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DivGradCurl
Sep18-04, 10:28 PM
Problem:

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

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

My answer:

(a) \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

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

My question:

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

Thanks a lot!!! :smile:

Gokul43201
Sep18-04, 11:42 PM
0.73908 looks right assuming a=1 rad.

phoenixthoth
Sep19-04, 12:52 AM
Problem:

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

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

My answer:

(a) \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

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

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.