MHB Proving $(x_n,y_n)$ Converge to $(20,20) for All $k$

  • Thread starter Thread starter alexmahone
  • Start date Start date
  • Tags Tags
    Limit Sequence
alexmahone
Messages
303
Reaction score
0
$y_0=k$ where $k$ is a constant.

$x_{n+1}=30-\dfrac{y_n}{2}$

$y_{n+1}=30-\dfrac{x_{n+1}}{2}$

Prove that $(x_n, y_n)$ converges to $(20, 20)$ for all values of $k$.

My attempt:

I wrote a computer program and verified this for a few values of $k$. But I don't know how to prove that $x_n$ and $y_n$ converge.
 
Physics news on Phys.org
The dependence of $y_n$ on $x_n$ is a red herring, the two variables are separable as $x_{n + 1}$ can just be substituted:
$$y_{n + 1} = 30 - \frac{30 - \frac{y_n}{2}}{2}$$
Then once you find that $y_n$ converges to 20 regardless of $k$, it's easy to show that $x_n$ must too. You can easily find the limit of $y_n$ by noting that this is a contraction mapping and using the Banach fixed-point theorem ;)
 
Last edited:
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
##\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...
Back
Top