MATH 221

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

Last updated: 6 September 2014

Lecture 31

Averages

Average of a bunch of numbers:

(a)	Add up the numbers
(b)	Divide by the number of numbers

The average of $1,2,3,\dots ,100$ $$\begin{array}{ccccccccccccccccc}1& +& 2& +& 3& +& 4& +& \cdots & +& 97& +& 98& +& 99& +& 100\\ 100& +& 99& +& 98& +& 97& +& \cdots & +& 4& +& 3& +& 2& +& 1\\ 101& +& 101& +& 101& +& 101& +& \cdots & +& 101& +& 101& +& 101& +& 101& =& 10100\end{array}$$ So $$1+2+3+\cdots +100=\frac{10100}{2}=5050\text{.}$$ So the average is $\frac{1+2+\cdots +100}{100}=\frac{5050}{100}=50.5\text{.}$

Compute the average of $$1,\frac{1}{3},\frac{1}{3^2},\frac{1}{3^3},\dots ,\frac{1}{3^{50}}$$ $$\begin{array}{ccc}{\displaystyle (1+x+x^2+\cdots +x^{50})(1-x)}& =& {\displaystyle 1+x+x^2+x^3+\cdots +x^{50}}\\ & & {\displaystyle \phantom{1}-x-x^2-x^3-\cdots -x^{50}-x^{50}}\\ & =& {\displaystyle 1-x^{51}}\end{array}$$ So $$1+x+x^2+\cdots +x^{50}=\frac{1-x^{51}}{1-x}\text{.}$$ So $$1+\frac{1}{3}+\frac{1}{3^2}+\cdots +{\left(\frac{1}{3}\right)}^{50}=\frac{1-{\left(\frac{1}{3}\right)}^{51}}{1-\frac{1}{3}}=\frac{1-\frac{1}{3^{51}}}{\frac{2}{3}}=\frac{3-\frac{1}{3^{50}}}{2}\text{.}$$ So $$\frac{1+\frac{1}{3}+\frac{1}{3^2+\cdots +\frac{1}{3^{50}}}}{51}=\frac{\frac{3-\frac{1}{3^{50}}}{2}}{51}=\frac{3-\frac{1}{3^{50}}}{102}\approx \frac{3}{102}\approx .03\text{.}$$

Estimating the average

What is the average of $1,\frac{1}{2},\frac{1}{3},\dots ,\frac{1}{100}\text{?}$ $$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{99}$$ is the area of the boxes in the picture $$\begin{array}{c} 1 2 3 99 100 x 1 y y=1x ⋯ \end{array}$$ So $$1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{99}\ge {\int }_1^{100}\frac{1}{x}dx=\text{ln}\hspace{0.17em}x{|}_{x=1}^{x=100}=\text{ln}\hspace{0.17em}100\text{.}$$ Next, $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots +\frac{1}{100}$ is the area of the boxes in the picture $$\begin{array}{c} 1 2 3 4 99 100 x 1 y y=1x \end{array}$$ So $$\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{100}\le {\int }_1^{100}\frac{1}{x}dx=\text{ln}\hspace{0.17em}x{|}_1^{100}=\text{ln}\hspace{0.17em}100\text{.}$$ So $$\frac{\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{100}}{100}\ge \frac{\text{ln}\hspace{0.17em}100+\frac{1}{100}}{100}=\frac{\text{ln}\hspace{0.17em}100}{100}+\frac{1}{{100}^2}$$ and $$\frac{\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{100}}{100}\le \frac{\text{ln}\hspace{0.17em}100+1}{100}=\frac{\text{ln}\hspace{0.17em}100}{100}+\frac{1}{100}\text{.}$$

Average value of a function

$$\begin{array}{c} a b x y y=f(x) \end{array}$$ $$\begin{array}{ccc}{\displaystyle \text{Average value of}\hspace{0.17em}f\hspace{0.17em}\text{from}\hspace{0.17em}a\hspace{0.17em}\text{to}\hspace{0.17em}b}& =& {\displaystyle \text{Average height of little boxes}}\\ & =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}\frac{f\left(a\right)\mathrm{\Delta }x+f(a+\mathrm{\Delta }x)\mathrm{\Delta }x+\cdots +f(b-\mathrm{\Delta }x)\mathrm{\Delta }x}{\left(\text{number of boxes}\right)\mathrm{\Delta }x}}\\ & =& {\displaystyle \underset{\mathrm{\Delta }x\to 0}{\text{lim}}\frac{f\left(a\right)\mathrm{\Delta }x+f(a+\mathrm{\Delta }x)\mathrm{\Delta }x+\cdots +f(b-\mathrm{\Delta }x)\mathrm{\Delta }x}{b-a}}\\ & =& {\displaystyle \frac{{\int }_a^{b}f\left(x\right)dx}{b-a}}\\ & =& {\displaystyle \frac{\left(\text{Area under}\hspace{0.17em}f\hspace{0.17em}\text{from}\hspace{0.17em}a\hspace{0.17em}\text{to}\hspace{0.17em}b\right)}{b-a}\text{.}}\end{array}$$ So $$\left(\text{Average value}\right)(b-a)=\frac{\text{Area under}\hspace{0.17em}f\left(x\right)\hspace{0.17em}\text{from}\hspace{0.17em}a\hspace{0.17em}\text{to}\hspace{0.17em}b}{b-a}\text{.}$$ $$\begin{array}{c} ⏟ b-a { A \end{array}\text{has the same area as}\begin{array}{c} a b x y y=f(x) \end{array}$$

Find the average value of $f\left(x\right)={\text{sin}}^{2}x,$ $0\le x\le \frac{\pi }{2}\text{.}$ $$\begin{array}{c} π 2 π x 1 y y=sin2x \end{array}$$ Average value is $$\begin{array}{ccc}{\displaystyle \frac{{\int }_0^{\frac{\pi }{2}}f\left(x\right)dx}{\frac{\pi }{2}-0}}& =& {\displaystyle \frac{{\int }_0^{\frac{\pi }{2}}{\text{sin}}^{2}xdx}{\frac{\pi }{2}}}\\ & =& {\displaystyle \frac{2}{\pi }{\int }_0^{\frac{\pi }{2}}\frac{1}{2}({\text{sin}}^{2}x+{\text{sin}}^{2}x)dx}\\ & =& {\displaystyle \frac{2}{\pi }{\int }_0^{\frac{\pi }{2}}\frac{1}{2}(1-{\text{cos}}^{2}x+{\text{sin}}^{2}x)dx}\\ & =& {\displaystyle \frac{1}{\pi }{\int }_0^{\frac{\pi }{2}}(1-({\text{cos}}^{2}x-{\text{sin}}^{2}x))dx}\\ & =& {\displaystyle \frac{1}{\pi }{\int }_0^{\frac{\pi }{2}}(1-\text{cos}\hspace{0.17em}2x)dx}\\ & =& {\displaystyle \frac{1}{\pi }{\int }_0^{\frac{\pi }{2}}(1-\text{cos}\hspace{0.17em}2x)dx}\\ & =& {\displaystyle \frac{1}{\pi }(x-\frac{\text{sin}\hspace{0.17em}2x}{2}){|}_0^{\frac{\pi }{2}}}\\ & =& {\displaystyle \frac{1}{\pi }(\frac{\pi }{2}-\frac{\text{sin}\hspace{0.17em}2\frac{\pi }{2}}{2})-\frac{1}{\pi }(0-\frac{\text{sin}\hspace{0.17em}0}{2})}\\ & =& {\displaystyle \frac{1}{\pi }(\frac{\pi }{2}-0)}\\ & =& {\displaystyle \frac{1}{2}\text{.}}\end{array}$$ So $$\begin{array}{c} π 2 x 1 2 y \end{array}\text{is the same area as}\begin{array}{c} π 2 x 1 1 2 y y=sin2x \end{array}$$

Notes and References

These are a typed copy of Lecture 31 from a series of handwritten lecture notes for the class MATH 221 given on November 22, 2000.

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