Proof of Taylor's Theorem
Derive an approximation result from the generalised mean value theorem.
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Study Proof of Taylor’s Theorem
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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