Rewrite power expansion and finding coefficients of $ f(x) = \sum_{i=0}^{\infty} x^i = \sum_{l=0}^{\infty} b_l y^l$

Question

Let $f(x) = \sum_{i=0}^{\infty} x^i$. And let $x= \sum_{j=0}^{k} a_j y^j$. I want to express $f(x)$ in terms of $y$ in a clever way. i.e.,
\begin{align} f(x) = \sum_{i=0}^{\infty} x^i = \sum_{i=0}^{\infty} \left( \sum_{j=0}^{k} a_j y^j \right)^i = \sum_{l=0}^{\infty} b_l y^l \end{align} Is there any clever way to rewrite $b_l$ in terms of $a_j$?

Answer 1

$$f(x) = \sum_{i=0}^{\infty} x^i = \sum_{i=0}^{\infty} \left( \sum_{j=0}^{k} a_j\, y^j \right)^i = \sum_{l=0}^{\infty} b_l \,y^l=\frac 1{1-\sum_{j=0}^{k} a_j\, y^j}$$ This gives $$b_0=\frac{1}{1-a_0}\qquad\qquad b_1=\frac{a_1}{(1-a_0)^2}\qquad\qquad b_2=\frac{a_1^2-a_0 a_2+a_2}{(1-a_0)^3}$$ $$b_3=\frac{a_1^3-2 a_0 a_1 a_2+2 a_1 a_2+a_0^2 a_3-2 a_0 a_3+a_3}{(1-a_0)^4}$$

The denominators are simple but numerators start to be quite messy