5.9 The proof of Taylor’s theorem

This section is nonexaminable.

We motivated Taylor’s theorem as a generalisation of the mean value theorem, replacing the constant polynomial $P_{0}^{f,a}(x):=f(a)$ with higher-degree approximating polynomials. Recall, the first step in the proof of the mean value theorem was to prove Rolle’s theorem. The first step to proving Taylor’s theorem is the following ‘higher-order’ Rolle’s theorem.

Lemma 5.85 (Higher-order Rolle’s theorem).

Let $I\subseteq\mathbb{R}$ be an open interval, $n\in\mathbb{N}_{0}$ and $f\colon I\to\mathbb{R}$ be $(n+1)$ -times differentiable. Let $a$ , $b\in I$ with $a<b$ and suppose

(5.36) (5.36)

f(a)=f^{\prime}(a)=\cdots=f^{(n)}(a)=f(b)=0.

Then there exists some $c\in(a,b)$ such that $f^{(n+1)}(c)=0$ .

The hypotheses of Lemma 5.85 are slightly different compared to the simple case of Rolle’s theorem as stated in Theorem 5.37. In particular, we assume $f$ is defined on an open interval $I$ rather than on some closed interval $[a,b]$ . This is just to avoid technicalities concerning defining the derivative $f^{(k)}(a)$ at the endpoint $a$ .⁹⁹ 9 One could formulate a version of Lemma 5.85 for functions $f\colon[a,b]\to\mathbb{R}$ , but then one would have to work with right-derivatives at the endpoint $a$ . In any case, Theorem 5.37 implies the $n=0$ case of Lemma 5.85 by applying Theorem 5.37 to the restricted function $f|_{[a,b]}\colon[a,b]\to\mathbb{R}$ .

Proof (of Lemma 5.85).

We prove the result by induction on $n\in\mathbb{N}_{0}$ . We have already observed that the $n=0$ follows from Theorem 5.37, which serves as the base case.

Let $n\in\mathbb{N}$ and suppose, as an induction hypothesis, that the lemma holds for the $n-1$ case. Let $f\colon I\to\mathbb{R}$ be $(n+1)$ -times differentiable, fix $a$ , $b\in I$ with $a<b$ and suppose (5.36) holds. Note that the restricted function $f|_{[a,b]}\colon[a,b]\to\mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$ and satisfies $f(a)=f(b)=0$ . By the simple case of Rolle’s theorem from Theorem 5.37 applied to $f|_{[a,b]}$ , there exists some $d\in(a,b)$ such that $f^{\prime}(d)=0$ . Now, the derivative $f^{\prime}\colon I\to\mathbb{R}$ is $n$ -times differentiable and satisfies

f^{\prime}(a)=(f^{\prime})^{\prime}(a)=\cdots=(f^{\prime})^{(n-1)}(a)=f^{% \prime}(d)=0.

We can therefore apply the induction hypothesis to $f^{\prime}\colon I\to\mathbb{R}$ to conclude that there exists some $c\in(a,d)$ such that $(f^{\prime})^{(n)}(c)=0$ . But in this case, $c\in(a,b)$ and $f^{(n+1)}(c)=(f^{\prime})^{(n)}(c)=0$ , as required. This closes the induction and completes the proof. ∎

Lemma 5.86 (Higher-order mean value theorem).

Let $I\subseteq\mathbb{R}$ be an open interval, $n\in\mathbb{N}_{0}$ and $f\colon I\to\mathbb{R}$ be $(n+1)$ -times differentiable. Let $a$ , $b\in I$ with $a<b$ . Then there exists some $c\in(a,b)$ such that

(5.37) (5.37)

f(b)-P_{n}^{f,a}(b)=\frac{f^{(n+1)}(c)}{(n+1)!}(b-a)^{n+1}.

We use the same idea to prove Lemma 5.86 as we did to prove the simple case of the mean value theorem from Theorem 5.41. Recall, for the proof of Theorem 5.41, we subtracted a linear polynomial from $f$ to reduce to a situation where Rolle’s theorem applies. Here we subtract a degree $n+1$ polynomial from $f$ to reduce to a situation where the higher-order Rolle’s theorem applies.

Proof (of Lemma 5.86).

First consider the function $g\colon I\to\mathbb{R}$ given by $g(x):=f(x)-P_{n}^{f,a}(x)$ for all $x\in I$ . We know from Exercise 5.73 that

(5.38) (5.38)

g^{(k)}(a)=f^{(k)}(a)-(P_{n}^{f,a})^{(k)}(a)=0\qquad\text{for all $0\leq k\leq n% $.}

So the function $g$ satisfies all the hypotheses of the higher-order Rolle’s theorem except possibly the condition $g(b)=0$ . We can fix this by considering the modified function

h\colon I\to\mathbb{R},\qquad h(x):=f(x)-P_{n}^{f,a}(x)-\frac{f(b)-P_{n}^{f,a}% (b)}{(b-a)^{n+1}}\cdot(x-a)^{n+1}\qquad\text{for all $x\in I$,}

since a simple computation shows that $h$ also satisfies $h(b)=0$ . Furthermore, observe that $\frac{\mathrm{d}^{k}}{\mathrm{d}x^{k}}(x-a)^{n+1}|_{x=a}=0$ for $0\leq k\leq n$ . Combining this with (5.38) we have

h^{(k)}(a)=0\qquad\text{for all $0\leq k\leq n$.}

Thus, $h$ satisfies all the hypotheses of the higher-order Rolle’s theorem, so we can apply Lemma 5.85 to deduce that there exists some $c\in(a,b)$ such that $h^{(n+1)}(c)=0$ . However,

0=h^{(n+1)}(c)=f^{(n+1)}(c)-(n+1)!\cdot\frac{f(b)-P_{n}^{f,a}(b)}{(b-a)^{n+1}}.

Rearranging, we obtain the desired identity (5.37). ∎

Exercise 5.87.

Let $I\subseteq\mathbb{R}$ be an open interval, $n\in\mathbb{N}_{0}$ and $f\colon I\to\mathbb{R}$ be $(n+1)$ -times differentiable. Let $a$ , $b\in I$ with $a<b$ . Show that there exists some $c\in(a,b)$ such that

f(a)-P_{n}^{f,b}(a)=\frac{f^{(n+1)}(c)}{(n+1)!}(a-b)^{n+1}.

This is a version of Lemma 5.86 with the roles of $a$ and $b$ flipped in the identity.

Hint: Let $J:=\{a+b-x:x\in I\}$ and apply Lemma 5.86 to the function $g\colon J\to\mathbb{R}$ given by $g(x):=f(a+b-x)$ for $x\in J$ .

Taylor’s theorem is an immediate consequence of the higher-order mean value theorem.

Proof (of Theorem 5.74).

We use the argument from Interpretation 3 of the mean value theorem.

Let $I\subseteq\mathbb{R}$ be an open interval, $f\colon I\to\mathbb{R}$ be $(n+1)$ -times differentiable and $a\in I$ . For any $x\in I$ with $x>a$ , the function $f$ satisfies the hypotheses of the higher-order mean value theorem with $b:=x$ . If $x\in I$ and $x<a$ , then the same holds with the roles of $a$ and $x$ swapped. From this, we deduce that for all $x\in I\setminus\{a\}$ there exists some $c_{x}\in I$ lying between $a$ and $x$ such that

f(x)=P_{n}^{f,a}(x)+\frac{f^{(n+1)}(c_{x})}{(n+1)!}(x-a)^{n+1}.

Indeed, for $x>a$ this follows directly from Lemma 5.86; for $x<a$ this follows from the ‘flipped’ form of Lemma 5.86 from Exercise 5.87. However, the above is exactly the statement of Taylor’s theorem! ∎