Mathematics - 2020-04-07 (page 1 of 2)

William Sun

00:17

Let $A$ be a commutative ring with identity. Is the following statement true?

For an additive subgroup $I$ of $A^+$, $I$ is an ideal if and only if the coset multiplication $(a+I)(b+I)=ab+I$ is well-defined.

In other words, is $I$ an ideal iff the quotient $A/I$ is well-defined?

In the case of groups we have the following: If $G$ is a group and $H$ a subgroup of $G$, then $H$ is normal iff the quotient $G/H$ is well-defined, i.e., $G/H$ is closed under the group operation.

I wonder if there is a general way of formulating this in the settings of universal algebra.

Ante

01:11

how is continuity globally approached?

Thorgott

continuity is a local property

Ante

yes it is, but everywhere continuous function has some pieces of information, other than considering uniform continuity

Thorgott

that sentence does not make much sense to me

Ante

what´s not clear?

Thorgott

which pieces of information you're talking about

Ante

01:21

mapping properties, when mapping various sets into various sets

it maps opens into opens

Thorgott

no

Ante

bounded into bounded

Thorgott

it does neither of that

at least not necessarily

Ante

not necessarily, but with some restrictions on the domain (the sets), it does

i am just thinking about, to take everywhere continuous function as a whole, and globally characterize its continuity, it could somehow be done i think

Heine does it with sequences tending to one point

Thorgott

what domains do you have in mind; this sounds wrong

Ante

01:30

i am not sure, $\mathbb Q \cap (a,b)$ seems good as a first step

Thorgott

a constant function on that domain is continuous, but surely doesn't map open sets to open sets

and you can easily construct a continuous function on this bounded domain with unbounded image using trig functions

Ante

none of that is a problem, continuity over Q could be defined, and , upon completion of interval by adding all limit points the continuous function over Q gives us unique continuous function over R, the extension

Thorgott

01:45

continuity over Q already is a defined property and not every continuous function on Q can be continuously extended to R

Ante

when it cannot be extended?

Thorgott

in either case, this has nothing to do with mapping open sets into open sets or bounded sets into bounded sets, which, in turn, has little to do with continuity

say, $1/(x^2-2)$

uniform continuity on the other hand suffices to guarantee the existence of a continuous extension

Masterphile

02:05

then how is continuity characterized?

by what you write it seems it is not characterized

only, for example, if f is defined on [a,b] and uniformly continuous there

but even then it is not

Thorgott

preimages of open sets are open

Masterphile

does that characterization goes on to metric spaces also, even topological spaces?

Thorgott

it is the definition of continuity

EnjoysMath

02:28

developer.ibm.com/callforcode

There's stuff in there about software for genomics research as well, I figured since you are all math people, maybe you can apply your skillset there

s2s-omxware.us-south.containers.appdomain.cloud/home

user777

02:53

Hi! I have a small question. Let A, B are different rings, then for A x B, then it is not Integral Domain. the key point here is that A x B have a zero divisor. To show this: (0,x)(y,0)=(0,0) even though (0,x) != (0,0) and (y,0) != (0,0) for any y \in A and x \in B. Is my proof correct.

Thorgott

What would happen if there is no $x\neq0$ in $A$ or no $y\neq0$ in $B$?

user777

sorry I should say for x!=0 and y!=0.

Thorgott

Right, but what if such elements don't exist?

Semiclassical

which definition of ring are you using

ring with identity or not necessarily

Thorgott

that doesn't really change the situation, does it?

Semiclassical

03:01

i mean, presumably 1 is not 0

Thorgott

the zero ring is a ring with identity

Semiclassical

huh. point

user777

@Semiclassical ring with not necessarily to have unity or commutative.

Thorgott

we just need to exclude the degenerate case and then both theorem and proof will be correct

Semiclassical

Let Z be the zero ring. Does that mean Z x Z would be an integral domain?

I've used those definitions so little that I genuinely dont' know.

looks like no, based on the definitions i see online

user777

03:08

@Thorgott what is the degenerate case here?

Thorgott

the definitions vary from place to place

Semiclassical

yuck

Thorgott

@user777 consider my previous question

Semiclassical

Yeah, so maybe it's best to specify the definitions being used

Edward Evans

04:52

M o r n i n '

feynhat

@EdwardEvans Hello.

Edward Evans

Hiya @feynhat

Balarka Sen

Suppose $M$ is a compact Riemannian manifold and $g_1, g_2$ are two metrics on $M$ such that the curvature tensors of $g_1, g_2$ "agree", in the sense that $R_1(X, Y, Z, W) = R_2(X, Y, Z, W)$ for all $X, Y, Z, W$ vector fields on $M$.

That should force $g_1 = g_2$, right?

How do you show this? The Taylor expansion of the metric in normal coordinates is completely determined by the curvature tensor, so if the metric was analytic, so if it's false the metric is necessarily non-analytic

Balarka Sen

05:19

Oh I should be careful. Scaling the metric doesn't change the curvature tensor

By a constant that is

Let's see, Koszul formula says $2g(\nabla_X Y, Z) = X g(Y, Z) + Y g(X, Z) - Z g(X, Y) + g([X, Y], Z) + g([Z, X], Y) - g([Y, Z], X)$

Edward Evans

ew

Balarka Sen

If $g' = \lambda g$ where $\lambda$ is a general function, how does $\nabla'$ relate to $\nabla$

$2 \lambda g(\nabla'_X Y, Z) = X \lambda + Y \lambda - Z \lambda + \lambda 2g(\nabla_X Y, Z)$

That's a complicated formula...

I mean $(X\lambda) g(Y, Z)$ etc etc oh god horrible I mean

Edward Evans

I enjoy your walls of incomprehensible Riemannian geometry

Balarka Sen

Lol

Edward Evans

I'm experiencing insomnia induced delirium, so that's probably why

Balarka Sen

05:27

Likely, my dude, likely

Robert Bryant says specifying the curvature tensor obviously specifies the conformal class of the metric

Why is it so obvious to him

Maybe some counting. The metric is a positive section of $T^*M \otimes T^*M$, curvature tensor is a section of $\Lambda^2 TM \otimes TM \otimes T^*M$

So the first has $n^2$ degrees of freedom, second has $n^3(n-1)/2$ degrees of freedom

Hopefully

The conformal class of the metric has $n^3$ degrees of freedom, but how does that help

idgi ill figure it out later lol

Ted Shifrin

06:01

@BalarkaSen Most definitely NOT!

Leaky Nun

@TedShifrin have you figured out multinomial theorem with non-integral number of terms

Balarka Sen

@TedShifrin Why?

Ok, I meant $g_1 = c g_2$ for a constant $c$

KReiser

06:57

Here's a dumb question that I just spent too much time not answering: let $R$ be a DVR with fraction field $K$. Let $X$ be the spectrum of $R$. Let $U$ be the unique proper nonempty open subset. Let $F$ be the non-quasicoherent sheaf on $X$ given by $F(X)=F(\emptyset)=0$ and $F(U)=K$. What's the cohomology of this sheaf?

The only question is what the 0th cohomology is. As the 0th cohomology is global sections, it vanishes, and by Grothendieck vanishing, the cohomology in degree >1 also vanishes. The cohomology in degree 1 has to be either $K$ or 0 - I tried computing the Godement resolution, which gave me that the 1st cohomology also vanished, but this seems very suspicious to me. Does this sound right?

Masterphile

09:36

does anyone of you has rights to endorse someone on arxiv?

Mats Granvik

09:51

I have
You have
He has
She has
It has
We have
They have

Edward Evans

Jag har

Astyx

Hi chat

Edward Evans

Yo

Balarka Sen

Hey

Masterphile

@EdwardEvans what would i need to prove that, for example, you endorse me? if you would ever do that?

Edward Evans

09:56

I wouldnt but idk

Astyx

Is there a deeper way to prove that SU_2 and SO_3 have isomorphic Lie algebra than computing and expliciting the isomorphism ?

Alessandro Codenotti

Hi

Edward Evans

Hey @Alessandro

Balarka Sen

There's a double cover SU(2) -> SO(3), I meant

Let's see, what's the best way to explain this

Identify SU(2) with the group of unit quaternions, and consider for any unit quaternion $q$ the action on $\Bbb R^3 = i \Bbb R \oplus j \Bbb R \oplus k \Bbb R$ by conjugation by $q$

That's the map $SU(2) \to SO(3)$

Astyx

10:13

So when you take the Lie Algebra you get the same thing

Right ?

Balarka Sen

Yeah

It's a group homomorphism which is also a local diffeomorphism

So isomorphism on the tangent space at identity

I keep learning the connection between rotations and quaternions but forget the story completely every time

I'll write something down on that note eventually

Knight

11:17

@BalarkaSen If $f$ is a continuous function on $[0,1] \rightarrow [0,1]$ with the property $$ f ( f( f(x) ) ) =x $$. Then, can I say that either $f$ is an identity function or the inverse of $f$ is $f(f(x))$

Balarka Sen

If $f$ is a function satisfying those properties then it is the identity function.

inverse of $f$ is $f\circ f$ tautologically; doesn't require any argument. it's just defn of inverse

Knight

I have drawn that conclusion because applying $f$ to $f \left ( f(x) \right) $ results in indentity function

Balarka Sen

sure that works

Knight

Can I say $f (f(x) )$ is inverse of $f$ without proving that $f$ is an indentity function?

It is not given that $f$ is invertible, but the problems doesn’t even the restricts the existence of invertibilty

Balarka Sen

that $f \circ f$ is inverse of $f$ is literally definition of inverse of a function. i dont understand the confusion

Knight

11:27

@BalarkaSen Really? I think that’s true only for identity function

Balarka Sen

no

Knight

It’s true for every function?

Balarka Sen

its true because you're given $f \circ f \circ f = id$

that's hypothesis

Knight

Okay.

Thanks

:)

CroCo

12:35

how to solve this equation

VJ123

@CroCo Subtract the cos term from both sides, then square both sides, and use the Pythagorean identity.

Express everything in terms of either sin or cos

And then you'll get a quadratic

MathStudent

13:08

Hi there,
I would like to look at a norm of a third derivative of a function $f: R^d \rightarrow R$. What would be a standard norm to use?

CroCo

@VJ123 I've done but ended up with large values.

VJ123

@CroCo What was your final quadratic?

CroCo

(b^2 +a^2) cos^2(th) -2bd cos(th) + (d^2-a^2) = 0
where a=275; b=158.75; d=300

do I need to normalize it?

VJ123

275sin(x)=300-158.77cos(x)

, squaring both sides and rearranging ==> 75625sin^2(x)-(300-158.77cos(x))^2=0

now if i use the pythag identity

the quadratic i get is

-100832.9129cos^2(x) + 95262cos(x) - 14375

i think you may have an error in your simplification/rearrangement to get to this quadratic?

CroCo

13:27

no error in the equation.

this is from the book

VJ123

hmmmmm the solutions u get from my equation are x = 79.134 degrees (plus or minus multiples of 360) and also x=40.866 degrees (plus or minus multiples of 360)

both of which work

by error in equation, I mean error in your quadratic

not the actual question

CroCo

I see but which angle should I choose for this problem

I've got the angles correctly but got stuck on choosing between the angles.

Abhas Kumar Sinha

VJ123

@CroCo So would i be right in saying the rope from B to A is different to the rope from A to C which is different to the rope from A to D

so three different ripes?

*ropes

CroCo

yes I guess

@VJ123 yes I guess

VJ123

13:45

do you have the answers to the question?

@CroCo

if so, do they pick a particular ange

*angle

oh wait hold up

I think you need to consider the actual weight forces, and not just the mass

so 275lb * gravitational field strength

and 300 lb* gravitational field strength

in your calculations

not just 275 and 300

MathStudent

14:45

I read some conditions for a theorem and it says that "We focus on the one-dimensional case, but
the exact same theorem, with obvious modications, holds for d-dimensional $\theta$"

One of the conditions is that the supremum of the absolute value of the third derivative of a function of $\theta$ is less or equal than a function $M(x)$

So when $\theta$ is d-dimensional, what norm would you use instead of taking the absolute value?

Astyx

What's the link between a Lie group's adjoint representation and its Lie algebra's adjoint representation ? Are they the same thing modulus integration by the exponential or something ?

feynhat

15:02

If $M$ is a compact, orientable Riemannian manifold, then, is $L^2(M)$ complete?

Alessandro Codenotti

What's your definition of $L^2(M)$?

Leaky Nun

chat: when's the last time you wrote $\because$ and $\therefore$?

Alessandro Codenotti

I'm pretty sure I've never written either of them

MathStudent

same

I wonder what the definition of "obvious" is when not even 1 of 20 in here can answer an "obvious" question

feynhat

@AlessandroCodenotti hmm... I don't know. How should I define it? Smooth functions $f$ on $M$ such that $\int_M |f|^2 < \infty$.

Oh, wait is that $< \infty$ redundant?

@LeakyNun You mean in chat or irl?

Leaky Nun

15:10

both

feynhat

I write both quite often in assignments and answer scripts.

I sometimes write $\therefore$ when I run out of my $\implies$ arrows.

Astyx

@LeakyNun 10 years ago I'd say

Actually no

5 years ago

Alessandro Codenotti

@feynhat I'm asking because it should be obviously complete by definition

Astyx

in philosophy class

Leaky Nun

integration course: spends the whole course proving integration lemmas and that L^2 is complete
Alessandro: L^2 is obviously complete

Alessandro Codenotti

15:18

You wouldn't call it $L^2$ if it weren't complete, qed

3

Got any more statements for me to prove?

Astyx

I have a question above

I think you'll solve it quickly enough using this approach

"
What's the link between a Lie group's adjoint representation and its Lie algebra's adjoint representation ? Are they the same thing modulus integration by the exponential or something ?"

feynhat

@AlessandroCodenotti How is it usually defined? Is smoothness required?

Alessandro Codenotti

When I say it should be obviously complete I really mean "it should be easy to show it's complete as long as you have a reasonable definition and know the Riesz-Fischer theorem"

Astyx

(I still don't know how to quote messages after 5 years in this chat)

feynhat

@Astyx Permalink

halirutan

15:21

Hey all. Would anyone know if the 3D shape of a cone section (the red part) has an official name?

Astyx

31 mins ago, by Astyx

What's the link between a Lie group's adjoint representation and its Lie algebra's adjoint representation ? Are they the same thing modulus integration by the exponential or something ?

Oh it's that easy

Thank you

Alessandro Codenotti

Density on a manifold

In mathematics, and specifically differential geometry, a density is a spatially varying quantity on a differentiable manifold that can be integrated in an intrinsic manner. Abstractly, a density is a section of a certain trivial line bundle, called the density bundle. An element of the density bundle at x is a function that assigns a volume for the parallelotope spanned by the n given tangent vectors at x. From the operational point of view, a density is a collection of functions on coordinate charts which become multiplied by the absolute value of the Jacobian determinant in the change o...

It's tricky

Balarka Sen

If $M$ is a Riemannian manifold it has a natural measure associated to it, the Riemannian volume form. $L^2(M)$ is simply $L^2$ functions wrt that measure

At least, if that's not what it is, I debunk functional analysis

Alessandro Codenotti

It'd better be that in the Riemannian case

I was thinking about metrizing $M$ and using $L^2$ wrt the Hausdorff measure for a general $M$, but this gets messy because it's metric dependent

(but in the Riemannian case integration with the volume form and with the Hausdorff measure should agree for nice functions so it's fine)

@BalarkaSen Hmm there might still be details to fix, probably a completion to take? Officially $\int_M f \mathrm{d}V_g$ is defined for smooth $f$, but to get completeness I'm pretty sure you want continuous functions

Balarka Sen

How is $\int_M f dV_g$ defined for only smooth $f$? $dV_g$ is a measure. Take the measurable, square integrable dudes wrt this measure

I am not integrating just forms

Alessandro Codenotti

15:26

Anyway I don't know geometry and I know even less analysis. I'm out

$\mathrm{d}V_g$ is a form, how do you get a measure out of it?

Balarka Sen

$dV_g(U) = \int_U dV_g$

That makes it a Borel measure

Alessandro Codenotti

Ah right of course

Makes sense

Mats Granvik

@halirutan Steradian angle or solid angle comes close, but since it is an ellips maybe ellipsoid steradian angle?

feynhat

Ok. Never mind $L^2$. Here's what I want. I am studying Hodge decomposition. At some points in its proof, we define $l : (H^k)^\perp \to \Bbb R$ and show that its a bounded functional. So, Hahn-Banach extends the functional to $\Omega^k(M)$. Now, for this to make sense $\Omega^k(M)$ should be complete.

Mats Granvik

ah never mind, not the name of the volume.

feynhat

15:32

$H^k$ is the space of harmonic k-forms.

Balarka Sen

@feynhat @Alessandro Any compact Riemannian manifold $M$ isometrically embeds in $\Bbb R^n$ for some large $n$ by Nash embedding theorem. That should tell you $L^2(M)$ sits as a closed subspace of $L^2(\Bbb R^n)$, which is complete.

feynhat

Wait.

Alessandro Codenotti

Hahn-Banach works everywhere, no need for completeness

Balarka Sen

Also yeah ^

halirutan

@MatsGranvik Yep, it's about the "body". A friend is implementing a function which visualizes these and was asking for a good name. So I guess we stick with "Upper half of the upright cone that Silvia cut" :)

Thanks anyway!

Mike Miller

15:34

Of course to use Hahn-Banach you need to assume the Weak Konig Lemma

Balarka Sen

Lol

Alessandro Codenotti

Where by everywhere I mean that any normed space is plenty enough

@MikeMiller Do you? I never remember how strong Hahn-Banach is exactly

feynhat

aaaaaaaaaaaaaa

Anyway. Is $\Omega_k(M)$ complete though?

Mike Miller

What is that precisely

And under what inner product

Alessandro Codenotti

With respect to which norm?

feynhat

15:37

The inner product is defined as $\langle \omega, \eta \rangle = \int_M \omega \wedge * \eta$.

Balarka Sen

Am I to respond with "with what metric"?

Alessandro Codenotti

Well that's actually irrelevant since it's finite dimensional to be fair

feynhat

$*$ = Hodge star.

Mike Miller

What kind of forms are $\Omega_k$

feynhat

@AlessandroCodenotti ?

Mike Miller

15:38

$L^2$? $C^\infty$?

feynhat

@MikeMiller Smooth.

Mike Miller

Then that is decidedly not complete

The completion is the space of $L^2$ sections of $\Lambda^k(T^*M)$

Balarka Sen

C^infty can never be complete!

Alessandro Codenotti

Ah, no, that was wrong, my bad

feynhat

of course. what garbage am i talking

Mike Miller

15:39

I suspect this is halfway into a discussion of something simpler

What's the main goal?

feynhat

There is no goal since Hahn-Banach doesn't require completeness.

I thought we needed completeness (maybe because it has 'Banach' in the name).

@MikeMiller To show that $(H^k)^\perp \subset \Delta(\Omega^k(M))$.

(the other inclusion is trivial)

Mike Miller

What do you know about $\Delta$ at this point in the argument?

I'm just curious.

MathStudent

15:57

Could anyone of you recommend a source with a form of the Wilks' theorem (the commonly used likelihood ratio statistic is asymptotically chi-square distributed) with conditions that are not that hard to check?

Or for a theorem of the MLE being asymptotically normal with not that hard to check conditions could work too

feynhat

@MikeMiller That it is self adjoint

Oh also, regularity

MathStudent

I'd like to check them for the logistic regression model if that information helps

feynhat

If $\Omega^k$ were complete, then the regularity theorem would have been useless because any weak solution $l : \Omega^k \to \Bbb R$ would have already been of the form $l(\phi) = \langle \phi, \omega \rangle$ for some smooth form $\omega$ (because Riesz representation).

(I hope we do require completeness for Riesz representation)

Balarka Sen

Yeah Riesz representation is strictly a theorem for Hilbert spaces

Those are complete

feynhat

16:12

Thanks.

CroCo

16:39

@VJ123 I've got it. It is 40 because the second angle doesn't satisfy another equation.

Abhas Kumar Sinha

Bro need help with matrix calculus

$$ \frac{\partial X^T}{\partial X} $$

what's it?

$X$ is a $n \times n$ matrix

feynhat

$X \mapsto X^T$ is a linear map.

Abhas Kumar Sinha

@feynhat what that means?

feynhat

I don't know what this notation $\dfrac{\partial X^T}{\partial X}$ means. But if you want to know the derivative of the map $X \mapsto X^T$, where we identify the space of matrices with $\Bbb R^{n^2}$, this it is the map itself (at each point).

Abhas Kumar Sinha

you mean answer is 1?

feynhat

16:47

What does that expression mean?

Abhas Kumar Sinha

@feynhat see for example $\frac{\partial X^T X}{\partial X} = 2X$ for matrix $X$ of $n \times n$ then what is that one^?

feynhat

How?

Abhas Kumar Sinha

@feynhat linear algebra.

Matrix calculus

In mathematics, matrix calculus is a specialized notation for doing multivariable calculus, especially over spaces of matrices. It collects the various partial derivatives of a single function with respect to many variables, and/or of a multivariate function with respect to a single variable, into vectors and matrices that can be treated as single entities. This greatly simplifies operations such as finding the maximum or minimum of a multivariate function and solving systems of differential equations. The notation used here is commonly used in statistics and engineering, while the tensor index...

feynhat

And, again what does $\frac{\partial X^T X}{\partial X}$ mean?

Abhas Kumar Sinha

@feynhat see wiki

Mike Miller

16:57

It would probably be smarter to be more patient with someone who wants to help you, I assure you feynhat knows both linear algebra and calculus.

feynhat

oh wow.

Mike Miller

I am certain anything on that page is a rephrasing of: "What is the derivative of the map $X \mapsto X^T X$, as a map $\text{Mat}_{n \times n}$ to itself?"

Call that map $F$; its derivative at a matrix $A$ is a linear map $DF_A$ from matrices to themselves.

Balarka Sen

@MikeMiller High schoolers suck

I can't write an intro to differential geometry for them

Why did I sign up for this

Mike Miller

Then the product rule gives that $DF_A(X) = X^T A + A^T X$

Abhas Kumar Sinha

@MikeMiller This is god level mathematics... Can you explain this to a kid like me ? ^_^

@feynhat wow...

feynhat

17:02

I am sorry about my ambiguous wow. I was wowing about that wikipedia page, nothing else.

@BalarkaSen I remember being one.

Abhas Kumar Sinha

sorry

Balarka Sen

I am sure you sucked, as did I

Abhas Kumar Sinha

@BalarkaSen I'm too...

what that $\rightarrow$ means?

The notation I use is that one^ or $\nabla_{X} X^T$ this one for gradient of matrix

Please some cool one night intro to Linear Algebra book. That'd ease my pain.

Balarka Sen

@feynhat Do we know each other outside of this chat by any chance? Some workshop or colloquium we have both attended maybe?

Out of curiosity

feynhat

@BalarkaSen Were you at Homotopy theory school at ISI, Kolkata last summer?

Balarka Sen

17:09

Ah no I wanted to be there but then didn't really attend

I know the people who organized that thing

feynhat

@BalarkaSen I really doubt that. It'd be really cool if we did.

@BalarkaSen You know Somnath Basu?

Balarka Sen

Yeah

I know him well

feynhat

How is he sooooo good... at everything.

Abhas Kumar Sinha

@feynhat You are too at ISI K?

Balarka Sen

He's a pretty good mathematician, yeah.

I don't know him outside of academia though

feynhat

17:13

@AbhasKumarSinha No. Who else is?

Abhas Kumar Sinha

I dun know, I only know Balarka, none other than him...

feynhat

@BalarkaSen I mean yeah he is a good mathematician. So are a lot of other people. But like, his... presentation is so on-point... the language he uses is laser-precise, his handwriting is calligraphic, his diagrams are so fucking neat.

okay... I am just fanboying at this point.

Balarka Sen

Hah, I have felt that way with both Samik and Somnath.

I tend to like people who speaks incoherently though, in contrast to precision

feynhat

@BalarkaSen Gromow?

Balarka Sen

Haha sure

He's the archetype of incoherence

It's funny that plenty of geometric group theorists I have met speak incoherently

All of them get glassy-manga-eyed when Gromov is mentioned

It runs in the field

Aight time for dinner

Lucas Henrique

17:45

hey chat

Astyx

hi

Lucas Henrique

I have a question (yup, it's homework, but I'm stuck and I've tried a few things)

I'm trying to solve $$ y^\prime + \frac y{x+x^2y^2} = \frac{xy^2}{x+x^2y^2}$$

I've multiplying the ugly denominator to get $(x+x^2y^2)y^\prime + (y-xy^2) = 0$

And the associated differential form $(x+x^2y^2)\mathrm dy + (y-xy^2)\mathrm dx = 0$

It would be nice if this equation was exact - but it isn't. So I've tried to multiply it by an integrating factor $\mu(x,y)$ such that it's exact. So I got the equation $$ \frac{\partial}{\partial x}(\mu M) = \frac{\partial}{\partial y}(\mu N)$$

Which is equivalent to $\mu (M_x - N_y) = \mu_y N - \mu_x M$

But now I'm stuck. :(

I think I can't just impose $\mu_x = 0$ or $\mu_y = 0$ since that, solving the equation, $\mu$ would still depend on both $x$ and $y$ so it's not the integrating factor we're looking for

Could anyone give me a clue?

Alessandro Codenotti

@BalarkaSen lmao

feynhat

18:03

@MikeMiller What else should I know?

Leaky Nun

@LucasHenrique voce tem tentado utilizar WolframAlpha?

Lucas Henrique

Oh, that's a nice suggestion.

Wolfram suggests $v(x) \overset{\mathrm{def}}{=} x^2y(x)^2$

I don't know why but... 'kay

feynhat

Why do you think it should depend on both x and y.

Leaky Nun

o que e v?

feynhat

18:19

@LucasHenrique Try $e^{-2y}/y^2$.

@LeakyNun You can speak Portuguese?

Leaky Nun

a bit

feynhat

Noice. How many languages can you speak a bit?

Lucas Henrique

@LeakyNun a substitution to solve the equation

Leaky Nun

ah, you looked at the steps, ok

Lucas Henrique

@feynhat well, from the partial derivatives equation, if $\mu_x = 0$, then $\frac 1\mu \mathrm d \mu = \frac{2xy+2x}{-xy^2+1} \mathrm dy$

which is x dependent

Mike Miller

18:27

@feynhat I realized as I tried to help that I'm pretty hazy on the details of the argument.

If you know the Fredholm property (which is equivalent to an inequality for $\Delta \big|_{(H^k)^\perp}$) then it is enough.

Write $P^k$ for $(H^k)^\perp$, and write $\Delta_P: P^k \hookrightarrow \Omega^k \to P^k \subset \Omega^k$, since you know that $\text{Im}(\Delta) \subset P^k$.

I'm going to stop writing about this before I give myself a refresher. This is a little embarassing. :)

Maybe I'll re-read Wells this weekend.

feynhat

@MikeMiller No worries.

I am not familiar with the Fredholm property.

Did you mean this inequality, $\omega \le c \|\Delta \omega\|$, c is a global constant.

for all $\omega$ orthogonal to harmonic forms.

@LucasHenrique There is an exact part in the form, forget about that. And multiply the remaining with what I wrote above. e: the exact part is $xdy + ydx$.

Mike Miller

I will be back with more details in about half an hour @feynhat.

feynhat

Sure.

Lucas Henrique

@feynhat what do you mean by "there is an exact part in the form"?

Ohh, so you mean to write that as $\left( x \mathrm dy + y \mathrm dx \right) + \left(x^2y^2\mathrm dy - xy^2\mathrm dx\right)$?

feynhat

18:43

You can rearrange the terms so that the form becomes (exact form) + (non-exact form).

yes.

Now, focus on the non-exact part.

Lucas Henrique

That's a nice trick. Thanks, @feynhat.

feynhat

Unpopular Opinion: Leibniz's notation is evil, and should be banned from MVC and ODE courses.

Lucas Henrique

So I want to make the non-exact form become an exact form.

But if I find that integrating factor, wouldn't that affect the exactness of $x\mathrm dy + y\mathrm dx$?

feynhat

hmm...

You're damn right it will.

Forget what I wrote lol

Transcript for

$Mathematics$ Mathematics