Unique solution of $x = \cos\left(\frac{x}{2}\right)$ How does one show that there is a unique solution to this equation?
$$x = \cos\left(\frac{x}{2}\right)$$
Furthermore how can we find it? 
 A: Note that the function $f\colon\mathbb{R}\to\mathbb{R}$ defined by $$f(x) = \cos\frac{x}{2}-x$$ is differentiable, with $f'(x) = -\frac{1}{2}\sin\frac{x}{2} -1 <0$ for all $x$, and $$\lim_{x\to-\infty} f(x)=+\infty, \qquad \lim_{x\to+\infty} f(x)=-\infty$$
By the intermediate value theorem (IVT), and strict monotonicity, there exists a unique $x^*\in\mathbb{R}$ such that $f(x^*)=0$.

With regard to finding the value of this $x^\ast$, however, I do not think there is any closed-form solution; besides numerical approaches to try and obtain close approximations (for instance, $f(0)=1 > 0$, $f(1)<0$, so $x^\ast\in(0,1)$; then one can for instance proceed by binary search: checking whether $f(1/2)>0$, etc.).
A: HINT
Let consider
$$f(x)=x-\cos\left(\frac x 2\right) \implies f'(x)=1+\frac12 \sin \left(\frac x 2\right)>0$$
then $f(x)$ is strictly increasing.
A: You may proceed as follows:


*

*$f(x) = \cos \frac{x}{2}$ is continuous and $f'(x) = -\frac{1}{2}\sin \frac{x}{2}$

*$f([0,\pi]) = [0,1] \subset [0,\pi]$

*According to the Fixed Point Theorem for continuous functions you get
$$\Rightarrow \exists x^{\star} \in [0,1]: f(x^{\star}) = x^{\star}$$
Only the uniqueness is missing. Assume there is a second fixpoint $x^{\star\star}$:
$$\Rightarrow \color{blue}{|x^{\star} - x^{\star\star}|} = |f(x^{\star}) - f(x^{\star\star})| = |f'(\xi)||x^{\star} - x^{\star\star}|\color{blue}{\leq \frac{1}{2}|x^{\star} - x^{\star\star}|}\quad (1)$$
$$\Rightarrow \boxed{\color{blue}{x^{\star} = x^{\star\star}}}$$
Besides this, a similar calculation as in (1) shows that the iteration $\color{blue}{x_{n+1}=f(x_n)\mbox{ converges to}}$ the fixpoint $\color{blue}{x^{\star}\mbox{ for any}}$  starting value $\color{blue}{x_0}$.
