Taylor Polynomials of a Function

Taylor Polynomials of a Function

 Definition: Let $f : A \to \mathbb{R}$ be $n$-times differentiable at $x_0 \in A$. The $n^{\mathrm{th}}$ Taylor Polynomial for $f$ at $x_0$ is the polynomial $\displaystyle{P_n(x) = f(x_0) + f'(x_0)(x - x_0) + \frac{f''(x_0)}{2!}(x - x_0)^2 + ... + \frac{f^{(n)}(x_0)}{n!}(x - x_0)^n}$.

For example, consider the function $f(x) = e^x$. The $4^{\mathrm{th}}$ Taylor polynomial for $f$ at $0$ is given by:

(1)
\begin{align} \quad P_4(x) &= e^0 + e^0(x - 0) + \frac{e^0}{2!}(x - 0)^2 + \frac{e^0}{3!}(x - 0)^3 + \frac{e^0}{4!}(x - 0)^4 \\ &= 1 + x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \frac{1}{4!}x^4 \end{align}
 Theorem 1: Let $f : A \to \mathbb{R}$ be $n$-times differentiable at $x_0 \in A$. Then $f^{(n)}(x_0) = P_n^{(k)}(x_0)$ for each $k \in \{ 0, 1, 2, ..., n \}$.
• Proof: We take the derivative of $P_n$ to get:
(2)
\begin{align} \quad P_n'(x) = f'(x_0) + f''(x_0)(x - x_0) + ... + \frac{f^{(n)}(x_0)}{(n-1)!} (x - x_0)^{n-1} \end{align}
• More generally, when we take the $k^{\mathrm{th}}$ derivative of $P_n$ we get:
(3)
\begin{align} \quad P_n^{(k)}(x) = f^{(k)}(x_0) + f^{(k+1)}(x_0)(x - x_0) + ... + \frac{f^{(n)}(x_0)}{(n - k)!} (x - x_0)^{n - k} \end{align}
• Plugging in $x = x_0$ gives us that $P_n^{(k)}(x_0) = f^{(k)}(x_0)$, $k \in \{ 0, 1, 2, ..., n \}$. $\blacksquare$