Okay, first off, we are grown ups. $u$ is a vector. so is $v$. We don't need no steenking arrows. Use $\mathbf{u},\mathbf{v}$ if you really, really must
Consider the space $\omega_1$ (the first uncountable ordinal) together with the order topology (a convenient base for which is $\{(\alpha, \beta] \mid \alpha, \beta \in \omega_1\} \cup \{0\}$ ). I wish to show:
Any closed interval of $\omega_1$ is compact under the order topology.
At first ...
Let $\textbf{u},\textbf{v}$ $\in$ $\mathbb{R}^{n}$. Describe the vectors $\textbf{x} = s\textbf{u} + t\textbf{v}$, where $s + t = 1$. 1) Assume $\textbf{u}$ and $\textbf{v}$ are nonparallel. Assume $0 < s,t < 1$. Then it seems like $\textbf{x}$ is a vector inside of the parallelogram made by $\textbf{u}$ and $\textbf{v}$ 2) If $s$ or $t$ $= 1$, then we just have either $\textbf{u}$ or $\textbf{v}$.
Beyond that, I've got some general intuition, but I'm not sure where to go.
The problem I mentioned right above. I'm just not sure how to "linearly" work through the problem. I have some general ideas, but there seem to be tons of special cases, so it doesn't seem too efficient to just sketch everything.
I may help, I am currently in ninth grade, however, do not fret, as I have mastered all branches of mathematics, and physics. These include: astrophysics, quantum mechanics, thermodynamics,
electrodynamics, fluid-dynamics, magneto-hydrodynamics, linear algebra, abstract algebra, abstract linear algebra, etc. I've even made my own type of physics pretty much infinitely harder than astrophysics, 'phantomphysics' the study of scientific-fictional things. I also had to invent my own branch of math, which has over 2000 unique symbols. It is incomprehensibly hard, but this is extraneous.
First, I would like to know from where are you getting this problem?
the equation $x = su + tv$ describes something geometrically (do you know what that object is?); the vectors $x$ are those vectors that terminate along that geometrical object
@xander yeah from sketching it, I've been able to see what happens when $s$ and $t$ are both positive at least. $\textbf{x}$ is all the points inside the parallelogram made by $\textbf{u}$ and $\textbf{t}$
Ok, so I have $\textbf{x} = s\textbf{u} + (1-s)\textbf{v} = s(\textbf{u}-\textbf{v}) + \textbf{v}$ Since we can choose any $s$ $\in$ $\mathbb{R}$, $s(\textbf{u} - \textbf{v})$ is a line. That means $\textbf{x}$ is the line through $\textbf{u}$ and $\textbf{v}$?
I made this formula in fifth grade: $\int_1^ax^ndx=\frac{a^{n+1}}{n+1}-\frac{1}{5}(\frac{a^{n+1}}{n+1})+(\frac{1}{5}(\frac{n^{n+1}}{n+1})-\frac{1}{n+1})$
First one is to find the probability of a random inscribed triangle to contain the origin, second is to do the same for a random tetrahedron with vertices on the sphere.
@Semiclassical It is one of those arguments that seems obvious when you first see it, then you spend a long time convincing yourself that yes, it really is that obvious
I had taken a look at two other sources in the meantime. One is the original published solution to A6 in the 53rd Putnam exam, available here: math.hawaii.edu/~dale/putnam/1992.pdf
I have not yet encountered a set of numbers that defies this, so I am almost positive that if it contains an error it was through the writing on the computer.
@Semiclassical Here's another silly little geometric problem with a slick solution: Suppose that 5 points are placed on a sphere. Is it always possible to divide the sphere in half such that one of the two (closed---include the boundary) halves contains four of the five points?
Someone answer this question: If you select four random points on a sphere, what is the probability that the tetrahedron formed by the points contains the center of the sphere?
electrodynamics, fluid-dynamics, magneto-hydrodynamics, linear algebra, abstract algebra, abstract linear algebra, etc. I've even made my own type of physics pretty much infinitely harder than astrophysics, 'phantomphysics' the study of scientific-fictional things. I also had to invent my own branch of math, which has over 2000 unique symbols. It is incomprehensibly hard, but this is extraneous.
> Any intelligent fool can make things bigger, more complex, and more violent. It takes a touch of genius—and a lot of courage to move in the opposite direction.
a character of $G$ is a homomorphism $\chi : G \to \mathbb{T}$, where $\mathbb{T}$ is the one-torus (i.e. the unit circle), with a group structure given by complex multiplication
...full disclosure, for anyone still listening: while I may disdain trolling in the usual sense---acting like an a** in order to get a reaction---there is a certain kind of trolling I do appreciate.
So for part (b), we have $\textbf{x} = r\textbf{u} + s\textbf{v} + t\textbf{w}$, where $\textbf{u},\textbf{v},\textbf{w}$ $\in$ $\mathbb{R}^{3}$. Rewriting $r = 1 - s - t$ gives us $\textbf{x} = \textbf{u} + s(\textbf{v}-\textbf{u}) + t(\textbf{w} - \textbf{u})$ The lines $s(\textbf{v}-\textbf{u})$ and $t(\textbf{w} - \textbf{u})$ form a plane right? So then $\textbf{x}$ is just the plane defined by $\textbf{v}$ and $\textbf{w}$ intersecting $\textbf{u}$
Consider the space $\omega_1$ (the first uncountable ordinal) together with the order topology (a convenient base for which is $\{(\alpha, \beta] \mid \alpha, \beta \in \omega_1\} \cup \{0\}$ ). I wish to show:
Any closed interval of $\omega_1$ is compact under the order topology.
At first ...
