The Result of Dividing 2 Power Series

Question

Is there a way to write a single series for the following division? $$\frac{\displaystyle\sum_{i=0}^{\infty}a_{i}x^{i}}{\displaystyle\sum_{j=0}^{\infty}(j-2)a_{j}x^{j}}$$

Thanks, Radz.

Answer 1

As told in answers and comments, a series exists but it is quite tedious to derive. Writing $$\frac{\displaystyle\sum_{i=0}^{\infty}a_{i}x^{i}}{\displaystyle\sum_{j=0}^{\infty}(j-2)a_{j}x^{j}}=\sum_{i=0}^{\infty}b_{i}x^{i}$$ the first coefficients are given by $$b_0=-\frac{1}{2}$$ $$b_1=-\frac{a_1}{4 a_0}$$ $$b_2=\frac{a_1^2-4 a_0 a_2}{8 a_0^2}$$ $$b_3=\frac{-a_1^3+4 a_0 a_2 a_1-12 a_0^2 a_3}{16 a_0^3}$$ $$b_4=\frac{a_1^4-4 a_0 a_2 a_1^2+8 a_0^2 a_3 a_1-32 a_0^3 a_4}{32 a_0^4}$$ $$b_5=\frac{-a_1^5+4 a_0 a_2 a_1^3-4 a_0^2 a_3 a_1^2+16 a_0^3 a_4 a_1-16 a_0^3 a_2 a_3-80 a_0^4 a_5}{64 a_0^5}$$ in which a few patterns seem to appear.

Answer 2

Hint: Note that $\displaystyle\sum_{j=0}^{\infty}(j-2)a_{j}x^{j}=(x\frac d{dx}-2)\sum_{i=0}^{\infty}a_{i}x^{i}$