Section 2.6 Proof of the remainder theorem

In this subsection, we will provide a proof of the remainder theorem.

Let \(f\) be a function that has \(n+1\) continuous derivatives on an open interval \(I\) containing \(a \text{.}\) We want to show that for each \(x\in I\) and for each positive integer \(n\text{,}\) there exists a number \(c\) between \(a\) and \(x\) such that

\begin{equation*} f(x) = p_n(x) + R_n(x), \end{equation*}

where

\begin{equation*} p_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!} (x-a)^k, \end{equation*}

and

\begin{equation*} R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{n+1}. \end{equation*}

To prove this, we will first prove the following lemma:

Lemma 2.25.

Let \(f(x)\) be a function that is \((n+1)\)-times differentiable. Also, suppose that \(f(a) = f'(a) = f''(a) = ... = f^{(n)}(a) = 0\) and \(f(x) = 0\text{,}\) where \(x\neq a\text{.}\) Then there exists a number \(c\) between \(a\) and \(x\) such that \(f^{(n+1)}(c) = 0\text{.}\)

Proof.

The proof of this lemma is based on the mean value theorem. Since \(f(a) = f'(a) = f''(a) = ... = f^{(n)}(a) = 0\text{,}\) we can apply the mean value theorem repeatedly to show that there exists a number \(c\) between \(a\) and \(x\) such that \(f^{(n+1)}(c) = 0\text{.}\) Below we will do so step by step.

First, since \(f(a) = f(x) = 0\text{,}\) by the mean value theorem, there exists a number \(c_1\) between \(a\) and \(x\) such that

\begin{equation*} f'(c_1) = \frac{f(x) - f(a)}{x - a} = 0 \end{equation*}

Second, since \(f'(a) = f'(c_1) = 0\text{,}\) by the mean value theorem, there exists a number \(c_2\) between \(a\) and \(c_1\) such that

\begin{equation*} f''(c_2) = \frac{f'(c_1) - f'(a)}{c_1 - a} = 0 \end{equation*}

Then by continuing this process, we can show that there exists a number \(c_n\) between \(a\) and \(c_{n-1}\) such that

\begin{equation*} f^{(n)}(c_n) = 0 \end{equation*}

Since \(f^{(n)}(a) = f^{(n)}(c_n) = 0\text{,}\) by the mean value theorem, there exists a number \(c_{n+1}\) between \(a\) and \(c_n\) such that

\begin{equation*} f^{(n+1)}(c_{n+1}) = 0 \end{equation*}

Therefore, we have shown that there exists a number \(c = c_{n+1}\) between \(a\) and \(x\) such that \(f^{(n+1)}(c) = 0\text{.}\)

Remark 2.26. The error function and its derivatives are zero at the point of expansion.

Note that for the error function \(R_n(x) = f(x) - p_n(x) \text{,}\) we have \(R_n(a) = R_n'(a) = R_n''(a) = ... = R_n^{(n)}(a) = 0\text{.}\)

Now, assume that \(x\neq a\text{.}\) We can construct the function \(g(t) = R_n(t) - \frac{R_n(x)}{(x-a)^{n+1}} (t-a)^{n+1}\) such that \(g(a) = g(x) = 0\) and \(g^{(k)}(a) = 0\) for \(k = 1, 2, ..., n\text{.}\) Therefore, by the lemma and considering Remark 2.26, there exists a number \(c\) between \(a\) and \(x\) such that \(g^{(n+1)}(c)=0\text{.}\) After computing \(g^{(n+1)}(c)\text{,}\) we have \(R_n^{(n+1)}(c) - \frac{R_n(x)}{(x-a)^{n+1}} (n+1)! = 0\text{.}\) Since \(R_n^{(n+1)}(c) = f^{(n+1)}(c)\text{,}\) we have \(R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{n+1}\text{.}\) This completes the proof of the remainder theorem.

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