Real Analysis

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

Last update: 8 July 2014

Lecture 2

$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$

Prove that $e^{x+y}=e^x{e}^y\text{.}$

Definitions

$\text{log}\hspace{0.17em}x$ is the expression that undoes $e^x\text{:}$ $$\text{log}\left(e^x\right)=x\text{and}{e}^{\text{log}\hspace{0.17em}x}=x,$$ $\sqrt{x}$ is the expression that undoes $x^2$ $$\sqrt{x^2}=x\text{and}{\left(\sqrt{x}\right)}^2=x$$ and $$\begin{array}{c}\text{arcsin}\left(\text{sin}\hspace{0.17em}x\right)=x\text{and}\text{sin}\left(\text{arcsin}\hspace{0.17em}x\right)=x,\\ \text{arccos}\left(\text{cos}\hspace{0.17em}x\right)=x\text{and}\text{cos}\left(\text{arccos}\hspace{0.17em}x\right)=x,\\ {\text{tan}}^{-1}\left(\text{tan}\hspace{0.17em}x\right)=x\text{and}\text{tan}\left({\text{tan}}^{-1}\hspace{0.17em}x\right)=x,\\ {\text{sinh}}^{-1}\left(\text{sinh}\hspace{0.17em}x\right)=x\text{and}\text{sinh}\left({\text{sinh}}^{-1}\hspace{0.17em}x\right)=x,\\ \text{arccosh}\left(\text{cosh}\hspace{0.17em}x\right)=x\text{and}\text{cosh}\left(\text{arccosh}\hspace{0.17em}x\right)=x\text{.}\end{array}$$

Note: $\text{arcsin}\hspace{0.17em}x$ and ${\text{sin}}^{-1}x$ have the same meaning. ${\text{sin}}^{-1}x$ does not mean $\frac{1}{\text{sin}\hspace{0.17em}x}$ (which would be written ${\left(\text{sin}\hspace{0.17em}x\right)}^{-1}\text{).}$

Derivatives - the definition

(a)	$\frac{d(f+g)}{dx}=\frac{df}{dx}+\frac{dg}{dx},$
(b)	$\frac{d\left(cf\right)}{dx}=c\frac{df}{dx}$ if $c$ is a constant,
(c)	$\frac{d\left(fg\right)}{dx}=f\frac{dg}{dx}+\frac{df}{dx}g,$
(d)	$\frac{dx}{dx}=1,$ and
(e)	$\frac{df\left(g\right)}{dx}=\frac{df}{dg}\frac{dg}{dx}\text{.}$

Consequences

Prove that $\frac{d{x}^2}{dx}=2x\text{.}$ $$\frac{d{x}^2}{dx}=\frac{d(x·x)}{dx}=x·\frac{dx}{dx}+\frac{dx}{dx}·x=x·1+1·x=2x\text{.}$$

Prove that $\frac{d{e}^x}{dx}=e^x\text{.}$

		Proof.
	Suppose we know that if $n\in {\mathbb{Z}}_{\ge 0}$ then $\frac{d{x}^n}{dx}=n{x}^{n-1}\text{.}$ Then $$\begin{array}{ccc}{\displaystyle \frac{d{e}^x}{dx}}& =& {\displaystyle \frac{d(1+x+\frac{1}{2!}{x}^2+\frac{1}{3!}{x}^3+\frac{1}{4!}{x}^4+\cdots )}{dx}}\\ & =& {\displaystyle 0+1+\frac{1}{2!}2x+\frac{1}{3!}3x^2+\frac{1}{4!}4x^3+\frac{1}{5!}5x^4+\cdots }\\ & =& {\displaystyle 1+x+\frac{1}{2!}{x}^2+\frac{1}{3!}{x}^3+\frac{1}{4!}{x}^4+\cdots }\\ & =& {\displaystyle e^x\text{.}}\end{array}$$ $\square$

Prove that $\frac{d\hspace{0.17em}\text{log}\hspace{0.17em}x}{dx}=\frac{1}{x}\text{.}$

		Proof.
	$$\begin{array}{ccc}1& =& {\displaystyle \frac{dx}{dx}}\\ & =& {\displaystyle \frac{d\hspace{0.17em}{e}^{\left(\text{log}\hspace{0.17em}x\right)}}{dx}}\\ & =& {\displaystyle \frac{d{e}^{\text{log}\hspace{0.17em}x}}{d\hspace{0.17em}\text{log}\hspace{0.17em}x}·\frac{d\hspace{0.17em}\text{log}\hspace{0.17em}x}{dx}}\\ & =& {\displaystyle e^{\text{log}\hspace{0.17em}x}·\frac{d\hspace{0.17em}\text{log}\hspace{0.17em}x}{dx}}\\ & =& {\displaystyle x\frac{d\hspace{0.17em}\text{log}\hspace{0.17em}x}{dx}\text{.}}\end{array}$$ So $\frac{d\hspace{0.17em}\text{log}\hspace{0.17em}x}{dx}=\frac{1}{x}\text{.}$ $\square$

Prove that $\frac{1}{1-x}=1+x+x^2+x^3+x^4+\cdots \text{.}$

		Proof.
	$$\begin{array}{ccc}{\displaystyle (1+x+x^2+x^3+\cdots )(1-x)}& =& {\displaystyle 1+x+x^2+x^3+x^4+\cdots }\\ & & {\displaystyle -x-x^2-x^3-x^4-\cdots }\\ & =& {\displaystyle 1\text{.}}\end{array}$$ Divide both sides by $1-x\text{.}$ So $$1+x+x^2+x^3+\cdots =\frac{1}{1-x}\text{.}$$ $\square$

Prove that $\frac{1}{1+x}=1-x+x^2-x^3+x^4+\cdots \text{.}$

		Proof.
	$$\begin{array}{ccc}{\displaystyle \frac{1}{1+x}}& =& {\displaystyle \frac{1}{1-(-x)}}\\ & =& {\displaystyle 1+(-x)+{(-x)}^2+{(-x)}^3+\cdots }\\ & =& {\displaystyle 1-x+x^2-x^3+x^4-x^5+\cdots \text{.}}\end{array}$$ $\square$

Prove that $\text{log}(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots \text{.}$

		Proof.
	Since $\frac{d\hspace{0.17em}\text{log}(1+x)}{dx}=\frac{1}{1+x},$ then $\int \frac{1}{1+x}dx=\text{log}(1+x)\text{.}$ So $$\begin{array}{ccc}{\displaystyle \text{log}(1+x)}& =& {\displaystyle \int \frac{1}{1+x}dx}\\ & =& {\displaystyle \int (1-x+x^2-x^3+x^4-x^5+\cdots )dx}\\ & =& {\displaystyle x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\cdots \text{.}}\end{array}$$ $\square$

Prove that $e^{ix}=\text{cos}\hspace{0.17em}x+i\text{sin}\hspace{0.17em}x\text{.}$ (By definition $i^2=-1\text{).}$

		Proof.
	$$\begin{array}{ccc}{\displaystyle \text{cos}\hspace{0.17em}x+i\text{sin}\hspace{0.17em}x}& =& {\displaystyle \frac{e^{ix}+e^{-ix}}{2}+i\frac{e^{ix}-e^{-ix}}{2i}}\\ & =& {\displaystyle \frac{e^{ix}+e^{-ix}+e^{ix}-e^{-ix}}{2}}\\ & =& {\displaystyle \frac{2e^{ix}}{2}}\\ & =& {\displaystyle e^{ix}\text{.}}\end{array}$$ $\square$

Prove that $\text{sin}\hspace{0.17em}2x=2\text{sin}\hspace{0.17em}x\hspace{0.17em}\text{cos}\hspace{0.17em}x$ and $\text{cos}\hspace{0.17em}2x={\text{cos}}^{2}x-{\text{sin}}^{2}x\text{.}$

		Proof.
	$$\begin{array}{ccc}{\displaystyle \text{cos}\hspace{0.17em}2x+i\hspace{0.17em}\text{sin}\hspace{0.17em}2x}& =& {\displaystyle e^{i2x}}\\ & =& {\displaystyle e^{i(x+x)}}\\ & =& {\displaystyle e^{ix}{e}^{ix}}\\ & =& {\displaystyle (\text{cos}\hspace{0.17em}x+i\hspace{0.17em}\text{sin}\hspace{0.17em}x)(\text{cos}\hspace{0.17em}x+i\hspace{0.17em}\text{sin}\hspace{0.17em}x)}\\ & =& {\displaystyle {\text{cos}}^{2}x+i^2{\text{sin}}^{2}x+2i\text{sin}\hspace{0.17em}x\hspace{0.17em}\text{cos}\hspace{0.17em}x}\\ & =& {\displaystyle ({\text{cos}}^{2}x-{\text{sin}}^{2}x)+i\left(2\text{sin}\hspace{0.17em}x\hspace{0.17em}\text{cos}\hspace{0.17em}x\right)\text{.}}\end{array}$$ So $\text{cos}\hspace{0.17em}2x={\text{cos}}^{2}x-{\text{sin}}^{2}x$ and $\text{sin}\hspace{0.17em}2x=2\text{sin}\hspace{0.17em}x\hspace{0.17em}\text{cos}x\text{.}$ $\square$

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100303Lect2.pdf and was given on 3 March 2010.

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