How does one get $\sum_{n=-\infty}^{\infty}\lvert n\rvert^2 \left (\lvert a_n\rvert^2+\lvert b_n\rvert^2\right)=1$, by applying Parseval's identity to $\frac 1 {2\pi}\int_0^{2\pi}(x'(s)^2+y'(s)^2)ds=1$
Hi. If I have 3 vectors a, b and u and a triple inner product ((a.u).b).a is it acceptable to say something like a.u = |a||u|cos(a,u) a real number so it becomes |a||u|cos(a,u) . a . b ?
Also, to whoever was doing the flagging, Sir Mix a Lot hasn't been offensive since the 80s
(also, it took me way too long to Google whether or not Sir Mix a Lot should be hyphenated---based on his official website, it is correct as I have written it above)
What is a wavicle? I learned in electronics class that electrons are little particles. In physics they even say that the electron orbits the nucleus thus exhibiting angular momentum. But in chemistry they were little electron clouds. And physics deals with position and momentum in a probabilistic...
> In my first year of college teaching I watched a freshman chemistry student asked a recent Harvard Ph.D. how the electron in one lobe of a p-orbital could get to the other lobe if there was zero electron density at the node? The brilliant professor was stumped, and I felt bad for him. But I used that story for years to encourage my students who were also stumped, and to help them realize: "You're thinking about it all wrong: they're not particles! And they're not waves! They're wavicles!"
@Abcd google taylor expansion, i think it can be understood at high school level , just computational stuff, it is just derivatives but not only first order, second order etc
I know that $\{e^{in}:n\in\Bbb N\}$ is dense in unit circle in complex plane. For what values of $\theta$, is the set \{e^{in\theta}:n\in\Bbb N\} dense in unit circle?
@Silent sorry I was out but the idea there is that if you rotate by a rational multiple of $\pi$ you end up being periodic
So yeah if you look at, say $e^{i\pi n\theta}$ for some $\theta = \frac{p}{q} \in \mathbb{Q}$, then for $n = 2qk$, you hit $1$
Then you just repeat
So there are only those values
Now, turns out the converse is true, if you are choosing $e^{i\pi n \theta}$ where $\theta\notin\mathbb{Q}$, the orbit should be dense
For this, note that you can identify the circle as $\mathbb{R}/\mathbb{Z}$, just using the exponential map, so you try to think about the stuff using addition mod 1
So okay, if you assume the orbit is a finite set, then you can find some $x$ such that $x = x + n\theta$ mod 1
But yeah since $\theta$ is irrational and $n > 0$ you're in trouble
So if your angle is irrational, your orbit is infinite, but by compactness there's a limit point, meaning you can find $x + n\theta$ and $x+m\theta$ within $\epsilon$ of each other, but then $(n-m)\theta$ is gonna be $\epsilon$-dense
I want to examine the convexity of the following set $\{(x_{1},x_{2}):x_{2} \le e^{x_{1}}\}$, checking the graph i can see that it is not convex, but how can i check it in theory (method?). Most questions i found online were about functions with equality (like $f(x,y)=$ something), but here it is an inequality, how can i solve this?
In equalities i can use hessian matrix, what to do here?
@TobiasKildetoft, Conway says: " An open set $G\subset \Bbb C$ is connected iff for any two points a, b in G there is a polygon from a to b lying entirely inside G. " Can't we say this about any set $G\subset \Bbb C$?
Hi. Does anybody have an idea how I can show that the closure of a set $Y$ is closed without using limit points? My definitions are $\partial Y= \{x \in X \mid \forall \varepsilon \gt 0: B_\varepsilon(x) \cap Y \neq \emptyset \neq B_\varepsilon(x) \cap (X \setminus Y) \}$ is the boundary of $Y$, and $\overline Y = Y \cup \partial Y$ is the closure.
Can some one confirme me that the limits of $(\frac1n,0)$ are all $(x,y)\in \mathbb{R}^2$ such that $$(x+2)^2+(y-2)^2\geq 8 $$? here : math.stackexchange.com/questions/2752199/…
A set is open if for every point in the set there is an $\varepsilon$-ball around that point which is contained in the set
and closed if complement open
Though I know the result that a set is open iff every convergent sequence whose limit is in the set has almost all of its elements in the set and a set is closed iff every convergent sequence which has all its elements in the set also has its limit in the set
hi guys, if $A \subset X$, $A$ convex and absorbing and $X$ vector space. Defining with $\mu_A(x) = \inf \left\{t : t^{-1} x \in A \right\}$ why is $\mu_A(sx) = s \mu_A(x)$, if $s \geq 0$?
This creature exists in an infinite world, a flat landscape that extends ad infinitum, where light rains from the sky from infinity during recurrent day-night cycles, and, similarly, the ground goes down continuously.
All kinds of critters populate the cosmos including the skies and the undergro...
An organism with the shape of a network is really the most reasonable choice, since an infinite organism has to have all its digestive system and whatnot to be everywhere in order to not starve to death
which is why fungus sounds natural
You cannot have a infinite but bounded organism as approaching the limit, the thickness will soon become so small that its physical effects no longer matter
Background
Theorem. Let $(X,\tau_X)$ and $(Y,\tau_Y)$ be two topological spaces and $f:(X,\tau_X)\to(Y,\tau_Y)$ be a homeomorphism between them. Then there exists a bijection $\varphi:\tau_X\to\tau_Y$.
Sketch of the Proof. Define $\varphi:\tau_X\to\tau_Y$ by $\varphi(U)=f(U)$. Then the m...
@user170039 ok - sometimes images can be disruptive in chat, so the least destructive thing I could think of was to un-one-box it. Everybody please be respectful of each other. :)