MATH 221

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

Last updated: 30 July 2014

Lecture 6

The exponential function is the function $e^x$ $$x⟶\begin{array}{c} ex \end{array}⟶x^2$$ such that $$\frac{d{e}^x}{dx}=e^x\text{and}{e}^0=1\text{.}$$ Figure out what $e^x$ is:

Suppose $$e^x=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+\cdots \text{.}$$ Then $$e^0=a_0+0+0+0+\cdots =1\text{.}$$ So $a_0=1\text{.}$ $$\begin{array}{ccc}{\displaystyle \frac{d{e}^x}{dx}}& =& {\displaystyle a_1+2a_{2}x+3a_3{x}^2+4a_4{x}^3+\cdots }\\ {\displaystyle e^x}& =& {\displaystyle a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+\cdots }\end{array}$$ So $$a_1=a_0,2a_2=a_1,3a_3=a_2,4a_4=a_3,\dots$$ So $$a_0=1,a_1=1,a_2=\frac{1}{2},a_3=\frac{1}{3·2},a_4=\frac{1}{4·3·2},\dots$$ So $$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3·2}+\frac{x^4}{4·3·2}+\frac{x^5}{5·4·3·2}+\cdots \text{.}$$ So $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\cdots \text{.}$$ Factorials $$\begin{array}{ccc}{\displaystyle 7!}& =& {\displaystyle 7·6·5·4·3·2·1=5040,}\\ {\displaystyle 5!}& =& {\displaystyle 5·4·3·2·1=120,}\\ {\displaystyle 3!}& =& {\displaystyle 3·2·1\text{.}}\end{array}$$ So $$\begin{array}{ccc}{\displaystyle e^1}& =& {\displaystyle 1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}+\frac{1}{720}+\cdots =2.7818\dots }\\ {\displaystyle e^{-3}}& =& {\displaystyle 1+(-3)+\frac{{(-3)}^2}{2}+\frac{{(-3)}^3}{3!}+\frac{{(-3)}^4}{4!}+\cdots }\\ & =& {\displaystyle 1-3+\frac{9}{2}-\frac{27}{6}+\frac{81}{24}-\frac{243}{120}+\cdots =}\end{array}$$ So we can evaluate $e^x$ for any number $x\text{.}$

By the chain rule $$\frac{d\left(e^{2+x}\right)}{dx}=e^{2+x}\frac{d(2+x)}{dx}=e^{2+x}·1=e^{2+x}$$ and $e^{2+0}=e^2\text{.}$ What could $e^{2+x}$ be? If $$e^{2+x}=b_0+b_{1}x+b_2{x}^2+b_3{x}^3+\cdots$$ then $$e^{2+0}=b_0+0+0+\cdots =e^2\text{.}$$ So $b_0=e^2\text{.}$ $$\frac{d{e}^{2+x}}{dx}=b_1+2b_{2}x+3b_3{x}^2+4b_4{x}^3+c.\dots =e^{2+x}\text{.}$$ So $$b_0=b_1,2b_2=b_1,3b_3=b_2,4b_4=b_3,\dots$$ So $$b_0=e^2,b_1=e^2,b_2=\frac{e^2}{2},b_3=\frac{e^2}{3·2},b_4=\frac{e^2}{4·3·2},\dots$$ So $$\begin{array}{ccc}{\displaystyle e^{2+x}}& =& {\displaystyle e^2+e^{2}x+\frac{e^2{x}^2}{2}+\frac{e^2{x}^3}{3·2}+\frac{e^2{x}^4}{4·3·2}+\cdots }\\ & =& {\displaystyle e^2(1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots )}\\ & =& {\displaystyle e^2{e}^x\text{.}}\end{array}$$ Similarly $e^{6342+x}=e^{6342}{e}^x$ and $$\begin{array}{|c|}\hline e^{y+x}=e^y{e}^x\\ \hline\end{array}$$ Since $e^{-x+x}=e^{-x}{e}^x$ and $e^{-x+x}=e^0=1$ we have $e^{-x}{e}^x=1\text{.}$ So $$\begin{array}{|c|}\hline e^{-x}=\frac{1}{e^x}\\ \hline\end{array}$$ $$\begin{array}{ccc}{\displaystyle e^{10x}}& =& {\displaystyle e^{x+x+x+x+x+x+x+x+x+x}}\\ & =& {\displaystyle e^x{e}^{x+x+x+x+x+x+x+x+x}}\\ & =& {\displaystyle e^x{e}^x{e}^{x+x+x+x+x+x+x+x}}\\ & =& {\displaystyle e^x{e}^x{e}^x{e}^{x+x+x+x+x+x+x}}\\ & =& {\displaystyle \cdots =e^x{e}^x{e}^x{e}^x{e}^x{e}^x{e}^x{e}^x{e}^x{e}^x}\\ & =& {\displaystyle {\left(e^x\right)}^{10}\text{.}}\end{array}$$ Similarly $$\begin{array}{ccc}{\displaystyle e^{6342x}}& =& {\displaystyle e^{x+x+x+x+x+\cdots +x}}\\ & =& {\displaystyle e^x{e}^x{e}^x{e}^x{e}^x{e}^x\cdots e^x}\\ & =& {\displaystyle {\left(e^x\right)}^{6342}\text{.}}\end{array}$$ In general, $$\begin{array}{|c|}\hline {\left(e^x\right)}^y=e^{xy}\\ \hline\end{array}$$

Logarithms

If $$x⟶\begin{array}{c} f \end{array}⟶f\left(x\right)$$ is a function then the inverse function to $f$ is the function that undoes $f\text{.}$

The logarithm $\text{ln}\hspace{0.17em}x$ is the inverse function to $e^x\text{.}$ $$\begin{array}{ccccc}x& ⟶& \begin{array}{c} ex \end{array}& ⟶& e^x\\ \text{ln}\hspace{0.17em}7& ⟶& ⟶& 7\end{array}\begin{array}{ccccc}{e}^x& ⟶& \begin{array}{c} ln x \end{array}& ⟶& x\\ 7& ⟶& ⟶& \text{ln}\hspace{0.17em}7\end{array}$$ So $$\begin{array}{c}{e}^{\text{ln}\hspace{0.17em}2}=2,\text{ln}\hspace{0.17em}{e}^{6762}=6762,\text{ln}\hspace{0.17em}{e}^{\sqrt{7}}=\sqrt{7},\\ e^{\text{ln}\hspace{0.17em}\sqrt{7}}=\sqrt{7},e^{\text{ln}(3+\frac{2}{\pi }i)}=3+\frac{2}{\pi }i,e^{\text{ln}\hspace{0.17em}J}=J\text{.}\end{array}$$ Since $$1=e^0,\begin{array}{|c|}\hline \text{ln}\hspace{0.17em}1=0\\ \hline\end{array}$$ In general, If $y=e^x$ then $\text{ln}\hspace{0.17em}y=\text{ln}\hspace{0.17em}{e}^x=x\text{.}$ Since $e^x{e}^y=e^{x+y}$ $$\text{ln}\left(ab\right)=\text{ln}\left(e^{\text{ln}\hspace{0.17em}a}{e}^{\text{ln}\hspace{0.17em}b}\right)=\text{ln}\left(e^{\text{ln}\hspace{0.17em}a+\text{ln}\hspace{0.17em}b}\right)=\text{ln}\hspace{0.17em}a+\text{ln}\hspace{0.17em}b\text{.}$$ Since $e^{-x}=\frac{1}{e^x},$ $$\text{ln}\left(\frac{1}{a}\right)=\text{ln}\left(\frac{1}{e^{\text{ln}\hspace{0.17em}a}}\right)=\text{ln}\left(e^{-\text{ln}\hspace{0.17em}a}\right)=-\text{ln}\hspace{0.17em}a\text{.}$$ Since $e^{nx}={\left(e^x\right)}^n,$ $$\text{ln}\left(a^n\right)=\text{ln}\left({\left(e^{\text{ln}\hspace{0.17em}a}\right)}^n\right)=\text{ln}\left(e^{n\text{ln}\hspace{0.17em}a}\right)=n\text{ln}\hspace{0.17em}a\text{.}$$

Find $\frac{dy}{dx}$ when $y=\text{ln}\hspace{0.17em}x\text{.}$

Well $e^y=x\text{.}$ So $$\frac{d{e}^y}{dx}=\frac{dx}{dx}\text{.}$$ So $$e^y\frac{dy}{dx}=1\text{.}$$ So $$\frac{dy}{dx}=\frac{1}{e^y}=\frac{1}{x}\text{.}$$ So $$\begin{array}{|c|}\hline {\displaystyle \frac{d\hspace{0.17em}\text{ln}\hspace{0.17em}x}{dx}=\frac{1}{x}}\\ \hline\end{array}$$

Notes and References

These are a typed copy of Lecture 6 from a series of handwritten lecture notes for the class MATH 221 given on September 18, 2000.

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