We consider $$\sigma (u,v)=(f(u)\cos v, f(u)\sin v, g(u))$$ Picking $u=\theta , v=\phi , f(\theta )=\cos \theta , g(\theta )=\sin \theta$ we get that the first fundamental form is $$d\theta^2+\cos^2 \theta \phi^2$$
I have found a smooth function $\psi$ such that the reparametrization $\tilde{\...
@DanielFischer @robjohn Could one of you delete my Springer post (in chat) that has been starred? All books have been retracted, and someone on Reddit states that the authors did not give permission to Springer to put those .pdfs out originally.
I know that Bellman operator, defined as $T(f(x))=\sup_y {\phi(x,y)+\beta f(y)}$, is a contraction provided that $\beta\in(0,1)$ and $\phi$ is a bounded function on $Gr\Gamma$.
In order to prove this result I was trying to use Blackwell Theorem which states that an operator $T:B(X)\rightarrow B...
Hello. I am confused about the following formula $\nabla_{X}fY= f\nabla_X Y + X(f)Y$. This is one of the properties for an affine connection. I don't understand what $X(f)$ stands for. See here math.stackexchange.com/questions/1084302/… for all relevant definitions of terms.
@StanShunpike $X(f)$ is the function you get by differentiating $f$ in direction $X$, equivalently, applying the vector field $X$ to the function $f$. If you are comfortable with differential forms, knowing that it can also be expressed as $df(X)$ may help. We have $X(f)(p) = X_p(f) = df_p(X_p)$.
@StanShunpike If you have the covariant derivative defined to act not only on vector fields but on functions too, then you can also write it as $\nabla_X f$. But $X(f)$ is the more primitive notion, hence often it's preferable to use that.
Doesn't the covariant derivative have to be able to act on functions in order to write $X(f)$? After all, you just stated $X(f)$ is obtained by differentiating $f$ in the direction $X$, so doesn't that imply the covariant derivative can act on functions?
@StanShunpike No, to define the covariant derivative, you need to know how vector fields act on functions. That's $X(f)$. In one definition of the tangent space of a manifold at a point [as a space of derivations], the action of tangent vectors on (germs of) functions is (part of) the definition of tangent vectors.
@DanielFischer: just out of curiousity, is it ok for SE users to use (frequently) multiple accounts? I assume it's ok unless one tries to use multiple accounts to circumvent suspensions and similar?
@MikeMiller Here's a minor modification of the proof of IVT in Ted's book I did. $f : U \subset \Bbb R^n \to \Bbb R^n$ be a $C^1$ function, $x \in U$ be a point such that $Df(x)$ is invertible. We want to prove there is a nbhd $V \subset U$ of $x$ such that $f$ has a $C^1$ inverse.
WLOG, assume $x = f(x) = 0$ and $Df(0) = I$ by scaling $f(x)$ by $Df(0)^{-1}f(x)$.
$f$ is $C^1$, hence there is an $r > 0$ such that $x \in B_r(0)$ implies $||Df(x) - I||\leq 1/2$. Fix $y$ with $||y|| \leq r/2$, and define $\varphi : \Bbb R^n \to \Bbb R^n$ by $\varphi(x) = x - f(x) + y$.
So a high-reputation, highly-active user has already been dinged once for a sockpuppet, and I just dinged him again. He probably has another sockpuppet as well.
What are the guidelines for handling users who game the system in this fashion? Specifically,
How can I be sure I'm looking at a soc...
Note that $||\varphi(x)|| \leq ||x - f(x)|| + ||y|| \leq r/2 + r/2 = r$, thus $\varphi$ restricted to $B_r(0)$ is a map $B_r(0) \to B_r(0)$. Also, if $x, y \in \text{cl} B_r(0)$, then we see $||\varphi(x) - \varphi(y)|| \leq 1/2||x - y||$, so $\varphi$ is actually a contraction mapping.
Ok, $\varphi$ is a contraction mapping for every $||y|| \leq r/2$, so there is a fixed point $x_y$ for $\varphi$. Thus, define $g : B_{r/2}(0) \to f^{-1}(B_{r/2}(0)) \cap B_r(0)$ (I hope my domains and codomains are right) by $f(y) = x_y$. I claim this is the $C^1$ local inverse for $f$.
The hardest part by hard is the set-level inverse. Once you get that, slight manipulation of formulae gives you that it's $C^1$, or $C^k$ if the original function is $C^k$.
@Balarka: Should be easier than that. But no worry.
I want you to look at the first part of this theorem and tell me precisely what you needed to prove it.
That is, if $F : U \times V \subset \Bbb R^{n+k} \to \Bbb R^n$ is a function such that $F(-, y)$ is a contraction mapping for all $y \in V$, then $g : V \to \Bbb R^n$ sending $Y$ to the fixed pt of $F(-, y)$, then $g$ is $C^1$. This is a direct application of Banach fixed point theorem.
Actually, I skipped some subtlety up there. $\varphi$ is a contraction mapping on $\text{cl} B_r(0)$, and it's not entirely clear why it has fixed points inside the interior of $B_r(0)$. But this can be done anyway.
@Balarka: I can either tell you what I mean (which answers the question) or you can try to figure it out. Given the questions you pose to others I figure you would want to do the latter.
I don't pose vague questions. But I'll try to figure out what you mean.
Oh, I think you mean how the function $f$ should be like. $f$ has to be continuously differentiable with continuous derivative. This is necessary and sufficient.
Well, $f : U \subset \Bbb R^n \to \Bbb R^n$ be a function, continuously differentiable with continuous derivative at $a \in U$, and $Df(a)$ is invertible. Then there is nbhd $V \subset U$ of $a$ where $f$ has a inverse which is continuously differentiable and has continuous derivative.
There are a few assumptions here. One is that $Df$ is invertible; relaxing this gets you implicit. There is an assumption here you have not brought up at all, yet, though.
I am not quite sure what other assumptions there are. The continuously differentiable with continuous derivative assumption is there because - intuitively - then $f$ would be monotonically increasing around $a$ which guarantees inverse (at least in one dimensional case).
Sorry, I have no idea what other condition there might be. I checked the book to make sure I am not being silly, but I am not seeing any more condition there.
@MikeMiller Inverse function theorem also says that if $g$ is that local inverse, then $Dg$ is the inverse for $Df$ at the tangent space level, but I guess that's not what you were referring to. For one, that's not an assumption on $f$, it's just a property of $g$ which is quite straightforward to prove.
This is an attempt to prove Theorem 1.5.3. in Casella and Berger. Recall for a random variable $X$, we define $$F_X(x) = \mathbb{P}(X \leq x)\text{.}$$
Theorem. $F$ is a CDF iff:
$\lim\limits_{x \to -\infty}F(x) = 0$
$\lim\limits_{x \to +\infty}F(x) = 1$
$F$ is nondecreasing.
Fo...
I don't know why you don't just shoot wildly. There are only so many words and I have been glad to tell you when you're wrong. You'd be right before long. But if you want I huess I can say.
If $f : U \subset \Bbb R^n \to \Bbb R^n$ is a $C^1$ function such that $Df(a)$ is invertible for some $a \in U$, then $f$ is a local homeomorphism at $a$, that's all the IFT says, right? I am not sure if there's any implicit assumption here.
Your function is defined on a subset of $\Bbb R^n$. That's what I was pointing out.
What do you need for your theorem? You need a vector space with a notion of differentiability. You need a norm. And, lastly, you need the unit ball to be complete.
Frechet means completely metrizable TVS. But there are points in the theorem where you need to pull a scalar out of the norm, or invoke the triangle inequality. IFT is false without serious modifications for Frechet spaces.
@MaryStar There are two kinds of orthogonal matrices in $\mathbb{R}^3$: rotations ($\det(P)=1$) and rotations and a mirror ($\det(P)=-1$). Since we know that if $\det(P)=1$, we have $P(a\times b)=P(a)\times P(b)$. Thus, $-P(a\times b)=-(-P(a)\times-P(b))$. Thus, if $\det(P)=-1$, we have $P(a\times b)=-P(a)\times P(b)$
@MaryStar I don't know what all your terms are. Is $P$ a matrix? If so, then I have no idea what $P\cdot P$ means. How did you get the equation above if you don't know what $P\cdot P$ means?
The dot product usually results in a scalar, so how does one take the dot product with a scalar?
The second fundamental form is $Ldu^2+2MNdudv+Ndv^2$, where $L=\sigma_{uu}\cdot \textbf{N}$, $M=\sigma_{uv}\cdot \textbf{N}$, $L=\sigma_{vv}\cdot \textbf{N}$, where $\textbf{N}=\frac{\sigma_u\times\sigma_v}{\|\sigma_u\times\sigma_v\|}$.
So do we get $P\sigma_{uu}\cdot \frac{\pm P (\sigma_u\times \sigma_v)}{\| \sigma_u \times \sigma_v\|}=\pm P P\sigma_{uu}\cdot \frac{\sigma_u\times \sigma_v}{\| \sigma_u \times \sigma_v\|}$, not dot product but just the product of matrices?
@Clarinetist I am very very slowly understanding differential entropy.. I think this is the next main hurdle to my understanding math.stackexchange.com/q/1590790/66307
@Clarinetist oh.. are you doing a phd? I am not sure what qualifying exams are
@Lembik Sigh, I wish it was. I'm not very familiar with the idea, although I do understand the question. Have you tried stats.stackexchange.com? I would imagine there'd be someone helpful there
@Clarinetist it is simply this.. consider a 3 by n matrix $M$ and a random vector $v$ . The elements of $M$ and $v$ are from $\{-1,1\}$. What is the pmf of $Mv$?
$M$ is not random
@Clarinetist I feel that understanding that will actually resolve everything :)
@Lembik Well, I'm assuming the elements of $v$ are Bernoulli, so that you end up (when you take $Mv$) with a multivariable linear combination of Bernoullis
@Lembik Nope, they're not. Without doing some $n$-dimensional summation and fiddling with CDFs, this question is pretty close to impossible to do, unfortunately. At least IMO. Someone might know something more
Let $X_1,\ldots,X_n$ be independent Bernoulli random variables such that $\mathbb{P}(X_i=\pm 1)=1/2$ and consider two collections of real numbers $a_1,\ldots,a_n, b_1,\ldots, b_n$. For the moment let us just take $a_i=1$ and $b_i=i$. Write $X=\sum_{i=1}^{n}a_iX_i$ and $Y=\sum_{i=1}^{n}b_iX_i$. Qu...
@Lembik I don't either (when it comes to what I posted), but the comments have a hint which involves material I don't know at all
@Lembik Well, here's the thing. You don't have independence. That is a problem. Because the first thing that you're gonna need when you want to find the PMF of a bunch of variables is the joint PMF (or PDF in the continuous case).
@Clarinetist Oh I see. The thing is that it's not even that I want to prove something true currently.. it's that I don't even know what is vaguely true about my problem
@MikeMiller: I'm never sure about whether or not he's trolling. The more I see his mathematical questions, the more I believe he's actually that kind of guy who just always barely learns the notions and tools to apply whatever he needs them for and doesn't learn anything else. He knows a lot of GR but would probably not be able to do a simple linear algebra or real analysis proof.
@DanielFischer Hi!!! I am evinda... I am trying to log in at my account but then at my profile it says MaryStar... MaryStar is my sister but we are not the same person... Could I have again my account , please ?
@MikeMiller: really frustrating to watch as a mathematician sometimes.
and there's those guys who are like "I've never seen functional analysis or measure theory being applied anywhere and I'm a professional physicist so it's just useless maths"
@evinda: I asked the other day if you were that new user and you said no. I never said anything about you and Mary Star here? Even though I did have that thought, yes.
@Huy I can somewhat sympathize with that statement, in that time is a finite thing and there's math which, for certain purposes, is therefore just not a useful investment
@Semiclassical: "useless for me" is something different than saying "useless entirely", plus that same guy was in the maths chat asking about a formula containing the delta distribution and probably still doesn't understand how to prove it
@Huy Me and MaryStar , we were doing the exercises together in differential geometry and that's why we both asked... lately I was just asking about optimization and MaryStar about dif... because I am studying now for the exams...
@Semiclassical: I just thought it was ironic that someone who says "measure theory and some other maths is useless" and then asks about a formula with the delta distribution because he doesn't manage to prove it using only basic analysis
partly that's because i've never actually taken measure theory---i went from a liberal arts college off to physics grad school---and partly because i've felt for a while that if i wanted to learn it properly i should learn about hyperfunctions :P
We consider $$\sigma (u,v)=(f(u)\cos v, f(u)\sin v, g(u))$$ Picking $u=\theta , v=\phi , f(\theta )=\cos \theta , g(\theta )=\sin \theta$ we get that the first fundamental form is $$d\theta^2+\cos^2 \theta \phi^2$$
I have found a smooth function $\psi$ such that the reparametrization $\tilde{\...
I know that Bellman operator, defined as $T(f(x))=\sup_y {\phi(x,y)+\beta f(y)}$, is a contraction provided that $\beta\in(0,1)$ and $\phi$ is a bounded function on $Gr\Gamma$.
In order to prove this result I was trying to use Blackwell Theorem which states that an operator $T:B(X)\rightarrow B...
So a high-reputation, highly-active user has already been dinged once for a sockpuppet, and I just dinged him again. He probably has another sockpuppet as well.
What are the guidelines for handling users who game the system in this fashion? Specifically,
How can I be sure I'm looking at a soc...
This is an attempt to prove Theorem 1.5.3. in Casella and Berger. Recall for a random variable $X$, we define $$F_X(x) = \mathbb{P}(X \leq x)\text{.}$$
Theorem. $F$ is a CDF iff:
$\lim\limits_{x \to -\infty}F(x) = 0$
$\lim\limits_{x \to +\infty}F(x) = 1$
$F$ is nondecreasing.
Fo...
