Ah... okay. This is precisely metric compatibility $X\langle Y, \nu \rangle = \langle \widetilde{\nabla}_X Y, \nu \rangle + \langle \widetilde{\nabla}_X \nu, Y \rangle$. The left hand side is zero because $Y$ is vector field on $M$ so it has no projection along $\nu$.
We get $\langle \widetilde{\nabla}_XY, \nu \rangle = -\langle \widetilde{\nabla}_X\nu, Y \rangle = -h(X,Y)$.
she complained about having to walk by herself. she's a little wobbly and we'll have to pick her up early from day care again. i think it's only 80% a plea for attention.
@o.9 My masters advisor thought that LaTeX was a bloated piece of garbage, and used Plain TeX exclusively. I learned on Plain TeX, but switched over to LaTeX when I started my PhD (my masters thesis is written in Plain TeX).
Michel is nuts, @Xander. I switched from AMS-TeX to LaTeX when I had to revise my first book (of four!) and realized that every time I reordered exercises or added a lemma I'd have to manually correct numbering throughout the whole document. Insanity.
what's nice about encyclopedias is you could have two encyclopedias deal with the same subject in an interesting manner from two separate article author's perspective
shintuku interesting point. another thing about paper encyclopedias is that at least some of the time, there is an effort to make cross-referenced material somewhat consistent. e.g. an entry on a math topic that cites another math topic might not point to something using completely different terms or notation. often because a small group of people would be responsible for editing multiple entries on a topic.
individual wikipedia pages in math can be good, but you click over to read more about some piece, and it could be a completely different set of people working on it, and mutual inconsistency in notation/definitions, etc.
latest math conundrum: if I average $f_1(\hat{n}\cdot \hat{a}) f_2(\hat{n}\cdot \hat{b})$ over the sphere, must the result be some function of $\hat{a}\cdot\hat{b}$?
i think the answer is yes, and yet my computation (for a specific special case) keeps laughing in my face
(in my case i'm getting $\cos(\theta/2)$ when I think i should be getting $\cos\theta=\hat{a}\cdot \hat{b}$, and it is driving me mental)
@TedShifrin Michel doesn't use either Tex or LaTeX. He has an administrative assistant who converts all of his handwriting into LaTeX. My masters advisor, on the other hand...
@o.9 He is, though I disagree with him on this one. LaTeX is a horrible pile of bloat and incompatible libraries and whatnot, but it has the advantage of being designed to converge semantics into pretty documents. TeX doesn't really understand semantics---the user is meant to have to think about layout themselves.
It is like the difference between html and css. css tells the browser what a page should look like (i.e. how it is formatted), while the html gives the semantic meaning (e.g. "this is a paragraph", "this is a header", "this is a table", etc).
Since I am neither an expert on typesetting, nor do I want to be, I choose to trust that the choices made by Leslie Lamport (and others) reflect good typesetting practice, so I use LaTeX.
On the other hand, my masters advisor also believed that C++ was a terrible, bloated piece of garbage, and insisted that I code in C when I had to code. I took an undergraduate C++ class, but I am much more comfortable in plain C than C++ (OO is nonsense).
Let $m$ be the Lebesgue measure on $[a,b]$.
What would be an example of a bounded linear functional $T:L^ \infty \to \mathbb{R}$ which cannot be expressed by an integral $\int_{[a,b]}fg$ for an integrable $f$?
I know that, if $1 \leq p < \infty$, then any bounded linear functional $T:L^p \to \m...
I am working on a similar problem, but I am trying to show that the linear functional defined by $L(f) = f(2/3)$ cannot be expressed as such an integral. Is it possible to modify the example used by Quoka to work with the linear functional I am dealing with?
The fact that his sequence works is because $f_n(0) = 1$, i.e., it is nonzero.
But for large enough $n$, $f_n(2/3) = 0$, which won't get me the same contradiction that Quoka obtained.
Basically, I just need to redefine $f_n$ such that $f_n(2/3) =0$ for all $n$ or for sufficiently large $n$, so that the limit in the end is nonzero---which is precisely where the contradiction arises.
And I still need the functions to converge to $0$.
I guess a spike at $2/3$ should work...
So, continuous functions approximating the indicator function $1_{\{2/3\}$...not sure if there is a simpler example. Let me write down the functions.
Oh, wait...that won't converge pointwise to zero, but to $1_{\{2/3\}}$
But that function is $0$ almost every...Does the dominating convergence theorem work when the pointwise convergence is almost everywhere?
Yeah, we should be able to replace pointwise convergence with pointwise convergence almost everywhere---that usually works.
Yes, I am defining a linear functional on $C[0,1]$, then using Hahn-Banach to extend it to $L^{\infty}[0,1]$, and then arguing it doesn't arise as an integral for some $g \in L^{1}[0,1]$.
And then from there I am supposed to conclude that $L^{\infty}[0,1]$ is not reflexive.
$\frac{1}{n(n+a)} = \frac{1}{an} - \frac{1}{a(a+n))}$ is a useful identity that is not immediately obvious. the idea might be silly, but is there any sort of compilation of algebraic identities like these somewhere?
user, seems you have the right idea in that you're trying to show that an integrable function, if it implements that functional, has to be zero a.e. (and thus can't implement that functional). how you formalize may be up to you.
one of the funny things about partial fraction decomposition is that it is sometimes taught before calculus, when there's very little use for it. it's just presented as something you can do. at least, i got it that way.
the problem with pre-calculus is that, if the only reason to learn it is "it'll be useful in calculus", then you don't have any motivation for it at that moment
that said, the thing which does come with partial fractions is the concept of "if two polynomials are equal for all inputs, then they're the same polynomial and you can identify coefficients"
and i do think that's not a bad lesson
i tossed this in chat earlier, but i do want a second opinion on this:
latest math conundrum: if I average $f_1(\hat{n}\cdot \hat{a}) f_2(\hat{n}\cdot \hat{b})$ over the sphere, must the result be some function of $\hat{a}\cdot\hat{b}$?
the practical version of this is now that $f_1(n\cdot a)=1$ if $(a+b)\cdot n<0$ and 0 otherwise. (yes, $f_1$ has implicit $b$ dependence as well. i think that can't be avoided in my problem of interest)
while $f_2(n\cdot b)=\frac12(1+n\cdot b)$
maybe the implicit $b$ dependence in $f_1(n\cdot a)$ is the issue
@Semiclassic Hermann Weyl's Theorem on Vector Invariants says that any $O(n)$-invariant function of a collection of vectors is a function of their mutual dot products. With $SO(n)$ one throws in determinants, as well (which wouldn't apply in your case, anyhow).
I'll steal a quote of the theorem from my thesis for you.
Representation theorists, invariant theorists, and integral geometers know this. I wouldn't be surprised if a certain cadre of physicists know it, along with the unitary analogue.
Do you think "to be genius" is more about DNA/biological process, that is, you need to have the biological advantage to be it. Or do you think it is more about persistence and train? Or maybe both?
Since you were a professor, you probably saw a lot of type of students.
I have seen a lot of innate talent in students and a lot of others in mathematics who succeeded quite well with very hard work and dedication. But I don't throw around the word genius. Even the mathematicians I know who were astonishing in their accomplishments, I never think about "genius."
Anyhow, @Semiclassic, because you're averaging, your result is of course orthogonally invariant and so your conjecture must be true :)
I think it's immediate from Weyl's main theorem. You have an $O(3)$ action on the unit sphere and your average value is invariant under that action. If $g\in O(3)$, it's clear (by moving the $g$ to $\vec n$) that $F(g\vec a,g\vec b) = F(\vec a,\vec b)$, no?
The only other method I know comes from the trick for integrating any polynomial over the sphere, which is to multiply by the Gaussian and use homogeneity. I haven't thought about whether that works for a hemisphere only.
Let me just scribble for a moment so I don't say nonsense.
I think I've written this down in some class notes or paper, but I don't remember where.
So I'm using spherical coordinates (inexplicitly) $(t,\sigma)$ on $\Bbb R^n-\{0\}$. You want to integrate $\int_{S^{n-1}} P(\sigma)\,d\sigma$. $P$ is a homogeneous polynomial of some degree here.
Rather than work it all out, I'll cheat and link you to this.
Say $P$ is homogeneous of degree $m$. Then we take spherical coordinates $(t,\sigma)$ on $\Bbb R^n$, with $t>0$, $\sigma\in S^{n-1}$. We want $\int_{S^{n-1}} P(\sigma)\,d\sigma$.
But if you write $P(x) = x_1^{\alpha_1}x_2^{\alpha_2} \dots x_n^{\alpha_n}$, you can now write this as a product of $n$ $1$-dimensional integrals, breaking the Gaussian up as the product of $n$ Gaussians.
Regardless, these integrals are well known in terms of Gamma.
It looks like you're integrating over a half-space instead, so one variable is different. Oh, these integrals should have been from $-\infty$ to $\infty$.
And then if $x_n\ge 0$, for the $x_n$ integral you just have $0$ to $\infty$.
Of course, if $\alpha_i$ is odd, the integral should be $0$, so I needed to have $-\infty$ to $\infty$ above.
Robjohn will no doubt come in later and show us the sneakiest way to do this, but I'm confident I've reconstructed the method. Every physicist should know this :P
I think I may have written this all up as an exercise for a grad geometry course once. I'm going to look for that.
Yeah, you do not have to do anything with specific spherical coordinates. You only need the volume element on $\Bbb R^n$ in terms of surface area of the sphere and $t^{n-1}dt$.
this method probably will help me if i ever try to understand the higher-dimensional aspects of this paper. i only needed the $d=2$ case but it's developed for arbitrary $d\geq 2$
looking for what's popular lately in mathematical modeling for social sciences in the scientific literature. is there any search engine of mathematics journals?
a lot of social sciences, even really quantitative stuff, is done by people who don't have any mathematical understanding. people who describe themselves as focused on 'methods' sometimes develop new models, but they are a tiny minority. maybe a few dozen people in the US.
everyone else just copies them with different data and less understanding.
applied mathematics might be better. something that's one step removed from particular data sets.
not necessarily. but i don't think they leverage mathematical understanding, or mathematical models very much, in what they do. even if their papers are full of what appears to be the output of R or stata or whatever.
so if someone with mathematical background were looking to model something in the social sciences, they might do well not to start with whatever they see, even in "top journals"
there's a tendency for people to duplicate the methods that other people use. editors who don't understand methods can't independently evaluate them, so feel better when people say they are using stuff they can cite as having been used in other papers.
i don't mean this as a diss on the social sciences. i wouldn't look to pure math journals to decide whether a physics paper was physically appropriate. pure math people don't have the language for discussing that.
she's begun backseat driving a little bit, but only with my wife. apparently the light will turn green and she'll shout "GO! IT'S GREEN!" even if there are cars in front of them.
i'm getting a flu vaccine tomorrow, so if anyone wonders why i'm in a mood, or missing something, that's my preemptive excuse for that. anyone know if it interferes with 5G?
Let $m$ be the Lebesgue measure on $[a,b]$.
What would be an example of a bounded linear functional $T:L^ \infty \to \mathbb{R}$ which cannot be expressed by an integral $\int_{[a,b]}fg$ for an integrable $f$?
I know that, if $1 \leq p < \infty$, then any bounded linear functional $T:L^p \to \m...
