Derive an approximation result from the generalised mean value theorem.

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Here, we will prove the theorem below.

**Theorem 1**. *Taylor’s formula Let
\(I\) be an interval and assume that
\(f:I\to\mathbb{R}\) is \(n+1\)-times differentiable. Let \(x_0\in I\) and \(h\in\mathbb{R}\) such that \(x_0+h\in I\). Then there exists some \(\theta\in(0,1)\) such that \[f(x_0+h)=\underbrace{\sum_{k=0}^n\frac{f^{(k)}(x_0)}{k!}h^k}_{\text{Taylor}\
\text{polynomial}}+\underbrace{\frac{f^{(n+1)}(x_0+\theta
h)}{(n+1)!}h^{n+1}}_{\text{remainder}\ \text{term}}.\] The number
\(x_0\) is called **expansion
point*.

*Proof:* Remember the generalised mean value
theorem \[\frac{F(x_1)-F(x_0)}{g(x_1)-g(x_0)}=\frac{F'(x_0+\theta(x_1-x_0))}{g'(x_0+\theta(x_1-x_0))}\]
for some \(\theta\in(0,1)\), which can
be reformulated as (with \(h=x_1-x_0\))
\[F(x_0+h)-F(x_0)=\frac{g(x_0+h)-g(x_0)}{g'(x_0+\theta
h)}\cdot F'(x_0+\theta h).\label{eq:genmean}\] Now consider
the functions \[F(x):=\sum_{k=0}^n\frac{f^{(k)}(x)}{k!}(x_0+h-x)^k,\qquad
g(x):=(x_0+h-x)^{n+1}.\] Then we have \[F'(x)=\sum_{k=0}^n\frac{f^{(k+1)}(x)}{k!}(x_0+h-x)^k-\sum_{k=1}^n\frac{f^{(k)}(x)}{(k-1)!}(x_0+h-x)^{k-1}=\frac{f^{(n+1)}(x)}{n!}(x_0+h-x)^{n}.\]
and \[g'(x)=-(n+1)(x_0+h-x)^{n}.\]
Moreover, we have \[F(x_0)=\sum_{k=0}^n\frac{f^{(k)}(x_{0})}{k!}h^k,\quad
F(x_0+h)=f(x_0+h),\quad g(x_0)=h^{n+1},\quad g(x_0+h)=0.\] Then
using the generalised mean value theorem, we obtain \[\begin{aligned}
&f(x_0+h)- \sum_{k=0}^n\frac{f^{(k)}(x_0)}{k!}h^k\\
=&F(x_0+h)-F(x_0)\\
=&\frac{g(x_0+h)-g(x_0)}{g'(x_0+\theta h)}\cdot
F'(x_0+\theta h)\\
=&\frac{-h^{n+1}}{-(n+1)((1-\theta) h)^{n}}\cdot
\frac{f^{(n+1)}(x_0+\theta h)}{n!}((1-\theta) h)^{n}\\
=&\frac{f^{(n+1)}(x_0+\theta h)}{(n+1)!}h^{n+1}.
\end{aligned}\] This implies Taylor’s formula.\(\Box\)

Note that, by the substitution \(h:=x-x_0\), Taylor’s formula can also be
written as follows:

**Theorem 2**. *Taylor’s formula,
alternative version Let \(I\) be
an interval and assume that \(f:I\to\mathbb{R}\) is \(n+1\)-times differentiable. Let \(x_0,x\in I\). Then there exists some \(\hat{x}\) between \(x_0\) and \(x\) such that \[f(x)=\underbrace{\sum_{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k}_{=:T_n(x,x_0)}+\underbrace{\frac{f^{(n+1)}(\hat{x})}{(n+1)!}(x-x_0)^{n+1}}_{=:R_n(x,x_0)}.\]*

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