Taylor Series

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
and

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 1 December 2009

Taylor series

Taylor's theorem.

1. If $f\left(x\right)=a_0+a_{1}x+a_2{x}^2+\cdots$ then $$a_l=\frac{1}{l!}\left({\frac{d^{l}f}{d{x}^l}|}_{x=0}\right)\text{.}$$
2. If $f:\left[a,b\right]\to ℝ$ and $N\in {\mathbb{Z}}_{>0}$ and $f^{\left(N\right)}:\left[a,b\right]\to ℝ$ is continuous and $f^{\left(N+1\right)}:\left(a,b\right)\to ℝ$ exists then there exists $c\in \left(a,b\right)$ such that $$f\left(b\right)=f\left(a\right)+f\prime \left(a\right)\left(b-a\right)+\frac{1}{2!}f″\left(a\right){\left(b-a\right)}^2+\cdots +\frac{1}{N!}{f}^{\left(N\right)}\left(a\right){\left(b-a\right)}^N+\frac{1}{\left(N+1\right)!}{f}^{\left(N+1\right)}\left(c\right){\left(b-a\right)}^{N+1}\text{.}$$

Remark. The following results follow from (a) after a tiny bit of algebra. $$\text{If}f\left(a+x\right)=a_0+a_{1}x+a_2{x}^2+\cdots \text{then}{a}_l=\frac{1}{l!}\left({\frac{d^{l}f}{d{x}^l}|}_{x=0}\right)\text{.}$$ $$\text{If}f\left(z\right)=a_0+a_1\left(z-a\right)+a_2{\left(z-a\right)}^2+\cdots \text{then}{a}_l=\frac{1}{l!}\left({\frac{d^{l}f}{d{z}^l}|}_{z=a}\right)\text{.}$$

Remark. The special case $N=1$ is the mean value theorem. The special case $N=1$ and $f\left(a\right)=f\left(b\right)$ is Rolle's theorem. The last term in $$f\left(a\right)+f\prime \left(a\right)\left(b-a\right)+\frac{1}{2!}f″\left(a\right){\left(b-a\right)}^2+\cdots +\frac{1}{N!}{f}^{\left(N\right)}\left(a\right){\left(b-a\right)}^N+\frac{1}{\left(N+1\right)!}{f}^{\left(N+1\right)}\left(c\right){\left(b-a\right)}^{N+1}$$ is Lagrange's form of the remainder.

		Proof of Taylor's theorem, part (a).
	Assume $f\left(x\right)=a_0+a_{1}x+a_2{x}^2+\cdots =\sum _{i=0}^{\infty }{a}_i{x}^i$ To show: ${\displaystyle a_l=\left({\frac{d^{l}f}{d{x}^l}|}_{x=0}\right)}$ for $l\in {\mathbb{Z}}_{>0}$. We have $$\begin{array}{rcl}\frac{d^{l}f}{d{x}^l}& =& \frac{d^l}{d{x}^l}\sum _{i=0}^{\infty }{a}_{\mathrm{i}}{x}^i\\ & =& \sum _{i=l}^{\infty }\left(i-\left(l-1\right)\right)\left(i-\left(l-2\right)\right)\cdots \left(i-1\right)i{a}_i{x}^{i-l}\\ & =& \sum _{i=l}^{\infty }\left(\prod _{j=1}^l\left(i-\left(l-j\right)\right)\right)a_i{x}^{i-l}\text{,}\end{array}$$ by repeated differentiation. So $$\begin{array}{rcl}{\frac{d^{l}f}{d{x}^l}|}_{x=0}& =& {\sum _{i=l}^{\infty }\left(\prod _{j=1}^l\left(i-\left(l-j\right)\right)\right)a_i{x}^{i-l}|}_{x=0}\\ & =& \left(\prod _{j=1}^l\left(l-\left(l-j\right)\right)\right)a_l\\ & =& l!a_l\text{.}\end{array}$$ Solving for $a_l$ givs the required result.proof changed

		Proof of the mean value theorem.
	To show: there exists $c\in \left(a,b\right)$ such that $$f\left(b\right)=f\left(a\right)+f\prime \left(c\right)\left(b-a\right)\text{.}$$ First do the case: $f\left(a\right)=f\left(b\right)$. Then, since $\left[a,b\right]$ is compact and connected and $f:\left[a,b\right]\to ℝ$ is continuous, $$f\left(\left[a,b\right]\right)\text{is compact and connected.}$$ So $f\left(\left[a,b\right]\right)$ is a closed interval. So $f\left(\left[a,b\right]\right)=\left[\min ,\max \right]$ for some $\min ,\max \in ℝ$. So there exists $c\in \left(a,b\right)$ such that $f\left(c\right)=\max$. Then, if $\epsilon \in {ℝ}_{>0}$ is small enough $f\left(c+\epsilon \right)\le f\left(c\right)$ and $f\left(c-\epsilon \right)\le f\left(c\right)$. So $f\prime \left(c\right)\ge 0$ and $f\prime \left(c\right)\le 0$. So $f\prime \left(c\right)=0$. Next do the case $f\left(a\right)\ne f\left(b\right)$. Let ${\displaystyle g\left(x\right)=-\left(\frac{f\left(b\right)-f\left(a\right)}{b-a}\right)\left(x-a\right)+f\left(x\right)}$. So $g\left(a\right)=f\left(a\right)$ and $g\left(b\right)=f\left(a\right)$ and ${\displaystyle g\prime \left(x\right)=-\left(\frac{f\left(b\right)-f\left(a\right)}{b-a}\right)+f\prime \left(x\right).}$ So by the first case, there exists $c\in \left(a,b\right)$ with $g\prime \left(c\right)=0$. So ${\displaystyle f\prime \left(c\right)-\left(\frac{f\left(b\right)-f\left(a\right)}{b-a}\right)=0}$. So ${\displaystyle f\prime \left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}}$ and $f\left(b\right)=f\left(a\right)+f\prime \left(c\right)\left(b-a\right)$.

References [PLACEHOLDER]

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)