Invert $2z-1$, one coefficient at a time

To compute the inverse of \(A(z) \da (2z-1)\), note \(a_0 = -1, a_1 = 2\), so

• \(b_0 = 1/a_0 = -1\)
• \(b_1 = -{1\over a_0}(a_1b_0) = 1(2\cdot -1) = -2\)
• \(b_2 = -{1\over a_0}(a_2 b_0 + a_1 b_1) = 1(0\cdot 1 + 2\cdot -2) = -4\)
• \(b_3 = -{1\over a_0}(a_3b_0 + a_2 b_1 + a_1 b_2) = 1(0 + 0 + 2\cdot -4) = -8\)

so \[ {1\over 2z-1} = -1 - 2z - 4z^2 - 8z^3 - \cdots = - \sum_{k\geq 0} (2z)^k .\]