12:15 AM
No, @johnny09. Plenty of diagonal matrices can be singular.
@schn: Sometimes you can, sometimes you can't. In your case, it's easy, because you have two sets of the same type of functions. You can see more examples on one of my YouTube videos.

@TedShifrin Okay. Yes, the lines given were ‘kind’. It seems like problems involving changing variables in double integrals requires a lot of testing what will work.

They all have to be made up to be "workable." A random one won't be.

True.

12:38 AM
@TedShifrin Ah okay. Thanks!

1:11 AM
When evaluating a double integral and one of the iterated integrals diverges, the double integral must diverge, correct?

@TedShifrin hello. I had a doubt
For the volume bounded between $x^2+y^2$=$z^2$ and plsnes z=0 and z=1
I thought of applying cylindrical coordinates
So your $/thets$ is (0.2π)
Your z varies from (0, 1)
I an not able to get limits of r .....
Maybe r varies from (0, z)..

draw a picture

I have drawn it
When I do the integration, my result doesn't match so I think my limits are wrong
Wait my z limits are wrong......

2:14 AM
@Aladdin: That's correct. Personally, I'd probably set up the problem in the order $dz\,dr\,d\theta$, with $0\le r\le 1$ and $r\le z\le 1$.

2 hours later…
4:28 AM
Hey.

@CaptainAmerica16 welcome back!

Sup Leaky

what happened when you were gone?

I kind of took a break over the summer and focused on music. I play in a band.

4:51 AM
cool
so are you in uni now?

No, last year of high school. I'm starting the application process tho
I'm going to be taking to some college math courses this school year

where will you take uni?
and are you going to study maths in uni?

One of my top options is PennState, so I'll be able to study CS and Math. The other college I'm considering is a STEM-only college, but they don't have a pure math degree for some reason.

5:10 AM
have you considered studying abroad?

no, not really

what do you think about humanities subject?

I think they are required in universities in USA
and this is something I wish someone would have explicitly told me when I was applying to uni
luckily UK unis don't have compulsory humanities
and I went to a UK uni

Yeah, general education courses. I'll just get through them as quickly as possible, that's why I'm taking some college courses while in high school.
I don't understand the GenEd courses. They seem like a waste of time.

5:15 AM
I'm just saying that studying abroad is an option if you don't want to face GEH courses

Yeah, idk. I literally never considered it

you can consider it

How have you been for the last couple of months?

ppl ihave a question : is the reeb graph of a function supposed to have it's own topology, or it's own subspace topology. I mean for those critical points to be clopen in the height function of a torus that I keep seeing, it shouldn't need an embedding but a half interval containing them

5:31 AM
@CaptainAmerica16 learning exciting maths :P

Niiccee
I'm getting back into doing math every day. My course load is way lighter this year

does anyone skateboard cus its my 20th years now, I like deck transforms and grothendeck etc

I skate a little, i'm not the best at it tho :P

i usually have to break for nearly two years each sprain that is badd I just now got 1 again it's healed up though ;!

woah, i don't think i've ever gotten a severe injury from skating. Just like busted lips and sore ankles

5:37 AM
the neuralgia is pretty bad except can remember a lot of texts that I was reading n I was shuffling cards 1 handed too isn't that odd lollll

I rembered it's a quotient topology

3 hours later…
9:00 AM
"A flat morphism is the algebraic analogue of a map whose fibres form a continuously varying family."

I don't understand how to make sense of 'continuity' here, any ideas?

1 hour later…
10:03 AM

1 hour later…
11:16 AM
@TedShifrin you're right that ||z||=z \bar z, but that equals x^2-y^2 in this case. I'm talking about a type of generalized complex numbers. $z=a+bj$ and $\bar z= a-bj,$ and $z \bar z= a^2- (bj)^2=a^2-b^2.$
In this system, the points that are 1 unit from the origin form a hyperbola
The polar form of such numbers is re^{j \theta}, and e^{jx}= cosh x + j sinh x

I'm looking for recommended reading on Topology but in the particular context of normed vector spaces (no abstract topological spaces or metric spaces, but I could get away with some reasonably concise generic introduction)

If someone could help me find the equation of circle with center at 0 with these conditions, I would be very grateful
If someone don't want to find the equation, but know how to find it, please tell me a method or even an hint. Thanks!

2 hours later…
1:07 PM
Given two quadratic forms [a b; b c] [A B; B C] what is the significance of Ac+aC-4Bb?

1:41 PM
If $x$ and $y$ denote elements in a Lie algebra, and $a$ is a scalar, what is $a[x,y]$? Is it $[ax,ay]$? Is it $[ax,y]$? Is it $[x,ay]$?
I'm trying to determine this from the axioms of a Lie bracket but I can't see it for some reason.
Wait! The axiom states that $[ax+by,z] = a[x,z] + b[y,z]$. If $y=0$, then we get $[ax,z] = a[x,z]$.

The Lie bracket is bilinear, that's all you need here

So $a[x_1,y_1] + b[x_2,y_2] = [ax_1 + bx_2,y_1 +y_2]$
?

2:10 PM
@user193319 Why would you think this is true?

2:29 PM
@GaloisintheField Because I thought it would follow from the axioms

I don't like subsets.

@user193319 So somehow because $[a,c]+[b,c] = [a+b,c]$ (noting that the second term in each bracket is the same), and the same for the other position, you think you can take $[a,b]+[c,d]=[a+c,b+d]$?

No, you're right. It should be $[a+c,b+d] = [a,b] + [a,d] + [c,b] + [c,d]$, right?

Indeed

Hmm....I'm trying to prove that if $I,J$ are Lie ideals, then so is $[I,J]$; but proving that it is a vector subspace is more difficult than I thought.

2:38 PM
What are some good math related jobs?

@user193319 the Lie Bracket is linear in both arguments, so its image $[V,W]$ of two vector subspaces $V,W$ is itself a vector subspace

@s.harp Yeah, I guess that's what I'm having trouble proving, that the image of $V \times W$ under a bilinear map is a vector subspace. Seems like a tricky calculation.

How do I find the surface area of a fork?

2:55 PM
thinking more clearly it might actually be false, you should probably be considering the span of $[I,J]$

Aw damn...That's exactly how my professor defined $[I,J]$...now I see it.
...I think...give me a second to do the calculation

@user193319 You can factor bilinear maps $V\times W\to X$ through a linear map $V\otimes W\to X$
@user193319 I don't think this is true, although I haven't thought of a counterexample

Isn't it trivial: $\alpha \sum [x_i,y_i] + \beta \sum [w_i, z_i ] = \sum [\alpha x_i,y_i] + \sum [\beta w_i,z_i]$.
$\alpha x_i \in I$ and $\beta w_i \in I$, because $I$ is a Lie Ideal, and the RHS is clearly a span of elements of the form $[x,y]$ for $x \in I$ and $y \in J$.

Once you allow spans, yes

Yeah, my professor defined $[I,J]$ as the span of such elements....
Wait...
The span of any set is by definition a subspace...
so I don't even need to prove specifically that $[I,J]$ is a vector subspace.

3:02 PM
Yes, with that definition

D'oh!
So all I need to show is that the second condition for a Lie ideal is satisfied.
And that should involve the Jacobi identity, I think.

@user193319 Yep, just use that $I,J$ are ideals, and the Jacobi identity, and that its a vector subspace
and anti-symmetry once
I.e. $[g,[i,j]] = -[i,[j,g]]-[j,[g,i]]$ by Jacobi, and by antisymmetry $=-[i,[j,g]]+[[g,i],j]$ and $[j,g]\in J$ and $[g,i]\in I$ since they're ideals, so that this is a sum of elements in $[I,J]$.

morning (in california, at least)

gday

3:20 PM
i came here to vent, but all the bile seems to have disappeared :-).

3:37 PM
Is it possible to evaluate the double integral over the triangular domain with vertices at (0,0),(1,0) and (1,1), bounded above by $(x^2+y^2)$?
Using a symmetry argument.

Because the function is symmetric in x and y, this is the same integral as the integral with domain the triangle whose vertices are (0,0), (0,1), (1,1). (this is your domain reflected across the line x = y; you send a point (x,y) to (y,x)).
So it suffices to halve the double integral over the unit square in the first quadrant
But it wouldn't be that bad to just do the integral over that triangle

just use polar coords :P

@MikeMiller Okay. Could one also evaluate the double integral as 4 times the integral over the triangle with vertices at (0,0),(1/2,1/2) and (1/2,0)?

@schn: Use a change of variables $(a,b)={1\over 2}(x,y)$. But what will that buy you?

Just trying to grasp the symmetry argument. Somehow when splitting the triangle into four pieces the calculations give a different result.

3:51 PM
of course it does, because the function has larger values further out

I'm not sure I see the four pieces in regards to symmetry. I only see two?
but i don't see how that symmetry makes things any simpler.

Okay. So in this case one could only half the triangle to still get the same result?
In general, isn’t a symmetric function and a symmetric domain enough of a condition to split the integral? Where is the symmetry lacking here?

You can split the $0,e_1,e_2$ triangle into two parts, but it doesn't simplify your life...

So the function probably needs to be constant over the domain as well.

@schn: not sure what you're getting at...

4:09 PM
@copper.hat For example the double integral over the rectangle $0\leq x\leq 1$ and $0\leq y\leq 1$, bounded above by $1/(x+y)$ can be evaluated by splitting the rectangle into a triangle...despite the fact that the function isn’t constant over the domain. The function is symmetric and so is the domain. Why wouldn’t it work splitting the triangle given above into four pieces?
How needs the symmetry between the function and the domain correlate?

Well, loosely, you need to split a domain $D$ into something like $D_1,...,D_m$ and have some maps $L_2,...,L_m$ such that their Jacobian is one, $D_k=L_k(D_1)$ and $f = f \circ L_k$.
I could quite easily have a major mistake above, but you get the idea.

4:36 PM
@schn no

5:32 PM
Hi @Sayan
Did you figure out that closed uncountable subset of the irrationals thing?

@AlessandroCodenotti Yeah, though I didn't get how to do it using cantor set, I did it by enumerating rationals since they are countable and forming a closed set by removing them. It turns out that my idea matched with the notion of rationals being a set of measure 0
And, Hi@AlessandroCodenotti

Well the construction I had in mind is similar to that of the Cantor, you enumerate the intervals and at each stage you remove a small open interval around one of them

Ah yeah

Here's a way to do it which I find is neater but less intuitive. Let $C$ be the standard Cantor set in $[0,1]$ and let $C+r=\{c+r\mid c\in C\}$ for $r\in \Bbb R$
Consider $K=\bigcup_{q\in\Bbb Q}(C+q)$. I claim that $K\neq\Bbb R$ and that if $x\in\Bbb R\setminus K$ then $C+x$ is a closed uncountable subset of the irrationals
The second claim is easy, while to show $K\neq\Bbb R$ you can either use the Baire category theorem or the fact that the Cantor set has measure $0$, you can think about the details

The $K \neq \Bbb{R}$ is just that since $C$ is of measure $0$, $C + r$ for some fixed $r$ is also of measure  as you're just translating the intervals. And then just union of measure 0 sets.
Intuitively for each rational you're considering a cantor set type neighborhood around it and since they are of measure $0$ whatever is outside has to be uncountable. Neat.
Is it right to think about it like that @AlessandroCodenotti

5:53 PM
Yes, it's important that this is a countable union of measure 0 sets

Hi, demonic @Alessandro, @Sayan

Hi @Ted

Hey@Ted
@AlessandroCodenotti Yes, ofcourse

6:17 PM
anyone know how to solve this
0

How can I find all involutions whose reciprocals are non trivial involutions? $f(x)=\frac{1}{x}$ does not satisfy the constraints because it's reciprocal is a trivial involution. Edit: as pointed out in the comments $f(x)= -\frac{1}{x}$ is one of the solutions.

6:35 PM
Randomness in a nutshell: Stuff that cannot be predicted one by one but can be predicted enmass (except for true randomness, which cannot be a mathematical object according to Ramsay Theory)

how does quantum uncertainty fit in with this?

It can be something between random and truly random, because it is clear we can still predict the probability of the outcomes

and it's Ramsey theory

Quantum randomness is actually not very unpredictable because the equation that governs its probability is actually deterministic
The schrodinger equation evolves the wavefunction deterministically

what if you're epsilon away from true randomness? Do you think mathematics can still be applied?
is there a geometric/topological flow that continously deforms a mathematical object into a more and more random object

7:43 PM
Schramm-Loewner Evolution

8:13 PM
Consider the intersection of three cylinders with equal radius each, and axes along the three coordinate axis in 3-space, respectively. In this question (math.stackexchange.com/q/2372904), finding the volume is solved by a symmetry argument involving finding the volume in an octant of the solid. What other symmetries are there to find this volume?

8:53 PM
@schn: That is in fact one of my favorite problems, but there is in fact further symmetry which makes the problem even easier. Consider symmetry about the planes $x=y$, $y=z$, etc. If you're curious, you can see a solution with no calculus whatsoever (the way Archimedes would have done it) in my MAA lectures posted on my webpage (linked in my profile).

I'm not exactly sure what the question is asking me to do. Could someone lend me a hand, please?

They're asking you to shorten the list to have every fraction represented once and once only.

What I have tried doing is putting all of these fractions under a common denominator of 10, and try finding the reminder. Is that a valid way to approach it?

What do you mean by finding the remainder? I don't think you have to be so fancy.
Start your list with the first fraction. Is the second fraction equivalent? If so, forget it. If not, put it on your list. Go on ...

@TedShifrin Sorry, I have recently started studying about equivalence of relations, so I wanted some sort of practice.

8:59 PM
Yes, I'm not quibbling about that. I suppose you can do it your way, but I still don't see what you mean about finding the reminder (remainder)?
If you don't have enough exercises, I can send you the exercises on equivalence relations I put in the appendix of my algebra book.

@TedShifrin Sorry, I had the wrong thought in mind. I overcomplicated the question. Sure, that would be great!

If you want to send me an email (at my email address in my profile), I'll be glad to send a few problems to you.

Will do, thanks so much!
So, regarding this question. Since a transversal is a set of elements with no more than two equivalent elements in that list, would that cause the number of elements of the largest transversal subset of that list to always be one?

No, not "no more than two." You want precisely one of each equivalence class.

I'm sorry, I meant that this set can not contain elements that are equivalent to each other at all. Is that the case here?

9:09 PM
Right. Each element of your list is equivalent only to itself.

So would that make the number of elements of the largest transversal subset of that list to only equal one since no two elements are equivalent to each other and can be in the same list by the definition of a transversal set?

No, no, you'll have a long list.
Try what I suggested. Put the first thing on your list. Put the second one on your list if it is not equivalent to the first. Look at the third. Is it equivalent to either of the first two?

I understand. So the list should contain only elements that are not equivalent to each other, which basically means that as long as the given fractions are not in the same equivalence class, they can go into the transversal list?

Right — you add it to the list provided it's not equivalent to anything on the list.

Okay, I understand. Out of curiosity, though, is there a different, perhaps more efficient way of determining this list? What would that list contain if we were given, for instance, over a thousand elements? Would we need to go over each and every element in that list, and determine whether it is equivalent to any other element that is already present in the list?

9:20 PM
I'm no computer science expert, but offhand that seems the most obvious way. Or you could sort by size, maybe.

I see. So regarding this exercise - when explaining how we know our answer is correct (after determining the transversal list) are we expected to justify it by explaining how we got that list? (By comparing each element to the other)

Seems reasonable to me. I haven't thought about other approaches.

@TedShifrin Cool. It’s an interesting problem...the cylinder intersection one.

Oh, it's fabulous. I've given it as a challenge problem every time I've taught multivariable calc. (And it's in my book. :P)
You can see neat pictures plus Greek mathematics if you look at that link.

@TedShifrin Okay, thank you! Would that be fine if I will try to solve this exercise, and share my results here? I am just not quite confident with this material yet, and would highly appreciate some feedback, if possible of course.

9:26 PM
@Harry: I have the two pages for you with more exercises. If you don't want to email me, I can put the .pdfs on here, but they'll be hard to read.
Sure, you can post your solution as a question on the main site, too, with a request for critique.

@TedShifrin Awesome, thanks again. I will send you an email right away.

Sure thing.

Consider the lines $y=0,y=1-x,x=0,x=1$. They bound a triangle with vertices (0,1),(0,0) and (1,0) in the xy-plane. With the change of variables $u=y-x, v=y+x$, the triangle gets mapped to the triangle in the uv-plane with vertices (-1,1),(1,1) and (0,0). The change of variables imply $0\leq v\leq 1$, but how does one find the range of $u$? Why isn't the triangle mapped to the triangle with vertices (-1,1),(0,1) and (0,0)?

@HarryBattersby: You could also reduce each element so that the numerator & denominator are coprime and the denominator is positive. This is an equivalent and unique representation for each, so you can count and compute equivalence classes easily.

Look at the triangle in the $uv$-plane, @schn, and set up the double integral by your usual algorithms. (For each fixed $v$, what does $u$ do?)

9:41 PM
@schn: The mapping is linear, so the range is the second set of vertices you listed?

@copper.hat I didn't know this was possible. That's interesting, thanks!

@Harry: Yeah, I should have suggested that, but somehow I was hoping you'd bump into it yourself :P

i mean the convex hull of the second set of vertices.
Note that any element of the domain can be written as $t_1 (0,1)+t_2 (0,0) + t_3(1,0)$ with $t_1+t_2+t_3 = 1$. Then the corresponding element after mapping is $t_1 (-1,1)+t_2 (0,0) + t_3(1,1)$ (with $t_1+t_2+t_3 = 1$, of course).

@TedShifrin Perhaps I would have after some time :P (By the way, I sent you an email).

@TedShifrin @copper.hat Okay. By looking at the given lines $y=0,y=1-x,x=0,x=1$, $u$ can take on the values $0$ and $-1$, but how does one find it also ranges to $1$? And how can one conclude from there that the triangle is $\textbf{not}$ the triangle with vertices (-1,1), (0,1) and (0,0)?

9:52 PM
@schn: I thought you told me the vertices of the triangle in the $uv$-plane.

Yes, they are (-1,1),(1,1) and (0,0), but couldn't they also be (-1,1), (0,1) and (0,0)?

The vertices in the u-v plane must be the mapping of the vertices in the x-y plane.
where are you getting the other vertices from?
It is a linear map. It comprises a scaling by $\sqrt{2}$ and a rotation by 45deg.

It's a one-to-one linear map.

think of it as multiplying by 1+i in the complex plane.

Okay, overthinking it maybe. So given $u=y-x, v=y+x$, one could simply plug in the vertices in xy-plane and determine thereby both the range of $u$ and $v$, as well as the new vertices in the uv-plane, or?

9:56 PM
well, plug in the source vertices, this gives you the u-v vertices. this gives you the range of the projections onto the u/v axes.

Yes

Sorry to interrupt guys, but is 0/1 formally equivalent (in terms of equivalence of relations) to 0/2 since the resultant is 0?

if you have vertices v_1,...,v_n then any point lying on or inside the vertices (convex hull) can be written as $\sum_k t_k v_k$ where $t_k \ge 0$ and $\sum_k t_k = 1$.

Sure.
@Harry: This is the usual stuff you've done with fractions ever since elementary school. Don't overcomplicate it.

@HarryBattersby: yes, 0 is a nuisance in terms of equivalence classes.
reduce 0 to 0/1, this is the only special case.

9:59 PM
If you're going to reduce, you also want a convention that the denominator is always a positive integer.

@copper.hat Okay. However, given the equations alone, i.e. the lines, would there be a way to tell the range of the projections? Given $u=y-x$ (and the lines above), how would one find $u$ can also equal $1$?

maybe the smallest positive integer that works. then it is unique.

I understand. I am indeed thinking way too much about it :P
So, -10/-5 could also be reduced to 2 since -10/-5 can be rewritten as 10/5, right?

@schn: i don't think it is straightforward, you need someway of knowing when to stop as it were. it is easier to work in terms of vertices.

Officially, 2/1.
2

10:02 PM
divide above & below by gcd(n,d)
which by convention is non negative.

Alright, cool. I really have to stop overthinking it :P

@copper.hat: Nah, I don't like that. I want a positive denominator in the unique representative.
@Harry: You should have an email.

@TedShifrin: sorry, that is what i meant to end up with

@copper.hat: But you don't with $-a/-b$ or $a/-b$. :P

i want the denominator to be positive and coprime

10:05 PM
@TedShifrin I am new to this platform, so I haven't really gone through my profile. I will do that.

No, no, I didn't mean that. I meant that I sent you the .pdf .... @Harry

ok, i should go to trader joes and do my weekly shopping. my son just took his psats.

@TedShifrin Haha my bad, I didn't understand it quite right. Yes, I have got the exercies, I am just finalizing the previous exercise, and then I will start working on them.

my daughter started college last monday.

@copper.hat@TedShifrin Thanks for the help.

10:07 PM
glad to be able to help

@schn: As before, I will recommend you watch a video or two of mine on this. :P

For the previous exercise that I have shared here, for part 1, I shortened the final list to the following list, {5/2, 2/5, 1/1, -2/5}. Does this seem reasonable?

@schn: what are you working on?

Calculus; A complete course.
Double integrals.

You're missing $0$, @Harry.

10:10 PM
But wouldn't 0/1 and 0/2 be equivalent to each other, thus they wouldn't be included in the final list?

One of them has to be!

i get 7 unique elements

I got a list with 6 numbers, so you're missing another.
Hmm, 7 ?

1/2 2/1 0/1 5/2 2/5 1/1 -2/5

Where did -1/5 come from?

10:11 PM
did i goof?
fixed

Yeah, you're right. I left one out.
I was doing it all mentally :P

i am incapable of working without paper or whiteboard

I try to ward off Alzheimers by using my brain as much as I can :P
Apparently I'm not being too successful :D

i have accepted some inevitable slowdown :-(.
but have much more difficultly accepting the physical slowdown :-)

Well, let's not discuss decrepitude. My vertebrae/disks in my neck are degenerating and the pain is no fun.

10:16 PM
sorry to hear it.

But, as my friends remind me, there are zillions of people way worse off than I am.

osteo is in my genes apparently. some n euro thing
as north euro

@TedShifrin I have got the following list {1/2, 2/1, 5/2, 2/5, 1/1, -2/5, 0/1} (7 elements in total) I have one extra than the list that you have. I am not exactly sure where I'm wrong, am I missing some key information?

Interestingly, my chiropractor told me osteoarthritis is not a genetic thing.
@Harry: No, @copper.hat and I already decided I had been wrong.

i will look it up and see if i am misremembering
something to do with iron in your blood

10:19 PM
Hmm ...

survival of the sickest.
sorry, i was wrong. related to vit. d, so environmental

sharon moalem

Just so long as you're right about mathematics ;P

rarely :-)

10:21 PM
@TedShifrin Great, thanks.

Sure thing.

my main strength in math, etc. has been a strong resilience to intellectual intimidation :-)
having has some fearsome advisors helped...
and some very smart & patient friends

10:38 PM
we all get by with a little help from our friends :-)

10:59 PM
just discovered my trump number is 3

-_-

gotta update my bio :-)

lol
not if he gets impeached

even if he is impeached, my number will remain the same :-)

true dat, but perhaps it will be interpreted as -3
or even -2.999...
:-)

11:09 PM
:-). i was just surprised to find out that a close friend has a Trump number of 2. i do have an obama number of 2, however :-).

coolio