Use this formulation to show that if \(A(z)= 1-z\) then \(1/A(z) = \sum z^k\).
Solution
Noting \(a_0 = 1, a_1 = -1\), we have
- \(b_0 = 1/a_0 = 1\)
- \(b_1 = -{1\over a_0}(a_1 b_0) = -1(-1\cdot 1) = 1\)
- \(b_2 = -{1\over a_0}(a_2 b_0 + a_1 b_1) = -1(0\cdot 1 + -1\cdot 1) = 1\),
and so on, so \[ {1\over A(z)} = 1 + 1\cdot z + 1\cdot z^2 + \cdots .\]