Can this equation be solved? $x+\sin(x)=\frac{11\pi}{48}$

Question

So I was revisiting an older problem and seeing if I could solve it in a different way. I boiled the equation down to this: $$x+\sin(x)=\frac{11\pi}{48}$$ I can't imagine how to isolate x, and a number of computer solvers also broke down in the effort. Desmos can solve it, but does not give the precise value. Is there a way to do it? Thanks.

Answer 1

$f(x)=x+\sin x$ is monotonically increasing, and without bounds, so there will be exactly one solution.

$y=\frac{11\pi}{48}$ is (relatively) small, so a first guess is obtained by setting $\sin x\approx x$ to get $x\approx \frac{11\pi}{96}$. An improved value is obtained by including the next term of the sine series, $$ 2x-\frac16x^3\approx y\implies x\approx \frac y2 + \frac{y^3}{96}. $$ This gives the numerical value $x=0.3638613210103829$ which is already close to the (more) exact value $0.363965532996313$.

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Instead of including more terms of the sine series, one could also directly iterate the fixed point equation $$ x=g(x)=\frac12(y+x-\sin(x)) $$

Answer 2

Hint.-The equation being trascendental we make numerical calculation and stop where we consider enough approximation. For this there are several ways. Here one.

First $a=\dfrac{11\pi}{48}\approx0.7194831$. Second in the neighborhood of $0$ one has $x\approx\sin(x)$ so $x+x=2x\approx0.7194831\Rightarrow x\approx0.359974$ which is a first approximation.

Now for $f(x)=x+\sin(x)$ we have successive values $$f(0.36)\approx0.712...\lt a\\f(0.365)\approx0.72...\gt a\\f(0.364)\approx0.72...\gt a\\f(0.363)\approx0.718...\lt a$$ Trying with $x$ from $0.3635$ till $0.3639$ we have $f(x)\lt a$ then with $x=0,36395$ till $0.363965$ we get $$f(0.363965)\approx 0.719947\approx a$$

We stop at this approximation $x\approx0.363965$

Answer 3

As said in comments and answers, finding the zero of function $$f(x)=x+\sin(x)-k$$ will require some numerical methods.

However, we can get approximations of the solution. For example, write $k=x+\sin(x)$, expand the rhs as a Taylor series and use series reversion to get $$x=\frac{k}{2}+\frac{k^3}{96}+\frac{k^5}{1920}+\frac{43 k^7}{1290240}+\frac{223 k^9}{92897280}+\frac{60623 k^{11}}{326998425600}+O\left(k^{13}\right)$$ Using it for $k=\frac{11\pi}{48}$ would give $x=0.3639655328$ while the "exact" solution would be $x=0.3639655330$.

Another way would be the use of $[1,n]$ Padé approximants and the succesive approximations would be given by $$\left( \begin{array}{cc} n & x_{(n)} \\ 1 & \frac{k}{2}\\ 2 & -\frac{24 k}{k^2-48}\\ 3 & \frac{k \left(k^2-48\right)}{4 \left(k^2-24\right)}\\ 4 & -\frac{80 \left(k^3-24 k\right)}{k^4-240 k^2+3840}\\ 5 & \frac{3 \left(k^5-240 k^3+3840 k\right)}{2 \left(11 k^4-960 k^2+11520\right)}\\ 6 & -\frac{28 \left(11 k^5-960 k^3+11520 k\right)}{k^6-1344 k^4+67200 k^2-645120} \end{array} \right)$$ which, for $k=\frac{11\pi}{48}$, would give the values $$\left( \begin{array}{cc} n & x_{(n)} \\ 1 & 0.3599741582 \\ 2 & 0.3639037546 \\ 3 & 0.3639471248 \\ 4 & 0.3639651460 \\ 5 & 0.3639655395 \\ 6 & 0.3639655330 \end{array} \right)$$

We could even recombine both approaches transforming the series expansion into a Padé approximant to get, as shorter expression, $$x=\frac k 2 \times \frac{1-\frac{541 }{6768}k^2+\frac{2221 }{2842560}k^4}{1-\frac{341 }{3384}k^2+\frac{697 }{379008}k^4}$$ which, for the worked case, would give again $x=0.3639655328$.