5

($x \in \mathbb{R}$) Graphically, it's obvious that the equation should have 3 solutions for x, but I can't think of any way to solve this without resorting to computation of [the Maclaurin series for $\sin(2x)$]$\div x$ or some cleverer computational trick.

I considered representing $\sin(2x)$ as $\left(1-\frac{x}{\frac{1}{2} \pi}\right)\left(1+\frac{x}{\frac{1}{2} \pi}\right)\left(1-\frac{x}{\frac{2}{2} \pi}\right)\left(1+\frac{x}{\frac{2}{2} \pi}\right)...$, but that seems too daunting to be of any use.

Edit: Note that I'm looking for a way to find the exact answer (i.e. not just an approximation), or a proof that it's impossible to find.

Michael Hardy
  • 1
Meow
  • 6,353
  • 3
    One of the solutions is exactly $0$. The other two are atractive fixed points of $\sin(2x)$, so just iterating the function gets you alternately upper and lower bounds on the solution, about $\pm0.94774713\ldots$. Newton's method would get it a lot faster. – Michael Hardy Dec 08 '12 at 20:04
  • What do you mean by "find the exact answer"? What sort of answer counts? – Chris Eagle Dec 08 '12 at 20:53
  • Exact in the way that $\sqrt{2}$ is, and $1.414$ is not; that is, an answer that, as written (in finite time) on paper, we know to infinite accuracy. – Meow Dec 08 '12 at 20:55
  • 2
    OK, the solutions are $0$, the unique positive real $a$ such that $\sin(2a)=a$, and $-a$. – Chris Eagle Dec 08 '12 at 21:00
  • I didn't specify that $sin(2x)=x$ and $sin(2x)=-x$, I was more looking for the 'exact' form of the number $\pm 0.94774713...$. – Meow Dec 08 '12 at 21:06
  • 1
    This is a transcendental equation and as I know it can't be solved exactly, except for trivia solutions. – m0nhawk Dec 08 '12 at 21:13
  • @monhawk: Ok, what are you using "solved exactly" to mean? – Chris Eagle Dec 08 '12 at 21:22
  • @ChrisEagle: Under "solved exactly" I mean finding such $x$ that is equals to finite number of any combination of elementary functions. – m0nhawk Dec 08 '12 at 21:43
  • 2
    Keep in mind "The positive solution to $\sin(2x) = x$" is an exact solution to your equation. It may even be useful for many applications. –  Dec 08 '12 at 22:30

2 Answers2

3

How about $$ x = \frac{\pi}{2} \mathrm{sinc}^{-1}\left(\frac{1}{2}\right) $$ where $\mathrm{sinc}$ is defined by $$ \mathrm{sinc}(u) = \frac{\sin(\pi u)}{\pi u} . $$

GEdgar
  • 111,679
  • I suppose given that there are few exact values of $\sin(2x)$, this is the best possible (I was asking too much for an exact solution to an equation including a function that doesn't have exact values for most $x$). – Meow Dec 08 '12 at 21:44
2

Three solutions exist, as is apparent by the graphs of $x$ and $\sin (2x)$.

One solution is trivially $0$.

The other two solutions, $S_1$ and $S_2=-S_1$, which don't have elementary closed forms, can be obtained to sufficient accuracy with Newton's method (or another root-finding algorithm).

$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} = x+\frac{x-\sin (2x)}{2\cos(2x)-1}$$

where the choice of $x_0$ will give one of the three different solutions.

Argon
  • 25,303