The form of $ \frac{\sin (2 x)}{x}=\sqrt{ (\frac{\sin (2 x)}{2 x}+2 \cos (2 x)-\frac 13 )^2-1}$ in terms of $\mathcal{O}(\frac 1x)$ as $x\to\infty$?

Question

I have this equation for $x\in(0,\infty)$ and over the real domain.

$$ \frac{\sin (2 x)}{x}-\sqrt{\left(\frac{\sin (2 x)}{2 x}+2 \cos (2 x)-\frac{1}{3}\right)^2-1}=0 $$

Can we say something about the approximate form of this equation as $x\to\infty$? I mean which terms may play the role as $x\to\infty$? What is the correct form of this equation in terms of $\mathcal{O}(\frac 1x)$?

Answer 1

Can we say something about the approximate form of this equation as $x \to \infty$?

It's a strange question. An equation is not a function, so it has not an "approximate form" as $x \to \infty$. Perhaps you mean its roots?

(Even then, I don't see how you could expect to write an expression of the roots that is a function of $x$!)

What you can readily see that, because $\sin$ and $\cos$ are bounded, for large $x$ (that is, for large roots) the equation should be asymptotically equivalent to

$$\left(2 \cos (z)-\frac{1}{3}\right)^2=1 \implies \cos z \in \{\frac{2}{3}, -\frac13\} \tag 1$$

where $z=2x$. Then the roots should be (asymptotically) around $ z=2 \pi k \pm a$ and $ z=2 \pi k \pm b$ where $a = \cos^{-1}(2/3)\approx 0.84107$ and $b=\cos^{-1}(-1/3)\approx 1.91063$

However, addionally we need to have $\sin(z)\ge 0$ so that the square root is defined, hence we are left with $ z=2 \pi k + a$ and $ z=2 \pi k + b$

This is a first order approximation, it could be refined.

Some graphs, for $k=15$: the blue line correspond to the approximate equation $(1)$, the red line to the exact equation. We are interested in the roots, ie, in the crossing with the horizontal axis. The first two graphs correspond to the valid roots , $z_1 \approx 30 \pi + a=95.08884828$ and $z_2 \approx 30 \pi + b=96.15841284$. The third graph correspond to one of "invalid" roots $30 \pi - a=93.40671094$.

Added - In terms of the original variable $x=z/2$, the $n-$th positive root (counting from $1$) would be at

$$ \pi k_n + a/2 = \pi \frac{(n-1)}{2} +a/2 = \frac{\pi}{2} n - 1.15026$$

for odd $n$, and

$$ \pi k_n + b/2 = \pi \frac{(n-2)}{2} +b/2 = \frac{\pi}{2} n -2.18628$$ for even $n$.

Answer 2

I shall not repeat what @leonbloy already wrote and commented and I shall just focus on the roots of the equation $$\frac{\sin (2 x)}{x}-\sqrt{\left(\frac{\sin (2 x)}{2 x}+2 \cos (2 x)-\frac{1}{3}\right)^2-1}=0$$ which are along two parallel straight lines correponding to the parity of $n$.

If $x_{(n)}$ denotes the $n^{\text{th}}$ root of the equation it seems that

• if $n$ is even :$\qquad(R^2 > 0.99999)$ $$x_{(n)}=a+ \frac \pi 2 n \qquad \text{with} \qquad a=-2.15907 \quad (\sigma=0.0164)$$
• if $n$ is odd $\qquad(R^2 >0.99999)$ $$x_{(n)}=b+ \frac \pi 2 n\qquad \text{with} \qquad b=-1.14697 \quad (\sigma=0.0079)$$

Could it be a coincidence that $(b \sim a+1)$ ?