)=f'(0)=0$ is it true that : $f''(1)+6f(1)\geq 4f'(1)$ ?

mick

2

Q: $a_2^{a_3^{... ^{a_n}}} = n $

$a_2 = 2$ $a_2^{a_3 } = 3$ So $a_3= \ln(3)/\ln(2)$. I wonder about all solutions $a_n$ such that $a_2^{a_3^{... ^{a_n}}} = n $ For all $n.$ How does $a_n$ behave ? What are the best asymptotics ? Ofcourse $a_n$ goes quickly towards values between $exp(1/e)$ and $1$ that is trivial. But I...

asymptotics tetration power-towers

Another annoying tetration question :)

s.harp

@Dattier $f$ is convex or $f''$ is convex?

Dattier

18:15

@s.harp f'' is convex

mick

0

Q: $f(x) = x^{{1/2}^{{x^{1/3}}^{x^{1/4}...}}}$ asymptotic?

Consider for positive real $x$ : $$ f(x) = x^{{1/2}^{{x^{1/3}}^{x^{1/4}...}}} $$ How does this function behave ? How fast does it grow ? Faster than any fixed iteration of exp sure, but How fast exactly ?? A brute estimate would be $tet( \ln(x+1))$ but that would likely be a bad estimate. W...

asymptotics roots tetration power-towers

Feel free to edit into sqrt symbols !!

Mona Jalal

18:29

do you know how I can solve this last integral here? imgur.com/0aLKKM8

Vrouvrou

Hello, who know the proof of this : " Let $(x_n)\subset \mathbb{R}^N$ , then for a subsequence, $|x_n|\to+\infty $ or $|x_n|\to x\in\mathbb{R}$ ?

Akiva Weinberger

@Vrouvrou You mean, then there exists a subsequence that either goes to $\infty$ or to a point?

Vrouvrou

yes

Akiva Weinberger

(A point in $\Bbb R^N$ rather than $\Bbb R$)

Vrouvrou

@AkivaWeinberger

Akiva Weinberger

18:39

Well, in compact sets, you know that all sequences have a convergence subsequence

Vrouvrou

no i put | x_n|

Akiva Weinberger

Oh

In any case, you can probably mimic the proof for compact sets

Or just apply it to the one-point compactificaion of $\Bbb R$

Vrouvrou

where i can found the proof please

Alessandro Codenotti

18:57

It's a standard exercise that every sequence in $\Bbb R$ admits a monotone (not strictly) subsequence so either it goes to infinity or it's bounded hence convergent

Dattier

@s.harp Do you think it's true ?

$f\in C^2([0,1])$ with $f''$ convex and $f(0)=f'(0)=0$ is it true that : $f''(1)+6f(1)\geq 4f'(1)$ ?

@Gérard comment cela se passe à la DGSE, ça va ton patron ne ta pas engueuler trop fort ?

@Mambo : hi, are you an AI ?

Mambo

That depends on whether or not you think this conversation is actually happening.

Dattier

19:14

@Mambo : answer yes or no is suffised

well, are you an AI ?

Mambo

No. It's all $\Bbb R$ intelligence here

Kasmir Khaan

19:28

Hi chat

Dattier

Hi

@KasmirKhaan : where do you came from ?

@Mambo : well, you are a RI lol

Bennett

@MonaJalal did you try completing the square and then subbing for whatever is under the square?

Mona Jalal

19:53

@Bennett the only way that wolfram alpha pro premium could solve this was by feeding the integral into two parts (basically breaking the integral based on absolute value), I am not aware how it has done the integral

additionally can you please have a look at math.stackexchange.com/questions/2639032/…

user193319

Does Sylvester's criterion hold for complex Hermitian matrices?

MatheinBoulomenos

yes

user193319

Thanks!

mick

20:18

0

Q: Conjecture reference : $\int_0^1 f(x) dx \int_0^1 g(x) dx = \int_0^1 h(x) dx$

Im looking for examples and references of unproved conjectures of type : $$\int_0^1 f(x) dx \int_0^1 g(x) dx = \int_0^1 h(x) dx$$ Where all integrals have positive value, but no closed form for any value from any of these 3 integrals. The equality is then conjectured but unproved.

calculus reference-request examples-counterexamples products conjectures

0celo7

20:33

@BalarkaSen help, I don't understand topology

Semiclassical

20:54

@EricSilva oh nooooes

To the extent that I rekt students on the HW they got back, it was mostly for not turning in the entire assignment

(It’s a bit weird: they have lecture MWF and turn in a few HW problems each time, but with all the problems in a week forming a single assignment)

(And this time it worked out so that not turning in the Monday assignment lost them half the points)

mick

Lol I barely understand my OWN questions

Semiclassical

@EricSilva what are you working on there atm ?

mick

How many unanswered question do I have ??

Who dares to downvote me ??

Downvoting without help or motivation should be illegal !! Vote Mick :)

Mike Miller

21:22

@s.harp I think the canonical filtration does answer your question. By the way it's built, you have a stratification on which the cohomology of the strata satisfies the desired vanishing conditions

Kasmir Khaan

22:12

@MatheinBoulomenos here? :)

@MatheinBoulomenos want to ask you about what book you using for intro to number theory

@LeakyNun Sup leaky :D

mick

22:24

I got a downvote spam ... someone hating me

Leaky Nun

@KasmirKhaan hi

Kasmir Khaan

leaky :D

what Courses are you taking?

been a while since we last talked :D

@anon anon text me when you see this , ill be here whole day :D

MatheinBoulomenos

22:56

@Kasmir I'm here now

We're not using a particular book, we're following a set of (German) lecture notes

Kasmir Khaan

Mathein my hero :D

aha okay =p I wanted to have a gentle book as intro to field extension

s.harp

@Mike do you mean something like this: Deligne construction on canonical stratification satifies AX_p[S] for any topological stratification S. Now this implies that it satisfies AX_p, a theorem says this must be unique up to canonical iso in D^b(X). This means that any sheaf $F$ satisfying AX_p is cannoical isomorphic to deligne sheaf, if $F$ is S-constructible it must satisfy also AX_p[S] since $F$ is isomorphic to the deligne sheaf, which satisfies AX_p[S] ...

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$Mathematics$ Mathematics