Continuous function sequences

  • #1
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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:
 

Answers and Replies

  • #2
Gokul43201
Staff Emeritus
Science Advisor
Gold Member
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0.73908 looks right assuming a=1 rad.
 
  • #3
1,605
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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.
 

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