Solutions to $x \sin x=1$ in the interval $0 < x \leq 2\pi$

Question

If I'm in an exam and do not have access to any sort of a calculator, how would I solve it? What method is applicable here or do I have to manually plot points??

Answer 1

The equation you're trying to solve is $$\sin x=\frac{1}{x}.$$ Plot both sides of the equation on the same set of axes for $x\in[0,2\pi]$. The maximum value of $\sin x$ is $1$ at $x=\tfrac{\pi}{2}\approx 1.57$, while $1/x$ takes the value $1$ at $x=1<\pi/2$ and continues to decrease from there. From the picture, it's clear that there are two solutions.

Answer 2

Consider the function $$f(x)=x \sin(x)-1$$ By inspection, you can notice that $$f(0)=-1$$ $$f(\frac{\pi }{2})=\frac{\pi }{2}-1 >0$$ $$f(\pi)=-1$$ $$f(\frac{3\pi }{2})=\frac{-3\pi }{2}-1 <0$$ $$f(2 \pi)=-1$$ So, you know that there is one root between $0$ and $\frac{\pi }{2}$ and another one between $\frac{\pi }{2}$ and $\pi$.

You could refine the procedure looking at the value of the function at some intermediate points for which you know the value of the sine. This will allow you to bracket better the solution.

If you need to go further, then use Newton method for a few iterations.

Answer 3

Presumably you have no access to a graphing calculator. Never fear! The question can still be solved. Firstly as $\sin x<0$ for $x \in [\pi,2\pi]$. we can forget that region. Now $x\sin x$ is increasing in $[0,\pi/2]$ and at $\pi/2$ it is bigger than 1. S, there is one and only one solution in that region. $\dfrac{dx\sin x}{dx}=\sin x+x\cos x=\sin x(1+x\cot x)$. Furthermore, $\sin x$ is positive in our region and after $1+x\cot x$ becomes negative in $[\pi/2,\pi]$ it stays negative. So the function increases and then decreases. At $\pi $ f is less than 1. So, there is one more solution in between and these are the only solutions.

Answer 4

I looked at your question and solved it by educated guess. I turned it to $\sin x=\frac{1}{x}$ to make familiar functions appear. I sketched an arch of a sinusoid by hand ($0$ to $\pi$ is enough as $\frac{1}{x}$ remains positive), and a branch of the hyperbola. There are visibly two intersections, one before $\frac{\pi}{2}$ and one after. Took me less than a minute.

If needed, this quick search can be rigorously confirmed by splitting the curves into monotonic parts ($0$ to $\frac{\pi}{2}$ and $\frac{\pi}{2}$ to $\pi$) and checking crossings of the curves: $0$ < $+\infty$, $1>\frac{2}{\pi}$ and $0<\frac{1}{\pi}$.

Directly working with the form $x.\sin x$ is much less convenient: you know (possibly ;-)) that it starts from the origin as the parabola $x^2$ (where $\sin x\approx x$), falls back to $0$ at $\pi$, and must have a maximum in between. But then you need to compute the height of the maximum to make sure it is higher than 1.

Answer 5

Most of these are the easiest to solve with simple iteration. Express

$$x=\frac{1}{\sin x}$$ start somewhere reasonable ($x=1$ for instance) and repeat the iteration (hit the enter key until the digits stop moving).

You just have to be careful that the derivative of the expression at the solution is $<1$ by absolute value. But if it isn't, just turn it around (for instance if this didn't work, you'd try $x=\arcsin\frac{1}{x}$). It's also easy to know if diverges when you do it with a calculator.

This method doesn't require almost any preprocessing or derivation (like the Newton method does) and works for most transcendental functions. I use it all the time.

EDIT: I kind of overlooked that even an ordinary calculator isn't available. In that case, I'd still do the same. Use something like $\sin x\approx x-\frac{x^3}{6}$ and do 2 iterations, you get $1.1$, close enough to $1.114$.

Answer 6

For general functions, there is no systematic procedure to find all roots. One of the reasons is that there can be an infinity of them (just think of $\sin \frac{1}{x}=0$).

What you can do is heuristically find changes of sign and prove that the function is monotonous between them.

If you suspect the presence of multiple roots (which can result in no change of sign), work with $\frac{f(x)}{f'(x)}$ instead.