Finding $f(x)$ for any $x$ by assuming $a_{n+1}$

Question

I'm trying to solve this problem in the implicit differentiation section of the book I'm going through:

The equation that implicitly defines $f$ can be written as

$y = \dfrac{2 \sin x + \cos y}{3}$

In this problem we will compute $f(\pi/6)$. The same method could be used to compute $f(x)$ for any value of $x$.

Let $a_1 = \dfrac{2\sin(\pi/6)}{3} = \dfrac{1}{3}$

and for every positive integer $n$ let

$a_{n+1} = \dfrac{2\sin(\pi/6) + \cos a_n}{3} = \dfrac{1 + \cos a_n}{3}$

(a) Prove that for every positive integer $n$, $|a_n - f(\pi/6)| \leq 1/3^n$ (Hint: Use mathematical induction) (b) Prove that $\lim_{n \to \infty} a_n = f(\pi/6)$

Now, I'm not sure how to solve $f(\pi/6)$ in the base case of the induction proof.

Also, does this above pattern of assuming $a_{n+1}$ and using that to compute $f(x)$ for any value of $x$ has any name ? I would like to read more about it.

Answer 1

It revolves around recognizing that $|\cos(x)-\cos(x)|\leq |x-y|$ for all $x,y\in\mathbb{R}$. This can be determined using the Mean Value Theorem. Then by induction we have $$\begin{align*} \Big|a_{n+1} - f\Big(\frac{\pi}{6}\Big)\Big| &= \Big|\frac{1 + \cos(a_n)}{3} - \frac{2\sin(\pi/6) + \cos(f(\pi/6))}{3}\Big| \\ &= \Big|\frac{1 + \cos(a_n)}{3} - \frac{1 + \cos(f(\pi/6))}{3}\Big| \\ &= \frac{|\cos(a_n) - f(\pi/6)|}{3} \leq \frac{1}{3^{n+1}}\end{align*}$$

Answer 2

For the induction base case, note that the RHS must be in the range $[0,2/3]$, so the LHS, and therefore the solution, must also be in that range. Letting our first guess be $1/3$, we are at most $1/3$ away.