Find \(1/\sin(z)\) using this method.
Solution
Note that the leading coefficient of the expansion for \(\sin(z)\) is 0, so this can’t be inverted directly. A standard trick: factor out the smallest power of \(z\) to get a piece with a nonzero leading coefficient, and invert that instead. Write \[ \sin(z) = z - {1\over 3!}z^3 + {1\over 5!}z^5 - \bigo(z^7) = z f(z) \da z(1 + c_2z^2 + c_4 z^4 + \bigo(z^6)) ,\] where
\(c_0 = 1\)
\(c_2 = -{1\over 3!}\)
\(c_4 = {1\over 5!}\)
\(c_i = 0\) for \(i\) odd.
Writing \(B(z) \da {1\over f(z)} = \sum b_k z^k\), we have
\(b_0 = \inverseof{c_0} = 1\)
\(b_2 = -(c_2 b_0 + c_1 b_1) = -c_2\)
\(b_4 = -(c_4 b_0 + c_2 b_2) = c_2^2 - c_4\)
Thus \(1/B(z) = 1 -c_2z^2 + (c_2^2-c_4)z^4 + \bigo(z^6)\), and \[ {1\over \sin(z)} &= {1\over z} {1\over B(z)} \\ &= {1\over z} - c_2 z + (c_2^2 - c_4)z^3 + \bigo(z^5) \\ &= {1\over z} + {1\over 3!} z + ({1\over 3! 3!} - {1\over 5!})z^3 + \bigo(z^5) \\ &= {1\over z} + {1\over 3!}z + {7\over 360}z^3 + \bigo(z^5) .\]