Informal Derivation of Taylor Series

We have a function $f(x)$ and we want to approximate it using an $n$th-order polynomial:

$$f(x) = f_0 + f_1 x + f_2 x^2 + \cdots + f_n x^n + R_{n+1}(x)$$

where $R_{n+1}(x)$, the approximation error, is called the remainder term. We may assume $x$ and $f(x)$ are real, but the following derivation generalizes unchanged to the complex case.

Our problem is to find fixed constants $\{f_i\}_{i=0}^{n}$ so as to obtain the best approximation possible. Let's proceed optimistically as though the approximation will be perfect, and assume $R_{n+1}(x)=0$ for all $x$ ( $R_{n+1}(x)\equiv0$), given the right values of $f_i$. Then at $x=0$ we must have

$$f(0) = f_0$$

That's one constant down and $n-1$ to go! Now let's look at the first derivative of $f(x)$ with respect to $x$, again assuming that $R_{n+1}(x)\equiv0$:

$$f^\prime(x) = 0 + f_1 + 2 f_2 x + 3 f_2 x^2 + \cdots + n f_n x^{n-1}$$

Evaluating this at $x=0$ gives

$$f^\prime(0) = f_1.$$

In the same way, we find

$$\begin{eqnarray*} f^{\prime\prime}(0) &=& 2 \cdot f_2 \\ f^{\prime\prime\prime}... ...cdot 2 \cdot f_3 \\ & \cdots & \\ f^{(n)}(0) &=& n! \cdot f_n \end{eqnarray*}$$

where $f^{(n)}(0)$ denotes the $n$th derivative of $f(x)$ with respect to $x$, evaluated at $x=0$. Solving the above relations for the desired constants yields

$$\begin{eqnarray*} f_0 &=& f(0) \\ f_1 &=& \frac{f^{\prime}(0)}{1} \\ f_2 &=& \... ...dot 2\cdot 1} \\ & \cdots & \\ f_n &=& \frac{f^{(n)}(0)}{n!}. \end{eqnarray*}$$

Thus, defining $0!\isdef 1$ (as it always is), we have derived the following polynomial approximation:

$$\zbox {f(x) \approx \sum_{k=0}^n \frac{f^{(k)}(0)}{k!} x^k}$$

This is the $n$th-order Taylor series expansion of $f(x)$ about the point $x=0$. Its derivation was quite simple. The hard part is showing that the approximation error (remainder term $R_{n+1}(x)$) is small over a wide interval of $x$ values. Another ``math job'' is to determine the conditions under which the approximation error approaches zero for all $x$ as the order $n$ goes to infinity. The main point to note here is that the Taylor series itself is simple to derive.