TITLE: Finding the power series for $y$ where $y + \sin(y) = x$
QUESTION [12 upvotes]: What do you do to find the power series for an inverse relationship such as for $y$ in $y + \sin(y) = x$?  I'm not sure where to begin.
(Similarly, the Lambert $W$ function has such a power series for $y$ in $y = W(x)$ ($y e^y = x$) which allows computation.)

REPLY [6 votes]: The Bell/Carleman-method, which J.M. mentions is my favorite method for tasks like this, which can be done by paper&pen for a truncation to coefficients at low index . First step is to list sufficiently many formal powers of the powerseries in question. So if
 $\small f(y)=y +  \sin (y) =  2y-y^3/3!+y^5/5!- \ldots + \ldots $  
then we list the first few formal powers of f(y):
$$\small
\begin{array} {lll} f(y)^0&=& 1 + O(x^8)\\
 f(y)^1&=&  2*x - 1/6*x^3 + 1/120*x^5 - 1/5040*x^7 + O(x^8)\\
 f(y)^2&=&  4*x^2 - 2/3*x^4 + 11/180*x^6 + O(x^8)\\ 
 f(y)^3&=&  8*x^3 - 2*x^5 + 4/15*x^7 + O(x^8)\\
 f(y)^4&=&  16*x^4 - 16/3*x^6 + O(x^8) \end{array} $$
 then the Carleman-matrix looks like
$$\small M_{f(y)}=
 \begin{bmatrix} 1&.&.&.&.&.&.&.\\  .&2&.&-1/6&.&1/120&.&-1/5040\\  .&.&4&.&-2/3&.&11/180&.\\  .&.&.&8&.&-2&.&4/15\\  .&.&.&.&16&.&-16/3&.\\  .&.&.&.&.&32&.&-40/3\\  .&.&.&.&.&.&64&.\\  .&.&.&.&.&.&.&128 \end{bmatrix} 
 $$
where the coefficients of the above powers of f(y) are filled rowwise into the matrix. This can be done using paper&pen for the first few rows/columns.
Because the matrix is triangular it is then again easy to invert it, by paper&pen at least for some leading terms and this gives:
$$\small M_{f^{[-1]}(y)}=
 \begin{bmatrix} 1&.&.&.&.&.&.&.\\  .&1/2&.&1/96&.&1/1920&.&43/1290240\\  .&.&1/4&.&1/96&.&29/46080&.\\  .&.&.&1/8&.&1/128&.&17/30720 \end{bmatrix} 
 $$
We need do this even only to the second row, because this contains the coefficients for the formal powerseries for the (compositional) inverse
 $$\small f(x)=f^{[-1]}(y) = 1/2 x + 1/96 x^3 + 1/1920 x^5 + \ldots $$