Proving f(L) = L for a sequence of iterates and finding fixed points

[DivGradCurl]

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!

 

[Gokul43201]

0.73908 looks right assuming a=1 rad.

 

[phoenixthoth]

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!

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.