For `f text[(] x text[)] = ln text[(]1 + xtext[)]` we have
`f^' text[(] x text[)] = text[(]1 + xtext[)]^(-1), f^('') text[(] x text[)] = -text[(]1 + xtext[)]^(-2), f^((3)) text[(] x text[)] = 2text[(]1 + xtext[)]^(-3), f^((4)) text[(] x text[)] = -3!text[(]1 + xtext[)]^(-4),`
and so on. When we substitute `x = 0`, we find
`f text[(] 0 text[)] = 0, f^(') text[(] 0 text[)] = 1, f^('') text[(] 0 text[)] = -1, f^((3)) text[(] 0 text[)] = 2, f^((4)) text[(] 0 text[)] = -3!`,
and so on. In general, except for the `0`th entry,
`f^((k))text[(] 0 text[)] = text[(]-1text[)]^(k+1)text[(]k-1text[)]!`.
When we divide by `k!` we find that
`c_k = (text[(]-1text[)]^(k+1))/k`.
These coefficients give the same terms in the polynomials as those we computed by integration.