convolution of sums

Question

I have to find an infinite power series $B(x)$ in accordance to an already given infinite power series $A(x)$ so that $B(x)A(x) = 1 $

$A(x)$ is defined as follows: $$1 + \sum_{n \ge 1 } a_n Z^n$$ when one considers the $0$. term as 1 then $$a_0Z^0 + \sum_{n \ge 1 } a_n Z^n = \sum_{n \ge 0 } a_n Z^n$$

so far I tried to define $B(x)$ as $$ B(x) := \sum_{n \ge 0 } b_n Z^n$$ with these limitations: $b_0 = 1 ; b_1= -1; b_i = 0$ for every $i > 1$

under the use of the convolution of sums $$\sum_{n \ge 0}^{} (\sum_{k=0}^{n} a_k b_{n-k})$$ I could cancel many terms, but not the problem is that I also cancel the first and the second $1$ with $-1$

$$ a_0b_0 + (a_0b_1 + a_1b_0) + (a_0b_2 + a_1b_1 + a_2b_0) + ... = 1 + (-1 +a_1) + (0 + -a_1+a_2) + ...$$

as you can see every last term is being canceled with the term before the last one in the next brackets but the two first cancel each other as well. How can one do it in this method so that it will fit nevertheless?

Thank you for your help

Answer 1

Trying to guess the $b_n$ coefficients in that manner will not work. There is a useful trick in problems like these though. Put $$C(z): = \sum_{n=1}^\infty a_nz^n,$$ so $A(z) = 1+C(z)$. In some sufficiently small neighborhood of $0$ we may assume $|C(z)|<1$, so by uniform convergence of geometric series $$\frac{1}{A(z)} = \frac{1}{1+C(z)} = \frac{1}{1-(-C(z))} = \sum_{j=0}^\infty (-1)^j(C(z))^j.$$ In other words, $$B(z) = 1-\bigg(\sum_{n=1}^\infty a_nz^n\bigg)+\bigg(\sum_{n=1}^\infty a_nz^n\bigg)^2-\bigg(\sum_{n=1}^\infty a_nz^n\bigg)^3+\dotsm.$$ Writing out an explicit formula for the coefficients would be quite tedious, but do note that $z^k$ only appears in the firt $k+1$ summands by the way the powers increase. We have $b_0=1$, and for $k\geq 1$ we have $$b_k = -a_k+(C^2)_k-(C^3)_k+\dotsm+(-1)^k(C^k)_k$$ where $(C^j)_k$ denotes the sum of all terms $a_{i_1}\dotsm a_{i_j}$ where $i_r\geq 1$ and $i_1+\dotsm+i_j = k$.

Answer 2

We can conveniently use the convolution formula to calculate the coefficients $b_n, n\geq 0$. We have the following situation \begin{align*} B(x)A(x)&=\left(\sum_{k=0}^\infty a_k x^k\right)\left(\sum_{j=0}^\infty b_j x^j\right)\\ &=\sum_{n=0}^\infty\left(\color{blue}{\sum_{k=0}^n a_kb_{n-k}}\right)x^n=1 \end{align*}

Using the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series we have \begin{align*} [x^0]B(x)A(x)&=a_0b_0=1\tag{1}\\ [x^n]B(x)A(x)&=\sum_{k=0}^na_kb_{n-k}=0\qquad n\geq 1\tag{2} \end{align*}

From (1) we derive with $a_0=1$: \begin{align*} [x^0]B(x)A(x)&=a_0b_0=\color{blue}{b_0=1} \end{align*} From (2) we derive \begin{align*} [x^n]B(x)A(x)&=\sum_{k=0}^n a_kb_{n-k}=a_0b_n+\sum_{k=1}^{n} a_kb_{n-k}=0\tag{3} \end{align*} With $a_0=1$ we obtain from (3) for $n\geq 1$

\begin{align*} \color{blue}{b_n}&\color{blue}{=-\sum_{k=1}^n a_kb_{n-k}}\\ \\ \color{blue}{b_1}&=-\sum_{k=1}^1 a_kb_{1-k}=-a_1b_0=\color{blue}{-a_1}\\ \color{blue}{b_2}&=-\sum_{k=1}^2 a_kb_{2-k}=-a_1b_1-a_2b_0\color{blue}{=a_1^2-a_2}\\ &\cdots \end{align*}