# 6.6: Taylor's Theorem

- Page ID
- 22676

## 6.6.1 Derivatives of Higher Order

Suppose \(f\) is differentiable on an open interval \(I\) and \(f^{\prime}\) is differentiable at \(a \in I .\) We call the derivative of \(f^{\prime}\) at \(a\) the second derivative of \(f\) at \(a,\) which we denote \(f^{\prime \prime}(a)\).

By continued differentiation, we may define the higher order derivatives \(f^{\prime \prime \prime},\) \(f^{\prime \prime \prime \prime},\) and so on. In general, for any integer \(n, n \geq 0,\) we let \(f^{(n)}\) denote the \(n\)th derivative of \(f,\) where \(f^{(0)}\) denotes \(f\).

Suppose \(D \subset \mathbb{R}, a\) is an interior point of \(D, f: D \rightarrow \mathbb{R},\) and \(f^{\prime \prime}(a)\) exists. Show that

\[\lim _{h \rightarrow 0} \frac{f(a+h)+f(a-h)-2 f(a)}{h^{2}}=f^{\prime \prime}(a).\]

Find an example to illustrate that this limit may exist even if \(f^{\prime \prime}(a)\) does not exist.

For any open interval \((a, b),\) where \(a\) and \(b\) are extended real numbers, we let \(C^{(n)}(a, b),\) where \(n \in \mathbb{Z}^{+},\) denote the set of all functions \(f\) with the property that each of \(f, f^{(1)}, f^{(2)}, \ldots, f^{(n)}\) is defined and continuous on \((a, b) .\)

### 6.6.2 Taylor's Theorem

(Taylor's Theorem).

Suppose \(f \in C^{(n)}(a, b)\) and \(f^{(n)}\) is differentiable on \((a, b) .\) Let \(\alpha, \beta \in(a, b)\) with \(\alpha \neq \beta,\) and let

\[\begin{aligned} P(x)=f(&\alpha)+f^{\prime}(\alpha)(x-\alpha)+\frac{f^{\prime \prime}(\alpha)}{2}(x-\alpha)^{2}+\cdots \\ &+\frac{f^{(n)}(\alpha)}{n !}(x-\alpha)^{n} \\=& \sum_{k=0}^{n} \frac{f^{(k)}(\alpha)}{k !}(x-\alpha)^{k}. \end{aligned}\]

Then there exists a point \(\gamma\) between \(\alpha\) and \(\beta\) such that

\[f(\beta)=P(\beta)+\frac{f^{(n+1)}(\gamma)}{(n+1) !}(\beta-\alpha)^{n+1}.\]

**Proof**-
First note that \(P^{(k)}(\alpha)=f^{(k)}(\alpha)\) for \(k=0,1, \ldots, n .\) Let

\[M=\frac{f(\beta)-P(\beta)}{(\beta-\alpha)^{n+1}}.\]

Then

\[f(\beta)=P(\beta)+M(\beta-\alpha)^{n+1}.\]

We need to show that

\[M=\frac{f^{(n+1)}(\gamma)}{(n+1) !}\]

for some \(\gamma\) between \(\alpha\) and \(\beta .\) Let

\[g(x)=f(x)-P(x)-M(x-\alpha)^{n+1}.\]

Then, for \(k=0,1, \ldots, n\),

\[g^{(k)}(\alpha)=f^{(k)}(\alpha)-P^{(k)}(\alpha)=0.\]

Now \(g(\beta)=0,\) so, by Rolle's theorem, there exists \(\gamma_{1}\) between \(\alpha\) and \(\beta\) such that \(g^{\prime}\left(\gamma_{1}\right)=0 .\) Using Rolle's theorem again, we see that there exists \(\gamma_{2}\) between \(\alpha\) and \(\gamma_{1}\) such that \(g^{\prime \prime}\left(\gamma_{2}\right)=0 .\) Continuing for \(n+1\) steps, we find \(\gamma_{n+1}\) between \(\left.\alpha \text { and } \gamma_{n} \text { (and hence between } \alpha \text { and } \beta\right)\) such that \(g^{(n+1)}\left(\gamma_{n+1}\right)=0 .\) Hence

\[0=g^{(n+1)}\left(\gamma_{n+1}\right)=f^{(n+1)}\left(\gamma_{n+1}\right)-(n+1) ! M.\]

Letting \(\gamma=\gamma_{n+1},\) we have

\[M=\frac{f^{(n+1)}(\gamma)}{(n+1) !},\]

as required. \(\quad\) Q.E.D.

We call the polynomial \(P\) in the statement of Taylor's theorem the Taylor polynomial of order \(n\) for \(f\) at \(\alpha .\)

Let \(f(x)=\sqrt{x} .\) Then the 4th order Taylor polynomial for \(f\) at 1 is

\[P(x)=1+\frac{1}{2}(x-1)-\frac{1}{8}(x-1)^{2}+\frac{1}{16}(x-1)^{3}-\frac{5}{128}(x-1)^{4}.\]

By Taylor's theorem, for any \(x>0\) there exists \(\gamma\) between 1 and \(x\) such that

\[\sqrt{x}=P(x)+\frac{105}{(32)(5 !) \gamma^{\frac{9}{2}}}(x-1)^{5}=P(x)+\frac{7}{256 \gamma^{\frac{9}{2}}}(x-1)^{5}.\]

For example,

\[\sqrt{1.2}=P(1.2)+\frac{7}{256 \gamma^{\frac{9}{2}}}(1.2-1)^{5}=P(1.2)+\frac{7}{256 \gamma^{\frac{9}{2}}}(0.2)^{5}=P(1.2)+\frac{7}{800000 \gamma^{\frac{9}{2}}},\]

for some \(\gamma\) with \(1<\gamma<1.2 .\) Hence \(P(1.2)\) underestimates \(\sqrt{1.2}\) by a value which is no larger than \(\frac{7}{80000} .\) Note that

\[P(1.2)=\frac{17527}{16000}=1.0954375\]

and

\[\frac{7}{800000}=0.00000875.\]

So \(\sqrt{1.2}\) lies between 1.0954375 and 1.09544625.

Use the 5th order Taylor polynomial for \(f(x)=\sqrt{x}\) at 1 to estimate \(\sqrt{1.2}\). Is this an underestimate or an overestimate? Find an upper bound for the largest amount by which the estimate and \(\sqrt{1.2}\) differ.

Find the 3rd order Taylor polynomial for \(f(x)=\sqrt[3]{1+x}\) at 0 and use it to estimate \(\sqrt[3]{1.1}\). Is this an underestimate or an overestimate? Find an upper bound for the largest amount by which the estimate and \(\sqrt[3]{1.1}\) differ.

Suppose \(f \in C^{(2)}(a, b) .\) Use Taylor's theorem to show that

\[\lim _{h \rightarrow 0} \frac{f(c+h)+f(c-h)-2 f(c)}{h^{2}}=f^{\prime \prime}(c)\]

for any \(c \in(a, b)\).

Suppose \(f \in C^{(1)}(a, b), c \in(a, b), f^{\prime}(c)=0,\) and \(f^{\prime \prime}\) exists on \((a, b)\) and is continuous at \(c .\) Show that \(f\) has a local maximum at \(c\) if \(f^{\prime \prime}(c)<0\) and a local minimum at \(c\) if \(f^{\prime \prime}(c)>0 .\)