What is $x$ in closed form if $2x-\sin2x=\pi/2$, $x$ in the first quadrant?


The solution is given by $$\displaystyle x = \pi/4 + D/2$$

where $\displaystyle D$ is the root of $\cos y = y$

The root of $\displaystyle \cos y = y$ is nowadays known as the Dottie Number and apparently has no known "closed form" solution. If you consider this number to be part of your constants, then the above can be considered a closed form solution.

For a proof:

If $\displaystyle y = \sin(2x)$

then we have that

$\displaystyle 2x = \pi/2 + y$

$\displaystyle y = \sin 2x = \sin (\pi/2 + y) = \cos y$.

The root of $$\displaystyle y = \cos y$$ is $\displaystyle y = 0.739085\dots$

Notice that $\displaystyle \pi/2 > x \gt \pi/4$ (as $\displaystyle f(x) = 2x - \sin 2x$ is increasing in $\displaystyle [0,\pi/2]$), so if $\displaystyle x = \pi/4 + z$ then

$\displaystyle \sin(2x) = \sin(\pi/2 + 2z) = \cos 2z = 0.739085\dots$

And thus $\displaystyle z = \dfrac{0.739085\dots}{2}$.

Thus $$\displaystyle x \sim \pi/4 + \dfrac{0.739085}{2} \sim 1.154940730005\dots$$

See Also: A003957.


Repeating the same thing as the other answers, but with a moderately more elegant one:

Let $y = 2x - \frac{\pi}{2}$, then substituting:

$y = sin(y+\frac{\pi}{2})$

$y = cos(y)$

$y = D$


$x = \frac{y+\frac{\pi}{2}}{2} = \frac{D}{2} + \frac{\pi}{4}$


An analytical form of x can be obtained solving Kepler equation:

$$M= E-\epsilon \sin(E)$$

with eccentricity=1 and mean anomaly = $\pi/2$ by means of Kapteyn series:

$$2x = \frac{\pi}{2}+\sum_{n=1} \frac{2J_n(n)}{n} \sin(\pi n/2)$$

where $J_n()$ are the Bessel functions. Simplifying:

$$2x = \frac{\pi}{2}+\sum_{n=0} \left( \frac{2J_{4n+3}(4n+1)}{4n+1} - \frac{2J_{4n+3}(4n+3)}{4n+3}\right)$$

$$x = \frac{\pi}{4}+\sum_{n=0} \left( \frac{J_{4n+1}(4n+1)}{4n+1} - \frac{J_{4n+3}(4n+3)}{4n+3}\right)$$

Such series can be numerically evaluated, but it converges slowly and n=10000 terms are required to obtain:

$$x = 1.154940317134$$



In order to improve the convergence, we can employ an acceleration series technique as Levin's acceleration. (See http://en.wikipedia.org/wiki/Series_acceleration)

With only 10 (ten!) terms we obtain:


A simple c++ code, based on gsl library is the following:

#include <iostream>
#include <fstream>
#include <iomanip>
#include "gsl_sf.h"
#include "gsl_sum.h"
using namespace std;

#include <cmath>

int main(int argc, char* argv[])
    double PIH = atan(1.)*2;
    double E=PIH;

    cout<<"raw series"<<endl;
    //raw series
    for( int i = 0 ; i < 1e4; i +=2 )
        double term = 2*gsl_sf_bessel_Jn( 2*i+1, 2*i+1 )/(2*i+1);
        double term2 = 2*gsl_sf_bessel_Jn( 2*i+3, 2*i+3 )/(2*i+3);

        E += (term-term2);
    cout<< E/2<<endl;

    cout<< "error: "<<E-sin(E)-PIH<<endl;

    cout<<"levin accelerated series"<<endl;
    const int N = 10;
    double t[N];
    double sum_accel=0, err;

    gsl_sum_levin_u_workspace* w =
        gsl_sum_levin_u_alloc( N );

    t[0] = PIH;
    for( int i = 1 ; i < N; i++ )
        double term = 2*gsl_sf_bessel_Jn( 4*i-3, 4*i-3 )/(4*i-3);
        double term2 = 2*gsl_sf_bessel_Jn( 4*i-1, 4*i-1 )/(4*i-1);

        t[i] = term-term2;

    gsl_sum_levin_u_accel( t, N, w, &sum_accel, &err );



    cout<<"error: "<<sum_accel-sin(sum_accel)-PIH<<endl;

  • $\begingroup$ By the way, the formula $$\sum_{n=1}^\infty \frac{2J_n(n)}{n} \sin(\pi n/2)$$ works for Dottier number. $\endgroup$
    – Anixx
    Mar 4 '15 at 10:19
  • $\begingroup$ terms for n=2,4,6,.. can be deleted (sin=0). Therefore we could rewrite the sum as: 2$\sum_{n=0} \left( \frac{J_n(4n+1)}{4n+1} - \frac{J_n(4n+3)}{4n+3}\right)$ Is it right? $\endgroup$ Mar 4 '15 at 10:27
  • $\begingroup$ Why the index of the Bessel functions is not equal to the argument in your expression? $\endgroup$
    – Anixx
    Mar 4 '15 at 10:39
  • $\begingroup$ I AGREE WITH YOU! TYPO! $\endgroup$ Mar 4 '15 at 10:41
  • $\begingroup$ I fixed the math. As you can see the code was consistent with the correct formula... I will fix the math also at the Dottie thread $\endgroup$ Mar 4 '15 at 10:43

To get a graphical impression of the solution have a look here:


Edit: That WA doesn't give a closed form is at least a hint that there might be none (very probable)

  • $\begingroup$ If you assume the Dottie number to be part of the "closed form" constants, then there is a closed form solution. See my answer. $\endgroup$
    – Aryabhata
    Nov 15 '10 at 19:35
  • $\begingroup$ @Moron: Well, I don't know if there is some "degree of closeness" but I guess this way you could define everything as "closed form" (which is often done btw, just think of Pi, e, Si(x), Li(x) asf... $\endgroup$
    – vonjd
    Nov 15 '10 at 19:42
  • $\begingroup$ Not sure what you mean. All I am saying is that it can be written in terms of well known constants. I guess we are just agreeing :-) $\endgroup$
    – Aryabhata
    Nov 15 '10 at 19:45
  • 2
    $\begingroup$ @Moron & @vonjd: You're both correct. As I said somewhere here previously, "closed-form" is just shorthand for "popular enough to be given a name and notation." $\endgroup$ Nov 15 '10 at 23:16

As others have noted, $ x = \pi/4 + D/2. $

Based on iteration $ \cos y = y $ there is a fun method to find D using a hand calculator:

Enter any real positive number and press Cos button some 10 or 15 times until none of last digits changes... this is the Dottie number.


use $$\sin (2 x)\simeq \sum _{n=0}^{\infty } \frac{i e^{-\frac{i}{2}} (-1)^n x^{n/2} \left((2 i)^n \Gamma \left(n+1,-\frac{i}{2}\right)-e^i (-2 i)^n \Gamma \left(n+1,\frac{i}{2}\right)\right) I_n\left(2 \sqrt{x}\right)}{2 \Gamma (n+1)}$$ few terms gives good polynomial approximating root or $$\sin (2 x)\simeq \sum _{n=0}^{\infty } \frac{e^{-\frac{i}{2}} (-1)^n \left(e^i E_{-n}\left(\frac{i}{2}\right)+E_{-n}\left(-\frac{i}{2}\right)\right) x^n \, _0\tilde{F}_1(;n+1;x)}{4 \Gamma (n+1)}$$


Not the answer you're looking for? Browse other questions tagged or ask your own question.