Proving that the coefficents of the inverse formal power series takes the form $b_{n} = \frac{-1}{a_{0}} (\sum_{i=1}^n a_{i} b_{n-i}). $ \
I found this proof online:
But I guess it is incorrect because it did not use the order of the power series $ a = \sum_{n}a_{n}x^n,$ am I correct? If so, below is my trial for the solution, If not please tell me how the argument used is the same as order argument.
My trial:
I got $a = a_{0}( 1 + xb(x))$ then the inverse $\frac{1}{1+xb(x)} = \sum (-xb(x))^k$ but then what happen to the first non-zero term I do not know, I got a hint that it starts further and furthur away but I do not know why and how that will help me in my proof? could any one help me in clarifying my confusions please?
Looking here we have the formula to multiply two series
$$f(x)=\sum _{n=0}^{\infty } a_n x^n;\;g(x)=\sum _{m=0}^{\infty } b_m x^n$$ $f$ and $g$ are inverse if $(fg)(x)=I(x)$, identity $$f(x)g(x)=\sum _{n=0}^{\infty } a_n x^n\sum _{m=0}^{\infty } b_m x^m=\sum _{k=0}^{\infty } c_k x^k$$ where coefficients $c_k$ are computed as follows $$c_k=\sum _{j=0}^{k } a_jb_{k-j}$$ In our problem we have the first coefficent $c_0=1$ and all $c_k=0$ for $k>0$.
Thus $a_0b_0=1\to b_0=\frac{1}{a_0}$
$$c_1=a_0b_1+a_1b_0= 0\to b_1=-\frac{1}{a_0}\left(\frac{a_1}{a_0}\right)$$ Coefficient $n$-th is zero, so we have, if we know the previous $(n-1)$ coefficients $$a_0b_n+a_1b_{n-1}+\ldots+a_{n-1}b_1+a_nb_0=0$$ $$b_n=-\frac{1}{a_0}\left(a_1b_{n-1}+\ldots+a_{n-1}b_1+a_nb_0\right)$$ and finally $$b_n=-\frac{1}{a_0} \sum_i^n a_ib_{n-i}$$
Hope it is useful