12:38 AM
@copper.hat thanks for replying, enjoy the match.

1 hour later…
1:39 AM
@user70432 you too :-)

2:25 AM
:-)

2 hours later…

6 hours later…
9:52 AM
still possible that an immersed submanifold might be homotopy equivalent to an embedded submanifold

10:37 AM
In proving formula (2.1), we first prove it for characteristic functions, then for simple functions and lastly for non-negative functions. In the last step, we need to use MCT; we assume there is a sequence of simple functions $\phi_n$ converging to $f$, then since it holds for simple functions $$\int f\, d\nu=\lim\int\phi_n\,d\nu=\lim\int\phi_n g\, d\mu.$$
Now, for any fixed $x$, $\phi_n g$ is increasing since $\phi_n$ is and $g$ is non-negative, however, what happens if for that specific $x$, $g(x)=\infty$ and $\phi_n\to 0$? I'm trying to verify that $\phi_n g$ is actually convering towards $fg$, so we can apply MCT again and finish proving (2.1).

11:17 AM
$\phi_n$ is increasing and non-negative, so if $\phi_n\rightarrow0$, we just have $\phi_n=0$ for all $n$

right :) but what if $g(x)=\infty$ on a set of nonzero measure?
the convention is that the integral of $0$ over a set of measure $\infty$ is $0\cdot\infty=0$, but here the situation is a bit different I think

Did I draw the tetrahedron figure correctly?
could someone check pls

why do you believe it's not correct?

11:33 AM
your question was whether $\phi_ng$ converges (pointwise) to $fg$ and the answer is yes. if $g(x)<\infty$, this is obvious. if $g(x)=\infty$ and $f(x)>0$, then $(\phi_ng)(x)=\infty=(fg)(x)$ for $n\gg0$ and if $g(x)=\infty$ and $f(x)=0$, then $(\phi_ng)(x)=0=(fg)(x)$ for all $n$, so the answer is yes in any case

@SineoftheTime It's the first time I've used geogebra in for 3D figures
I'm following an exercise I found on mathexchange

so you have the coordinates of the points and you have to draw the tetrahedron?

yes, I wanted to have a graphical representation

looks good

@Thorgott ok, but it seems to me you are using a convention here in asserting that $(\phi_ng)(x)=0$ when $g(x)=\infty$, aren't you?

11:37 AM
yes, $0\cdot\infty=0$ in measure theory always

@SineoftheTime to determine the volume, I need to find the plane equations for all faces of this tetrahedron, right?

what formula do you use to find the volume of the tetrahedron?

I wanted to use the method: Direct triple integral
we started seeing it 2 days ago

are you sure it's the best way?
can you send the source of the ex?

wait
im following this
7

Please help me with the problem below. Find the volume of a tetrahedron with vertices: $O(0,0,0)$, $A(1,2,3)$, $B(-2,1,5)$, $C(3,7,1)$ by using triple integral. Hint: First find the the equations of the planes. Note: My professor told me that I can use change of variables (which I need to cal...

I was looking at the answer and there were various methods provided to use

11:45 AM
the direct computation seems very messy

What is mean by messy
maybe I understood

I mean the bounds of $x,y,z$ in the tetrahedron
hard to find them

Plane defined by O, A and B. $ax + by + cz + d_1 = 0$
- For $x$: $ax = -by - cz - d_1$, then $x = -\frac{by}{a} - \frac{cz}{a} - \frac{d_1}{a}$.
- For $y$: $by = -ax - cz - d_1$, then $y = -\frac{ax}{b} - \frac{cz}{b} - \frac{d_1}{b}$.
- For $z$: $cz = -ax - by - d_1$, then $z = -\frac{ax}{c} - \frac{by}{c} - \frac{d_1}{c}$.
@SineoftheTime so it's better to change method?

yeah
I don't think this exercise is too relevant, maybe it's better if you choose one from your textbook

@SineoftheTime what would you go on?
@SineoftheTime oh ok
I'll take something from the teacher pdfs

11:57 AM
good idea

yes, I'm trying to solve a direct triple integral
I'll let you know if i can find the sol
ution

12:21 PM
$\int_A(x+z)dxdydz, A=[0,2] \times [1,2] \times [1,4]$
$\int_0^2\int^2_1\int_1^4 (x+z) \ dzdydx$
$\int_1^4 (x+z) \, dz = \left. \left(xz+\frac{z^2}{2}\right) \right|_1^4 = 3x+\frac{15}{2}$
$\int^2_1 3x+\frac{15}{2}dy = \left. \left(3xy+\frac{15}{2}y\right)\right|_1^2 = 3x+\frac{15}{2}$
$\int^2_13x+\frac{15}{2}dx=\left. \left(\frac{3x^2}{2}+\frac{15}{2}x\right)\right|^2_0 = 21$
Ok this was pretty simple
the solution is 21
However, I write the steps on the book otherwise it would take too long on the PC

1:20 PM
@Pizza you could have solve it using the properties of the center of mass also
in fact $\int_A xdV=x_G V(A)$ where $x_G$ is the $x$ coordinate of the center of mass and $V(A)$ is the volume of the region.
$\int_A xdV+\int_A z dV=V(A)*(x_G+z_G)=2\cdot 1\cdot 3\cdot(1+2.5)=21$

1:44 PM
$x_G = \frac{1}{V} \int_A x \, dV$
$z_G = \frac{1}{V} \int_A z \, dV$
you did that right?
to find xg and zg

nope, the other way around
$x_G$ and $z_G$ are easy to find without computations
and I used them to compute the original integral

wait
to find $V = (2-0) (2-1) (4-1) = 6$

@SineoftheTime how

they're the mean of the respective intervals
try to draw $[0,2]\times[1,2]$ in $\Bbb R^2$

1:52 PM
@SineoftheTime ok
@SineoftheTime ok
@SineoftheTime get a rectangle

correct
what are the coordinates of $G$?

$x$ = 1
$y$ = 1.5 so (1,1.5)

This method is so fast

can you find $z_G$ right?
@Pizza it's a trick that works when you integrate for example $x$ over a region that is symmetric somehow

2:01 PM
@SineoftheTime in theory in $\Bbb R^2$ i cant?

yes, you can't in $\Bbb R^2$
but you can consider now the whole region $A$ in $\Bbb R^3$

$z_G = \frac{1 + 4}{2} = \frac{5}{2}$

nice
so now that you have $x_G$ and $z_G$, the integral is straighforward

@SineoftheTime but there is no integral?
i mean
V(A)*(x_G+z_G)
you did that

this is equivalent to $\int (x+z) dV$, which is what you were trying to solve

2:06 PM
yes but if I use V(A)*(x_G+z_G), won't I finish the exercise?

yes
this is the value of the integral $\int_A(x+z)dV$

i can just say $\int (x+z) dV = 21$

yes
do you understand why?

I was going to ask you how I can tell if a region is somehow symmetrical

well, this is a parallelepiped
so in this case it's easy
in general you can use this trick where $A$ is a known geometric shape

2:14 PM
mm ok
however
I was trying to solve this cauchy problem before
i have a question
\begin{cases}
2y' - \frac{y}{\sqrt{x}} = \sqrt{x} \\
y(1) = 0
\end{cases}
I have to divide everything by 2, right?
i mean
$y' - \frac{y}{2\sqrt{x}} = \frac{\sqrt{x}}{2}$

if you want to apply the general formula, you divide by $2$

@SineoftheTime now can i multiply by $2\sqrt{x}$?

no?
why would you do that?

I used to think math was different from bio because memorization is really important in biology but now i've realized they're not so different :P
So many things to remember in math although it is technically possible to derive anything at any point from axioms usually

@SineoftheTime $2\sqrt{x}y' - y = x$

2:29 PM
personally, I don't memorize things

but why?

then how do you remember the definitions of all these algebraic objects and concepts :P
I get that it's not as much memorization as biology but there's still a bit of a language we build

@SineoftheTime ah yes, it's actually useless, I'll return to my previous form

Oh I remembered some random thing you said @Thorgott about how if we don't require a ring to be 2 binary operations in which one of them forms an abelian group, something follows anyway?
I forgot exactly but I'm curious what the ring axioms you were talking about were

2:39 PM
@SineoftheTime $y' + P(x)y = Q(x)$ so i want this form
$P(x) = -\frac{1}{2\sqrt{x}}$
$Q(x) = \frac{\sqrt{x}}{2}$
oh well but therefore the multiplicative integral is Px

wdym?

wait im wrting
$\mu(x) = e^{\int P(x)dx} = e^{\int-\frac{1}{2\sqrt{x}}dx} = e^{-\sqrt{x}}$
@SineoftheTime

@Obliv I just remember what intuitively makes sense to me
@Obliv the point was that being abelian actually follows from the other axioms

yes procede

3:39 PM
Let $S=[0,1]^2$. Ignoring issues having to do with boundaries and corners, a chart is a diffeomorphism $\varphi \colon S \to S$. Let $g_0$ denote the flat (euclidean) metric. Given a chart $\varphi \colon (S, g_0) \to (S, g_0)$, which need not be an isometry, consider the curves $\alpha_u(t)=\varphi^{-1}(u,t)$ and $\beta_v(t)=\varphi^{-1}(t,v)$ in $S$ which are generally not geodesics.
Does there exist a harmonic map $F: (S,g_0)\to T^2$ for which $F(\alpha_u(t))$ and $F(\beta_v(t))$ are all geodesics?
$T^2$ is the torus
I believe the standard Cartesian chart works?
however I'm more interested in the case where $\alpha, \beta$ are not geodesics on $S$

3:58 PM
$e^{-\sqrt{x}} \cdot y' - \frac{ e^{-\sqrt{x}} y}{2\sqrt{x}} = \frac{\sqrt{x}}{2} \cdot e^{-\sqrt{x}}$

substitute $\frac{e^{-\sqrt{x}}}{2\sqrt{x}} = \frac{d}{dx}e^{-\sqrt{x}}$

$e^{-\sqrt{x}} y' + \frac{d}{dx}\left(e^{-\sqrt{x}}\right)y = \frac{\sqrt{x}}{2} \cdot e^{-\sqrt{x}}$

$\frac{d}{dx}(e^{-\sqrt{x}} \cdot y) = \frac{\sqrt{x}}{2} \cdot e^{-\sqrt{x}} \Longrightarrow \frac{d}{dx}\left(e^{-\sqrt{x}}y\right) = \frac{\sqrt{x}}{2} \cdot e^{-\sqrt{x}}$
I wrote the same thing twice, sorry
(last line)
i procede
ill back

4:17 PM
@Thorgott me neither
I think that's just the normal state of things

1 hour later…
5:26 PM
$e^{-\sqrt{x}}y = -e^{-\sqrt{x}}(x+2\sqrt{x}+2)+C$

1 hour later…
6:50 PM
@SineoftheTime are you studying differential geometry?

7:10 PM
@SoumikMukherjee yes, an introduction. Basically curves and surfaces

7:42 PM
@SineoftheTime From which book?

my professors follows Do Carmo's book
"Differential geometry of curves and surfaces"

8:11 PM
Okay

9:01 PM
In Cohn's book on measure theory and on Wikipedia, I'm reading about the image measure and change of variables formula. I understand mostly everything in proving this theorem, i.e. showing the formula displayed. However, what I don't understand is how to show $g$ is integrable wrt to the image measure iff $g\circ f$ is integrable with respect to $\mu$. Maybe I'm missing the point of the proof?
It says that "In that case, the integrals coincide..." i.e. when say $g$ is integrable with respect to the image measure. But how does one show that $g\circ f$ is then integrable with respect to $\mu$? This confuses me.

9:21 PM
well, I don't know what you're looking at, but typically the proof should establish both the claim about integrability and the value of the integrals simultaneously

ok, strange, because as far as I can see, this is not proved in Cohn's book, although he writes the integrability claim in the proposition, but only shows the integral formula. Folland doesn't talk much about image measures at all. I need to find another source maybe.

I would assume it's implicit in Cohn's proof

2 hours later…
11:20 PM
do you guys use overleaf for LaTex?

i don't, but the last time i wanted to do something in latex, the only stuff i could find was (a) 20 years old or (b) had a bunch of dependencies with stuff that i assume is part of, or also assumed by, overleaf
so it seems quite popular

I am using that as well.
but I installed MacTex it is weird.
maybe I am a noob in LaTex LOL.