MATH 221

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

Last update: 28 July 2014

Lecture 1

Calculus is the study of

(a)	Derivatives
(b)	Integrals
(c)	Applications of Derivatives
(d)	Applications of Integrals

A derivative is a creature you put a function into, it chews on it and spits out a different function $$f⟶\begin{array}{c} ddx \end{array}⟶\frac{df}{dx}\text{.}$$ The integral is the derivative backwards: $$f⟵\begin{array}{c} ∫dx \end{array}⟵\frac{df}{dx}\text{or}\frac{df}{dx}⟶\begin{array}{c} ∫dx \end{array}⟶f\text{.}$$ A function is one down on the food chain $$\text{input number}\hspace{0.17em}x⟶\begin{array}{c} f \end{array}⟶\text{output number}\hspace{0.17em}f\left(x\right)\text{.}$$ Functions take a number as input, chew on it a bit and spit out a number.

The inverse function to $f$ is $f$ backwards $$x⟵\begin{array}{c} f-1 \end{array}⟵f\left(x\right)\text{or}\begin{array}{ccccc}f\left(x\right)& ⟶& \begin{array}{c} f-1 \end{array}& ⟶& x\\ z& ⟶& ⟶& f^{-1}\left(z\right)\end{array}\text{.}$$

$$\begin{array}{c} ⟶ ⟶ ⟶ ⟶ ⟶ ⟶ ⟶ ⟶ ⟶ ⟶ ⟶ ⟶ ⟶ ⟶ x 1 2 3 -3 π 7 x2 1 4 9 9 π2 7 f(x)=x2 \end{array}$$ The inverse function is $$\begin{array}{c} ⟶ ⟶ ⟶ ⟶ ⟶ ⟶ ⟶ ⟶ ⟶ ⟶ ⟶ ⟶ ⟶ ⟶ x2 1 4 9 9 π2 7 x 1 2 3 -3 π 7 f-1(x)=x \end{array}$$ The inverse function is not always a function because there might be some uncertainty about what the inverse function will spit out $$\begin{array}{c} ⟶ ⟶ 9 3 f-1(x)=x \end{array}\text{or}\begin{array}{c} ⟶ ⟶ 9 -3 f-1(x)=x \end{array}\text{.}$$

Numbers are at the very bottom of the food chain.

Numbers

At some point humankind wanted to count things and discovered the positive integers $$1,2,3,4,5,6,\dots$$ Great for counting something BUT what if you don't have anything i.e. $$\text{nothing, nulla, zip}$$ and so we discovered the nonnegative integers $$0,1,2,3,4,5,\dots$$ GREAT for adding $1+3=4,$ $5+0=5,$ $9+16=25$ BUT not so great for subtracting $1-3=\text{???}$ and so we discovered the integers $$\dots ,-4,-3,-2,-1,0,1,2,3,4,\dots$$ GREAT for adding, subtracting and multiplying BUT not so great if you only want part of the sausage... and so we discovered the rational numbers $$\frac{a}{b},a\hspace{0.17em}\text{an integer,}$ b$\hspace{0.17em}\text{an integer,}b\ne 0\text{.}$$ GREAT for addition, subtraction, multiplication and division, BUT not so great for finding $\sqrt{2}\dots$ and so we discovered the real numbers $$\text{all finite and infinite decimal expansions.}$$

$$\begin{array}{ccc}{\displaystyle \sqrt{2}}& =& {\displaystyle 1.414\dots }\\ {\displaystyle e}& =& {\displaystyle 2.71828\dots }\\ {\displaystyle \pi }& =& {\displaystyle 3.1415926\dots }\\ {\displaystyle \frac{1}{6}}& =& {\displaystyle .16666\dots }\\ {\displaystyle \frac{1}{8}}& =& {\displaystyle .125=.1250000\dots }\end{array}$$ GREAT for addition, subtraction, multiplication and division BUT not so great for finding $\sqrt{9}\dots$ and so we discovered the complex numbers $$a+bi,a\hspace{0.17em}\text{a real number,}b\hspace{0.17em}\text{a real number,}i=\sqrt{-1}\text{.}$$

$$\begin{array}{ccc}3+4i& & 0+10i=10i\\ 7+9i& & \pi +0i=\pi \\ 3.2+6.7i& & {\displaystyle \frac{1}{3}+\frac{2}{6}i=\frac{1}{3}+\frac{1}{3}i}\\ 5+0i=5& & \sqrt{7}+\sqrt{2}i\end{array}$$ and $${\left(3i\right)}^2=3^2{i}^2=9i^2=-9\text{.}\text{So}\sqrt{-9}=3i$$ GREAT.

Addition: $$(3+4i)+(7+9i)=10+13i\text{.}$$ Subtraction: $$(3+4i)-(7+9i)=3-7+4i-9i=-4-5i\text{.}$$ Multiplication: $$\begin{array}{ccc}{\displaystyle (3+4i)(7+9i)}& =& {\displaystyle 3(7+9i)+4i(7+9i)}\\ & =& {\displaystyle 21+27i+28i+36i^2}\\ & =& {\displaystyle 21+27i+28i-36}\\ & =& {\displaystyle -15+55i\text{.}}\end{array}$$ Division: $$\begin{array}{ccc}{\displaystyle \frac{3+4i}{7+9i}}& =& {\displaystyle \frac{\left(3+4i\right)}{\left(7+9i\right)}\frac{\left(7-9i\right)}{\left(7-9i\right)}}\\ & =& {\displaystyle \frac{21-27i+28i+36}{49-63i+63i+81}}\\ & =& {\displaystyle \frac{57+i}{130}}\\ & =& {\displaystyle \frac{57}{130}+\frac{1}{130}i\text{.}}\end{array}$$ Square roots: $$\sqrt{-3+4i}=\pm (1+2i)$$ since $${(1+2i)}^2=1+2i+2i+4i^2=1+4i-4=-3+4i$$ and $${(-(1+2i))}^2={(1+2i)}^2=-3+4i\text{.}$$ Another way is: $$\sqrt{-3+4i}=a+bi\text{.}$$ So $$\begin{array}{ccc}{\displaystyle -3+4i={(a+bi)}^2}& =& {\displaystyle a^2+abi+abi+b{i}^2}\\ & =& {\displaystyle a^2-b^2+2abi\text{.}}\end{array}$$ So $$a^2-b^2=-3\text{and}2ab=4\text{.}$$ Solve for $a$ and $b\text{.}$ $$b=\frac{4}{2a}=\frac{2}{a}\text{.}$$ So $$a^2-{\left(\frac{2}{a}\right)}^2=-3\text{.}$$ So $$a^2-\frac{4}{a^2}=-3\text{.}$$ So $$a^4-4=-3a^2\text{.}$$ So $$a^4+3a^2-4=0\text{.}$$ So $$(a^2+4)(a^2-1)=0\text{.}$$ So $$a^2=-4\text{or}{a}^2=1\text{.}$$ So $$a=\pm 1\text{.}$$ So $$b=\frac{2}{\pm 1}=2\text{or}-2\text{.}$$ So $$a+bi=1+2i\text{or}a+bi=-1-2i\text{.}$$ So $$\sqrt{-3+4i}=\pm (1+2i)\text{.}$$ Graphing $$\begin{array}{c} 3 4 3+4i \end{array}\begin{array}{c} -1 1 -1 1 -1+34i \end{array}$$ Factoring $$\begin{array}{ccc}{\displaystyle x^2+5}& =& {\displaystyle (x+\sqrt{5}i)(x-\sqrt{5}i),}\\ {\displaystyle (x^2+x+1)}& =& {\displaystyle (x-\left(\frac{-1+\sqrt{-3}}{2}\right))(x-\left(\frac{-1-\sqrt{-3}}{2}\right))}\\ & =& {\displaystyle (x-(-\frac{1}{2}+\frac{\sqrt{3}}{2}i))(x-(-\frac{1}{2}-\frac{\sqrt{3}}{2}i))\text{.}}\end{array}$$ The fundamental theorem of algebra is one reason why the complex number system is "the right" number system to use. It says that any polynomial can be factored completely as $$(x-u_1)(x-u_2)(x-u_3)\cdots (x-u_n)$$ where $u_1,u^2,\dots ,u_n$ are some complex numbers.

Notes and References

These are a typed copy of Lecture 1 from a series of handwritten lecture notes for the class MATH 221 given on September 6, 2000.

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