@Rememberme He gave me a pdf of something you've unlikely yo find by yourself on the internet, and in the Mendley database a significant part of the ones who owned that pdf were from Switzerland
Maybe I forgot my derivation, but if I wanted to find the derivative of $y = f(x)$ with respect to $x$ (assuming derivatives do exist for $f(x)$, $x \in \mathbb{R}$), then I would notationally represent it as
I'm asking because I came across the $\\nabla$ notation, which represents a gradient. Maybe I read wrong, but that is used to express the "derivative" of some $f:\mathbb{R}^n \to \mathbb{R}^n$, e.g. the derivative of $f(\mathbf{x})$, $\mathbf{x} \in \mathbb{R}^n$.
Did I get it correct so far?
By $\\nabla$ I meant $\nabla$.
I'm asking because I came across this equation, and I'm asked to solve for $\mathbf{w}$, in the following equation:
@Rememberme He gave me a pdf of something you've unlikely yo find by yourself on the internet, and in the Mendley database a significant part of the ones who owned that pdf were from Switzerland
@SalehenRahman Basically gradient of a $f, f:\Bbb{R^m}\to \Bbb{R^n}$ is a vector field at each point $a$ where partial derivatives $D_1(f(a),....D_n(f(a))$ exist. To get an idea of this in $\Bbb{R^2}$ $\nabla f(a) = \frac{{\partial}(a)}{{\partial}{x}}\hat{i} + \frac{{\partial}(a)}{{\partial}{y}}\hat{j}$
@Hippalectryon Firstly I have showed that $r(t)=(\cos^2 t-\frac{1}{2}, \sin t \cos t, \sin t)$ satisfies both of the given surfaces $x^2+y^2=\frac{1}{4}$ and $\left( x+\frac{1}{2}\right)^2+y^2+z^2=1$ and hence their intersection, that is $x^2+x=\frac{1}{2}$.
It remains to show that any point on that intersection is of the form $(\cos^2 t-\frac{1}{2}, \sin t \cos t, \sin t)$.
If $(x,y,z)$ lies on the sphere $\left( x+\frac{1}{2} \right)^2+y^2+z^2=1$ then $z^2=1-\left( x+\frac{1}{2} \right)^2-y^2 \leq 1$. Therefore $|z| \leq 1$. It follows that $z$ must be of the form $z= \sin t$ for s…
@evinda You know that any point on the intersection is either $(\cos^2t-1,\cos t\sin t,\sin t)$ or $(\cos^2t-1,-\cos t\sin t,\sin t)$. Now check whether both those cases actually work.
@evinda In your reasoning, you talk of $r(t)$. Where does it come from ? When you later say 'it remains to show', what is that based on ? WHat's the more global goal of the exercise ?
Show that $r(t)=\left (\cos^2 t-\frac{1}{2}, \sin t\cos t, \sin t\right )$ is a parametrization of the curve of intersection of the circular cylinder of radius $\frac{1}{2}$ and axis the $z$-axis with the sphere of radius $1$ and centre $\left (-\frac{1}{2}, 0, 0\right )$. This is called Viviani’s Curve. @Hippalectryon
For any $c_1(t)= \sin t \cos t$ there exists $t_2$ auch that $c_2(t_2)=- \sin{t_2} \cos{t_2}= \sin t \cos t$. We can pick $t_2=-t$. Then $c_2(t_2)=-\sin (-t) \cos (-t)= \sin t \cos t$. @Hippalectryon
let's say that I have a number C which will be hypotenuse of pythogorean triple. I need to find a,b such that a^2 + b^2 = c^2 and both a and b are integers.
So is the following justification right and sufficient? Or could we improve something? For any $c_1(t)= \sin t \cos t$ there exists $t_2$ auch that $c_2(t_2)=- \sin{t_2} \cos{t_2}= \sin t \cos t$. We can pick $t_2=\pi-t$. Then $c_2(t_2)=-\sin (\pi-t) \cos (\pi-t)= - \sin t (- \cos t)= \sin t \cos t$, $\cos^2(\pi-t)-\frac{1}{2}= \cos^2 t-\frac{1}{2}, \sin(\pi-t)=\sin t$. @Hippalectryon
@evinda I just realized I told you something that was wrong before. The existence of $t_2$ is not enough. However, here we have shown that for any $t$, $c_2(t)=c_1(\pi-t)$. As a result, the set of images of both curves are exactly identical. Hence the curves are the same (as in, their 3D representation is the same).
@Hippalectryon So can we justify it as follows? Or can we say it better? Since $- \sin t \cos t=\sin (\pi-t) \cos (\pi-t)$, $x(t)=x(\pi-t)$, $z(t)=z(\pi-t)$ we have that $y=\sin t \cos t$.
@hippa The solution is$$x}\right)\ln^{2}\left(x+1\right)+2\operatorname{Li}_{2}\left(x+1\right)\ln\left(x+1\right){-2\operatorname{Li}_{3}\left(x+1\right)}+C$$
@Chris'ssistheartist Basically $\int_0^1\ln^2(1+x)/xdx=\int_0^1\sum_{i,j=1}^\infty\dfrac{(-x)^{i+j-1}}{ij}=\sum_{i,j=1}^\infty(-1)^{k+j}$... which doesn't converge in the usual sense, but we might be able to do something with it
@Gato Hmm.... pas vraiment sur les groupes, mais ça me fait penser à l'exo suivant : soit $C_n$ l'ensemble des matrices de $\mathcal{M}_n(\mathbb{R})$ tq $^tAA=I_n$, on note $d_n$ la dimension maximale d'un sous espace vectoriel de $C_n$. donner $d_1,d_2,d_3$.
@Addem I'm working more on a computation than a proof. A function and its partition is given and we are supposed to find the upper and lower sums for that partition
So it sounds like for the lower sum you would use the lowest end-points of each partition, plug them into the function and collect a bunch of values. Pair them with the partition lengths and multiply pairwise. This results in a new list of values which you then sum. @Paradox101
@Paradox101 Yes, which would obviously not converge, which just means that this particular sum is not well-defined. That's not surprising since the area under the curve to infinity is not defined for this function. But that would be basically the procedure.
@Addem I just entered riemann sum followed by the function and the interval. In the detailed answer they mentioned whether it was using left or right endpoints
I'm not exactly sure what you mean by entering the Riemann sum but there's a chance Wolfram is just assuming that you want the exact integral, not an approximation by this particular partition.
I can't speak to what Wolfram is telling you without knowing what you entered, but the pattern you gave above, f(1.5)*0.5+f(2.4)*0.9+.., is correct.
What I'm looking for is some simple formula $\phi(\mathbb S)$ such that $$\phi(\mathbb S)\iff\left[\forall X, Y \in \Bbb S: X \ne Y \implies X \cap Y = \varnothing\right].$$
If a finite set $\mathbb{S}$ is disjoint, then $\left|\bigsqcup\mathbb{S}\right|=\left|\bigcup\mathbb{S}\right|$ holds. Does this also hold for infinite case?
