📂Calculus

Taylor Series and Maclaurin Series

Build-Up¹

Assume that the given function $f$ is expressed as a power series.

$$f(x) = c_{0} + c_{1}(x - a) + c_{2}(x - a)^{2} + c_{3}(x - a)^{3} + \cdots \qquad |x - a| \lt R \tag{1}$$

In this context, finding the power series representation of the function $f$ is equivalent to determining the coefficients of each term $c_{n}$. By substituting $x = a$ into both sides, we can obtain $c_{0}$.

$$f(a) = c_{0} + c_{1}(a - a) + c_{2}(a - a)^{2} + c_{3}(a - a)^{3} + \cdots = c_{0}$$

Differentiating $(1)$, we obtain the following:

$$f^{\prime}(x) = c_{1} + 2c_{2}(x - a) + 3c_{3}(x - a)^{2} + \cdots \qquad |x - a| \lt R \tag{2}$$

Substituting $x = a$ into both sides, we can find $c_{1}$.

$$f^{\prime}(a) = c_{1}$$

Differentiating $(2)$ again and substituting $x = a$,

$$f^{\prime\prime}(x) = 2c_{2} + 3 \cdot 2c_{3}(x - a) + \cdots \qquad |x - a| \lt R \tag{3}$$

$$f^{\prime\prime}(a) = 2c_{2}$$

Repeating once more,

$$f^{\prime\prime\prime}(x) = 3 \cdot 2c_{3} + 4 \cdot 3 \cdot 2c_{4}(x - a) + \cdots \qquad |x - a| \lt R$$

$$f^{\prime\prime\prime}(a) = 3 \cdot 2c_{3}$$

Following this pattern, we find that $c_{n}$ can be expressed as follows:

$$f^{(n)}(a) = n! \cdot c_{n} \implies c_{n} = \frac{f^{(n)}(a)}{n!}$$

Therefore, $(1)$ is expressed as follows:

$$f(x) = f(a) + \frac{f^{\prime}(a)}{1!}(x - a) + \frac{f^{\prime\prime}(a)}{2!}(x - a)^{2} + \frac{f^{\prime\prime\prime}(a)}{3!}(x - a)^{3} + \cdots = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^{n}$$

Here, the series on the right is defined as the Taylor series of $f$.

Definition

The Taylor series of a function $f$, which is infinitely differentiable, at $a$ is defined as follows:

$$\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^{n} = f(a) + \frac{f^{\prime}(a)}{1!}(x - a) + \frac{f^{\prime\prime}(a)}{2!}(x - a)^{2} + \frac{f^{\prime\prime\prime}(a)}{3!}(x - a)^{3} + \cdots$$

Explanation

Specifically, the series at $a = 0$ is called the Maclaurin series.

If $f$ can be expressed as a power series, then the power series of $f$ and the Taylor series of $f$ are identical. However, the Taylor series of $f$ does not always match $f$ itself. It has been proven that the power series of $f$ and the Taylor series of $f$ are the same under certain conditions.

An Example Where the Taylor Series of $f$ Differs from $f$¹

The following function $f$ is given.

$$f(x) = \begin{cases} e^{-1/x^{2}} & x \neq 0 \\ 0 & x = 0 \end{cases}$$

Let us find $f^{\prime}(0)$ to determine the coefficients of the series.

$$\lim\limits_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \lim\limits_{h \to 0} \frac{e^{-1/h^{2}} - 0}{h} = \lim\limits_{h \to 0} \frac{1}{h} \cdot e^{-1/h^{2}}$$

By organizing the equation and using L’Hôpital’s rule,

$$\lim\limits_{h \to 0} \frac{1}{h} \cdot e^{-1/h^{2}} = \lim\limits_{h \to 0} \dfrac{1/h}{e^{1/h^{2}}} = \lim\limits_{h \to 0} \dfrac{-1/h^{2}}{-2e^{1/h^{2}}/h^{3}} = \lim\limits_{h \to 0} \dfrac{h}{2e^{1/h^{2}}} = 0$$

Similarly, it can be shown that $n \in \mathbb{N}$ corresponds to $f^{(n)} = 0$. Therefore, the Maclaurin series of $f$ is as follows:

$$\sum\limits_{n = 0}^{\infty} \frac{f^{(n)}(0)}{n!}x^{n} = \sum\limits_{n = 0}^{\infty} \frac{0}{n!}x^{n} = 0$$

However, it is evident that $x \ne 0$ leads to $f(x) \ne 0$, indicating that the Maclaurin series of $f$ and $f$ are not the same.