How can I relate this question to Series expansion of functions?

Question

if $\,x^3+y^3+xy-1=0\,$ then show that $y=1-\frac{x}{3}-\frac{26}{81}x^3+....$

Answer 1

There are two main ways that I can think of. One is to insert $y = a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots$, multiply out, sort by degree and solve. you get $$ x^3 + (a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots)^3 + x(a_0 + a_1x + a_2x^2 + a_3x^3 + \cdots) - 1 = 0\\ a_0^3 - 1 + (3a_1a_0^2 + a_0)x + (3a_0^2a_2 + 3a_0a_1^2 + a_1)x^2 + \cdots = 0 $$ (I might have made some mistakes here, it is messy.) Degree $0$ says that $a_0^3 - 1 = 0$, so $a_0 = 1$. Inserting that, degree $1$ says that $3a_1 + 1 = 0$, giving $a_1 = -\frac13$, and so on.

The other approach I would consider would be to implicitly differentiate. Insert $x = 0$ to get $y(0) = 1$. Differentiating your equation with respect to $x$ once gives the equation $$ 3x^2 + 3y'y^2 + xy' + y = 0 $$ Now inserting $x = 0$ and $y(0) = 1$ gives you $3y'(0) + 1 = 0$, showing that $y'(0) = -\frac13$. Differentiate once more to get $y''(0)$, and so on.

These two approaches are very closely related, as differentiating $n$ times and inserting $x = 0$ in the second approach is the same as isolating the degree $n$ coefficient from the first approach (multiplied by $n!$).