Infinitely nested inverse

Calculus Level 4

Let $f : [-1, 1] \to [0, \pi]$ be defined as $f(x) = \arccos(x)$ . Define $f_{n}(x)$ as $\underbrace{f \circ f \circ f \circ \cdots \circ f}_{\text{n times}}$ if it's defined.

For some $x \in [-1, 1]$ , $\displaystyle \lim_{n \to \infty}{f_{n}(x)} = A$ , a real number. Evaluate $A$ to $3$ decimal places.

The answer is 0.739.

1 solution

Akeel Howell
Apr 22, 2017

Let $\displaystyle \lim_{n \to \infty}{f_n{(x)} = L} \implies \cos{L} = \lim_{n \to \infty}{\cos{ \left( f_n{(x)} \right) }}.$ So $L = \cos{L}$ .

But $L = \arccos{(\cos{L})} \implies \arccos{L} = L \\ \therefore \cos{L} = \arccos{L}.$

Using the fact that $\cos{ \left( \dfrac{\pi}{4} \right) } = \dfrac{1}{\sqrt{2}}$ and $\dfrac{\pi}{4} - \dfrac{1}{\sqrt{2}} = 0.07829...$ , we can perform a linearization at $x = \dfrac{\pi}{4}$ to estimate $L$ $\left( \text{since } \cos{\dfrac{\pi}{4}} = \dfrac{1}{\sqrt{2}} \text{ and } \dfrac{\pi}{4} \approx \dfrac{1}{\sqrt{2}} \right)$ .

The linearization formula is $f(x) \approx f(x_0) + f'(x_0)(x-x_0)$ .

Hence, $L \approx \dfrac{1}{\sqrt{2}} - \dfrac{1}{\sqrt{2}} \left( L - \dfrac{\pi}{4} \right) = \dfrac{1}{\sqrt{2}} - \dfrac{L}{\sqrt{2}} + \dfrac{\pi}{4\sqrt{2}} \\ \sqrt{2} L + L \approx 1 + \dfrac{\pi}{4} \implies L \approx \dfrac{4 + \pi}{4(\sqrt{2} + 1)}$

$\text{So } \ L \approx 0.739$ ( $3$ decimal places).

Note that it is not true " we can proceed by iterating f until it converges".

As pointed out in the reports, we have a repelling fixed point. You have to start with the exact value in order to get back to it.

As an explicit example, if the question was: $g(x) = 2 x$ , $\lim g^{(n) } (x)$ converges, find the limit.
Then, no amount of "iterating g" will let it converge unless we started out with $x = 0$ .

Calvin Lin Staff - 4 years, 1 month ago

Thanks. I've removed that line from the solution.

Akeel Howell - 4 years, 1 month ago

Infinitely nested inverse

The answer is 0.739.

1 solution

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