What is the power series expansion at $x=0$ of the algebraic function defined by $(27x-4)y^3 + 3y + 1 = 0$?

Question

Let $y$ denote the complex-valued algebraic function defined implicitly near $x=0$ by $(27x - 4)y^3 + 3y + 1=0$ and such that $y(0)=1$. What is the power series expansion of this function at $x=0$? The first few terms are $$ y = 1 + 3x + 15x^2 + 84x^3 + 495x^4 + \cdots $$

I have a couple of ideas about how to approach this problem, but none of them have panned out.

• Use implicit differentiation to compute $y^{(n)}(0)$ recursively. You'll end up with the equation $$ (27x-4)\sum_{i+j+k=n}\binom{n}{i,j,k}y^{(i)}(0)\cdot y^{(j)}(0)\cdot y^{(k)}(0) + 27\sum_{i+j+k=n-1}\binom{n}{1,i,j,k}y^{(i)}(0)\cdot y^{(j)}(0)\cdot y^{(k)}(0) + 3y^{(n)} = 0, $$ which could then be used to compute $y^{(n)}(0)$ recursively. But this equation looks prohibitively complicated.
• Use the cubic formula to compute an explicit formula for $y$ near $x=0$. I don't like this method because it wouldn't generalize to algebraic functions of higher degree. Also, I'm not that familiar with the cubic equation.
• Substitute an indeterminate power series into the equation defining $y$ and solve recursively for the coefficients. But it seems too difficult to compute the power series expansion of $y^3$ from the power series expansion of $y$. It also seems like this won't generalize to the higher-degree case.

More broadly, does the general version of this problem -- compute the power series expansion of a given algebraic function about a given point -- have an algorithmic solution? What if we work over a different field than $\mathbb C$?

I'd prefer a closed-form expression for the coefficients of the power series, but a recurrence relation is almost as good. For the general problem, I expect that finding a recurrence relation is the best we can do.

Answer 1

By the Lagrange inversion theorem, the solution of $w^3-w=-x$ has the following Taylor series: $$ w(x)=\sum_{k\geq 0}\binom{3k}{k}\frac{x^{2k+1}}{2k+1} $$ whose radius of convergence is $\frac{2}{3\sqrt{3}}$. If we set $y(x)=\sqrt{\frac{3}{4-27 x}}\,w(x)$, by Lagrange inversion:

$$ y(x)=\sum_{k\geq 0}\binom{3k}{k}\frac{3(-1)^k}{(2k+1)(4-27x)^{k+1}}\tag{1}$$

and we may apply stars and bars to recover the coefficients of the Taylor series as convolutions.

If you show that your function $y$ fulfills $y=w'(\sqrt{x})$, with $w$ being the solution of $w^3-w=-x$, you prove Robert Israel's extraordinary simple claim:

$$ y(x)=\sum_{k\geq 0}\binom{3k}{k}x^k .\tag{2}$$

It is probably easier to go in the opposite direction: take $w(x)$ as the solution of $w^3-w=-x$, define $y$ as $w'(\sqrt{x})$ and show by differentiation that $y$ is a solution of $(27x-4)y^3+3y+1=0$.

Still another way is to show that

$$ y(x)=\frac{2\cdot\cos\left(\frac{1}{3} \arcsin\left(\frac{3 \sqrt{3x}}{2}\right)\right)}{\sqrt{4-27 x}}\tag{3}$$

by solving the original third-degree equation through Chebyshev polynomials, then finding the Taylor coefficients of $(3)$ through Euler's beta function, leading to $(2)$.

Answer 2

$$ \sum_{k=0}^\infty {3k \choose k} x^k$$

See OEIS sequence A005809