How to find the inverse of $f(x) = x+\sin(x)$, analytically?

Well how should I proceed to find the inverse of $f(x)$? Basically I have applied graphical approach to solve the equation, but I want to know the inverse equation by analytical method.

  • 3
    $\begingroup$ See this. $\endgroup$ Dec 18, 2014 at 9:41
  • $\begingroup$ You are unlikely to be able to achieve much more than a series expansion which might start $\dfrac12 y + \dfrac1{96} y^3 +\dfrac{1}{1920} y^5+\cdots$ $\endgroup$
    – Henry
    Dec 18, 2014 at 9:55

1 Answer 1


Let $y = x + \sin x = f(x)$; So, $x = f^{-1}(y)$. Differentiating the first expression according to $y$ leads to $x' + x'\cos(x) = 1$, or $x'(1+\cos x) = 1$. Let us put $x=\arccos t$, so $x' = -t'/\sqrt{1-t^2}$. This leads to $-t'(1+t)/\sqrt{1-t^2} = 1$, or $t'=-\sqrt{\frac{1-t}{1+t}}$. I believe that this is a standard differential equation. I don't know if it is solvable. If it is not, then the problem has no analytical closed expression.

Another way: Using the formula $\cos x = \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$, we are lead to the differential equation $x' = \frac{1+\tan^2(x/2)}{2}$. We can put $u = \tan(x/2)$, so $u'=\frac{1}{2}x'(1+u^2)$, and $x' = \frac{2u'}{1+u^2}$. So, the equation becomes $u' = \frac{(1+u^2)^2}{4}$. If I'm not wrong, there is a method to handle somehow this kind of equations. I will try tomorrow.

  • $\begingroup$ Ooops, it's true. Let me say that it was to test the vigilance of the student :-) . My answer remains exact, with some care regarding imaginaries (I believe). $\endgroup$
    – MikeTeX
    Dec 18, 2014 at 11:01
  • $\begingroup$ This is insufficient as an answer. Are you sure it leads to an explicit solution ? Can you exhibit it ? $\endgroup$
    – user65203
    Dec 18, 2014 at 12:08
  • $\begingroup$ Indeed, the hint was wrong. I have replaced it with another hint which is hopefully correct. $\endgroup$
    – MikeTeX
    Dec 18, 2014 at 13:39
  • $\begingroup$ Sooo... $x=-i\ln t$ AND $x=\ln t$? $\endgroup$
    – Did
    Dec 18, 2014 at 16:13
  • $\begingroup$ You pointed to a minor trailing assertion from a previous incorrect solution. This minor bug was corrected, but anyway, the solution is strange ; I don't find where is the problem (checking the solution directly shows that $x$ satisfies the differential equation $2x'+e^x + e^{-x} = 2$). $\endgroup$
    – MikeTeX
    Dec 18, 2014 at 16:40

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