12:00 AM
so this means I have to take the closure: $\overline{(0,1) \cup (2,3)} = ((0,1) \cup (2,3)) \cup \{0,1,2,3\}$

And are the interior points in $S$?

I would say yes......

@TedShifrin, after watching lecture 2 I think I know how to solve q1A-13 in this pset, so i take back all that stuff last week about lectures being a wasre of time

Right, so this example hasn’t quite been the right one. Can you think of a clever modification, @dc3rd?

My solution prior to watching the lecture was purely pictoral, i drew the vectorz and concluded they produce the same triangle. Now i think using the strategy of explicitly referencing vectors from origin I can say why that is so (but cant say more now because i need to get back to work)

12:10 AM
Achieving understanding is not measured just by problem sets, but they're helpful if good problems.

$\overline{(0,1) \cup (2,3]} = ((0,1) \cup (2,3]) \cup \{0,1,2,3\}$ I closed one of the sides of an interval

I said open sets. So not clever at all.
You're just going back to your original failed attempt.
You do need to learn not to fall back always in your same ruts.
It’s a real weakness.

I'll discuss that with you in a minute, because you see a recurring pattern. But back to this....a modification you say...hmm

Hello everyone
How much math is too much math to learn at once?

So I want to do something along the lines of

$\overline{(0,1) \cup (1, 3)} = ((0,1) \cup (1,3)) \cup \{0,1,3\}$, but then I'm including everything from $(1,2)$ in the set

12:21 AM
I'm currently seeing differential geometry, data structures (as in computer science) and grad linear algebra. I also would like to start measure theory by myself and algebra (with aluffi's algebra 0 which teaches with cat theory). Should I forget about MT and algebra by now?

@dc3rd Good. Keep analyzing.
What level diff geo?

@Tangoed I think it depends on the person but I think there may be a general tendency to try to do too much. I fell into this trap in undergrad. I took a lot of upper level math classes at once and I didn't learn them as well as I needed to and I've been going back and filling in ever since. I just wish I had slowed down a little and really learnt the stuff well the first time. But that's just me.

In general, that is good advice, @Quin.

@Quin....you're not alone and I'm going through this in the middle of my undergrad, took a year off from it to do just that

@Quin I feel like I have a lot of thing to learn, I can't lose anymore time!! I've found Rieffel's notes on measure theory very interesting to self-learn

12:26 AM
Text?

do Carmo M., Differential Geometry of Curves and Surfaces
Duistermaat J. J., Kolk J. A. C., Lie Groups
Lee J., Introduction to Smooth Manifolds
Spivak M., Calculus on Manifolds
Wolf, Spaces of constant curvature
Analysis was the only proof-based course I did
Not sure if this is too much too

Geez. Completely mixed bag. I can’t tell from that what your prof is actually teaching.

@TedShifrin From his website it seems that the two first books are the main ones, the others are only references or to be read only some annexes

This could beca super demanding course, especially without multivariable analysis and point-set topology.

Sorry, only the first one*

12:29 AM
The first one is a standard undergrad text, but demanding.

@TedShifrin Ok... but feasible?

Spivak is a hard version of the multivariable analysis you should learn. The others are graduate texts. Wolf assumes a lot.

the Spivak pamphlet I call it.......

My book replaces Spivak.

rieffel's notes on measure theory are interesting. almost any approach to measure theory involves a good number of arbitrary choices. i think of it as more of a vibe or a perspective than a body of theorems, although obviously there are some core ones.

12:31 AM
@TedShifrin Do you know Portuguese? His website
@leslietownes So to a first exposure to the theme it would be alright?

Nope, I don't. You have so many basic subjects to learn.

Thank god, because that book is waaayyy too terse. Why would he make it so terse @TedShifrin ? I ask because you know him.

Groups, rings, multivariable anslysis ...

it seems like a strange place to start if only because measure theory isn't a preliminary to a lot of other stuff. you can do so much useful integration without it.

That was his first book and he was asked to write it as part of a short book series.
He was just out of graduate school.

12:33 AM
but as a treatment of measure theory it is a fine one.

also in tweaking my example I just have to somehow manipulate the objects we already stated right? I don't need to include anything else?

@TedShifrin Ok... So maybe it isn't a good idea to take dif geo right now?

No, diff geo is fine. Your choice of reading material is my complaint.

before you begin going crazy with measure theory or even constructing haar measure on locally compact groups it is helpful to have a rich familiarity with a body of examples that that theory intends to abstract from.

What is the interior of $\bar S$ with this example?

12:36 AM
@TedShifrin You mean the closure of any set S?
The interior of the closure of S?

@Tangoed Assuming your geom prof is competent and not doing those advanced books.
That was for dc3rd, Tangoed.

@TedShifrin LOL sorry

No prob. I get too many conversations.

@TedShifrin When I asked him if I could audit the course he only sent me to his notes and asked if I felt ok with it. It's a first year master's course, and obvsly they expect undergrad knowledge
Anyway, thanks @TedShifrin @leslietownes and @Quin, I have this year to prepare myself as best as I can, so I may be panicking a little

interior of $\bar{S} = \overline{(0,1) \cup (2, 3)} = ((0,1) \cup (2,3))$

12:41 AM
Really?
Did you forget which example we’re doing?

well then this may be where my issue is......

Scroll back. You had $(0,1)\cup (1,3)$.

oh that one...I binned it because I thought you were being sarcastic as in "good, don't think in that direction and go back to analyzing the first example"

Huh? The good was not sarcastic. The rest was getting you to recognize your bad patterns of behavior.
You do get stuck in ruts and ignore what we tell you, over and over.

so then in this case the interior of $\bar{S} = \overline{(0,1) \cup (1,3)}$ is $(0,1) \cup (1,3)$.

12:48 AM
Draw a damn picture. You are seriously thinking in symbols and not thinking.

DrW?

Well, I just found this data visualization from UC-Berkeley of all places...

Why of all places, and what is MIDS?

MIDS = Masters in Data Science
And two, because UC-Berkeley's name brand is extremely well respected in DS

Who the hell puts the I in there?
I don’t understand your point.

12:56 AM
Don't ask me, that's how they've chosen to initialize it

MPA = masters in public administration
MIS = masters in information science

oooh... I think I may have what you mean Ted, so the interior of $\overline{(0,1) \cup (1,3)}$ is $(0,3)$, that is because of the definitions of interior point being satisfied, but $1$ is not in $S$ itself, but is a frontier point.

Finally! Learn from this torture! :)

@dc3rd i was on the edge of my seat for so long with that one!

My takeaway....the closure of a set and the set are not the same

1:00 AM
Learn from the process...

:)

No, wrong takeaway.
You have to learn to try really different sorts of scenarios... not get stuck in ruts that get you nowhere.

examples, examples, examples.........I also read this in '"How to think mathematically"..........."specialize your conjectures with concrete examples, then worry about generalizing to proof"
well you answered my next question with regards to these ruts
I have to live what I preach "Always be open minded and experiment"
looks like I have to get better at it in practice in this environment

You have a tendency to revert to familiar territory that got you nowhere — so you are not learning from your missteps.
Stubbornness can be good, but often it is the plague.
That said, this is a difficult example for many people, but you need a mental list of novel approaches.

novel approaches and thinking is why I chose this arderous subject........for better or for worse.

1:05 AM
It did just come up in a more advanced setting in a question I commented on on main. Very similar thought.
But you will frustrate me if you keep repeating the same thing we've explained to you is wrong!
So expect me to get bitchy if you do that.

@StudySmarterNotHarder nothing much, I was just busy when you pinged me

One of my issues I've found is I have a "fear/hesitancy" to manipulate/reconstruct certain objects. So for instance I was really hesitant of using $(0,1) \cup (1,3)$ because it wasn't the same thing I started with, i.e $(0,1) \cup (2,3)$ and was different
I'll take the lesson to heed about novel approaches though Ted...

Thanks :)
Of course I pushed you to make it different, so don't be scared.

time for dinner

Here too, soon. Bon appétit.

1:16 AM
merci pour tout.

de rien.

@dc3rd you need to make sure you have the closure, interior, frontier of the various types of intervals and simple combinations down without thinking.
@TedShifrin Thanks for the wishes! Much appreciated. No fog today, spectacular, nice clear day, brisk breeze. Delightful.

@copper I'm looking forward to a trek “homeward” in the near future.

How's it going?

1:42 AM
fog reminds me of berkeley too.
and a certain kind of wind in the fall.

2:02 AM
Howdy, demonark.

2:17 AM
If Q is to find Range of f(x) = $x^2 + x +1$. Then , We can use D greater than or equal to zero. Why can’t it be D = 0 here.

a standard way of computing the range of a polynomial (when the domain is not in some way specially restricted) is completing the square. it makes things clearer. wolframalpha.com/input/?i=complete+the+square+x%5E2%2Bx%2B1
of a quadratic polynomial, i meant to say. it's not always that easy.

Aah k
@leslietownes Thnx

2:49 AM
when i was younger, springtime was windsurfing season in the berkeley area, got my endorphins on the bay...

@Tedshifrin I take it back. I dont get how to prove these vectors sum to 0. There is clearly a question I need to ask myself that I am missing...
Will need to ponder tgis after work. Methinks

3:16 AM
This is a different idea from what appeared in my lecture, @Andrew. Draw all the vectors (all 6) with a common origin.

1 hour later…
4:43 AM
@leslietownes I saw this post just now, and it seems easiest to see how this works for two caterpillars.

5 hours later…
10:11 AM
@TedShifrin I worked on the question a little more this evening, I still havent solved the second part (and your hint has given me more than enough to work with), but feel like the first part is a way better argument now that the one I came up with before watching your lecture!

https://gofile.io/d/jdrdSC

(for reference my first attempt to answer the same question 2 months ago: https://gofile.io/d/B2OklM)

10:31 AM
ohh I getit
Vectors are arrows with no fixed location, so In the first part I can think of the vectors as originating at the verticies, OR I can think of them coming from an arbitrary point in space, wherever they are they sum to 0. So if I rotate them all 90degrees they still all sum to 0
isn't geometry fun!

10:47 AM
Say I have two lines defined with two 3d vectors : e for axis and o for origin. How do I compute the distance between them?

1 hour later…
11:57 AM
@Eminem take general point on one if the line and find the perpendicular distance from it to other line.

12:40 PM
Hello, can anyone please explain how can I get the results in the last paragraph? I am able to get the result if in addition I use 2nd Kirchhoff' law.
The text is from the book Randomized algorithm by Motwani, Raghavan.

Which result are you struggling with?

The part after "a simple calculation using Kirchoff's Law and Ohm's Law yields the following: ...".
I guess that If I find the ampere through the branches, then the rest about voltage is ok

The point is that i1 + i2 = 1A where i1 is the current throught the top branch and i2 through the bottom branch
Now U = R1i1 = R2i2, but R1 = 1+1 = 2 = R2
Thus i1 =i2, thus i1=i2=1/2

How do you know that U = R1i1 = R2i2?

12:58 PM
That's Ohm's law IIRC

This should be right. But I do not see how from this point I can get i1 = i1 = 1/2.

Why is i3=1 ?

It should be i1=i3.

You need to see that u1 + u3 = u2
@Kapur that I agree with

I see that u1 + u3 = u2 from the second Kirchhoff's law, but in the textbook is not mentioned. And I do not see any other way how to get u1 + u3 = u2. Do you?

1:10 PM
How is u defined for you ?

@Astyx, do you happen to have time?

I am not sure, so for that reason I search for other derivation. But what I remember from the old studies at high school, that it is something like that if I go through the "circle", then the sum of voltages on resistances is equal to 0.
Thus, in this case u1 + u3 - u2 = 0, which is equivalent to u1 + u3 = u2.

@Kapur It's something like that. You might be right that the book isn't too accurate by saying you only need Kirchoff and Ohm's laws. But it doesn't really matter in the end
@ShaVuklia Sure

@Astyx Super, thank you for your time.

I'm trying to check the well-definedness of a definition, and I'm failing ;x it's about this one:

1:18 PM
@Kapur Glad to help

I'm trying to show that if $[\phi_0,\dots,\phi_n]$ is regular at $P$
and $[g\phi_0,\dots,g\phi_n]$ is also regular at $P$
then $[g\phi_i(P)]=[\phi_i(P)]$ for each $i$
and so, what I did so far was:

Your phi's are the f's of the pictures right?

yes
what I did so far was:
if $\phi_i$ is regular at $P$, then we have $\phi_i=f_i/g_i$ where $g_i(P)\neq 0$
and likewise $g\phi_i=f_i'/g_i'$ for some $g_i'(P)\neq 0$
if we write $g=r/s$
then we have: $rf_i/(sg_i)=f_i'/g_i'$
so $rg_i'f_i=sg_if_i'$
so that's getting close to what I want to show
I don't know if I can factor out stuff
I'm a bit unsure in this multivariate case
I think the following could maybe help (if this is correct):
if we have $f/g$ where $f(P)=0$ and $g(P)=0$, then we can factor out enough so that we get some $f'/g'$, where either $f'(P)\neq 0$ or $g'(P)\neq 0$, and $f/g=f'/g'$
I would think this is allowed, because we can always factor out roots, right? and so you keep doing this until you get the fraction you want

@ShaVuklia What do you mean "for each i" ?

@Astyx ah, that wasn't phrased correctly
I should have written:
$[g\phi_0(P),\dots,g\phi_n(P)]=[\phi_0(P),\dots,\phi_n(P)]$

1:33 PM
One of the $\phi_i(P)$ is nonzero otherwise the point is not well defined
Look at $g\phi_i$ at $P$. It's either zero or nonzero, it cannot have a pole there, otherwise it's not regular at $P$
I think, my brain is kinda fried rn
If $g(P) = 0$, then it's not well defined at $P$ because you get $0$ on the LHS
Thus $g(P)\ne 0$ and you get your equality

oml
let let lelteltleltellet me see :''xx
@Astyx you can't necessarily evaluate $g$ rite. I think you should say that "if $f_i'/g_i'=0$..." instead of $g$

That's why I established just before

okok, but I can say the following:
oh wait
wait

1:49 PM
:)

x'D
Right, so we have the equality $g=f_i'/g_i'\cdot g_i/f_i$, from which it's also clear that $g(P)\neq 0$
ok, massive thanks. I think I wanted to check a few more things, lemme see if I can figure them out

2:16 PM
If a vector field $b(t,x):\mathbb{R}^+ \times \mathbb{R}^d \to \mathbb{R}^d$ to preserves a measure $\mu$? Say the lebesgue measure, then what does this say about the Jacobian of b(t,\cdot) ?

2:38 PM
If the Jacobian is $1$, then areas are preserved.

Say I push forward a measure $\mu$ by a vector field $X$, what happens to its density call it $f$? is it just $f \circ X^{-1}$
assuming $\mu$ was absolutely cont. w.r.t lebesgue measure

@Monty do you mean following the flow of $X$?
considering $\exp(X)$?

no time involed here
just push forward of a measure by a vector field.
trying to work out how the density changes after a push forward

I am not sure about what you mean by "push forward". Just move every point by the vector in the field at that point?

Is it always true that for a rational function $f/g$ and a point $P$, either $f/g$ or $g/f$ is defined at $P$? I would think it is, but I don't know how to show it. In the case of smooth points on curves, I can work with the order at $P$, and then the proof is simple - but is it also true in general? (and why?)

2:49 PM
what does it mean to push forward a measure by a vector field

@ShaVuklia unless $f$ and $g$ vanish together.

I'd assume you don't mean the literal pushforward measure on the tangent bundle

@robjohn Defined at $P$ means that there exists a representative of $f/g$, say $f'/g'$, such that $g'(P)\neq 0$
in my case
so it could still be that $f$ and $g$ both vanish at $P$, while $f/g$ is defined at $P$
maybe I should add that I'm looking at rational functions on an irreducible variety $V$
So if I denote by $I(V)$ the irreducible ideal corresponding to $V$, we would have $f,g\in K[X]/I(V)$
($V$ is an irreducible projective variety btw)

In measure theory, a discipline within mathematics, a pushforward measure (also push forward, push-forward or image measure) is obtained by transferring ("pushing forward") a measure from one measurable space to another using a measurable function. == Definition == Given measurable spaces ( X 1 , Σ 1 ) {\displaystyle (X_{1},\Sigma _{1})} and ( X 2...
just wondering what happens to the density of the push forward

meh, if that's really what you want, I don't think there's a meaningful answer, but good luck

3:15 PM
is it true that if $X$ is a CW complex and $Y \subset X$ is a subcomplex such that $X \setminus Y$ only contains cells of dimension at least $n$ then $Y$ contains $X^{n-1}$ , and therefore $Y^{n-1} = X^{n-1}$?
I have an argument for it but am a bit doubtful

How can I integrate $$sin(x^2+2)ln(1+sqrt{x})dx$$? I don't seem to recognise what to substitute.

Can anyone help me understand mathoverflow.net/questions/327781/…

@Liang it might not have an antiderivative expressible in terms of elementary functions
if you have a range you are integrating over it that would make the problem easier somewhat

Yes, it does not seem to be integrable in elementary functions.

3:57 PM
@ShaVuklia so you mean there is a continuous extension at $P$.

Can I post coding decoding questions here?

NO.
@Thor: Ready for the talk?

Why are you so angry in me @TedShifrin
I am an unemployed youth seeking some help and guidance for employment..

@robjohn I'm only thinking algebraically at this point I'm afraid

Because you repeatedly (like ten times) ignored my polite requests and kept bothering us with things that do not belong in this room.

4:06 PM
but yea, that would boil down to a continuous extension probs

@Monty: Do you know that the divergence theorem tells you that the rate of change of the volume of a region as you flow by a vector field is the integral of divergence over the region?

@TedShifrin I think I got it now actually,
Ill shout back if Im stuck, but I think it makes sense

It's quite intuitive, because the change of volume will be given by the flux of the vector field across the boundary, and by the divergence theorem this is in turn the integral of divergence over the inside.
(This is proved quite rigorously in the last section of my multivariable math book, but it is a classic result.)

It is very intuitive that divergence free vector field preserves lebesgue measure
since divergence free means I can take any small set and the mass flow in is equal to the mass flow out, and since the lebesgue measure has "distributed mass evenly" then the total change will be zero
or is this intuition bad?
silly question given any vector field will there always exist some distribution which is preserved under the flow of that vector field ?

could I reask my question? it was buried

4:18 PM
@porridgemathematics this question?

yup thats the one
how did you link to that by the way?

@porridgemathematics if you click on the downarrow that appears to the left of the text of the comment, there is a link to a "permalink". I used that link in a [this question](<link address>) reference.

4:42 PM
@Monty I'm not sure I know what you mean, but if divergence is everywhere positive, every region expands under the flow.

hey chat

Hey Lucas

sup, Ted?
so I'm solving my analysis problem set and I had to prove this as a step:
1

I suppose your definition of a "null set" is a set $A\subseteq\mathbb R$ such that, given any $\varepsilon\gt0,$ we can find a sequence $\{I_n\}$ of open intervals such that $A\subseteq\bigcup_{n=1}^\infty I_n$ and $\sum_{n=1}^\infty|I_n|\lt\varepsilon$ where $|I_n|$ is the length of $I_n.$ Assu...

where did his idea of the characteristic functions come from?

Huh? It's totally natural and shows up all over analysis, probability, and more.

@LucasHenrique That is a standard definition.

4:48 PM
If you want the length of a subset or the area of a region, you integrate the function that is $1$ on that subset/region and $0$ outside.
In probability, it's called an indicator function.

@robjohn: I know the definition. I'm just not used to use them :p

@Astyx do you have time for one more thing? ;o basically I had two definitions, and you helped me with well-definedness of one, and I finally proved well-definedness of the other, and now I want to show that they are 'equivalent', and I'm only lacking one direction ;x

@LucasHenrique Sorry, I was confused by the question.

In manifold theory, we use smooth version of this. We use a bump function that is $1$ on some neighborhood of a point of interest and $0$ far outside.

sure

4:49 PM
(it's a lot to write out, hence I'm checking first \w u)

Hi Ted

oh cool! ;o

Hi, @Astyx. :)

And Lucas and robjohn and others

Thank goodness we're not "others."

4:50 PM
lol Ted

Hell is others
Or however that's translated

@TedShifrin just barely... I was included at the very end.

ok so, this is the alternative definition:

L'existence devant autrui?

4:51 PM
@robjohn Be careful not to get omitted next time!

Well, @robjohn, I wasn't going to brag about getting top billing.

I want to show now that if I start with a rational map $\phi=[\phi_0,\dots,\phi_n]$, where $\phi_i$ are rational, then by the procedure they describe, I will get an 'equivalent' rational map
This was the first definition, btw

@TedShifrin Glory hog! ;-p

Let $\phi$ be as in the first definition, and $\Phi$ as in the second one (where we've cleared denominators)
I've already shown that on each point $P$ where $\phi$ is regular, $\phi$ and $\Phi$ coincide
I want to show now that if $\Phi$ is regular at some $P$, then $\phi$ is regular there too, and the values coincide
if I write: $\phi_i=f_i/g_i$ for each $i$
then $\Phi_j=\prod_{i\neq j} g_i f_j$
since $\Phi$ is regular at $P$, we have some poly's $h_i$ s.t. $H=[h_0,\dots,h_n]$ and $h_i\Phi_j=h_j\Phi_i$
assume $h_i(P)\neq 0$
then I would have to multiply $\phi$ by some rational function
so that things work out
hm, I see a mistake I made now... I was multiplying each component $i$ in $\phi$ by $h_i/h_j$, but that's wrong, because then for each component I get a different rational function that I multiply by
hmm, I guess my question is:
would you know what rational function $r$ to choose to multiply $\phi$ with, such that $r\phi(P)$ would be regular (and equal to $\Phi(P)$)

4:57 PM
Remember things only have to be equal up to a (common) nonzero scalar at each point.

Yes

So you can always pull nonvanishing functions in and out of the vector of homogeneous coordinates.
Not that I'm following your details, because I'm too lazy.

Lol, that's fine x')
My problem right now seems to be that I don't know which non-vanishing function to choose. Maybe I could pick $X_k^d/h_j$ for some $k$ s.t. the $k$-th coordinate of $P$ doesn't vanish, and $d$ = degree of $h_j$