Taylor's theorem

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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

\[ \begin{eqnarray} f(x) &=& f(a) + f'(a)(x-a) + \frac{1}{2}f''(a)(x-a)^2 + \cdots\nonumber\\ && + \frac{1}{(n-1)!}f^{(n)}(a)(x-a)^{n-1} + \frac{1}{n!}f^{(n)}(c_x)(x-a)^{n}.\nonumber\\ \label{eq:taylor} \end{eqnarray} \]

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

Suppose \(b \in I\), \(b\neq a\). We need to show

\[ \begin{eqnarray} f(b) &=& f(a) + f'(a)(b-a) + \frac{1}{2}f''(a)(b-a)^2 + \cdots\nonumber\\ && + \frac{1}{(n-1)!}f^{(n-1)}(a)(b-a)^{n-1} + \frac{1}{n!}f^{(n)}(c_b)(b-a)^{n}.\nonumber\\ && \label{eq:taylorb} \end{eqnarray} \]

for some \(c_b\). Let us define the constant \(A\) by

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

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

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

\(g(x)\) 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 \(g'(c_b) = 0\).

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

Let us calculate \(g'(x)\). Noting that, for \(k \geq 1\),

\[\left[\frac{1}{k!}f^{(k)}(x)(b-x)^k\right]' = \frac{1}{k!}f^{(k+1)}(x)(b-x)^k - \frac{1}{(k-1)!}f^{(k)}(x)(b-x)^{k-1},\]

the first several terms of \(g'(x)\) are

\[-f'(x) +f'(x) - f''(x)(b-x) + f''(x)(b-x) - \frac{1}{2}f'''(x)(b-x)^2 + \cdots\]

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

\[ g'(x) = -\frac{1}{(n-1)!}f^{(n)}(x)(b-x)^{n-1} + nA(b-x)^{n-1}. \]

Since \(g'(c_b) = 0\), 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

\[ P_n(x) = f(a) + f'(a)(x-a) + \frac{1}{2}f''(a)(x-a)^2 +\cdots + \frac{1}{n!}f^{(n)}(a)(x-a)^{n}. \]

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

\[ \begin{eqnarray} P_1(x) &=& x,\\ P_3(x) &=& x - \frac{1}{6}x^3,\\ P_5(x) &=& x - \frac{1}{6}x^3 + \frac{1}{120}x^5,\\ &\vdots& \end{eqnarray} \]

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

Approximation of \(sin x\) 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!) □

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