@Ted: brief question. If $M$ is parametrized by $\vec x(u,v) = \vec\alpha(u) + v\vec\beta(u)$ with $\vec\alpha$ regular and $\|\vec\beta(u)\| = 1$ for all $u$, if we let $(\vec\alpha' \cdot \vec\beta)(u) = f(u)$, then $\vec\alpha(u) + F(u)\vec\beta(u)$ (where $F$ is an antiderivative of $f$) gives the desired "WLOG $\vec\alpha' \cdot \vec\beta = 0$", right?
@manooooh Yes and no. In principle, you wouldn't be able to find exactly what $f(x)$ is---your answer can differ from $f(x)$ by an arbitrary constant. But by definition, indefinite integrals are antiderivatives, so the integral you wrote evaluates to $f(x) + C$.
@manooooh No, this is just an expression---the only thing you can really learn is that, for the expression to be defined, $f$ had better be differentiable on its domain.
This question is based on proving this implication: $$f(x+h)-f(x)=hf'(x)\implies f(x)=ax+b,\;a,b,x,h\in\Bbb R,\text{$f$ is continuous and differentiable on $\Bbb R$}$$
@manooooh This isn't saying "if this is true for a particular $h$", because what you just said---$f(x) - f(x) = 0$---is always true for any function, so it wouldn't imply $f(x) = ax + b$ on its own. It's saying "if this is true for every $h$, then $f(x) = ax + b$".
@TedShifrin what I was trying to do is to start from the left member of the thesis, and arrive through the hypothesis to the right member. It is right?
@TedShifrin I am not sure why: 1) We are forced to take $x=0$ (I know it holds for all $h$, but why is it impossible to solve it with for example $x=1$?), 2) We have considered the function $g(h)=f(x+h)-f(x)-hf'(x)$, why should we equal it to $0$?
I'm reading through an opencv example that performs camera alignment.
More specifically the homography is being estimated in such code, but there are these few lines that aren't quite clear to me:
//! [compute-plane-normal-at-camera-pose-1]
Mat normal = (Mat_<double>(3, 1) << 0, 0, 1);
Mat norma...
@traducerad your last equation is equivalent to $$\frac{gR}{V^2} = sin(2\phi)$$, comparing this with your first equation leads to $\sin^2 (\phi) = sin(2\phi)$, which isn't true in general
Hey, I'm trying to find a solved example how to develop a function using haar wavelets without success.. Do you maybe have some suggestion, where I could look into?
@user8469759 difficult question... They are different. What I love about London is that it is extremely active, I don't like the US English accent and apparently I should be able to make quite a lot of money there. Berlin is more "rebelious" and has an amazing underground (party) scene
and staff at your workplace should help you to cope with the bureaucracy, that's really only thing you will need german for... and then you can pick it up. its not that much different from english
@user8469759 yep, i get that, but i'd like to see a worked example first. i'm trying to find one since learning from examples aids understanding the theory
and you want to project, for example, the function $f(t) = e^t$ in the subspace $$V = \left\{ \alpha_0 \psi_{0,0}(t) + \alpha_1 \psi_{0,1}(t) + \alpha_2 \psi_{0,2}(t)\right\} = Span(\psi_{0,0},\psi_{0,1},\psi_{0,2})$$
according to that wiki page, as standard anyway, with $$<f_1,f_2> = \int_{\mathbb{R}} f_1(x) f_2(x) dx $$
well now once you do this, if you assume (though this isn't true) that $f(t) \in V$ you'll have an equation like
@traducerad yep, Munich seemed quite calm to me as well.. but then again it depends where in Munich(?) The same goes for Berlin.. If you go to Marzan, it is definitely something else than in Dahlem or in Kreuzberg.
yeah, sadly we only covered basics maths needed for psychology and computer science.. so if i want to develop let's say $e^x$ with up to 3rd order Haar functions, I should set n to 3?
or have 3 terms in the what would be "series expansion" in fourier?
@user8469759 yeah but in the context of splines, so that's where confusion comes from.. they's just an overview slides of what methods exist to approximate functions and i'd like to work some examples to understand what one or another method does
I'm reading through an opencv example that performs camera alignment.
More specifically the homography is being estimated in such code, but there are these few lines that aren't quite clear to me:
//! [compute-plane-normal-at-camera-pose-1]
Mat normal = (Mat_<double>(3, 1) << 0, 0, 1);
Mat norma...
It does because if $V$ is open in $X$ (topological vector space) then for every $x \in X$ we have $x + V$ is also open. So if $p \in X$ and $\gamma$ is a local base at $p$ you can construct a local base at $0$ by simply translating the members of $\gamma$.
I think so: Let $V$ be an open neighborhood of $0$. Since $0 + 0 = 0 \in V$, by continuity of addition there are open neighborhoods $B_1,B_2$ of $0$ such that $0 \in B_1 + B_2 \subseteq V$. Let $\mathcal{B}$ be the collection of all such neighborhoods. Note that it is not empty, because $0$ has at least one open neighborhood (namely, $X$), and by construction it is a local base.
@AlessandroCodenotti $\mathcal{B}$ is a local base of $0$ if, for any open neighborhood of $V$ of $0$, there is some open set $B \in \mathcal{B}$ such that $0 \in B \subseteq V$.
Suppose $f:[0,+\infty)\to [0,+\infty)$ is a continuous function such that: (1) $f(0)=0$; (2) $f(t)>0$ for $t>0$; (3) there is some $t_0>0$ such that $f$ is constant on $[t_0,+\infty)$. Question: can I find a continuous function $g:[0,+\infty)\to [0,+\infty)$ which is: (1) strictly increasing; (2) $g(0)=0$; and (3) $g(t)\leq f(t)$ for all $t\in [0,+\infty)$. This seems to be true, but I don't know how to prove it.
