How can I effectively prove that f(L) = L in this problem?

Click For Summary
SUMMARY

The discussion focuses on proving that for a continuous function f, if the sequence defined by a_n = f(a_{n-1}) converges to L, then f(L) = L. The user illustrates this with the function f(x) = cos(x) and starts with a = 1, estimating L to be approximately 0.73908. The user confirms the correctness of their approach and seeks alternative methods for finding L. Feedback suggests that presenting L explicitly on the left side of the equation enhances clarity in the proof.

PREREQUISITES
  • Understanding of limits and convergence in calculus
  • Familiarity with continuous functions
  • Basic knowledge of trigonometric functions, specifically cosine
  • Proficiency in numerical estimation techniques
NEXT STEPS
  • Study the properties of continuous functions and their fixed points
  • Learn about iterative methods for finding roots, such as the fixed-point iteration method
  • Explore numerical methods for estimating limits, including Newton's method
  • Investigate the implications of the Banach fixed-point theorem in real analysis
USEFUL FOR

Students of calculus, mathematicians interested in fixed-point theory, and anyone looking to deepen their understanding of continuous functions and numerical estimation techniques.

DivGradCurl
Messages
364
Reaction score
0
Problem:

(a) Let [tex]a_1 = a, a_2 = f(a), a_3 = f(a_2) = f(f(a)), \ldots, a_{n+1} = f(a_n),[/tex] where [tex]f[/tex] is a continuous function. If [tex]\lim _{n \to \infty} = L,[/tex] show that [tex]f(L) = L[/tex].

(b) Illustrate part (a) by taking [tex]f(x) = \cos x , a = 1,[/tex] and estimating the value of [tex]L[/tex] to five decimal places.

My answer:

(a) [tex]\lim _{n \to \infty} a_{n+1} = \lim _{n \to \infty} f(a_n) = f \left( \lim _{n \to \infty} a_n \right) = f(L) = L[/tex]

(b) I have used my calculator to get this one. First, I plugged in: [tex]\cos 1[/tex]. I took the cosine of the result. Then, I kept on taking the cosine until my results agreed to five decimal places. I got [tex]L \approx 0.73908[/tex].

My question:

Did I get it right? Are there other ways to find answer (b)?

Thanks a lot! :smile:
 
Physics news on Phys.org
0.73908 looks right assuming a=1 rad.
 
thiago_j said:
Problem:

(a) Let [tex]a_1 = a, a_2 = f(a), a_3 = f(a_2) = f(f(a)), \ldots, a_{n+1} = f(a_n),[/tex] where [tex]f[/tex] is a continuous function. If [tex]\lim _{n \to \infty} = L,[/tex] show that [tex]f(L) = L[/tex].

(b) Illustrate part (a) by taking [tex]f(x) = \cos x , a = 1,[/tex] and estimating the value of [tex]L[/tex] to five decimal places.

My answer:

(a) [tex]\lim _{n \to \infty} a_{n+1} = \lim _{n \to \infty} f(a_n) = f \left( \lim _{n \to \infty} a_n \right) = f(L) = L[/tex]

(b) I have used my calculator to get this one. First, I plugged in: [tex]\cos 1[/tex]. I took the cosine of the result. Then, I kept on taking the cosine until my results agreed to five decimal places. I got [tex]L \approx 0.73908[/tex].

My question:

Did I get it right? Are there other ways to find answer (b)?

Thanks a lot! :smile:
I think it would be more convincing if you wrote L= first on the left hand side and then came to f(L) on the right hand side. You basically have f(L)=L in your argument which is what you want to prove. Right idea though.
 

Similar threads

Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
8
Views
3K