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12:17 AM
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.
 
1:11 AM
how is continuity globally approached?
 
continuity is a local property
 
yes it is, but everywhere continuous function has some pieces of information, other than considering uniform continuity
 
that sentence does not make much sense to me
 
what´s not clear?
 
which pieces of information you're talking about
 
1:21 AM
mapping properties, when mapping various sets into various sets
it maps opens into opens
 
no
 
bounded into bounded
 
it does neither of that
at least not necessarily
 
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
 
what domains do you have in mind; this sounds wrong
 
1:30 AM
i am not sure, $\mathbb Q \cap (a,b)$ seems good as a first step
 
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
 
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
 
1:45 AM
continuity over Q already is a defined property and not every continuous function on Q can be continuously extended to R
 
when it cannot be extended?
 
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
 
2:05 AM
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
 
preimages of open sets are open
 
does that characterization goes on to metric spaces also, even topological spaces?
 
it is the definition of continuity
 
2:28 AM
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
 
2:53 AM
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.
 
What would happen if there is no $x\neq0$ in $A$ or no $y\neq0$ in $B$?
 
sorry I should say for x!=0 and y!=0.
 
Right, but what if such elements don't exist?
 
which definition of ring are you using
ring with identity or not necessarily
 
that doesn't really change the situation, does it?
 
3:01 AM
i mean, presumably 1 is not 0
 
the zero ring is a ring with identity
 
huh. point
 
@Semiclassical ring with not necessarily to have unity or commutative.
 
we just need to exclude the degenerate case and then both theorem and proof will be correct
 
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
 
3:08 AM
@Thorgott what is the degenerate case here?
 
the definitions vary from place to place
 
@user777 consider my previous question
 
Yeah, so maybe it's best to specify the definitions being used
 
 
2 hours later…
4:52 AM
M o r n i n '
 
@EdwardEvans Hello.
 
Hiya @feynhat
 
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
 
5:19 AM
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 $2g(\nabla_X Y, Z) = X g(Y, Z) + Y g(X, Z) - Z g(X, Y) + g([X, Y], Z) + g([Z, X], Y) - g([Y, Z], X)$
 
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
 
I enjoy your walls of incomprehensible Riemannian geometry
 
Lol
 
I'm experiencing insomnia induced delirium, so that's probably why
 
5:27 AM
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
 
6:01 AM
@BalarkaSen Most definitely NOT!
 
@TedShifrin have you figured out multinomial theorem with non-integral number of terms
 
@TedShifrin Why?
Ok, I meant $g_1 = c g_2$ for a constant $c$
 
6:57 AM
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?
 
 
3 hours later…
9:36 AM
does anyone of you has rights to endorse someone on arxiv?
 
9:51 AM
I have
You have
He has
She has
It has
We have
They have
 
Jag har
 
Hi chat
 
Hey
 
@EdwardEvans what would i need to prove that, for example, you endorse me? if you would ever do that?
 
9:56 AM
I wouldnt but idk
 
Is there a deeper way to prove that SU_2 and SO_3 have isomorphic Lie algebra than computing and expliciting the isomorphism ?
 
Hey @Alessandro
 
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 $SU(2) \to SO(3)$
 
10:13 AM
So when you take the Lie Algebra you get the same thing
Right ?
 
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
 
 
1 hour later…
11:17 AM
@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 $f(f(x))$
 
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
 
I have drawn that conclusion because applying $f$ to $f \left ( f(x) \right) $ results in indentity function
 
sure that works
 
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
 
that $f \circ f$ is inverse of $f$ is literally definition of inverse of a function. i dont understand the confusion
 
11:27 AM
@BalarkaSen Really? I think that’s true only for identity function
 
It’s true for every function?
 
its true because you're given $f \circ f \circ f = id$
that's hypothesis
 
Okay.
Thanks
:)
 
 
1 hour later…
12:35 PM
how to solve this equation
 
@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
 
1:08 PM
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?
 
@VJ123 I've done but ended up with large values.
 
@CroCo What was your final quadratic?
 
(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?
 
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?
 
1:27 PM
no error in the equation.
this is from the book
 
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
 
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.
 
@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
 
yes I guess
@VJ123 yes I guess
 
1:45 PM
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
 
2:45 PM
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 modi cations, 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?
 
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 ?
 
3:02 PM
If $M$ is a compact, orientable Riemannian manifold, then, is $L^2(M)$ complete?
 
What's your definition of $L^2(M)$?
 
chat: when's the last time you wrote $\because$ and $\therefore$?
 
I'm pretty sure I've never written either of them
 
same
I wonder what the definition of "obvious" is when not even 1 of 20 in here can answer an "obvious" question
 
@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?
 
3:10 PM
both
 
I write both quite often in assignments and answer scripts.
I sometimes write $\therefore$ when I run out of my $\implies$ arrows.
 
@LeakyNun 10 years ago I'd say
Actually no
5 years ago
 
@feynhat I'm asking because it should be obviously complete by definition
 
in philosophy class
 
integration course: spends the whole course proving integration lemmas and that L^2 is complete
Alessandro: L^2 is obviously complete
 
3:18 PM
You wouldn't call it $L^2$ if it weren't complete, qed
3
Got any more statements for me to prove?
 
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 ?"
 
@AlessandroCodenotti How is it usually defined? Is smoothness required?
 
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"
 
(I still don't know how to quote messages after 5 years in this chat)
 
@Astyx Permalink
 
3:21 PM
Hey all. Would anyone know if the 3D shape of a cone section (the red part) has an official name?
 
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
 
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
 
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
 
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
 
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
 
3:26 PM
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?
 
$dV_g(U) = \int_U dV_g$
That makes it a Borel measure
 
Ah right of course
Makes sense
 
@halirutan Steradian angle or solid angle comes close, but since it is an ellips maybe ellipsoid steradian angle?
 
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.
 
ah never mind, not the name of the volume.
 
3:32 PM
$H^k$ is the space of harmonic k-forms.
 
@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.
 
Wait.
 
Hahn-Banach works everywhere, no need for completeness
 
Also yeah ^
 
@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!
 
3:34 PM
Of course to use Hahn-Banach you need to assume the Weak Konig Lemma
 
Lol
 
Where by everywhere I mean that any normed space is plenty enough
@MikeMiller Do you? I never remember how strong Hahn-Banach is exactly
 
aaaaaaaaaaaaaa
Anyway. Is $\Omega_k(M)$ complete though?
 
What is that precisely
And under what inner product
 
With respect to which norm?
 
3:37 PM
The inner product is defined as $\langle \omega, \eta \rangle = \int_M \omega \wedge * \eta$.
 
Am I to respond with "with what metric"?
 
Well that's actually irrelevant since it's finite dimensional to be fair
 
$*$ = Hodge star.
 
What kind of forms are $\Omega_k$
 
@AlessandroCodenotti ?
 
3:38 PM
$L^2$? $C^\infty$?
 
@MikeMiller Smooth.
 
Then that is decidedly not complete
The completion is the space of $L^2$ sections of $\Lambda^k(T^*M)$
 
C^infty can never be complete!
 
Ah, no, that was wrong, my bad
 
of course. what garbage am i talking
 
3:39 PM
I suspect this is halfway into a discussion of something simpler
What's the main goal?
 
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)
 
What do you know about $\Delta$ at this point in the argument?
I'm just curious.
 
3:57 PM
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
 
@MikeMiller That it is self adjoint
Oh also, regularity
 
I'd like to check them for the logistic regression model if that information helps
 
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)
 
Yeah Riesz representation is strictly a theorem for Hilbert spaces
Those are complete
 
4:12 PM
Thanks.
 
4:39 PM
@VJ123 I've got it. It is 40 because the second angle doesn't satisfy another equation.
 
Bro need help with matrix calculus
$$ \frac{\partial X^T}{\partial X} $$
what's it?
$X$ is a $n \times n$ matrix
 
$X \mapsto X^T$ is a linear map.
 
@feynhat what that means?
 
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).
 
you mean answer is 1?
 
4:47 PM
What does that expression mean?
 
@feynhat see for example $\frac{\partial X^T X}{\partial X} = 2X$ for matrix $X$ of $n \times n$ then what is that one^?
 
How?
 
@feynhat linear algebra.
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...
 
And, again what does $\frac{\partial X^T X}{\partial X}$ mean?
 
@feynhat see wiki
 
4:57 PM
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.
 
oh wow.
 
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.
 
@MikeMiller High schoolers suck
I can't write an intro to differential geometry for them
Why did I sign up for this
 
Then the product rule gives that $DF_A(X) = X^T A + A^T X$
 
@MikeMiller This is god level mathematics... Can you explain this to a kid like me ? ^_^
@feynhat wow...
 
5:02 PM
I am sorry about my ambiguous wow. I was wowing about that wikipedia page, nothing else.
@BalarkaSen I remember being one.
 
I am sure you sucked, as did I
 
@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.
 
@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
 
@BalarkaSen Were you at Homotopy theory school at ISI, Kolkata last summer?
 
5:09 PM
Ah no I wanted to be there but then didn't really attend
I know the people who organized that thing
 
@BalarkaSen I really doubt that. It'd be really cool if we did.
@BalarkaSen You know Somnath Basu?
 
Yeah
I know him well
 
How is he sooooo good... at everything.
 
@feynhat You are too at ISI K?
 
He's a pretty good mathematician, yeah.
I don't know him outside of academia though
 
5:13 PM
@AbhasKumarSinha No. Who else is?
 
I dun know, I only know Balarka, none other than him...
 
@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.
 
Hah, I have felt that way with both Samik and Somnath.
I tend to like people who speaks incoherently though, in contrast to precision
 
@BalarkaSen Gromow?
 
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
 
5:45 PM
hey chat
 
hi
 
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?
 
@BalarkaSen lmao
 
6:03 PM
@MikeMiller What else should I know?
 
@LucasHenrique voce tem tentado utilizar WolframAlpha?
 
Oh, that's a nice suggestion.
Wolfram suggests $v(x) \overset{\mathrm{def}}{=} x^2y(x)^2$
I don't know why but... 'kay
 
Why do you think it should depend on both x and y.
 
o que e v?
 
6:19 PM
@LucasHenrique Try $e^{-2y}/y^2$.
@LeakyNun You can speak Portuguese?
 
a bit
 
Noice. How many languages can you speak a bit?
 
@LeakyNun a substitution to solve the equation
 
ah, you looked at the steps, ok
 
@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
 
6:27 PM
@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.
 
@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 $xdy + ydx$.
 
I will be back with more details in about half an hour @feynhat.
 
Sure.
 
@feynhat what do you mean by "there is an exact part in the form"?
Ohh, so you mean to write that as $\left( x \mathrm dy + y \mathrm dx \right) + \left(x^2y^2\mathrm dy - xy^2\mathrm dx\right)$?
 
6:43 PM
You can rearrange the terms so that the form becomes (exact form) + (non-exact form).
yes.
Now, focus on the non-exact part.
 
That's a nice trick. Thanks, @feynhat.
 
Unpopular Opinion: Leibniz's notation is evil, and should be banned from MVC and ODE courses.
 
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 $x\mathrm dy + y\mathrm dx$?
 
hmm...
You're damn right it will.
Forget what I wrote lol
 
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