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Given a transcendental equation $f(x,y)=0$, is there a way for Mathematica to automatically solve the equation as a series? I already know that I can use
NSolve[f(x,y)==0,y]
if I substitute a value for $x$. However, what I'm specifically wondering is if Mathematica can output something that looks like:
$$y=c_0+c_1x+c_2x^2+\cdots$$
or
$$y=c_0+c_1x^{-1}+c_2x^{-2}+\cdots$$
where $c_i$ is a numerical constant.
Possibly, expand f to first order in y
Series[f[x, y], {y, y0, 1}] // Normal;
(Solve[(% /. y - y0 -> z) == 0, z] /. z -> y - y0)[[1, 1]]
(* y - y0 -> -(f[x, y0]/Derivative[0, 1][f][x, y0]) *)
Then, the right side of the last expression can be expanded in x to the desired power. For instance,
Series[%[[2]], {x, x0, 2}] // Normal
(* -(f[x0, y0]/Derivative[0, 1][f][x0, y0]) +
((x - x0)*(-(Derivative[0, 1][f][x0, y0]*Derivative[1, 0][f][x0, y0]) +
f[x0, y0]*Derivative[1, 1][f][x0, y0]))/Derivative[0, 1][f][x0, y0]^2 +
((x - x0)^2*(2*Derivative[0, 1][f][x0, y0]*Derivative[1, 0][f][x0, y0]*
Derivative[1, 1][f][x0, y0] - 2*f[x0, y0]*Derivative[1, 1][f][x0, y0]^2 -
Derivative[0, 1][f][x0, y0]^2*Derivative[2, 0][f][x0, y0] +
f[x0, y0]*Derivative[0, 1][f][x0, y0]*Derivative[2, 1][f][x0, y0]))/
(2*Derivative[0, 1][f][x0, y0]^3) *)
yielding a power series, here to third order, in x - x0.
Incidentally, an illustration of the approach is given in my answer to question 94663.
f, so you'd have to be more specific. If, e.g., you can solve forx, then the series forycan be obtained directly usingInverseSeries. $\endgroup$ – Jens Sep 15 '15 at 2:56