Is it possible to solve sin(x) - $\frac{1}{x} = 0$ for x analytically?

Question

I was wondering how I might go about finding the roots of this equation? Can it only be done using root approximation methods like newton's method?

Answer 1

The answer is no. You need some sort of numerical method (fixed point iteration, perhaps).

Answer 2

This is a transcendental equation which does not show any closed form solution and, then, numerical methods are required.

Considering the problem of the first and second positive roots, I cannot resist the pleasure to show once more the beautiful approximation $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ whcih was proposed, more than 1400 years ago, by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician (see here).

If we use it, the problem reduces to $$16 x^3-4(4 \pi -1) x^2-4 \pi x+5 \pi ^2=0$$ Using Cardano's method, it is easy to show that there are three real roots and that one of them is negative. For sure, the analytical expressions are quite ugly but perfectly workable. The numerical values are $$x_1=1.11495\qquad x_2=2.77389$$ while the exact solutions, obtained using Newton method, are$$x_1=1.11416\qquad x_2=2.77260$$

As Henry commented, the next roots appear to be closer and closer to $n\pi$. As he proposed to do it, use Taylor expansions around $x=n\pi$. This will give $$\sin (x)-\frac{1}{x}=-\frac{1}{\pi n}+(x-\pi n) \left(\frac{1}{\pi ^2 n^2}+(-1)^n\right)+O\left((x-\pi n)^2\right)$$ from which $$x_n\approx\frac{\pi n \left(\pi ^2 (-1)^n n^2+2\right)}{\pi ^2 (-1)^n n^2+1}\tag 1$$ Just for illustration purposes, I wrote down the values generated using $(1)$ and the exact solutions obtained using Newton method. $$\left( \begin{array}{cc} n & \text{from } (1) & \text{Newton} \\ 1 & 2.78740 & 2.77260\\ 2 & 6.43841 & 6.43912\\ 3 & 9.31747 & 9.31724\\ 4 & 12.6454 & 12.6455\\ 5 & 15.6440 & 15.6440\\ 6 & 18.9025 & 18.9025\\ 7 & 21.9456 & 21.9456\\ 8 & 25.1725 & 25.1725\\ 9 & 28.2389 & 28.2389 \end{array} \right)$$

Just for the fun, using $(1)$ $$x_{1000}=3141.5929718996\color{red}{472} $$ while Newton would lead to $$x_{1000}=3141.5929718996\color{red}{525}$$