Real Analysis

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

Last update: 13 July 2014

Lecture 18

Derivatives by limits

Let $f:[a,b]\to ℝ\text{.}$ Let $c\in [a,b]\text{.}$

The derivative of $f$ at $x=c$ is $$f\prime \left(c\right)=\underset{x\to c}{\text{lim}}\frac{f\left(x\right)-f\left(a\right)}{x-c}\text{.}$$ Alternatively, $$f\prime \left(c\right)=\underset{\mathrm{\Delta }x\to 0}{\text{lim}}\frac{f(c+\mathrm{\Delta }x)-f\left(x\right)}{\mathrm{\Delta }x}\text{.}$$

Let $f:[a,b]\to ℝ$ and $g:[a,b]\to ℝ$ and let $\beta ,\gamma \in ℝ\text{.}$ Assume that $f\prime \left(c\right)$ and $g\prime \left(c\right)$ exist. Then

(a)	$(\beta f+\gamma g)\prime \left(c\right)=\beta f\prime \left(c\right)+\gamma g\prime \left(c\right),$
(b)	$\left(fg\right)\prime \left(c\right)=f\prime \left(c\right)g\left(c\right)+f\left(c\right)g\prime \left(c\right),$
(c)	Assume $f:[a,b]\to ℝ$ is given by $f\left(x\right)=x\text{.}$ then $f\prime \left(c\right)=1\text{.}$
(d)	If $f\prime \left(c\right)$ exists then $f$ is continuous at $x=c\text{.}$

Define $$f″\left(c\right)=\left(f\prime \right)\prime \left(c\right),f^{\left(3\right)}\left(c\right)=\left(f″\right)\prime \left(c\right)\text{and}{f}^{\left(N\right)}\left(c\right)=\left(f^{(N-1)}\right)\prime \left(c\right)\text{.}$$

(Taylor's theorem with Lagrange's remainder). If $f:[a,b]\to ℝ$ and $N\in {\mathbb{Z}}_{\ge 0}$ and $f^{\left(N\right)}:[a,b]\to ℝ$ is continuous and $f^{(N+1)}:[a,b]\to ℝ$ exists then there exists $c\in (a,b)$ such that $$f\left(b\right)=f\left(a\right)+f\prime \left(a\right)(b-a)+\frac{1}{2!}f″\left(a\right){(b-a)}^2+\cdots +\frac{1}{N\prime }{f}^{\left(N\right)}\left(a\right){(b-a)}^N+\frac{1}{(N+1)!}{f}^{(N+1)}\left(c\right){(b-a)}^{N+1}\text{.}$$

Remarks:

(1)	The last term in $f\left(b\right)=f\left(a\right)+\cdots +\frac{1}{(N+1)!}{f}^{(N+1)}\left(c\right){(b-a)}^{N+1}$ is Lagrange's form of the remainder.
(2)	The special case $N=0$ is the Mean Value Theorem. If $f:[a,b]\to ℝ$ and $f^{*}:[a,b]\to ℝ$ is continuous and $f\prime :[a,b]\to ℝ$ exists then there exists $c\in (a,b)$ such that $$f\left(b\right)=f\left(a\right)+f\prime \left(c\right)(b-a)$$ i.e. $$f\prime \left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\begin{array}{c} a b c x y \end{array}$$
(3)	The special case $N=0$ and $f\left(a\right)=f\left(b\right)$ is Rolle's theorem If $f:[a,b]\to ℝ$ is continuous and $f\prime :[a,b]\to ℝ$ exists then there exists $c\in (a,b)$ such that $f\prime \left(c\right)=0$ $$\begin{array}{c} a b x f ( a ) = f ( b ) y \end{array}$$
(4)	The proof of these theorems uses Intermediate value theorem (a) If $f:[a,b]\to ℝ$ is continuous and $w$ is between $f\left(a\right)$ and $f\left(b\right)$ then there exists $c\in (a,b)$ such that $f\left(c\right)=w\text{.}$ (b) If $f:[a,b]\to ℝ$ is continuous then there exist $m,M\in ℝ$ such that $$f\left([a,b]\right)=[m,M]$$ $$\begin{array}{c} a b f ( a ) f ( b ) m M \end{array}$$
(5)	Let $f:[0,2\pi ]\to ℂ$ be given by $$f\left(x\right)=\text{cos}\hspace{0.17em}x+i\text{sin}\hspace{0.17em}x\text{.}$$ Then $f\left(0\right)=f\left(2\pi \right)$ but $f\prime \left(x\right)$ is never $0\text{.}$ Why is not a contradiction to Rolle's theorem?
(6)	The first $N$ terms in (all but the remainder term) $$f\left(b\right)=f\left(a\right)+\cdots +\frac{1}{N!}{f}^{\left(N\right)}\left(c\right){(b-a)}^N+\frac{1}{(N+1)!}f\prime \left(c\right){(b-a)}^{N+1}$$ are the Taylor approximation to $f$ at $x=a$ of order $N\text{.}$
(7)	$f:[a,b]\to ℝ$ is differentiable at $x=c$ if $f\prime \left(c\right)$ exists.

Approximate ${28}^{1/3}$ to $5$ decimal places.

Let $f\left(x\right)={(27+x)}^{1/3}\text{.}$ Then $${(27+x)}^{1/3}=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\cdots$$ So $$\begin{array}{ccc}{\displaystyle a_0}& =& {\displaystyle {(27+0)}^{1/3}=3,}\\ {\displaystyle a_1}& =& {\displaystyle \left(\frac{d}{dx}{(27+x)}^{1/3}\right){|}_{x=0}=1/3{(27+x)}^{-\frac{2}{3}}{|}_{x=0}=1/3\frac{1}{3^2}=\frac{1}{27},}\\ {\displaystyle a_2}& =& {\displaystyle \frac{1}{2!}\left(\frac{d^2}{d{x}^2}{(27+x)}^{1/3}\right){|}_{x=0}=\frac{1}{2}1/3(-\frac{2}{3}){(27+x)}^{-\frac{5}{3}}{|}_{x=0}=\frac{1}{2}·\frac{(-2)}{3^2}\frac{1}{3^5}=-\frac{1}{3^7},}\\ {\displaystyle a_3}& =& {\displaystyle \frac{1}{3!}\left(\frac{d^3}{d{x}^3}{(27+x)}^{1/3}\right){|}_{x=0}=\frac{1}{2·3}1/3\frac{(-2)}{3}\frac{(-5)}{3}{(27+x)}^{-\frac{8}{3}}{|}_{x=0}=\frac{5}{3^4}·\frac{1}{3^8}=\frac{5}{3^{12}},}\\ {\displaystyle \frac{1}{4!}{f}^{\left(4\right)}\left(c\right){(28-27)}^4}& =& {\displaystyle \frac{1}{4!}1/3\frac{(-2)}{3}\frac{(-5)}{3}\frac{(-8)}{3}{(27+c)}^{-\frac{11}{3}}}\\ & =& {\displaystyle \frac{-2^4·5}{2^3·3^5}{(27+c)}^{-\frac{11}{3}}=\frac{-2·5}{3^5}\frac{1}{{(27+c)}^{\frac{11}{3}}}\text{.}}\end{array}$$ So ${(27+x)}^{1/3}=3+\frac{1}{27}x-\frac{1}{3^7}{x}^2+\frac{5}{3^{12}}{x}^3+\cdots$ and $$\begin{array}{ccc}{\displaystyle {28}^{1/3}}& =& {\displaystyle {(27+1)}^{1/3}}\\ & \approx & {\displaystyle 3+\frac{1}{27}-\frac{1}{3^7}+\frac{5}{3^{12}}}\end{array}$$ with error equal to $$\frac{2·5}{3^5}\frac{1}{{(27+c)}^{\frac{11}{3}}}\text{for some}c\in (0,1)\text{.}$$ So the error is less than $$\frac{2·5}{3^5}\frac{1}{{27}^{\frac{11}{3}}}=\frac{2·5}{3^5·3^{11}}=\frac{2·5}{3^{16}}\text{.}$$

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100419Lect18.pdf and was given on 19 April 2010.

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