Find intersection of $|\cos(x)/2|$ and $|\arctan(x)|$

Question

I want to find the intersection for: $$\left|\frac{\cos(x)}{2}\right| = |\arctan(x)|, \ \forall x> 0$$

my attempt: I tried to find the value of $x$ as follows; $$\frac{\cos(x)}{2} = \frac{\cos(x)}{\sin(x)}$$

Then

$$\frac{1}{\sin(x)} = \frac{1}{2}$$

$$\csc(x)=\frac{1}{2}$$ Then $x$ does not have any answer. Is it correct or I made a mistake?

Answer 1

As Parcly Taxel already answered, you need a numerical metod to solve the equation, that is to say to find the zero of function $$y=\cos (x)-2\tan^{-1}(x)$$

Remember that is already the case for the "simple" $x=\cos(x)$.

Using Newton method, and being lazy, I shall start with $x_0=0$ and the iterates would be $$\left( \begin{array}{cc} n & x_n \\ 0 & 0.00000000000000000000 \\ 1 & 0.50000000000000000000 \\ 2 & 0.47609308186884695044 \\ 3 & 0.47614701448567249514 \\ 4 & 0.47614701474806253971 \end{array} \right)$$ which is recognized by inverse symbolic calculators.

You could get approximate solutions expanding $y$ as Taylor series around, say, $x= 0$ and get for example $$y=1-2 x-\frac{x^2}{2}+\frac{2 x^3}{3}+\frac{x^4}{24}-\frac{2 x^5}{5}-\frac{x^6}{720}+\frac{2 x^7}{7}+\frac{x^8}{40320}-\frac{2 x^9}{9}-\frac{x^{10}}{3628800}+O\left(x^{11}\right)$$ Now, using series reversion, this would give $$x=t-\frac{t^2}{4}+\frac{11 t^3}{24}-\frac{91 t^4}{192}+\frac{1141 t^5}{1920}-\frac{18937 t^6}{23040}+\frac{372539 t^7}{322560}-\frac{8694019 t^8}{5160960}+\frac{33483283 t^9}{13271040}-\frac{793236041 t^{10}}{206438400}+O\left(t^{11}\right)$$ where $t=\frac{1-y}2$.

Making $y=0$, that is to say $t=\frac 12$, you should end with $$x=\frac{902783482711}{1902536294400}\approx 0.474516$$ I suppose that you see how simple is Newton method for this problem.

Edit

There is another possible way to approximate the solution at the price of a simple linear equation.

Instead of using Taylor series, use the $[1,n]$ Padé approximant. As a function of $n$, the numerator and the solution of it are listed below $$\left( \begin{array}{cccc} n & \text{numerator} & x_{(n)}=& x_{(n)}\approx \\ 1 & 1-\frac{9 x}{4} & \frac{4}{9} & 0.444444 \\ 2 & 1-\frac{56 x}{27} & \frac{27}{56} & 0.482143 \\ 3 & 1-\frac{67 x}{32} & \frac{32}{67} & 0.477612 \\ 4 & 1-\frac{704 x}{335} & \frac{335}{704} & 0.475852 \\ 5 & 1-\frac{20711 x}{9856} & \frac{9856}{20711} & 0.475882 \\ 6 & 1-\frac{913376 x}{434931} & \frac{434931}{913376} & 0.476180 \\ 7 & 1-\frac{15344777 x}{7307008} & \frac{7307008}{15344777} & 0.476189 \\ 8 & 1-\frac{96681728 x}{46034331} & \frac{46034331}{96681728} & 0.476143 \\ 9 & 1-\frac{553781219 x}{263677440} & \frac{263677440}{553781219} & 0.476140 \end{array} \right)$$

Answer 2

$\left|\frac{\cos x}2\right|$ is periodic with period $\pi$ and attains a maximum of $\frac12$, while $|\tan^{-1}x|$ is strictly increasing over all $\mathbb R^+$. Furthermore, we note that $|\tan^{-1}\pi/2|>\frac12$ already, whereas $\left|\frac{\cos x}2\right|$ is strictly decreasing over $(0,\pi/2)$.

The original equation thus has exactly one positive solution, which does not have an elementary expression; it is the solution of $\cos x=2\tan^{-1}x$, and is approximately $0.476147$.