Yes, but they all treat the point that "the $g_i$ should agree on $f(Z)$ when defined" as self-evident and focus on more difficult issues. But I don't think that claim which everybody takes as self-evident is true
i'd love to read an essay about what it is about children and serving tea. she really, really loves this. i don't think she's absorbed this from some external source.
the exercise in Hirsch feels like it's wrong to me, though, but I can't prove my counter-example
verbatim, the exercise states "A C1 immersion f:M ~ N which is injective on a closed subset K c M is injective on a neighborhood of K." This is well-known for $K$ compact. If we instead assume that $f\vert_K$ is a topological embedding, I also know this to be true (under even weaker hypothesis, from which I can prove an inverse function theorem even more general than what I've seen any textbook state). But just injectivity feels insufficient. Consider the projection $\mathbb{R}^2\rightarrow T^2$. This is injective on a line with irrational slope (which is closed), but I don't think it shoul…
i don't know about the fonction de reparittion. if that's the cumulative distribution function it would just be 1 - F_X(a) give or take some static about the endpoints
@leslietownes hmm, that's interesting. text before that exercise section mentions limits. i'd probably also express that second term with cdf, for the sake of it.
it's also different from the claim that Hirsch would actually need to construct tubular neighborhoods for non-closed submanifolds (and he makes no such restriction in the statement of the tubular neighborhood theorem)
but this is not the first time I find one part of this book claiming it follows from another part of the book, just for chasing through the book to reveal that what's actually proven doesn't have sufficient generality
A Generalized function theorem should be a statement of the following sort: If $f\colon M\rightarrow N$ is a smooth map, $X\subseteq M$, $df\vert_x$ is an isomorphism for each $x\in X$, $f\vert_X$ is a topological embedding and <additional hypotheses>, then $f$ maps an open neighborhood of $X$ diffeomorphically onto an open neighborhood of $f(X)$.
The G&P exercise has the additional assumption be that $X$ is a smooth submanifold.
The Hirsch exercise has the additional assumption be that $X$ is closed.
non-closed submanifolds ought to have tubular neighborhoods too
they're just not $\varepsilon$-neighborhoods
closed submanifolds are better, of course
I realized only today that it's actually a subtle matter to choose a tubular neighborhood of a closed submanifold such that its disc bundle is closed in the ambient manifold
Here's my site (very rough): https://abstract-spacecraft.herokuapp.com/ Note that if you create diagrams right now they could be deleted any time because I'm gonna change the DB model, and the site isn't even featured out enough yet to backup the database.
Also I'm running in DEBUG=True mode because DEBUG=False (production) is causing Server 500 error
But yeah, you can create an account and check out what I've done with Quiver editor
Also your accounts will probably also get deleted :), but you can easily recreate them later
Suppose that $y_1,y_2$ are linearly independent on interval I and are also solutions of the ODE: $y’’+p(x)y’+q(x)=0$. If $a$ and $b$ are two consecutive zeroes (suppose that a<b) of $y_1$ in $I$ then does there exist a zero of $y_2$ in $(a,b)$?
It is given that p and q are continuous on I.
We can say for sure that the the function $y_1y_2’-y_1’y_2$ never vanishes on $I$.
draw a picture and it will be clear. since $y'(a)>0$ there is some $a'>a$ (and $a'<b$) such that $y(a')>y(a)=0$ and similarly there is some $b'<b$ (and $b'>a'$) such that $y(b')<y(b) = 0$, now use the intermediate value theorem.
@Thorgott Okay is there a reference for that properness and nets condition you mentioned? The thing is that if that limit condition holds then I can show that $f$ actually attains a minimum, which is what I require to show the Kempf-Ness theorem(equivalence of symplectic and algebraic quotients)
Okay sanity check: The function $f = -2x + \log(\cosh (x))$ is strictly convex ($f'' = (sech(x))^2$) and is proper (for |x|>>1, it is -2x + |x| - log(2)) but it still does not have a critical point. This works right?
But then as Thorgott claims that if f is proper, it implies $\lim_{|x| \to \infty} f(x) = \infty$ happens then $f$ cannot be proper as it will go to $-\infty$
But this is weird still. I want to show that if $f$ is proper and strictly convex on geodesics it attains a critical point. So reducing to $\Bbb{R}$ essentially proper and strictly convex should imply there is a critical point. But that $f$ is a counterexample
Could somebody give some short hint on how to manipulate with the series I obtained for the generating functions here in order to prove they are equal: math.stackexchange.com/q/4353557/823281
Okay so here is what is happening: Let $X \subset \Bbb{P}^n$ be a projective submanifold and let $G$ be a reductive algebraic group action on $X$ and consider a representation of $G$ on $GL(n+1,\Bbb{C})$. Let $K \subset G$ be its maximal compact subgroup. A point $p \in X$ is said to be polystable under $G$ (Mumford) if $G \cdot \tilde{p}$ is closed where $G$ acts on a lift of $p$ in $\Bbb{C}^{n+1}$ (this is called a linearization of the action to a line bundle (in this case O(1)). Now the Kempf-Ness theorem states the following:
The first and second implications are fine, the only issue is the last one, for which the idea I was asked to think about was convexity on lines + properness of a smooth function and it's relation to critical points
To note: There is a question on site with same diagram but the questions I ask regarding this diagram , are different.
Find $ S1$ in the parallelogram below
The Question:
Given:
$S_1 = 10\ \mathrm{ m^2}$ , $S_2=3\ \mathrm{ m^2}$. Find $S_1$.
$S$ in the diagram is the shaded region. Answer is $...
Suppose a polynomial $P$ of degree $n$ has $n$ distinct real roots then $P'$ (the derivative of $P$) has $n-1$ distinct real roots.
Proof by Induction:
Base case: For $n=1$, $P_1 (x)=a_0+a_1x, a_1\neq 0$ has $1$ real (distinct) root, $x=\frac{-a_0}{a_1}$. Then $P_1 ' (x)=a_1$ has $1-1=0 $ ...
@LalitTolani An important fact is that if the polynomial has degree $n$ and it has $n$ distinct roots, all of its roots are simple (no repeated roots).
Can we use the theorem of schwarz in iterative manner ? Say as long as the requirements apply, we can exchange partial derivations even higher than rank "2"
@LalitTolani If it is truly duplicate, the system will pick it up and flag it. If it is essentially a duplicate, you can leave a comment saying so. However, if it is not a true duplicate, then perhaps the latter answer has better exposition or is easier to understand. If the user has a habit of plagiarism, then you can flag for moderator attention.
i have never seen plagiarism but i do think it is fairly common for people to post answers without checking if someone else has posted something similar (particularly if one answer goes up while someone is composing another). there is usually something to recommend both answers, some difference in exposition. even if it looks like noise, it is probably an OK kind of noise.
i guess some of this behavior could be meta-gamed by tweaking the rank in which the platform displays answers. i assume brighter minds than mine have thought about it. and of course one cannot design away that kind of problem.
i tend not to answer at all if i have a feeling that someone else is also likely to answer right away. that is usually also a sign that a question might be a duplicate.
i like a question to sit unanswered for a few days. of course, then there's maybe less of a chance that the asker ever reads the answer.
@MadSpaces Again, with all due respect to the effort you put into the question, it is, at the end of the day, not a good question for the SE format. Numerous users have explained the problems.
@MadSpaces If no one has answered, you can edit the question to make it better. It would be best if the question remains the same so that if anyone is in the process of writing an answer, they don't have to scrap all their work.
@TedShifrin Hello Prof.Ted. Are you referring to that of Schwarz? In our lecture, we have showed it, or say more accurately, stated this theorem for functions of order $k=2$. But it is only logical to be iterative if the conditions apply, as Prof.Rob pointed out to me.
@MadSpaces: The authors of MathJax disapprove of using MathJax to add color to text. That is why some of your text extends beyond the standard column bounds.
@TedShifrin Thanks for the motivation Prof! Alot of my fellow students look down upon me for being very rigorous and say i make myself extra work, which is true. It is not easy, but i am not doing it because i am sadistic. I do it because i actually like what i am doing! it is very nice to hear words of encourgement :)
@MadSpaces: There was a time when we used to use MathJax to center images on the page, but the MathJax authors raised such a fuss, that we just leave images left-aligned these days.
@robjohn I believe studying in General, any topic for that matter, in a good manner, requires that Prof. And as a note, i am not actually a Mathematics student. I am a physics student. But i will eventually get a Bachelor degree in mathematics after i finish my Physics degree. I am intentionally picking out all the mathematic courses that i can pick so i can transfer points later when applying to math department.
I believe so far, i am well more known in the mathematics department "and for that matter liked" in my university than in my physics department... I have constantly expressed criticisim of the rigor of the physics lectures. And Sadly, flunk Lab four times. Not because i am dumb, but i was not interested and writing pages of already known theoretical results and why my ice did not achieve its theoretical temperature in the laboratory.
Suppose that we have a parametrization of a surface $r$ (a map from R^2 to R^3, say) and if $r_u$ and $r_v$ are partial derivatives then the area of the surface is $\int||r_u\times r_v|| du dv$ if $r_u\times r_v$ never vanishes on the surface. :(
Oh. Yea. If the crossproduct is zero, it means the partial derivatives are parallel to each other. Geometrically, that would look as leslie said, crinckly ( i think atleast)
But then who cares about such a rectangle. Why is that rectangle important? I mean since the cross product is zero there anyways, the goofy-ness is taken care of by multiplication by zero.
i'm imagining something where half of a square is mapped to some surface, and the other half of the square is mapped to the same surface. there's gonna be some singularity in there, maybe reflected in r_u x r_v being zero, or not defined, or something. the thing you get out of the integral won't relate to the surface. it will be an artifact of parametrization.
you can still evaluate the integral. i don't think it's a question of making sense of the integral, which is kind of a separate issue. it's a question of what the integral represents.
i should teach math again just so a classroom of people who are very unsure of the drawing abilities of their instructor can hear me say "this classroom is a NO GOOFY CRINKLES ZONE."
our day care has helpfully given us 30 minutes of 'remote learning', a concept with no obvious relation to day care. my daughter is dancing and singing in front of my laptop.
i should film this and blackmail her with it when she is a teenager.
the song is about a snake going through the woods looking for pieces of his tail. it's really annoying. i think in a classroom environment a kind of conga line or something would form, but it's not working remotely.
@leslietownes I think I need a revision of multiple integrals and Green's theorem etc. Then probably I'll understand more the intricacies of the cross product being non-zero.😬
it's something you want geometrically. changes in parameters u and v should give you changes in different directions on the surface. i'm going to throw myself down a well now because of all the geometry. i can't take it.
@leslietownes wow! getting children used to remote learning is such a post pandemania thing to do at such an early age, no? I'm not saying that's a bad thing, just a new thing
it's a constant adjustment. i suspect that it's driving her a little bit crazy, but it's all she knows. it's hard for me to romanticize my own experiences (i didn't do day care but did do preschool) because i don't remember them. which suggests that maybe everything will be fine.
they're gonna try to go back to in person next week assuming no other staff come down with covid.
does anybody in the chat have it now? i suspect my daughter might, she has a funny cough. but seems otherwise fine. if i've got it this time i am asymptomatic.
it's a strange disease, it seems there are basically two outcomes, either you're kinda fine fairly quickly, or it really messes your life up. they will probably need to analyze years of data to figure it all out.
It is truly, at the begining of Corona, i devoloped Psychosomatic symptoms. And have had health issues since that time, that still presist to this day.
Anyways, nice chatting to you all. I am going to make one last push for one more hour of good Analysisstudy before i end my day. Have a good evening everybody :)
I had what I thought was a bad flu in November 2019. Some of my friends think it may have been an early case of covid, but I don't think I had any symptoms that would indicate it wasn't simply the flu.
@robjohn Of course, but this is circular reasoning. The conditiion we're discussing is the "easy" way to see the vectors are linearly independent. It's not a determinant.
To note: There is a question on site with same diagram but the questions I ask regarding this diagram , are different.
Find $ S1$ in the parallelogram below
The Question:
Given:
$S_1 = 10\ \mathrm{ m^2}$ , $S_2=3\ \mathrm{ m^2}$. Find $S_1$.
$S$ in the diagram is the shaded region. Answer is $...
Suppose a polynomial $P$ of degree $n$ has $n$ distinct real roots then $P'$ (the derivative of $P$) has $n-1$ distinct real roots.
Proof by Induction:
Base case: For $n=1$, $P_1 (x)=a_0+a_1x, a_1\neq 0$ has $1$ real (distinct) root, $x=\frac{-a_0}{a_1}$. Then $P_1 ' (x)=a_1$ has $1-1=0 $ ...
