geocalc, possibly? it's certainly the union of four quadrants, assuming the quadrants are defined to include all of the axes and the origin appropriately. whether this corresponds to a 'gluing' of components might depend on what you have in mind by 'gluing' and what objects can serve as inputs to that. note e.g. that $\{(x,y): x > 0, y > 0\}$ is a manifold but $\{(x,y): x \geq 0, y \geq 0\}$ is not.
this is not an insurmountable thing, but does at least raise abstract technical questions about how you'd generally put a manifold structure on a 'glued' result of pieces that might not themselves be manifolds where you want to 'glue' them.
Good points. Well what if you take the four manifolds $\{(x,y): x > 0, y > 0\}$ and $(-x,y)$ and $(-x,-y)$ and $(x,-y)$ and glue $(x,y)$ to $(-x,y)$ at their common boundary $(0,y)$ continuing like this for the other ones?
well, that's the issue. as you've defined them (the open quadrants? looks like) they don't share any points that you can glue together. they do sit next to each other in R^2 and there's something in the ambient R^2 that fills those gaps. which is extra structure you wouldn't have with four random manifolds that you want to glue together. so i don't know if it's helpful to think of it as 'gluing,' if it's just 'unioning stuff together inside a larger manifold that we already know is there.'
@leslietownes it is a manifold with boundary, however
and gluing manifolds with boundaries along pieces of their boundaries is a very well-understood construction, both in the topological and the smooth category
or, more precisely, topologically it's a manifold with boundary
xander that marker technique is so good it's almost distracting. it might be too good. i guess students get used to it.
i do like using color or boldface, in moderation, to draw focus to particular things, particularly pieces of things that change in a calculation. i always thought of this as something you could only do in typeset documents. with this marker trick you could actually do it at the board.
When I gave guest lectures and had to use markers, they without exception caused me great distress. I stuck with blackboards and plenty of colored chalk.
euclid is surprisingly clear, by modern standards, at least in the dover translation with footnotes and context, although it's still far from ideal as a modern math book.
In the Squeeze Theorem proof of $\lim\limits_{\theta\to 0}\frac{sin{\theta}}{\theta} $, the appeal is made to the obvious triangle inclusion in the image on the unit circle below to say that
$$\frac{\sin{\theta}}{2} \leq \frac{\theta}{2} \leq \frac{\tan{\theta}}{2} $$
While it is indeed obvious t...
i met someone who said they were reading newton's principia once, and my only question was 'why?' i've skimmed it and the impression i get is that you'd need either to be newton, or to already understand both calculus and physics, to get much out of it.
@leslietownes This guy named Thomas Heath heavily annotated Euclid with some of his thoughts regarding the books. I've been slowly making my way through the first one.
well, not meaningless, but usually not defined at any higher level of 'rigor' (whatever that is) than the area argument. there's no starting point for what a real number is, for heaven's sake.
i see ted already made this point or something equivalent in connection with pi and 2pi.
The way I have understood, $0/0$ is undefined or indeterminate because, if $c=0/0$ then $c\cdot 0=0$, where $c$ can be any finite number including $0$ itself.
If we also observe a fraction $F=a/b$ where $a,b$ are positive real numbers, the value of $F$ increases with the decrements of $b$. Bein...
a lot of books introduce trig functions and need that limit well before any mean value theorem. you need lim sin(t)/t just to see that sin is differentiable, for example.
under: another thing in this area, similar vibe, the calc book will assume that we all understand what 2^x is for real x, but we kinda don't, really? most people go to a calculator for 2^(1/2) and at least that's something whose square is 2. something like 2^pi doesn't have an interpretation as an nth root of an integer, what the heck is it.
these aren't at the level of mysteries of the cosmos, but they're very much not the focus of attention. so if you dig into it it's easy for people to get into arguments about what the words mean, or what they're assuming vs. proving.
i think it's more than enough to know that this stuff can be made rigorous, in any number of different ways, and so it kinda doesn't matter how. i feel the same about trig functions and trig limits. i like the power series definitions but they are definitely not what i would start with.
it's just not a very well-founded question. if someone pops up and says, here are my axioms, here's my definition of sin(t), is this is a proof from those axioms and definition that some limit exists, OK, you can answer that yes or no. and sometimes the answer would be no for something involving triangles.
but there's such a longstanding tradition of not working at that level of granularity when these limits are first introduced, it seems silly to fault people for not playing a logic game that they didn't even know they were playing, or not following rules that they didn't have.
it's like the galaxy brain meme. normal brain is whatever you think a real number is in 9th grade. the glowy brain is maybe a real number being something in a complete ordered field. the slightly more glowy brain is crap about set theory axioms and dedekind cuts or cauchy sequences. then the galaxy brain is a real number is whatever you thought it was in 9th grade.
some folks use 'hilbert space' to explicitly mean infinite dimensional (+ complex scalars), or even infinite dimensional + countable orthonormal basis (+ complex scalars).
real infinite dimensional hilbert spaces get a big thumbs down from me.
@leslietownes hi. I didn't understand. I think since $x\mapsto e^{a\sin x}\cos^{2}x$ is well defined and continuous on $[-\pi/2,\pi/2]$ so the integral convergent.
I was trying with integration by parts but it doesn't work.
It is convergent for sure but having a closed form is not guaranteed. So it may not have a closed form. That's why I ask if you have a reason to believe that the closed form exists.
In fact if you compute $(a\partial_a^2+3\partial_a-a)e^{a\sin x}\cos^2(x)$ you'll find the $x$-derivative of $-a^2\cos^3(x)e^{a\sin(x)}$ @Alex
The differential operator is the annihilating operator of $I_1(a)/a$, but you can derive it directly from the integrand, assuming you have the time/a computer to do it for you
Then you want to compute it and its derivative for specific values of a and compare that to a basis of solutions of $a\partial_a^2+3\partial_a-a$ (ie $I_1(a)/a$ and $K_1(a)/a$ if I'm not mistaken)
you can use the Beta Integral $$ \begin{align} \int_{-1}^1x^{2n}\sqrt{1-x^2}\,\mathrm{d}x &=\int_0^1x^{n-1/2}\sqrt{1-x}\,\mathrm{d}x\\ &=\frac{\Gamma\!\left(n+\frac12\right)\Gamma\!\left(\frac32\right)}{\Gamma(n+2)}\\ &=\frac\pi2\frac{(2n)!}{2^n2^nn!(n+1)!} \end{align} $$
along with the series for $e^{ax}$ to get $$ \begin{align} \int_{-\pi/2}^{\pi/2}e^{a\sin(x)}\cos^2(x)\,\mathrm{d}x &=\int_{-\pi/2}^{\pi/2}e^{a\sin(x)}\cos(x)\,\mathrm{d}\sin(x)\\ &=\int_{-1}^1e^{au}\sqrt{1-u^2}\,\mathrm{d}u\\ &=\frac\pi2\sum_{k=0}^\infty\frac{a^{2n}}{2^{2n}n!(n+1)!} \end{align} $$
Oh, son of a b---! I just figured out why I couldn't get the right formula for the determinant of a 2x2 matrix in lecture yesterday. Arg... I am so f'kin' dyslexic. In the last step of the Gauss-Jordan elimination, I flipped a fraction upside down. ARG!
But why did you flip the fraction? Did you think the fraction $\frac pq$ to be such that p is in $i-$th row and $q$ to be in $i+1$th row of the same column? And all this was an accident because the bar between p and q was overlooked.
Ahh. One of my teachers would do such things to test if students are paying attention to him. I remember once he drew a container C1 with liquid in it with volume V and temperature T and then he transferred (in the picture) half the liquid into another container C2 and then wrote: So temperature and volume of liquid in C2 are T/2 and V/2 respectively.
One of my classmates nodded yes to that :P
Transferring a liquid to another container shouldn't change temperature under usual situations so he was declared as not paying attention in the class.
I mean, I am the kind of asshole who would do something like that, but I also have moderate dyscalculia, so I make those kinds of mistakes all by myself with great frequency.
Which is why I work out my notes ahead of time with great care.
But I lecture in a more improvisational style, so generally ignore my notes. And then I get into trouble.
@Koro Yes and no. I work at a place which serves a large, geographically disparate area. In the before-fore times, most classes were taught remotely (e.g. an instructor would be in a classroom on one campus, and there would be students in three or four other classrooms on other campuses).
So "offline" has not really been a thing here in 15+ years.
And before that, we had classes taught over the radio (the college owns four or five radio transmission towers).
@XanderHenderson over radio 😯. I never heard about that :)
One of my teachers makes it very clear before starting his lecture (online) that if he asks a question -everyone has to respond even if the response is wrong. If someone doesn't respond he'll make them exit class. So everyone responds in his class.
@Koro Yeah, the problem that I have with that is there there is a lot of mixed modality going on. I have some students in WebEx (who can respond in chat), but other students in classrooms, who have difficulty responding individually (they are generally too small for me to distinguish on the screen).
In the spring, I am thinking about requiring those students to come to class with cleverphones, then setting up something like a discord server for that kind of instant feedback.
Hi math lovers. I have a small question about quotient vector space $V/W$. I know the defs and thms. But when I look to it closely I don't understand some things. Suppose $v_1,v_2\in V$ then by definition $v_1\sim v_2\iff v_1-v_2\in W$. I want to know is $v\sim 2v$? seemingly the answer is no, but it is weird to me that why these two very similar vectors are not in same class and produce different co-sets.
CFG: if you haven't seen the inner product interpretation it might be a little bit of a detour, but even without the formalism, the idea is that in V/W, everything that's in W becomes equivalent to 0, and things that are 'perpendicular' to W (think non-rigorously if necessary) do not become identified with other things perpendicular to W
CFG: for maybe a concrete example, try V = R^2 and W = {(0,y): y in R}
@leslietownes I know that and that example is easy to handle. my question is why $v$ and $2v$ are in different classes. I am asking this because of I have some issue in De Rham cohomology. i.e. why a diff form $\omega$ and $2\omega$ may fall into two distinct cohomology classes? isn't this produce many classes and make that group a infinite dim?
CFG: in general, v and 2v are in the same class in V/W if only if v is in W. otherwise, they are different. but the fact that they might represent different vectors doesn't change the fact that they're scalar multiples of each other
R contains infinitely many different numbers and is still a 1-dimensional vector space over R
@C.F.G Two answers. One is that cohomology classes are characterized by integration over oriented compact submanifolds of the correct dimension. If $\int_Z \omega = \int_Z 2\omega$ for every $Z$, then $\int_Z \omega = 0$ for every $Z$. Do you know what this says? Alternatively, if $\omega\sim 2\omega$, then $2\omega-\omega = \omega = d\eta$ for some form $\eta$, which says that $\omega$ had to be exact to start with, i.e., represents the zero class.
@TedShifrin I am good in high math but so bad in basics. !! How is that possible I don't know! but this building will be destroyed as it has weak basic. isn't?
that's a good analogy. a prof i had once said something like, if there is a crack in the foundation, eventually something will fall right through it. and maybe not eventually but sooner rather than later.
Well, you can go back and understand quotient groups, quotient rings, quotient vector spaces. It's all the same notion — groups being the hardest because of non-commutativity. But if you really understand differential forms and duality, then my comments are pretty direct.
Yeah, my personal belief is that one cannot actually understand the "advanced" stuff without having mastered the basics.
i haven't heard otherwise, but i haven't heard from him in a while. i invited him to my wedding a few years ago and he said that he no longer travels in the summer (he used to summer in the LA area). which might have just been a nice way of saying no thanks. i hate weddings.
Yes, Wu-Yi is Wu-Yi. Different person for sure. Here's your problem, @C.F.G. You wrote $\langle[\omega]\rangle = \Bbb R$. You should be talking about $[\omega]$ and $[2\omega]$ as elements of $\Bbb R$.
Wu-Chung used to be at Princeton. I assume long since dead.
Anyhow, get back to the basics here. Both $\omega$ and $2\omega$ generate the same one-dimensional vector subspace. That does not make them the same class.
No obviously. we are talking about equivalent not equal. in this case [(1,1)] and [(2,2)] are in one class. but our [v] and [2v] are not in one class because $v-2v\notin V$
The word class is here because we're looking at equivalence classes. You're talking about the classes AND about the subspace generated by the classes. Think about our concrete example.
CFG: in any vector space, the span of a vector is a set of vectors. in a quotient space V/W, an equivalence class of an element v in V is a single vector in V/W.
the span of an element in a quotient space is generally a set of equivalence classes, not a single one. ted's example, where equivalence classes are lines parallel to something, is a good one. the span of one of those equivalence classes is a set of lines, not one line.
@TedShifrin Something with enough interesting features to make the problem interesting, but without having to get into a lot of really gross and tedious computation.
@TedShifrin Yeah, I don't want to give them a rational function, because they are going to mess up the algebra, but I feel like giving them the derivatives is too much of a hint.
Well, I got to the point of giving the derivative and second derivative so that I could test them on the analysis and synthesis. That is the point, after all.
@C.F.G i find it best to have a concrete model of the equivalence space. thinking of the span of equivalence classes is a bit much for my simple mind. in the above case you can take a representative point where the equivalence class intersects the $x$ axis.
