(Theorem) Taylor approximation

Taylor’s theorem §

Definition: Taylor polynomial

If $f:[a,b]\to \mathbb{R}$ is $n$-times differentiable at $x_0\in [a,b]$, then its $n$-th order Taylor polynomial centered at $x_0$ is the series defined as

$$T_{n,x_0}(x):=\sum _{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0{)}^k.$$

Notice that for all $0\le k\le n$, we have $T_{n,x_0}(x_0)=f^{(k)}(x_0)$. Further, denote the approximation error or remainder as $x\to x_0$ by

$$R_n(x)=f(x)-T_{n,x_0}(x)=o(|x-x_0{|}^n),$$

where the final expression uses little-o notation.

#wip Taylor of order 1 is the tangent?

Theorem: Taylor approximation

Let $f:[a,b]\to \mathbb{R}$ be $(n=k+1)$-times continuously differentiable at some point $x_0\in [a,b]$, where $k>0$, and write $T_{n,x_0}$ for the $n$-th order Taylor polynomial of $f$ and $R_n(x)$ for the approximation error as $x\to x_0$. Then:

• (i) Peano form of the remainder. There exists a function $h_n:[a,b]\to \mathbb{R}$ defined by $h_n(x):=f(x)-(T_{n,x_0})(x)$ which satisfies

$$\underset{x\to x_0}{\lim }\frac{h_n(x)}{(x-x_0{)}^n}=0.$$

Thus, Taylor’s theorem says that the error term $o(|x-x_0{|}^n)$ decays as fast as some multiple of $|x-x_0{|}^{n+1}$, and in particular faster than any multiple of $|x-x_0{|}^n$.

• (ii) Cauchy form of the remainder. There exists some ${\xi }_C\in (x_0,x)$ such that

$$R_{k,C}(x)=\frac{f^{(k+1)}({\xi }_C)}{k!}(x-x_0)(x-{\xi }_C{)}^k.$$

• (iii) Lagrange form of the remainder. There exists some ${\xi }_L\in (x_0,x)$ such that

$$R_{k,L}(x)=\frac{f^{(k+1)}({\xi }_L)}{(k+1)!}(x-x_0{)}^{k+1}.$$

• (iv) Integral form of the remainder.

$$R_k(x)={\int }_{x_0}^x\frac{f^{(k+1)}(t)}{k!}(x-t{)}^{k}dt.$$

Proof (i) from Modern Analysis I.

By (Theorem) L’Hopital’s rule,

$$\begin{array}{rl}\underset{x\to x_0}{\lim }\frac{h_n(x)}{(x-x_0{)}^n}& =\underset{x\to x_0}{\lim }\frac{h_n^{\prime }(x)}{n(x-x_0{)}^{n-1}}=\underset{x\to x_0}{\lim }\frac{h_n^{(n)}(x)}{n!}\\ & =\frac{h^{(n)}(x_0)}{n!}=\frac{f^{(n)}(x_0)-T_{n,x_0}^{(n)}(x_0)}{n!}=0.\end{array}$$

#wip

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Estimating the remainder §

Theorem: Estimating the remainder of Taylor’s approximation

If $f$ is $C^{(k+1)}$ on an interval $[x-h,x+h]$ and $M={\text{max}}_{t\in [0,h]}|f^{(k+1)}(x+t)|$, then the remainder can be estimated by

$$|R_k(x)|=|f(x+h)-P_k(h)|\le \frac{M}{(k+1)!}{h}^{k+1}$$

where

is Taylor’s polynomial.