Hi @robjohn! If my toothache gets fixed by my dentist (it was supposed to already be on the mend :() I'm planning a quick trip to LA Thursday to Saturday.
So you can only do it approximately. There's no way to parametrize the path explicitly, so your only hope is to have a conservative vector field. You know the Fundamental Theorem of Calculus for Line Integrals?
According to Wikipedia,
Hilbert space [...] extends the methods of vector algebra and calculus from the two-dimensional Euclidean plane and three-dimensional space to spaces with any finite or infinite number of dimensions
However, the article on Euclidean space states already refers to
...
@Mockingbird: OK ... Basically, then the answer will only depend on the potential function (the $f$ so that $\nabla f = \vec F$) and the endpoints of the path. But the endpoints you can only get approximately (e.g., Newton's method).
@nbro: There's nothing more to say. $\Bbb R^n$ or $\Bbb C^n$ is a finite-dimensional Hilbert space (when you equip them with inner products).
You need $(x,y)$ for both, @Mockingbird, and that equation can only be solved numerically ā if you specify an $x$, it should give us only a few different $y$'s ... but unlikely a unique one!!
Interesting that you raise this question, @Mockingbird. I spent about a week figuring out a solution of a similar, very hard question on MSE a few years ago.
If I have a torus and I want to calculate the "volume of its interior" is that the volume or the surface area? The torus is just a shell, of course. Because I have a question asking for this. Asks for the volume. As a hint says, compute this integral. I know for sure the integral is the surface area ???
Yeah, they want the volume of the interior. How can you modify the parametrization of the torus (the surface) to get a parametrization of the interior?
At one point at Mathcamp, a professor looked at her notes, said "I used both phi and varphi? That's evil!" and then proceeded to use both $\phi$ and $\varphi$ on the blackboard
@Daminark a few things. for one the priors on what is taken as elementary is just totally different. Cartan clearly expects readers to be comfortable pulling of ridiculous linear algebra computations that modern students probably wouldn't be able to do. The other thing is just lack of modern language making like everything a little obtuse to a modern reader, im sure there's still a wealth of insight to be gleaned but my melted fever brain can't do it rn
I'm still not understanding where you got n=4m+k from, since rewriting the modulo would give you $(7^{ n })(mod\quad 10)=7,9,3,1\Rightarrow 7^{ n }-1=10k$, right?
@Daminark there's a particular thing which happens a lot where he says you have something which hasn't been defined anywhere but he can compute what it should be (but he doesn't, so i think he assumes the reader either knows or should figure it out) so i guess he doesn't need to come up with a formal construction or something
also, I can see exactly why $$\int_{\delta}^{2\pi - \delta} |D_N(t)|\,\mathrm{d}t \ge C \log(N)$$ for some constant $C$ (where $D_N$ is the Dirichlet kernel), but the mother ****ing rigor refuses to behave!
Given a regular pentachoron ($ABCDE$), find the cross section when it is cut along a hyperplane which is equidistant from and parallel to line $AB$ and plane $CDE$.
Here, a regular pentachoron means a $5-$ cell, i.e., a $4D$ figure with $5$ vertices A,B,C,D,E all at equal distances from o...
Is a morphism of sheafs uniquely defined by the induced stalk morphisms?
I mean, yes, $f=g$ if and only if $f_x=g_x$ for all $x$.
But given morphisms $f_x$ on the stalks, can I say that they induce a morphism of sheafs $f$ such that the stalk morphisms induced by $f$ are exactly the $f_x$?
Let $(X,\mathcal{A},\mu)$ be a measure space. \\ $(a)$ If $(A_k)$ is an increasing sequence of sets that belong to $\mathcal{A},$ then $\mu(\bigcup_{k}A_k)=\lim_{k}\mu(A_k).$\\ $(b)$ If $(A_k)$ is a decreasing sequence of sets that belong to $\mathcal{A},$ and if $\mu(A_n)<+\infty$ holds for some $n,$ then $\mu(\cap_k A_k)=\lim_k \mu(A_k).$
If $f: X\rightarrow [0,+\infty[$ is a measurable function, there exists subsets $A_1,\ldots, A_n\subset X$ not necessarily disjoint and $(\alpha_n)\subset\mathbb{R}_+$ such that $$f=\sum_{n=1}^{\infty} \alpha_n \chi_{A_n},$$ with $$\sum_{n=1}^k\alpha_n\chi_{A_n}(x)\leq f(x),~x\in \Omega ~\text{and}~ k\in \mathbb{N}.$$
$$\textbf{1/ }\exists f \in F([-1,1],[-1,1]), \forall x\in [-1,1], f^3(x)+f^2(x)+f(x)+x=1 \text{ and } 3f^2(x)+2f(x)+x=0\text{ ?}$$
$$\textbf{2/ }\exists f \in C([-1,1],[-1,1]), \forall x\in [-1,1], f^3(x)+f^2(x)+f(x)+x=0 \text{ and } 3f^2(x)+2f(x)+x=0 \text{ ?}$$
$$\textbf{3/ } P_1,..,P_n \in \mathbb C[z], \text{ find a NCS on the }(P_i)_i \text{ for that system (S) have a solution in } F(\mathbb C,\mathbb C),$$ $$\text{(S) : }\forall i=1..n, \forall z \in \mathbb C, P_i(f)(z)=0$$
Hi all, I have a differential equation with a Bessel function as a solution (obtained from WolframAlpha) Rewriting it appropriately does not appear to me instantly, so any good methods or books I can look at to squeeze out the right answer?
> There are two very different types of ultrafilter: principal and free. A principal (or fixed, or trivial) ultrafilter is a filter containing a least element. Consequently, principal ultrafilters are of the form Fa = {x | a ā¤ x} for some (but not all) elements a of the given poset. In this case a is called the principal element of the ultrafilter. Any ultrafilter that is not principal is called a free (or non-principal) ultrafilter.
In other words, they look like onions with no core
@MatheinBoulomenos Well that's true, so I guess the best we can only say is that it is a subset that has no least element and is upward closed
Recipe for making non measurable sets: 1. DC but no inaccessibles 2. Ugly functions (what does that mean in the minimalisitc sense?) 3. Existence of a free ultrafilter on the naturals 4. (I don't recognise this variety of AC)
To get a feeling for how weird non-prinicpal ultrafilters are: if you choose a non-principal ultrafiter $\mathcal F$ on the set of primes $\Bbb P$ , then the set $I \subset \prod_{p \in \Bbb P} \overline{\Bbb F_p}$ given by $(x_p)_{p \in \Bbb P} \in I \Leftrightarrow \{p \in \Bbb P \mid x_p = 0\} \in \mathcal F\}$ is a maximal ideal (okay that part is not so weird) and $(\prod_{p \in \Bbb P} \overline{\Bbb F_p})/I$ is isomorphic to $\Bbb C$
okay, you need a second application of choice for the isomorphism to $\Bbb C$, but you can at least say that it's an algebraically closed field of characteristic $0$ with the cardinality of the continuum
here $\overline{\Bbb F_p}$ is the algebraic closure of $\Bbb F_p$
$\prod_{p \in \Bbb P} \overline{\Bbb F_p}$ is countable, $I$ is also countable, $\mathcal{F}$ is countable(?), and $(\prod_{p \in \Bbb P} \overline{\Bbb F_p})/I$ is of size continuum...
when is the last time I saw two countable things quotiented together give an uncountable thing...?
o wait my mistake, $\prod_{p \in \Bbb P} \overline{\Bbb F_p}$ is uncountable since $\overline{\Bbb F_p}$ is countable each
If $f: X\rightarrow [0,+\infty[$ is a measurable function, there exists subsets $A_1,\ldots, A_n\subset X$ not necessarily disjoint and $(\alpha_n)\subset\mathbb{R}_+$ such that $$f=\sum_{n=1}^{\infty} \alpha_n \chi_{A_n},$$ with $$\sum_{n=1}^k\alpha_n\chi_{A_n}(x)\leq f(x),~x\in \Omega ~\text{and}~ k\in \mathbb{N}.$$
@AkivaWeinberger Right so I pinged you because of this: Remember that I defined $\mathcal{L}_X Y$ as $\partial_t (\varphi_{-t})_* Y_{\varphi_t(p)}|_{t = 0}$ where $\varphi_t(x) : \Bbb R \times M \to M$ is the flow of $X$ defined by solution to the ODE $\partial_t \varphi_t(x) = X_{\varphi_t(x)}$ for all time $t$ and $x$ (assume $M$ is compact for now so the flow is everywhere defined), and $(\varphi_{-t})_* : T_{\varphi_t(p)} M \to T_p M$ is the derivative
You should think of this as, let $p \in M$ and $\gamma$ be the integral curve of $X$ starting at $p$, then take $X_{\gamma(t)}$, "slide it back along the path" to $X_{\gamma(0)}$, taking the difference quotient and let $t \to 0$ - like we discussed earlier.
I meant $Y_{\gamma(t)}$ and $Y_{\gamma(0)}$ there.
@AkivaWeinberger There's actually a similar interpretation to $\nabla_X Y$, where instead of sliding $Y_{\gamma(t)}$ to $T_{\gamma(0)}M$ along $\gamma$ using the flow $(\varphi_{-t})_*$, you parallel transport it to $T_{\gamma(0)}M$. Let me explain what that is.
@AkivaWeinberger So let's assume that you already have a connection $\nabla_{(-)}(-)$ eating two vector fields on $M$ and spitting another vector field, linear on bottom component ($\nabla_{fX} Y = f \nabla_X Y$) and Leibniz on front component ($\nabla_X (fY) = f \nabla_X Y + D_X(f) Y$).
If $\gamma$ is a path in $M$, a vector field $X$ along $\gamma$ is said to be parallel if $\nabla_{\gamma'} X = 0$. You should think of this as, if an ant walks along $\gamma$ then under the force of $X$ it would NOT feel ANY acceleration on $M$.
Ted's notes has a picture of a parallel vector field along a meridian of $S^2$ (under a natural connection). It twists as it moves along the meridian. Notice that $\gamma$ is a geodesic iff $\gamma'$ is a parallel vector field.
Parallelism gives a natural way to identify two tangent spaces $T_p M$ and $T_q M$ where $\gamma$ is a path on $M$ with $\gamma(0) = p$ and $\gamma(1) = q$. Namely, pick a basis $(e_1, \cdots, e_n)$ of $T_p M$, construct parallel vector fields $X_i$ along $\gamma$ starting at $e_i$ at $p$, and define the map $\mathscr{T} : T_p M \to T_q M$ by $\mathscr{T}(e_i) = X_i(q)$. (I think that's it)
The upshot of all the formalism is that you can recover $\nabla$ from the notion of parallel transport: $\nabla_X Y$ is $\partial_t \mathscr{T}_{-t} Y_{\varphi_t(p)}|_{t =0}$ where $\mathscr{T}_{-t}$ is the parallel transport map $T_{\varphi_t(p)} \to T_{\varphi_0(p)}$ along the integral curve of $X$ starting at $p$.
I think this is an exercise in do Carmo. It's not hard to prove in any case.
To give a not-so-excited but formal example consider the fact that if $X, Y$ are compact manifolds and $Z \subset Y$ is a submanifold then (1) given a map $f : X \to Y$ transverse to $Z$ any homotopy $f_t : X \to Y$ starting at $f_0 = f$ stays transverse to $Z$ for all $t < \epsilon$ for some $\epsilon > 0$ (2) given a map $f : X \to Y$ there is an arbitrarily small homotopy $f_t : X \to Y$ such that $f_1$ is transverse to $Z$
Let $C^\infty(X, Y)$ be the space of smooth maps from $X$ to $Y$ and $C_{\pitchfork}^\infty(X, Y) \subset C^\infty(X, Y)$ be the subset of maps transverse to $Z \subset Y$ (I'm not going to describe the topology)
(1) says $C_{\pitchfork}^\infty(X, Y)$ is an open subset (2) says $C_{\pitchfork}^\infty(X, Y)$ is a dense susbet
You can also define it purely formally, by describing the topology on $C^\infty(X, Y)$. (This is done in the first page of chapter 2 Hirsch if you want a reference)
You want an elaborate version of compact-open topology, more or less.
What is the usual way to denote $O(2,\mathbb{R})$ using set notation? Do you just say, determinant must have absolute value of 1 or is there an easier way?
According to Wikipedia,
Hilbert space [...] extends the methods of vector algebra and calculus from the two-dimensional Euclidean plane and three-dimensional space to spaces with any finite or infinite number of dimensions
However, the article on Euclidean space states already refers to
...
Given a regular pentachoron ($ABCDE$), find the cross section when it is cut along a hyperplane which is equidistant from and parallel to line $AB$ and plane $CDE$.
Here, a regular pentachoron means a $5-$ cell, i.e., a $4D$ figure with $5$ vertices A,B,C,D,E all at equal distances from o...
