This function appears to be monotonic but interestingly Wolfram alpha and other symbolic math utilities I've tried haven't been able to invert this (aka solve for x=).

Is it possible to invert this function? If not, why is it so difficult?

  • 5
    $\begingroup$ It is a transcendental equation and there does not exist any analytic form of inverse. You could try to evaluate it using numerical methods $\endgroup$ Jun 3, 2018 at 15:54
  • $\begingroup$ nothing is difficult if there exists one (for this question) and if there exists no one (no analytic form), should we call it as difficult? $\endgroup$ Jun 3, 2018 at 15:59

3 Answers 3


The inverse function could be represented as an infinite series.

From $$y=x+\sin x$$ the inverse function satisfies, $$ y+\sin y = x $$

Let $$y=y(0)+ y'(0)x + y''(0) x^2/2+.....$$

We can find the derivatives using the equation $$ y+\sin y = x $$

We have $y(0)=0$.

Differentiation of $$ y+\sin y = x $$ implies $$ y'+\cos y y'=1$$

Evaluating at $x=0$, we have $ y'(0)=1/2 $

Similarly we can find higher derivatives and find the power series for $y$.


The function $f(x) = \sin(x) + x$ is in fact invertible. You have $f'(x) = \cos(x) + 1 \ge 0$ so that it is increasing. Since $f'(x) = 0$ only at the isolated points $x_k = (2k+1)\pi$, it is strictly increasing. Moreover $lim_{x\to \pm \infty} f(x) = \pm \infty$ so that $f$ is a bijection $\mathbb{R} \to \mathbb{R}$.

Unfortunately the inverse of $f$ cannot be expressed in terms of "elementary functions". However, as Mohammad Riazi-Kermani has explained, you can find a power series which allows to approximate $f^{-1}$.

  • $\begingroup$ You have a power series for $f(x)$ since you have one for $\sin(x)$. Concerning inverting a power series see math.stackexchange.com/q/478281. $\endgroup$
    – Paul Frost
    Jun 3, 2018 at 17:39

A way to get a finite approximation that can work is to write the inverse recursively.

$$y+\sin y = x$$ $$y = x-\sin y$$ $$y = x - \sin(x-\sin(y))$$ $$y = x-\sin(x-\sin(x-\sin(\cdots)))$$ At this point you can choose to go arbitrarily as many recursive steps as you like to try to improve on this.

Now to show this works, I'll show that for this recurrence, $a_{n+1}=a_0 - \sin(a_n)$ the difference $|a_{n+1}-a_n|$ gets smaller with each iteration from any arbitrary starting $a_0$. If $a_{n+1}=a_n$ then it's converged, otherwise we can start from this fact,

$$\left|\sin\left(\frac{a_{n+1}-a_n}{2}\right)\right| < \left|\frac{a_{n+1}-a_n}{2}\right|$$ Cosine can only make this term to the left smaller as it's bounded between -1 and 1, so this doesn't affect the inequality,

$$\left|\sin\left(\frac{a_{n+1}-a_n}{2}\right)\cos\left(\frac{a_{n+1}+a_n}{2}\right)\right| < \left|\frac{a_{n+1}-a_n}{2}\right|$$

A trig identity allows us to simplify the left side,

$$\left|\frac{\sin(a_{n+1})-\sin(a_n)}{2}\right| < \left|\frac{a_{n+1}-a_n}{2}\right|$$ Multiplying both sides by 2 and writing 0 in the fancy way $a_0-a_0$ we have,

$$|-a_0+\sin(a_{n+1})+ a_0-\sin(a_n) | < |a_{n+1}-a_n|$$ These are the next terms in the recurrence, and this difference is strictly less than

$$|a_{n+2}-a_{n+1}| < |a_{n+1}-a_n|$$

So it does converge. The actual reasoning to work this out was to start from this end and work backwards to the start.


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