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 1

Location
Announcements
Assessment – The plan is to take exam questions directly from the problem sheets. Assignment 1 due 15 March.
Consultation hours: 2 – 4:30 in Old Geology 1.
Labs run in weeks 2, 3, 4, and 6.
Text: I highly recommend the book of W. Chen.

Language of Mathematics

Assume
If $A$ then $B .$

The foundation is definitions. The explanation is proofs. It is impossible to do a proof without knowing the definitions.

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

The definition of $e^{x}$ is $e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots$ where $k! = k (k - 1) (k - 2) \dots 2 \cdot 1, if k \in ℤ_{> 0} .$ The definition of $ℤ_{> 0}$ is $ℤ_{> 0} = {1, 1 + 1, 1 + 1 + 1, 1 + 1 + 1 + 1, \dots}$ which is often written $ℤ_{> 0} = {1, 2, 3, 4, \dots} .$ Example inside the example: $\begin{matrix} 1! = 1, & 4! = 24, & 7! = 5040 . \\ 2! = 2, & 5! = 120, \\ 3! = 6, & 6! = 720, \end{matrix}$ So $e^{1} = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + \dots .$ Welcome to Wolfram alpha. $e^{1} \approx 2.7182818284590452353602874713 .$ Prove that $e^{x} e^{y} = e^{x + y} .$

Proof.

Righthand side: $\begin{matrix} e^{x + y} & = & 1 + (x + y) + \frac{{(x + y)}^{2}}{2!} + \frac{{(x + y)}^{3}}{3!} + \frac{{(x + y)}^{4}}{4!} + \dots \\ = & \begin{matrix} 1 \\ + x + y \\ + \frac{1}{2} (x^{2} + 2 x y + y^{2}) \\ + \frac{1}{6} (x^{3} + 3 x^{2} y + 3 x y^{2} + y^{3}) \\ + \frac{1}{4!} (x^{4} + 4 x^{3} y + 6 x^{2} y^{2} + 4 x y^{3} + y^{4}) \\ + \frac{1}{5!} (x^{5} + 5 x^{4} y + 10 x^{3} y^{2} + 10 x^{2} y^{3} + 5 x y^{4} + y^{5}) \\ + \dots \end{matrix} \\ = & \begin{matrix} 1 \\ + x + y \\ + \frac{1}{2!} x^{2} + x y +12!>y^{2} \\ + \frac{1}{3!} x^{3} + \frac{1}{2!} x^{2} y + \frac{1}{2!} x y^{2} + \frac{1}{3!} y^{3} \\ + \frac{1}{4!} x^{4} + \frac{1}{3!} x^{3} y + \frac{1}{2^{1}} \frac{1}{2!} x^{2} y^{2} + \frac{1}{3!} x y^{3} + \frac{1}{4!} y^{4} \\ + \frac{1}{5!} x^{5} + \frac{1}{4!} x^{4} y + \frac{1}{2!} \frac{1}{3!} x^{3} y^{2} + \frac{1}{3!} \frac{1}{2!} x^{2} y^{3} + \frac{1}{4!} x y^{4} + \frac{1}{5!} y^{5} \\ + \dots \end{matrix} \\ = & e^{x} + e^{x} y + e^{x} \frac{1}{2!} y^{2} + e^{x} \frac{1}{3!} y^{3} + e^{x} \frac{1}{4!} y^{4} + \dots \\ = & e^{x} (1 + y + \frac{y^{2}}{2!} + \frac{y^{3}}{3!} + \frac{y^{4}}{4!} + \frac{y^{5}}{5!} + \dots) \\ = & e^{x} e^{y} . \end{matrix}$

$□$

Some more definitions: $\begin{matrix} sin x = \frac{e^{i x} - e^{- i x}}{2 i} with i^{2} = - 1, \\ cos x = \frac{e^{i x} + e^{- i x}}{2}, tan x = \frac{sin x}{cos x}, \\ cot x = \frac{1}{tan x}, sec x = \frac{1}{cos x}, csc x = \frac{1}{sin x}, \\ sinh x = \frac{e^{x} - e^{- x}}{2}, cosh x = \frac{e^{x} + e^{- x}}{2}, tanh x = \frac{sinh x}{cosh x}, \\ coth x = \frac{1}{tanh x}, sech x = \frac{1}{cosh x}, csch x = \frac{1}{sinh x} . \end{matrix}$ log undoes $e^{x} :$ $log (e^{x}) = x and e^{log x} = x .$

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100301Lect1.pdf and was given on 1 March 2010.

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