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JH
Numerade Educator

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Problem 84 Medium Difficulty

(a) Let $ a_1 =a, a_2 = f(a), a_3 = f(a_2) = f( f(a)), . . . , a_{n + 1} = f(a_n), $ where $ f $ is a continuous function. If $ lim_{n \to\infty} a_n = 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.

Answer

(a) $$
f(L)=f\left(\lim _{n \rightarrow \infty} a_{n}\right)=\lim _{n \rightarrow \infty} f\left(a_{n}\right)=\lim _{n \rightarrow \infty} a_{n+1}=L
$$
(b) $$
L \approx 0.73909
$$

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Video Transcript

So notice our sequence here we're starting off. Our first term is a constant and then the general term is given by Ryker. Shin or F is continuous. So we're trying to show a f of l equals l. So let's start with the left hand side Now, using the definition of El, I can rewrite Ellis this limit over here and now. Here's the key step. Using continuity of f, we can take the limit from inside and put it outside. So this is key here. This is given information that we're using f his continuous allows youto pull the limit outside. Now half of a n buy up. This this expression up here we know that it's just an plus one. And if you notice over here the limit of a gazelle, if you add one to the end, it's and it's still going to infinity. So this expression is also equal to help. And that's what we wanted to show f of l equals l and part B. They're giving you a constant A and a specific function f and we want to find the value l In our case, we could just go ahead and put a plus one. And this will be limit as n goes to Infinity Co signed to the end power of one. And so at this point, I would just go into a calculator, preferably a scientific. You have co sign of one and then insert that is co sign of the previous answer co sign of the previous answer and so on. And you want to do this until the first five decimal places stopped moving and the first time that should happen. The decimal is point seven three nine o nine. So this is a estimation of l two five decimal places, and that completes part B.