The second derivative of `p text[(] x text[)]` has the form
`p'' text[(] x text[)] = 2c_2+6c_3x+12c_4x^2`.
Thus `2c_2 = p'' text[(] 0 text[)]`, so `c_2 = p'' text[(] 0 text[)/]2`. Differentiating again, we obtain
`p^((3)) text[(] x text[)] = 6c_3 + 24c_4x`,
which implies that `c_3 = p^((3))text[(] 0 text[)] text[/] 6`. (The prime notation for derivatives quickly gets out of hand for higher derivatives. Instead of adding another prime for each new derivative, we put the order of the derivative in parentheses.) Differentiating one more time, we find that `c_4 = p^((4)) text[(] 0 text[)] text[/] 24`.
See the discussion following Activity 1 for the general rule.