# Application of Taylor's Theorem

Calculate an approximation via Taylor's Theorem

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Taylor's Theorem | An approximation method for differentiable functions. |

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

## Study *Application of Taylor’s Theorem* #

The application of Taylor’s formula is twofold: First, it gives a polynomial that approximates a given function quite fine in some neighbourhood. The second application is the computation of values of “complicated functions”. We will present examples for both kinds of application.

**Example 1**. Consider the function \(f(x)=\sin(x)\). We want to determine the
Taylor polynomial of degree 6 with expansion point \(x_0=\frac\pi2\). Since we have \[\sin'(x)=\cos(x),\quad\sin''(x)=-\sin(x),\quad\sin^{(3)}(x)=-\cos(x),\]
\[\sin^{(4)}(x)=\sin(x),\quad\sin^{(5)}(x)=\cos(x),\quad\sin^{(6)}(x)=-\sin(x)\]
and \[\sin(x_0)=\sin\left(\frac\pi2\right)=1,\quad
\cos(x_0)=\cos\left(\frac\pi2\right)=0,\] the Taylor polynomial
of degree 6 is given by \[T_6(x)=1-\frac12\left(x-\frac{\pi}{2}\right)^2+\frac{1}{24}\left(x-\frac{\pi}{2}\right)^4-\frac1{720}\left(x-\frac{\pi}{2}\right)^6.\]
The remainder term reads \[R_6(x,x_0)=\frac{\sin^{(7)}(\hat{x})}{7!}\left(x-\frac{\pi}{2}\right)^{7}
=\frac{-\cos(\hat{x})}{5040}\left(x-\frac{\pi}{2}\right)^{7}.\]
Taking into account that \(|\cos(x)|\leq1\) for all \(x\in\mathbb{R}\), we have that \[|R_6(x,x_0)|\leq\frac{\left|x-\frac{\pi}{2}\right|^{7}}{5040}.\]
This leads to the estimate \[|\sin(x)-T_6(x)|=|R_6(x,x_0)|\leq\frac{\left|x-\frac{\pi}{2}\right|^{7}}{5040}.\]

**Example 2**. We want to compute \(\log(1.2)\) up to 3 digit precision. A nice
expansion point is \(x_0=1\) since we
know the precise values of \(\log^{(k)}(1)\). Consider \[\log'(x)=\frac1x,\quad
\log''(x)=-\frac{1!}{x^2},\] \[\log^{(3)}(x)=\frac{2!}{x^3},\quad\log^{(4)}(x)=-\frac{3!}{x^4}\]
and therefore, the Taylor polynomial of degree 3 is given by \[T_3(x)=(x-1)-\frac12(x-1)^2+\frac13(x-1)^3.\]
In particular, we have \[T_3(1.2)=0.2-\frac12\cdot 0.2^2+\frac13 \cdot
0.2^3=\frac{137}{750}.\] Now we estimate \(|\log(1.2)-\frac{137}{750}|\): The
remainder term is given by \[R_3(x,x_0)=-\frac{3!}{\hat{x}^4}\frac{(x-x_0)^{4}}{4!}=-\frac{(x-x_0)^{4}}{4\hat{x}^4}\]
for some \(\hat{x}\) between \(x\) and \(x_0\). For \(x=1.2\), \(x_0=1\) we have \(1<\hat{x}<1.2\) and therefore \[| R_3(1.2,1) |
=\frac{(0.2)^{4}}{4\hat{x}^4}=4\cdot10^{-4}\frac{1}{\hat{x}^4}\leq4\cdot10^{-4}.\]
This leads to \[|\log(1.2)-0.182\overline{6}|=|R_3(1.2,1)|\leq4\cdot10^{-4},\]
so we have determined \(\log(1.2)\) up
to three digits.

**Theorem 3**. *Let \(I\subset \mathbb{R}\) be an open interval,
\(n\in\mathbb{N}\) and \(f:I\rightarrow\mathbb{R}\) an \(n\)-times continuously differentiable
function. Suppose that for \(a\in I\)
holds \[f'(a)=f''(a)=\dots=f^{(n-1)}(a)=0\quad
\text{and}\quad f^{(n)}(a)\neq 0 \ .\] If \(n\) is odd, then \(a\) is not a local extremum. If \(n\) is even and \(f^{(n)}(a)> 0\), then \(a\) is a local minimum. If \(n\) is even and \(f^{(n)}(a)< 0\), then \(a\) is a local maximum.*

*Proof:* By assumption the Taylor expansion of
\(f\) of degree \(n-1\) in the expansion point \(a\) reads: \[\label{eq:locext1}
f(x)=f(a)+\frac{f^{n}(z)}{n!}(x-a)^n\] where \(z=z(x)\) lies between \(x\) and \(a\). Since \(f^{(n)}\) is continuous and \(f^{(n)}(a)\neq 0\), there is a
neighbourhood \(U:=~(a-\varepsilon,a+\varepsilon)~\subset
I\), \(\varepsilon>0\), of
\(a\) such that \(f^{(n)}(x)\neq 0\) for all \(x\in U\). This means that for all \(x\in U\), \(f^{(n)}(x)\) and \(f^{(n)}(a)\) have the same sign. Then for
any \(x_l \in~(a-\varepsilon,a)\) and
any \(x_r \in~(a,a+\varepsilon)\)
Theorem 3 implies that for \(z_l:=z(x_l)\in~(x_l,a)\) and \(z_r:=z(x_r)\in~(a,x_r)\) holds \[f(x_l) = f(a)+\frac{f^{n}(z_l)}{n!}(x_l-a)^n
,\] \[f(x_r) =
f(a)+\frac{f^{n}(z_r)}{n!}(x_r-a)^n\] and \(0\neq f^{n}(a), f^{n}(z_l),f^{n}(z_r)\)
have the same sign. If \(n\) is odd,
then \((x_l-a)^n < 0 <
(x_r-a)^n\) and therefore either \(f(x_l)<f(a)<f(x_r)\) or \(f(x_l)>f(a)>f(x_r)\) so that \(a\) is not a local extremum. If \(n\) is even and \(f^{(n)}(a)>0\), then \((x_l-a)^n,(x_r-a)^n >0\) and \(f(x_l),f(x_r)>f(a)\) so that \(a\) is a local minimum. Finally, if \(n\) is even and \(f^{(n)}(a)<0\), then \((x_l-a)^n,(x_r-a)^n >0\) and \(f(x_l),f(x_r)<f(a)\) so that \(a\) is a local maximum. \(\Box\)

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