4.5 Differentiating Taylor polynomials

Proof Each time we differentiate $p_{n}$, we reduce the degree by one, since

$${{d}\over{dx}}(x-a)^{n}=n(x-a)^{n-1}.$$

Hence we find that

$$p_{n}(a)=a_{0}=f(a)$$

$$p_{n}^{\prime}(a)=a_{1}=f^{\prime}(a),$$

$$p_{n}^{\prime\prime}(a)=2a_{2}=f^{\prime\prime}(a),$$

$$\vdots\quad\quad\vdots$$

$$p_{n}^{(n)}(a)=n!a_{n}=f^{(n)}(a),$$

so the first $n$ derivatives of $p_{n}(x)$ and $f(x)$ are equal at $x=a$.