Strictly decreasing sequence (analysis)

Click For Summary
SUMMARY

The discussion focuses on the analysis of the recursive sequence defined by x_{n+1} = f(x_n), where f(x) = (3 sin(x))/(2 + cos(x)). It is established that the sequence {x_n} is strictly decreasing in the interval (0, (2/3)*pi] and converges to 0. The proof requires demonstrating that f(x) < x for all x in the specified interval and ensuring that the sequence remains within the domain. Additionally, the discussion addresses finding an integer k such that {n^(1/k) x_n} converges to a finite, non-zero real number.

PREREQUISITES
  • Understanding of recursive sequences and their properties
  • Knowledge of limits and convergence in calculus
  • Familiarity with trigonometric functions and their behavior
  • Proficiency in mathematical induction and inequalities
NEXT STEPS
  • Study the properties of recursive sequences in calculus
  • Learn about convergence criteria for sequences and series
  • Explore the behavior of trigonometric functions within specific intervals
  • Review techniques for proving inequalities and using induction in mathematical proofs
USEFUL FOR

Mathematics students, particularly those studying calculus and analysis, as well as educators looking for examples of recursive sequences and convergence proofs.

riskandar
Messages
3
Reaction score
0

Homework Statement



f(x) = (3 sin(x))/(2 + cos(x))

let x_0 be a number in (0, (2/3)*pi], and define a sequence recursively by setting
x_n+1 = f(x_n)

(1) prove that the sequence {x_n} is strictly decreasing sequence in (0, (2/3)*pi] and that lim x_n =0

(2) Find an integer k greater or equal to 1 such at {n^(1/k) x_n} is convergent to a finite, non-zero real number and evaluate lim n^(1/k) x_n

Homework Equations


The Attempt at a Solution



I tried the first problem by induction but I don't know how to prove it for P(n+1) (assuming P(n) is true). Any help is appreciated thank you.
 
Physics news on Phys.org
Don't think about this as an inductive argument at the bottom, but as an inequality.

Because your sequence is defined by iterating the function [tex]f[/tex], the simplest way to prove that [tex]x_{n+1} = f(x_n) < x_n[/tex] for all [tex]n[/tex] is to prove that [tex]f(x) < x[/tex] for all [tex]0 < x \leq \frac{2\pi}3[/tex].

(You also need to show that [tex]f(x)[/tex] still lies in [tex]\left(0, \frac{2\pi}3\right][/tex]; otherwise you can't conclude [tex]f(x_{n+1}) < x_{n+1}[/tex], because you don't know that [tex]x_{n+1}[/tex] is in the domain to which your inequality applies. This part of the argument is an induction.)

To show that [tex]x_n \to 0[/tex], you need a little more than that. When you prove the fact above, try to show not just that [tex]f(x) < x[/tex], but that [tex]f(x) < x[/tex] in some kind of systematic or predictable way. Think about examples of simple sequences that converge to zero, and try to establish a comparison with one of those.
 
Last edited:

Similar threads

  • · Replies 13 ·
Replies
13
Views
2K
Replies
1
Views
1K
Replies
34
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K