If ${x_{n + 1}} = \cos {x_n}$, prove that $\mathop {\lim }\limits_{n \to \infty } {x_n}$ exists.
Note 1: As we want to prove that a limit of a sequence exists, some methods can be used.
- $\left\{ {{x_n}} \right\}$ is bounded and monotonic, then $\left\{ {{x_n}} \right\}$converges.
- $\left\{ {{x_n}} \right\}$ converges if and only if $\left\{ {{x_{2n}}} \right\}$ and $\left\{ {{x_{2n+1}}} \right\}$ both converge to the same limit.
- In any metric space $X$, every convergent sequence is a Cauchy sequence. And every Cauchy sequence converges in a complete metric space.
Definiton of Cauchy sequence: A sequence ${x_n}$ in a metic space $X$ is said to be a Cauchy squence if \[\forall \varepsilon > 0,\exists N > 0{\rm{ s}}{\rm{.t}}{\rm{. }}\forall n,m > N \Rightarrow \left| {{x_m} - {x_n}} \right| < \varepsilon. \]
It's obvious that in Cauchy sequence, $x_n$'s are very closed to each other whereas $n$ is sufficiently large.
Since the sinusoid is bounded by 1, we only need to consider $\left| {{x_0}} \right| \le 1$. Otherwise, we will omit ${{x_0}}$ and it does nothing to the convergence of this sequence.
Proof 1: We are easy to check that whatever ${{x_0}}$ is, $\left\{ {{x_{2n}}} \right\}$ and $\left\{ {{x_{2n+1}}} \right\}$ both are monotonic with different monotonicity. Thus they both converges to some nonnegative numbers, say $a$ and $b$, respectively. However, from the iterative equation ${x_{2n}} = \cos \left( {\cos {x_{2n - 2}}} \right)$ we get that $a = \cos \left( {\cos a} \right)$. Similarly, $b = \cos \left( {\cos b} \right)$. If $a = b$, we complete the proof by method 2 in the Note 1 above.
Now let $f\left( x \right) = \cos \left( {\cos x} \right) - x$, $x \in \left[ { - 1,1} \right]$. Differentiating $f\left( x \right)$ we can easily conclude that $f\left( x \right)$ is monotone-decreasing. Since $f\left( { - 1} \right) > 0$, $f\left( { 1} \right) < 0$ and $f\left( x \right)$ is continuous on a closed interval, by the Intermediate Value Theorem of continuous function, the equation $f\left( x \right) = 0$ has a unique solution. Thus $a = b$. Q.E.D.
Proof 2: Now we will use the Cauchy criterion to prove this proposition. Let $m > n$, then \[\begin{array}{l}
\left| {{x_m} - {x_n}} \right| = \left| {\cos {x_{m - 1}} - \cos {x_{n - 1}}} \right|\\
= 2\left| {\sin \frac{{{x_{m - 1}} + {x_{n - 1}}}}{2}\sin \frac{{{x_{m - 1}} - {x_{n - 1}}}}{2}} \right|\mathop \le \limits^{{\rm{(a)}}} 2\left( {\sin 1} \right)\left| {\sin \frac{{{x_{m - 1}} - {x_{n - 1}}}}{2}} \right|\\
\le \left( {\sin 1} \right)\left| {{x_{m - 1}} - {x_{n - 1}}} \right| \le {\left( {\sin 1} \right)^n}\left| {{x_{m - n}} - {x_0}} \right| \le {\left( {\sin 1} \right)^n}\left( {1 + \left| {{x_0}} \right|} \right).
\end{array}\]
Where inequality (a) follows form the fact that $\left| {{x_n}} \right| < 1,\forall n \Rightarrow \left| {\sin \frac{{{x_{m - 1}} + {x_{n - 1}}}}{2}} \right| < \sin 1$.
Since $\sin 1 < 1$ strictly, $\forall \varepsilon > 0$, $\exists N > 0$ s.t. $\forall n > N \Rightarrow {\left( {\sin 1} \right)^n} < \frac{\varepsilon }{{1 + \left| {{x_0}} \right|}}$. Thus $\forall n,m > N \Rightarrow \left| {{x_m} - {x_n}} \right| < \varepsilon $. We have prove that $\left\{ {{x_n}} \right\}$ is a Cauchy sequence, it then convergs. Q.E.D.
Note 2: As in proof 2, we recognize that $f(x)= \cos {x}$ is a contraction since $\left| {\cos x - \cos y} \right| < \theta \left| {x - y} \right|$ where $\theta = \sin 1 < 1$. The following contraction principle tells us that there exist one and only one $x$ such that $\cos x = x$.
Theorem (Contraction Principle): If $X$ is a complete metric space, and if $\varphi $ is a contraction of $X$ into $X$, then there exist one and only one $x \in X$ such that $\varphi \left( x \right) = x$. (cf: Principles of Mathematical Analysis Rudin, Page220)