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Continuous function sequences

  1. Sep 18, 2004 #1

    (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:
  2. jcsd
  3. Sep 18, 2004 #2


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    0.73908 looks right assuming a=1 rad.
  4. Sep 19, 2004 #3
    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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