Taylor's theorem

The essence of differentiation is to approximate an arbitrary function by a linear function. We can extend this idea by using higher-order derivatives to obtain better approximations.

Theorem (Taylor's theorem)

Let $f(x)$ be a function of class $C^n$ on an open interval $I$. Let $a\in I$. Then for all $x\in I$, there exists $c_x$ between $x$ and $a$ such that

$$\begin{array}{rcl}f(x)& =& f(a)+f^{\prime }(a)(x-a)+\frac{1}{2}{f}^{″}(a)(x-a{)}^2+\cdots \\ & & +\frac{1}{(n-1)!}{f}^{(n)}(a)(x-a{)}^{n-1}+\frac{1}{n!}{f}^{(n)}(c_x)(x-a{)}^n.\end{array}$$

Proof. If $x=a$, then we can set $c_x=a$ and Eq. (eq:taylor) clearly holds.

Suppose $b\in I$, $b\ne a$. We need to show

$$\begin{array}{rcl}f(b)& =& f(a)+f^{\prime }(a)(b-a)+\frac{1}{2}{f}^{″}(a)(b-a{)}^2+\cdots \\ & & +\frac{1}{(n-1)!}{f}^{(n-1)}(a)(b-a{)}^{n-1}+\frac{1}{n!}{f}^{(n)}(c_b)(b-a{)}^n.\\ & & \end{array}$$

for some $c_b$. Let us define the constant $A$ by

$$A(b-a{)}^n=f(b)-[f(a)+f^{\prime }(a)(b-a)+\cdots +\frac{1}{(n-1)!}{f}^{(n-1)}(a)(b-a{)}^{n-1}]$$

and the function $g(x)$ on $I$ by

$$g(x)=f(b)-[f(x)+f^{\prime }(x)(b-x)+\cdots +\frac{1}{(n-1)!}{f}^{(n-1)}(x)(b-x{)}^{n-1}+A(b-x{)}^n].$$

$g(x)$ is a differentiable function on $I$ and $g(a)=g(b)=0$. Thus, by Rolle's theorem, there exists $c_b$ between $a$ and $b$ such that $g^{\prime }(c_b)=0$.

[See also: Mean Value Theorem for Rolle's theorem.]

Let us calculate $g^{\prime }(x)$. Noting that, for $k\ge 1$,

$${[\frac{1}{k!}{f}^{(k)}(x)(b-x{)}^k]}^{\prime }=\frac{1}{k!}{f}^{(k+1)}(x)(b-x{)}^k-\frac{1}{(k-1)!}{f}^{(k)}(x)(b-x{)}^{k-1},$$

the first several terms of $g^{\prime }(x)$ are

$$-f^{\prime }(x)+f^{\prime }(x)-f^{″}(x)(b-x)+f^{″}(x)(b-x)-\frac{1}{2}{f}^{‴}(x)(b-x{)}^2+\cdots$$

so that most terms are canceled. After all, we have

$$g^{\prime }(x)=-\frac{1}{(n-1)!}{f}^{(n)}(x)(b-x{)}^{n-1}+nA(b-x{)}^{n-1}.$$

Since $g^{\prime }(c_b)=0$, we have

$$0=-\frac{1}{(n-1)!}{f}^{(n)}(c_b)(b-c_b{)}^{n-1}+nA(b-c_b{)}^{n-1}$$

or

$$A=\frac{1}{n!}{f}^{(n)}(c_b).$$

Substituting this $A$ into $g(a)=0$, we have Eq. (eq:taylorb). ■

Let's see what Taylor's theorem means. Let us define the polynomial of degree $n$ by

$$P_n(x)=f(a)+f^{\prime }(a)(x-a)+\frac{1}{2}{f}^{″}(a)(x-a{)}^2+\cdots +\frac{1}{n!}{f}^{(n)}(a)(x-a{)}^n.$$

Taylor's theorem states that the difference between $f(x)$ and $P_n(x)$ is given by

$$f(x)-P_n(x)=\frac{1}{(n+1)!}{f}^{(n+1)}(c_x)(x-a{)}^{n+1}$$

for some $c_n$. If $n$ is sufficiently large and $x$ is close to $a$ (i.e., $|x-a|$ is small), then $|(x-a{)}^{n+1}|$ is very small so that the above difference is very small. Thus, Taylor's theorem indicates that the function $f(x)$ can be approximated by polynomial functions.

Example. Consider $f(x)=\sin x$. With $a=0$, we have

$$\begin{array}{rcl}{P}_1(x)& =& x,\\ P_3(x)& =& x-\frac{1}{6}{x}^3,\\ P_5(x)& =& x-\frac{1}{6}{x}^3+\frac{1}{120}{x}^5,\\ & ⋮& \end{array}$$

By plotting the graphs of these polynomials, we can see that they approximate $\sin x$ with increasingly better accuracies.

Approximation of $sinx$ by Taylor expansions.

□

Definition (finite Taylor expansion and finite Maclaurin expansion)

The right-hand side of Eq. (eq:taylor) in Taylor's Theorem is called the finite Taylor expansion and the last term $\frac{1}{n!}{f}^{(n)}(c_x)(x-a{)}^n$ is called the remainder or residual. The finite Taylor expansion can be also expressed as

$$f(x)=\sum _{k=0}^{n-1}\frac{1}{k!}{f}^{(k)}(a)(x-a{)}^k+\frac{1}{n!}{f}^{(n)}(a+\theta (x-a))(x-a{)}^n$$

where $\theta \in (0,1)$ is a constant dependent on $x$.

In particular, the finite Taylor expansion with $a=0$ is called the finite Maclaurin expansion. 

Example. The finite Maclaurin expansion of $e^x$ is given by

$$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\cdots +\frac{x^{n-1}}{(n-1)!}+\frac{e^{\theta x}{x}^n}{n!}.$$

(Exercise!) □

Example. The finite Maclaurin expansion of $\log (x+1)$ is given by

$$\log (x+1)=x-\frac{1}{2}{x}^2+\frac{1}{3}{x}^3-\frac{1}{4}{x}^4+\cdots +\frac{(-1{)}^n}{(n-1)}{x}^{n-1}+\frac{(-1{)}^{n+1}{x}^n}{n(\theta x+1{)}^n}.$$

(Exercise!) □