Finding the power series for $y$ where $y + \sin(y) = x$ 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.)
 A: 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
 $$\small f(x)=f^{[-1]}(y) = 1/2 x + 1/96 x^3 + 1/1920 x^5 + \ldots $$
A: lhf already brought up Lagrangian inversion (a special case of the more general Lagrange-Bürmann series):
$$f^{(-1)}(x)=\sum_{k=0}^\infty \frac{x^{k+1}}{(k+1)!} \left(\left.\frac{\mathrm d^k}{\mathrm dt^k}\left(\frac{t}{f(t)}\right)^{k+1}\right|_{t=0}\right)$$
There are a number of nice series reversion algorithms that make use of coefficients from the series to be inverted. Here is a Mathematica implementation of an algorithm due to Henry Thacher (also used here):
a = Rest[CoefficientList[Series[(x + Sin[x])/2, {x, 0, 20}], x]];
n = Length[a];
Do[
    Do[
      c[i, j + 1] = Sum[c[k, 1]c[i - k, j], {k, 1, i - j}];
      , {j, i - 1, 1, -1}];
    c[i, 1] = Boole[i == 1] - Sum[a[[j]] c[i, j], {j, 2, i}]
    , {i, n}];
Table[c[i, 1]/2^i, {i, n}]

{1/2, 0, 1/96, 0, 1/1920, 0, 43/1290240, 0, 223/92897280, 0, 
60623/326998425600, 0, 764783/51011754393600, 0, 
107351407/85699747381248000, 0, 2499928867/23310331287699456000, 0, 
596767688063/63777066403145711616000}

Thacher's method expects the series to be inverted to take the form $x+\cdots$ (and the series for $x+\sin\,x$ starts out $2x+\cdots$), so some rescaling is necessary before feeding the series coefficients to the algorithm; the division by powers of $2$ at the end recovers the coefficients of the original function to be inverted.
There a lot more methods (e.g. Carleman matrices, Bell polynomials); search around for more information on them.
A: You should compute the $n$th derivative of the given expression. So,
$$
  y(x)+\sin(y(x))=x
$$
and then $y(0)=0$. The first derivative will give
$$
 y'(x)(1+\cos(y(x))=1
$$
and so
$$
  y'(0)=\frac1{2}.
$$
Proceeding in a similar way you will get $y''(0)=0$ and $y'''(0)=\frac{1}{16}$ and eventually you will have
$$
  y(x)=\frac1{2}x+\frac1{16}\frac{x^3}{3!}+\frac1{16}\frac{x^5}{5!}+\frac{43}{256}\frac{x^7}{7!}+\frac{223}{256}\frac{x^9}{9!}+O(x^{11}).
$$
