What is the inverse of
$$f(x)=\sin(x)+x.$$
I thought about it for a while but I couldn't figure it out and I couldn't find the answer on the internet.
What about
$$f(x)=\sin(a \cdot x)+x$$
where $a$ is a known real constant.
Thank you for taking the time to read this question! Sorry if this has been asked before...
On
Why not give another “closed form” since @GEdgar’s solution for the Kepler equation technically has $E(a,M)$ as the solution in the Wikipedia link. Here is a solution to the inverse of $y=\sin(x)+x$ using mathematica function using Inverse Beta Regularized which is a standard function introduced in $1996$. The answer is from:
where
$$M=E-e\sin(E)\iff x=y-a\sin(y)\implies y=\text E(a,x)$$
and
$$\text E(-1,x)=2\sin^{-1}\sqrt{\text I^{-1}_{\frac x\pi}\left(\frac12,\frac32\right)}=\text{hav}^{-1}\left(\text I^{-1}_{\frac x\pi}\left(\frac12,\frac32\right)\right)$$
Here is a numerical tester for the inverse function
Here is a plot of:
$$\boxed{x=\sin(y)+y\implies y=\text E(-1,x) = \text{hav}^{-1}\left(\text I^{-1}_{\frac x\pi}\left(\frac12,\frac32\right)\right),0\le x\le \pi}: $$
The code is inversehaversine(inversebetaregularized(x/pi,1/2,3/2))
where appears the Inverse Haversine, a transformed inverse sine function.
The differential equation for the inverse of $y=\sin(x)+x$ is:
$$y’\text{vercos}(y))=1$$
where vercos is the versed cosine, a transformed cosine function. The inverse function’s domain is extended by using a series expansion of $\text I^{-1}_z(a,b)$. Please correct me and give me feedback!
Unfortunately, it seems that no more solutions of the inverse of $x-a\sin(x)=y$ can be solved for using this method. The only exception is that the inverse of $x-\sin(x)$ can be given in closed form by switching the $\frac32$ and $\frac12$ in the boxed answer.
On
Using the fixed-point technique seems to be the simplest solution I found:
$y = x - \sin(y)$
$y = x - \sin(x - \sin(y))$
$y = x - \sin(x - \sin(x - \sin(y)))$
…
$y = x - \sin(x - \sin(x - \sin(x - \sin(x - ... \sin(y) ...))))$
You can assign any value to $y$ on the right side since the effect is eventually eliminated after numerous iterations of the sine operation.
Here are the plotted functions, where the $z$ slider represents the $y$ value on the right side of the equation:
As observed, the source and inverse functions exhibit symmetry with respect to the line $y = x$.
On
That's a kind of Kepler's equation. Already Liouville proved that the inverse isn't an elementary function:
$$x+\sin(ax)=f(x)$$ $$x-(-1)\sin(ax)=f(x)$$ $x\to\frac{t}{a}$: $$t-(-a)\sin(t)=af(x)$$
see How to solve Kepler's equation $M=E-\varepsilon \sin E$ for $E$?
1) Lambert W, Generalized Lambert W
$$\sin(ax)+x=y$$
$$-\frac{1}{2}i\left(e^{aix}-e^{-aix}\right)+x=+y$$
$$-\frac{1}{2}i\left(e^{aix}\right)^2+\frac{1}{2}i+e^{aix}x-e^{aix}y=0$$
$$e^{aix}=-ix\pm\sqrt{-x^2+2yx-y^2+1}+iy$$
We see, the inverse cannot be represented in closed form in terms of Lambert W either, and not by generalized Lambert W of Mezö et. al. But possibly by other generalized Lambert W - see the references below.
2) "Leal-functions"
$$\sin(x)+x=y$$ because $\sin(it)=i\sinh(t)$: $x\to it$: $$\sin(it)+it=y$$ $$i(\sinh(t)+t)=y$$ $$\sinh(t)+t=-iy$$ $$t=\text{Lsin}_2(-iy)$$ $$x=i\ \text{Lsin}_2(-iy)$$
[Mezö 2017] Mezö, I.: On the structure of the solution set of a generalized Euler-Lambert equation. J. Math. Anal. Appl. 455 (2017) (1) 538-553
[Mezö/Baricz 2017] Mezö, I.; Baricz, A.: On the generalization of the Lambert W function. Transact. Amer. Math. Soc. 369 (2017) (11) 7917–7934 (On the generalization of the Lambert W function with applications in theoretical physics. 2015)
[Castle 2018] Castle, P.: Taylor series for generalized Lambert W functions. 2018
Kepler's equation ... its solution is known not to be an elementary function.