For example, if $$ f(x)^5 + f(x) = x $$
i.e. the solution $ s $ to the quintic equation $ s^5 + s - x = 0 $, how can we find a Maclaurin/Taylor series expansion for $ f(x) $?
According to WolframAlpha, the solution in this case is a hypergeometric function, which does have a series approximation, so how can it be found?
My attempt was to use implicit differentiation:
$$ \text{Let } y = f(x) \rightarrow y^5 + y = x \rightarrow 5y^4 + 1 = \frac{\text{d} x}{\text{d} y} \rightarrow \frac{\text{d} y}{\text{d} x} = \frac{1}{5y^4 + 1} $$
so $ f(x) \approx f(0) + x \cdot \frac{1}{5f(0)^4 + 1} \approx x $ where $ f(0) = 0 $ by solving the original.
However this seems to stop working after the linear term as I cannot do a simple reciprocal of the $ \frac{\text{d}^2 x}{\text{d} y^2} $ to get the desired $ \frac{\text{d}^2 y}{\text{d} x^2} $.
This function (up to a sign convention) is called the Bring radical, and its Taylor series expansion can be computed using Lagrange inversion. The answer turns out to be unexpectedly nice:
$$f(x) = \sum_{k=0}^{\infty} (-1)^k {5k \choose k} \frac{x^{4k+1}}{4k+1}.$$
This can be interpreted in terms of counting $5$-ary trees as explained e.g. here. (That post considers inverting a slightly different function but the two are related by a suitable change of coordinates.)