Find \(1/\cos(z)\) using this method.
Solution
We have \(\cos(z) = c_0 +c_2^2 + c_4 z^4 + \bigo(z^6)\) where
\(c_0 = 1\)
\(c_2 = -1/2!\)
\(c_4 = 1/4!\)
Note that \(c_i\) for \(i\) odd all vanish. Write \(1/\cos(z) = \sum b_k z^k\), then
\(b_0 = \inverseof{c_0} = 1\).
\(b_1 = -\inverseof{c_0}(c_1 b_0) = -(0) = 0\).
\(b_2 = -\inverseof{c_0}(c_2 b_0 + c_1b_1) = -(c_2 + 0) = -c_2 = 1/2!\)
\(b_3 = -\inverseof{c_0}(c_3b_0 + c_2b_1 + c_1b_2) = -(0 +0 + 0)= 0\)
\(b_4 = -\inverseof{c_0}(c_4b_0 + c_3b_1 + c_2 b_2 + c_1b_3)=-(c_4 + 0 + (-1/2!)(1/2!) + 0) = 5/24\).
So \[ {1\over \cos(z)} = 1 + {1\over 2!}z^2 + {5\over 24}z^4 + \bigo(z^6) .\]