B Define Vector on x-Axis: -d/2 to d/2

  • B
  • Thread starter Thread starter Philosophaie
  • Start date Start date
  • Tags Tags
    Vector
Philosophaie
Messages
456
Reaction score
0
How do I define a vector starting on the x-Axis at -d/2 ending at d/2 on the x-Axis?
 
Physics news on Phys.org
Philosophaie said:
How do I define a vector starting on the x-Axis at -d/2 ending at d/2 on the x-Axis?
You define it as ##(d)##.

A vector notation doesn't say anything about where it is applied at. It simply has a length - here ##d## - and a direction - here only in one coordinate, so ##(*)\,##. If you want to describe a vector as its point of application and its length and direction, then you need to use pairs ##(p;\vec{v})##. Here it would be ##(p;\vec{v})=(-\frac{d}{2};d)##.
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top