Question

If $y=ax+b{x}^2+c{x}^3+\cdots$, express $x$ in ascending powers of $y$ as far as the term involving $y^3$.

Answer 1

Assume $x=py+q{y}^2+r{y}^3+\cdots$,
and substitute in the given series; thus
$y=a(py+q{y}^2+r{y}^3+\cdots )+b{(py+q{y}^2+\cdots )}^2+c{(py+q{y}^2+\cdots )}^3+\cdots$
Equating coefficients of like powers of $y$, we have
$ap=1$; whence $p=\frac{1}{a}$.
$aq+b{p}^2=0$; whence $q=-\frac{b}{a^3}$.
$ar+2bpq+c{p}^3=0$; whence $r=\frac{2b^2}{a^5}-\frac{c}{a^4}$.
Thus $x=\frac{y}{a}-\frac{b{y}^2}{a^3}+\frac{(2b^2-ac)y^3}{a^5}+\cdots$.