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 19

Areas

$$\begin{array}{c}\begin{array}{c} ⏟ Δx ℓ r f(ℓ) \end{array}\text{has area}f\left(\ell \right)\mathrm{\Delta }x\\ \begin{array}{c} ⏟ Δx ℓ r f(ℓ) f(r) \end{array}\text{has area}\begin{array}{c}f\left(\ell \right)\mathrm{\Delta }x+\frac{1}{2}\mathrm{\Delta }x(f\left(r\right)-f\left(\ell \right))\\ =\frac{\mathrm{\Delta }x}{2}(2f\left(\ell \right)+f\left(r\right)-f\left(\ell \right))\\ =\frac{\mathrm{\Delta }x}{2}(f\left(\ell \right)+f\left(r\right))\end{array}\\ \begin{array}{c} ⏟ ⏟ Δx Δx ℓ m r f(ℓ) f(m) f(r) \end{array}\text{has area}\frac{\mathrm{\Delta }x}{3}(f\left(\ell \right)+4f\left(m\right)+f\left(r\right))\end{array}$$

Riemann Integral

$$\begin{array}{ccc}{\displaystyle {\int }_a^{b}f\left(x\right)dx}& =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}\left(\text{sum of the areas of the little rectangles}\right)}\\ & =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}\mathrm{\Delta }x(f\left(a\right)+f(a+\mathrm{\Delta }x)+\cdots +f(b-\mathrm{\Delta }x))}\\ \multicolumn{3}{c}{\begin{array}{c} a b \end{array}}\end{array}$$

Trapezoidal integral

$$\begin{array}{ccc}{\displaystyle {\int }_a^{b}f\left(x\right)dx}& =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}\left(\text{sum of the areas of the little trapezoids}\right)}\\ & =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}\frac{\mathrm{\Delta }x}{2}\left(\begin{array}{c}f\left(a\right)+f(a+\mathrm{\Delta }x)+f(a+\mathrm{\Delta }x)+f(a+2\mathrm{\Delta }x)\\ +f(a+2\mathrm{\Delta }x)+f(a+3\mathrm{\Delta }x)\\ +\cdots +f(b-2\mathrm{\Delta }x)+f(b-\mathrm{\Delta }x)\\ +f(b-\mathrm{\Delta }x)+f\left(b\right)\end{array}\right)}\\ & =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}\frac{\mathrm{\Delta }x}{2}(f\left(a\right)+2f(a+\mathrm{\Delta }x)+2f(a+2\mathrm{\Delta }x)+\cdots +2f(b-\mathrm{\Delta }x)+f\left(b\right))}\end{array}$$

Simpson's Integral

$$\begin{array}{c} ⏟ ⏟ Δx Δx ℓ m r f(ℓ) f(m) f(r) \end{array}\text{has area}\frac{\mathrm{\Delta }x}{3}(f\left(\ell \right)+4\left(m\right)+f\left(r\right))$$ so that $$\begin{array}{ccc}{\displaystyle {\int }_a^{b}f\left(x\right)dx}& =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}\left(\text{sum of the areas of the little camel humps}\right)}\\ & =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}\frac{\mathrm{\Delta }x}{3}\left(\begin{array}{c}f\left(a\right)+4f(a+\mathrm{\Delta }x)+f(a+2\mathrm{\Delta }x)\\ +f(a+2\mathrm{\Delta }x)+4f(a+3\mathrm{\Delta }x)+f(a+4\mathrm{\Delta }x)\\ +\cdots \\ +f(b-2\mathrm{\Delta }x)+4f(b-\mathrm{\Delta }x)+f\left(b\right)\end{array}\right)}\\ & =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}\frac{\mathrm{\Delta }x}{3}(f\left(a\right)+4f(a+\mathrm{\Delta }x)+2f(a+2\mathrm{\Delta }x)+4f(a+3\mathrm{\Delta }x)+2f(a+4\mathrm{\Delta }x)+\cdots +f\left(b\right))\text{.}}\end{array}$$

Let $N\in {\mathbb{Z}}_{>0}$ and $f:[a,b]\to ℝ$ be a function and $M\in {ℝ}_{>0}\text{.}$

(a)	Assume that $f^{(N+1)}:[a,b]\to ℝ$ exists and $\left|f^{(N+1)}\left(c\right)\right|<M$ for $c\in [a,b]\text{.}$ Let $$\text{TaylorErr}\left(N\right)=f\left(b\right)-(f\left(a\right)+f\prime \left(a\right)(b-a)+\frac{1}{2!}f″\left(a\right){(b-a)}^2+\cdots +\frac{1}{N!}{f}^{\left(N\right)}\left(a\right)(b-a))\text{.}$$ Then $$\left|\text{TaylorErr}\left(N\right)\right|<\frac{1}{(N+1)!}{(b-a)}^{N}M\text{.}$$
(b)	Assume that $f″:[a,b]\to ℝ$ exists and $\left|f″\left(c\right)\right|<M$ for $c\in [a,b]\text{.}$ Let $\mathrm{\Delta }x=\frac{b-a}{N}$ and $$\text{TrapErr}\left(N\right)={\int }_a^{b}f\left(x\right)dx-\left(\frac{b-a}{N}\right)\frac{1}{2}\left(\begin{array}{c}f\left(a\right)+2f(a+\mathrm{\Delta }x)\\ +2f(a+2\mathrm{\Delta }x)+\cdots \\ \dots +2f(b-\mathrm{\Delta }x)+f\left(b\right)\end{array}\right)\text{.}$$ Then $$\left|\text{TrapErr}\left(N\right)\right|<\frac{{(b-a)}^3}{12N^2}·M\text{.}$$
(c)	Assume that $f^{\left(4\right)}:[a,b]\to ℝ$ exists and $\left|f^{\left(4\right)}\left(c\right)\right|<M$ for $c\in [a,b]\text{.}$ Let $\mathrm{\Delta }x=\frac{b-a}{N}$ and $$\text{SimpErr}\left(N\right)={\int }_a^{b}f\left(x\right)dx-\frac{(b-a)}{3N}\left(\begin{array}{c}f\left(a\right)+4f(a+\mathrm{\Delta }x)+2f(a+2\mathrm{\Delta }x)\\ +4f(a+3\mathrm{\Delta }x)+2f(a+4\mathrm{\Delta }x)+\cdots \\ \cdots +2f(b-2\mathrm{\Delta }x)+4f(b-\mathrm{\Delta }x)+f\left(b\right)\end{array}\right)\text{.}$$ Then $$\left|\text{SimpErr}\left(N\right)\right|<\frac{{(b-a)}^5}{180N^4}·M\text{.}$$

Find $\text{log}\left(2\right)$ to within $.01\text{.}$ $$\text{log}(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots +\frac{x^N}{N}+\mathrm{Error}\left(N\right)\text{.}$$ In fact, if $f\left(x\right)=\text{log}(1+x)$ then $$\begin{array}{c}{\displaystyle f\prime \left(x\right)=\frac{1}{1+x},f″\left(x\right)=\frac{-1}{{(1+x)}^2},f^{\left(3\right)}\left(x\right)=\frac{(-1)(-2)}{{(1+x)}^3},}\\ {\displaystyle f^{\left(4\right)}\left(x\right)=\frac{\left(-1\right)(-2)(-3)}{{(1+x)}^4},\dots ,f^{(N+1)}\left(x\right)=\frac{(-1)(-2)\cdots (-N)}{{(1+x)}^{N+1}}\text{.}}\end{array}$$ So $$\text{TaylorError}\left(N\right)=\frac{(-1)(-2)\cdots (-N)}{{(1+c)}^{N+1}}·\frac{1}{(N+1)!}·{(2-1e)}^{N+1}\text{with}c\in (0,1)\text{.}$$ So $$\left|\text{TayErr}\left(N\right)\right|=\frac{1}{(N+1){(1+c)}^{N+1}}<\frac{1}{N+1}\text{.}$$ So, to get an approximation within $\frac{1}{100}=.01$ we should let $N=99\text{.}$

So $\text{log}\left(2\right)\approx 1-\frac{1}{2}+1/3+\cdots +\frac{1}{97}-\frac{1}{98}+\frac{1}{99},$ to within $.01\text{.}$

Find $\text{log}\hspace{0.17em}2$ to within $.01\text{.}$ $$\text{log}\hspace{0.17em}2={\int }_1^2\frac{1}{x}dx$$ Use a trapezoidal approximation with $$f\left(x\right)=\frac{1}{x},a=1\text{and}b=2\text{.}$$ How many slices (i.e. what should $N$ be)? $$\begin{array}{c} 1 2 x y y=1x \end{array}$$ Then $f\prime \left(x\right)=\frac{-1}{x^2}$ and $f″\left(x\right)=\frac{(-1)(-2)}{{𝓍}^3}=\frac{2}{x^3}\text{.}$

So $\left|f″\left(c\right)\right|<2$ for $c\in [1,2]\text{.}$

So $$\left|\text{TrapError}\left(N\right)\right|<\frac{{(b-a)}^3}{12N^2}·2=\frac{1}{6·N^2}\text{.}$$ If $N=5$ then $\frac{1}{6·N^2}=\frac{1}{6·25}=\frac{1}{150}<.01\text{.}$

So $5$ slices will do. $$\begin{array}{ccc}{\displaystyle \text{log}\left(2\right)}& =& {\displaystyle {\int }_1^2\frac{1}{x}dx}\\ & \approx & {\displaystyle \frac{(b-a)}{5}\frac{1}{2}\left(\begin{array}{c}f\left(a\right)+2f(a+\mathrm{\Delta }x)+2f(a+2\mathrm{\Delta }x)\\ +2f(a+3\mathrm{\Delta }x)+2f(a+4\mathrm{\Delta }x)+f\left(b\right)\end{array}\right)}\\ & =& {\displaystyle \frac{1}{10}(\frac{1}{1}+2\frac{1}{1+\frac{1}{5}}+2\frac{1}{1+\frac{2}{5}}+2\frac{1}{1+\frac{3}{5}}+2\frac{1}{1+\frac{4}{5}}+\frac{1}{2})}\\ & =& {\displaystyle \frac{1}{10}(1+2·\frac{5}{6}+2·\frac{5}{7}+2·\frac{5}{8}+2·\frac{5}{9}+\frac{1}{2})}\\ & =& {\displaystyle \frac{1}{10}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{20}\text{.}}\end{array}$$

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100421Lect19.pdf and was given on 21 April 2010.

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