$ l(x)g = l(x)\frac{dv}{dx} + v^2$ becomes $ l(t)g=l(t)\frac{dv}{dt} + v^2 $ with the clause that $ v=\frac{dl}{dt} $ , the question as you said does become totally different from the one I posted initially ,though. Sorry for that
@user193319 relatively late to this party, but i'm not sure this needs the invariance condition. if you take e.g. any positive f then 0 <= f.(1 - 1_e) <= ||f||.(1 - 1_e) and positivity of m implies 0 <= m(f.(1 - 1_e)) <= ||f|| (m(1) - m(1_e)), and if m(1_e) = m(1) = 1, this implies that m(f.(1-1_e)) = 0, but the LHS of this last is also m(f.1) - m(f.1_e) = m(f) - m(f(e) 1_e) = m(f) - f(e) m(1_e) = m(f) - f(e).
i'm not sure i'm using anything too groupy, in fact, just the commutativity of the cstar algebra sturcture and i guess this distinguished element (1_e) that multiplies on algebra elements to give scalar multiples of itself back. which maybe could be derived from any atom in the measure space playing the role that G does here.
i always get nervous when there's lotsa structure that's not being used, but maybe that's why you encountered difficulty.
Let $M$ be a smooth manifold. I am trying to prove that $M$ admits a vector field with only finitely many zeros.
This will follow if we can find a function $f : M\rightarrow \mathbb R$ such that $df$ has only finitely many zeros, but I cannot find such a function with this property either. My i...
I have many images, with many points selected on them (measured) with real-world coordinates (2D).
Measured points are matched between images (tie points).
Now, I have to estimate Helmert 2D parameters in a few iterations using set of points matched together.
Previously I calculated average coord...
Does anyone know something about it? For now fo each detected object I'm calculating the average position, and after that, for each image, I'm calculating Helmert parameters separately, where I'm trying to transform image points to the calculated average positions.
I'm pretty sure that should be a better method which will be calculating all parameters in once and the final objects positions would be more accure
Makes sense, all compact sets of Q are nowhere dense and so it is not locally compact.
*Subsets
mathworld.wolfram.com/ConvergentSequence.html claims that "Every bounded monotonic sequence converges". But isn't this not true, for example: the sequence 2,4,6,8,10..., is boundless. So is it wrong to say "Every"?
@Ajay I'm not sure how it proves the statement. This may be part of some theorem that I don't know yet.
The sequence statement is true or false depending upon the definitions you're taking. I suppose that in the link you shared, they are taking the sequence to be in $\mathbb R$ or in $\mathbb C$ so 'the sequence converges' means that there is some r in R or in C, which is a limit of the sequence.
@Ajay this is usually referred to as a divergent sequence.
The proof is not correct as it stands, because you don't know beforehand that the compact neighbourhood of $x = (x_i)$ is of the form $\prod_i N_{x_i}$.
So you do know that there exists a compact set $C \subseteq \prod_i X_i$ such that $(x_i)$ is in the interior of $C$ (i.e. $C$ is a neighbourho...
Can anyone please explain to me that in this answer why $X_i$ for i in F is locally compact?
@Ajay fortunately, I have studied these theorems :).
but not sure how you want to use the theorems in the link to prove that Q is not locally compact.
@Koro This is because if one takes $x_i\in X_i$, then $x_i\in \pi_i(O)\subset \color{blue}{\pi_i(C)}$, the blue colored set is compact, $\pi_i(O)$ is open containing $x_i$, therefore $X_i$ is compact.
and the Heine Borel equivalent in metric spaces theorem also that you mentioned but I never had to use ‘totally bounded’ before and I don’t know its proof.
@Ajay how exactly?
If Q were locally compact, then there is a compact subspace of Q containing a nbd of 0. We can write every interval $(-d,d) \cap Q$ as $((-d,r) \cap (r,d))\cap Q$, where r is an irrational number in the interval.
Similarly we can get infinitely many open sets which along with some more open sets form an open cover of the compact subspace but it has no finite subcover which contradicts the assumption of local compactness.
@AlessandroCodenotti No, no no empty open set in Q is complete.
Can we always find an infinite sequence of irrationals $r_n$ with $a<r_1<r_2<r_3<\dotsb<b$?
with $\lim r_n=b$
If so, we can write $(a,b)=(a,r_1)\cup(r_1,r_2)\cup(r_2,r_3)\cup\dotsb$
@Koro Actually, how about this: did you prove compact implies sequentially compact?
If a subset of $\Bbb Q$ has a subset that's open in $\Bbb Q$, you should be able to find a sequence of rationals in there that approaches an irrational
f(x,y) = 2x^2 - y^2 +xy - 8x + 10 g(x,y) = 2x + y - 2 = 0 P=(1/4, 3/2) is a relative maximum point of f constrained to g. Fine. But I wanted to use Lagrange multipliers. As P is a relative maximum, I should get negative eigenvalues. But here's what I get: https://www.wolframalpha.com/input?i2d=true&i=det%7B%7B4-x%2C1%2C-2%7D%2C%7B1%2C-2-x%2C-1%7D%2C%7B-2%2C-1%2C-x%7D%7D What did I do wrong?
@Curio When you test constrained extreme points, you need to use what's called a bordered Hessian. If you use the regular second derivative test, you're forgetting the constraint.
I think I understood what was going on. I was wrongly looking for the eigenvalues of the bordered Hessian, forgetting that I just needed to look at the determinant. Thanks!
To prove that $\lim_{x \to 0^+}(\sin x-x^{2021})=0^+$, is the following reasoning valid? For $x \ge 0$ it is $x-x^{2021}\ge \sin x-x^{2021}\ge x-x^3/6-x^{2021}=x\left(1-x^2/6-x^{2020}\right)$. Since $-x^2/6-x^{2020} \to 0$ when $x \to 0^+$, there exists $\delta>0$ such that $-x^2/6-x^{2020}>-1/2$ for any $0<x<\delta$. Hence $x\left(1-x^2/6-x^{2020}\right)>x/2$ for any $0<x<\delta$.
Let $M$ be a smooth manifold. I am trying to prove that $M$ admits a vector field with only finitely many zeros.
This will follow if we can find a function $f : M\rightarrow \mathbb R$ such that $df$ has only finitely many zeros, but I cannot find such a function with this property either. My i...
