Calculus/Algebra

The purpose of this section is for readers to review important algebraic concepts. It is necessary to understand algebra in order to do calculus. If you are confident of your ability, you may skim through this section.

Rules of arithmetic and algebra

The following laws are true for all $a,b,c$ in $\mathbb {R}$ whether these are numbers, variables, functions, or more complex expressions involving numbers, variable and/or functions.

Addition

• Commutative Law: $a+b=b+a$ .
• Associative Law: $(a+b)+c=a+(b+c)$ .
• Additive Identity: $a+0=a$ .
• Additive Inverse: $a+(-a)=0$ .

Subtraction

• Definition: $a-b=a+(-b)$ .

Multiplication

• Commutative law: $a\times b=b\times a$ .
• Associative law: $(a\times b)\times c=a\times (b\times c)$ .
• Multiplicative identity: $a\times 1=a$ .
• Multiplicative inverse: $a\times {\frac {1}{a}}=1$ , whenever $a\neq 0$
• Distributive law: $a\times (b+c)=(a\times b)+(a\times c)$.

Division

• Definition: ${\frac {a}{b}}=r+nb$, where r is the remainder of a when divided by b, and n is an integer.
• Definition: ${\frac {a}{b}}=a\times {\frac {1}{b}}$ , whenever $b\neq 0$ .

Let's look at an example to see how these rules are used in practice.

${\frac {(x+2)(x+3)}{x+3}}$	$=\left[(x+2)\times (x+3)\right]\times \left({\frac {1}{x+3}}\right)$ (from the definition of division)
	$=(x+2)\times \left[(x+3)\times \left({\frac {1}{x+3}}\right)\right]$ (from the associative law of multiplication)
	$=((x+2)\times (1)),\qquad x\neq -3$ (from multiplicative inverse)
	$=x+2,\qquad x\neq -3$ (from multiplicative identity)

Of course, the above is much longer than simply cancelling $x+3$ out in both the numerator and denominator. However, it is important to know what the rules are so as to know when you are allowed to cancel. Occasionally people do the following, for instance, which is incorrect:

${\frac {2\times (x+2)}{2}}={\frac {2}{2}}\times {\frac {x+2}{2}}=1\times {\frac {x+2}{2}}={\frac {x+2}{2}}$ .

The correct simplification is

${\frac {2\times (x+2)}{2}}=\left(2\times {\frac {1}{2}}\right)\times (x+2)=1\times (x+2)=x+2$ ,

where the number $2$ cancels out in both the numerator and the denominator.

Interval notation

There are a few different ways that one can express with symbols a specific interval (all the numbers between two numbers). One way is with inequalities. If we wanted to denote the set of all numbers between, say, 2 and 4, we could write "all $x$ satisfying $2<x<4$". This excludes the endpoints 2 and 4 because we use $<$ instead of $\leq$. If we wanted to include the endpoints, we would write "all $x$ satisfying $2\leq x\leq 4$ ."

Another way to write these intervals would be with interval notation. If we wished to convey "all $x$ satisfying $2<x<4$" we would write $(2,4)$. This does not include the endpoints 2 and 4. If we wanted to include the endpoints we would write $[2,4]$. If we wanted to include 2 and not 4 we would write $[2,4)$; if we wanted to exclude 2 and include 4, we would write $(2,4]$.

Thus, we have the following table:

Endpoint conditions	Inequality notation	Interval notation
Including both 2 and 4	all $x$ satisfying $2\leq x\leq 4$	$[2,4]$
Not including 2 nor 4	all $x$ satisfying $2<x<4$	$(2,4)$
Including 2 not 4	all $x$ satisfying $2\leq x<4$	$[2,4)$
Including 4 not 2	all $x$ satisfying $2<x\leq 4$	$(2,4]$

In general, we have the following table, where $a,b\in \mathbb {R}$.

Meaning	Interval Notation	Set Notation
All values greater than or equal to $a$ and less than or equal to $b$	$[a,b]$	$\{x:a\leq x\leq b\}$
All values greater than $a$ and less than $b$	$(a,b)$	$\{x:a<x<b\}$
All values greater than or equal to $a$ and less than $b$	$[a,b)$	$\{x:a\leq x<b\}$
All values greater than $a$ and less than or equal to $b$	$(a,b]$	$\{x:a<x\leq b\}$
All values greater than or equal to $a$	$[a,\infty )$	$\{x:x\geq a\}$
All values greater than $a$	$(a,\infty )$	$\{x:x>a\}$
All values less than or equal to $a$	$(-\infty ,a]$	$\{x:x\leq a\}$
All values less than $a$	$(-\infty ,a)$	$\{x:x<a\}$
All values	$(-\infty ,\infty )$	$\{x:x\in \mathbb {R} \}$

Note that $\infty$ and $-\infty$ must always have an exclusive parenthesis rather than an inclusive bracket. This is because $\infty$ is not a number, and therefore cannot be in our set. $\infty$ is really just a symbol that makes things easier to write, like the intervals above.

The interval $(a,b)$ is called an open interval, and the interval $[a,b]$ is called a closed interval.

Intervals are sets and we can use set notation to show relations between values and intervals. If we want to say that a certain value is contained in an interval, we can use the symbol $\in$ to denote this. For example, $2\in [1,3]$ . Likewise, the symbol $\notin$ denotes that a certain element is not in an interval. For example $0\notin (0,1)$ .

Exponents and radicals

There are a few rules and properties involving exponents and radicals. As a definition we have that if $n$ is a positive integer then $a^{n}$ denotes $n$ factors of $a$ . That is,

$a^{n}=a\cdot a\cdot a\cdots a\qquad (n~{\mbox{times}})$

If $a\neq 0$ then we say that $a^{0}=1$ .

If $n$ is a negative integer then we say that $a^{-n}={\frac {1}{a^{n}}}$ .

If we have an exponent that is a fraction then we say that $a^{\frac {m}{n}}={\sqrt[{n}]{a^{m}}}=({\sqrt[{n}]{a}})^{m}$ . In the expression ${\sqrt[{n}]{a}}$ , $n$ is called the index of the radical, the symbol ${\sqrt {\;\;}}$ is called the radical sign, and $a$ is called the radicand.

In addition to the previous definitions, the following rules apply:

Rule	Example
$a^{n}\cdot a^{m}=a^{n+m}$	$3^{6}\cdot 3^{9}=3^{15}$
$(a^{n})^{m}=a^{n\cdot m}$	$(x^{4})^{5}=x^{20}$
$(ab)^{n}=a^{n}b^{n}$	$(3x)^{5}=3^{5}x^{5}$

Simplifying expressions involving radicals

We will use the following conventions for simplifying expressions involving radicals:

1. Given the expression $a^{\frac {b}{c}}$, write this as ${\sqrt[{c}]{a^{b}}}$
2. No fractions under the radical sign
3. No radicals in the denominator
4. The radicand has no exponentiated factors with exponent greater than or equal to the index of the radical

Example: Simplify the expression $\left({\frac {1}{8}}\right)^{\frac {1}{2}}$ Using convention 1, we rewrite the given expression as (1) $\left({\frac {1}{8}}\right)^{\frac {1}{2}}={\sqrt[{2}]{\left({\frac {1}{8}}\right)^{1}}}={\sqrt {\frac {1}{8}}}$ The expression now violates convention 2. To get rid of the fraction in the radical, apply the rule $\left({\frac {a}{b}}\right)^{n}={\frac {a^{n}}{b^{n}}}$ and simplify the result: (2) ${\sqrt {\frac {1}{8}}}={\frac {\sqrt {1}}{\sqrt {8}}}={\frac {1}{\sqrt {8}}}$ The resulting expression violates convention 3. To get rid of the radical in the denominator, multiply by ${\frac {\sqrt {8}}{\sqrt {8}}}$: (3) ${\frac {1}{\sqrt {8}}}={\frac {1}{\sqrt {8}}}\cdot {\frac {\sqrt {8}}{\sqrt {8}}}={\frac {\sqrt {8}}{8}}$ Notice that $8=2^{3}$. Since the index of the radical is 2, our expression violates convention 4. We can reduce the exponent of the expression under the radical as follows: (4) ${\frac {\sqrt {8}}{8}}={\frac {\sqrt {2^{3}}}{8}}={\frac {\sqrt {2^{2}\cdot 2}}{8}}={\frac {2\cdot {\sqrt {2}}}{8}}={\frac {\sqrt {2}}{4}}$

Exercise

$144^{\frac {5}{3}}=$ ${\sqrt[{3}]{}}$

Logarithms

Consider the equation

(5) $y=b^{x}$

$b$ is called the base and $x$ is called the exponent. Suppose we would like to solve for $x$ . We would like to apply an operation to both sides of the equation that will get rid of the base on the right-hand side of the equation. The operation we want is called the logarithm, or log for short, and it is defined as follows:

Definition: (Formal definition of a logarithm)

$\log _{b}y=x$ exactly if $y=b^{x}$ and $x>0$, $b>0$, and $b\neq 1$.

Logarithms are taken with respect to some base. What the equation is saying is that, when $x$ is the exponent of $b$, the result will be $y$.

Example

Example: Calculate $\log _{10}100000$ $\log _{10}100000$ is the number $x$ such that $10^{x}=100000$. Well $10^{5}=100000$, so $\log _{10}100000=5$

Common bases for logarithms

When the base is not specified, $\log$ is taken to mean the base 10 logarithm. Later on in our study of calculus we will commonly work with logarithms with base $e\approx 2.718282$ . In fact, the base $e$ logarithm comes up so often that it has its own name and symbol. It is called the natural logarithm, and its symbol is $\ln$ . In computer science the base 2 logarithm often comes up.

Properties of logarithms

Logarithmic addition and subtraction

Logarithms have the property that $\log _{b}x+\log _{b}y=\log _{b}(x\cdot y)$ . To see why this is true, suppose that:

$\log _{b}x=r$ and $\log _{b}y=s$

These assumptions imply that

$x=b^{r}$ and $y=b^{s}$

Then by the properties of exponents

$x\cdot y=b^{r}\cdot b^{s}=b^{r+s}$

According to the definition of the logarithm

$\log _{b}(x\cdot y)=r+s=\log _{b}x+\log _{b}y$

Similarly, the property that $\log _{b}x-\log _{b}y=\log _{b}\left({\frac {x}{y}}\right)$ also hold true using the same method.

Historically, the development of logarithms was motivated by the usefulness of this fact for simplifying hand calculations by replacing tedious multiplication by table look-ups and addition.

Logarithmic powers and roots

Another useful property of logarithms is that $\log _{b}a^{c}=c\cdot \log _{b}a$ . To see why, consider the expression $\log _{b}a^{c}$ . Let us assume that

$\log _{b}a^{c}=x$

By the definition of the logarithm

$a^{c}=b^{x}$

Now raise each side of the equation to the power ${\frac {1}{c}}$ and simplify to get

$a=b^{\frac {x}{c}}$

Now if you take the base $b$ log of both sides, you get

$\log _{b}a={\frac {x}{c}}$

Solving for $x$ shows that

$x=c\cdot \log _{b}a$

Similarly, the expression $\log _{b}{\sqrt[{c}]{a}}={\frac {\log _{b}a}{c}}$ holds true using the same methods.

Converting between bases

Most scientific calculators have the $\log$ and $\ln$ functions built in, which do not include logarithms with other bases. Consider how one might compute $\log _{b}a$, where $b$ and $a$ are given known numbers, when we can only compute logarithms in some base $\beta$. First, let us assume that

$\log _{b}a=x$

Then the definition of logarithm implies that

$a=b^{x}$

If we take the base $\beta$ log of each side, we get

$\log _{\beta }a=\log _{\beta }b^{x}=x\cdot \log _{\beta }b$

Solving for $x$ , we find that

$x={\frac {\log _{\beta }a}{\log _{\beta }b}}$

For example, if we only use base 10 to calculate $\log _{2}45$, we get $\log _{2}45={\frac {\log 45}{\log 2}}\approx {\frac {1.653212514}{0.301029996}}\approx 5.491853096$ .

Identities of logarithms summary

A table is provided below for a summary of logarithmic identities.

	Formula	Example
Product	$\log _{b}(x\cdot y)=\log _{b}x+\log _{b}y$	$\log _{2}32=\log _{2}(4\cdot 8)=\log _{2}4+\log _{2}8=2+3=5$
Quotient	$\log _{b}\!{\frac {x}{y}}=\log _{b}x-\log _{b}y$	$\log _{2}16=\log _{2}\!{\frac {64}{4}}=\log _{2}64-\log _{2}4=6-2=4$
Power	$\log _{b}\left(a^{c}\right)=c\log _{b}a$	$\log _{2}64=\log _{2}\left(2^{6}\right)=6\log _{2}2=6$
Root	$\log _{b}{\sqrt[{c}]{a}}={\frac {\log _{b}a}{c}}$	$\log _{10}{\sqrt {1000}}={\frac {1}{2}}\log _{10}1000={\frac {3}{2}}=1.5$
Change of base	$\log _{b}a={\frac {\log _{\beta }a}{\log _{\beta }b}}$	$\log _{9}243={\frac {\log _{3}243}{\log _{3}9}}={\frac {5}{2}}=2.5$

Factoring and roots

Given the expression $x^{2}+3x+2$ , one may ask "what are the values of $x$ that make this expression 0?" If we factor we obtain

$x^{2}+3x+2=(x+2)(x+1)$

.

If $x=-1,-2$ , then one of the factors on the right becomes zero. Therefore, the whole must be zero. So, by factoring we have discovered the values of $x$ that render the expression zero. These values are termed "roots." In general, given a quadratic polynomial $px^{2}+qx+r$ that factors as

$px^{2}+qx+r=(ax+c)(bx+d)$

then we have that $x=-{\frac {c}{a}}$ and $x=-{\frac {d}{b}}$ are roots of the original polynomial.

A special case to be on the look out for is the difference of two squares, $a^{2}-b^{2}$ . In this case, we are always able to factor as

$a^{2}-b^{2}=(a+b)(a-b)$

For example, consider $4x^{2}-9$ . On initial inspection we would see that both $4x^{2}$ and $9$ are squares of $2x$ and $3$, respectively. Applying the previous rule we have

$4x^{2}-9=(2x+3)(2x-3)$

The AC method

There is a way of simplifying the process of factoring using the AC method. Suppose that a quadratic polynomial has a formula of

$x^{2}+qx+r$

If there are numbers $a$ and $b$ that satisfy both

$a\cdot b=r$ and $a+b=q$

Then, the result of factoring will be

$x^{2}+qx+r=(x+a)(x+b)$

The quadratic formula

The quadratic formula
Given any quadratic equation $ax^{2}+bx+c=0\ ,\ a\neq 0$, all solutions of the equation are given by the quadratic formula:

$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$

Note that the value of $b^{2}-4ac$ will affect the number of real solutions of the equation.

If	Then
$b^{2}-4ac>0$	There are two real solutions to the equation
$b^{2}-4ac=0$	There is only one real solution to the equation
$b^{2}-4ac<0$	There are no real solutions to the equation

Example: Find all the roots of $4x^{2}+7x-2$ Finding the roots is equivalent to solving the equation $4x^{2}+7x-2=0$ . Applying the quadratic formula with $a=4\ ,\ b=7\ ,\ c=-2$ , we have: $x={\frac {-7\pm {\sqrt {7^{2}-4(4)(-2)}}}{2(4)}}$ $x={\frac {-7\pm {\sqrt {49+32}}}{8}}$ $x={\frac {-7\pm {\sqrt {81}}}{8}}$ $x={\frac {-7\pm 9}{8}}$ $x={\frac {2}{8}}\ ,\ x={\frac {-16}{8}}$ $x={\frac {1}{4}}\ ,\ x=-2$

The quadratic formula can also help with factoring, as the next example demonstrates.

Example: Factor the polynomial $4x^{2}+7x-2$ We already know from the previous example that the polynomial has roots $x={\frac {1}{4}}$ and $x=-2$ . Our factorization will take the form $C(x+2)\left(x-{\tfrac {1}{4}}\right)$ All we have to do is set this expression equal to our polynomial and solve for the unknown constant C: $C(x+2)\left(x-{\tfrac {1}{4}}\right)=4x^{2}+7x-2$ $C\left(x^{2}+\left(-{\tfrac {1}{4}}+2\right)x-{\tfrac {2}{4}}\right)=4x^{2}+7x-2$ $C\left(x^{2}+{\tfrac {7}{4}}x-{\tfrac {1}{2}}\right)=4x^{2}+7x-2$ You can see that $C=4$ solves the equation. So the factorization is $4x^{2}+7x-2=4(x+2)\left(x-{\tfrac {1}{4}}\right)=(x+2)(4x-1)$

Vieta's formulae

Vieta's formulae relate the coefficients of a polynomial to sums and products of its roots. It is very convenient because under certain circumstances when the sums and products of the quadratic's roots are provided, one does not require to solve the whole quadratic polynomial.

Vieta's formulae in quadratic polynomials
Given any quadratic equation $ax^{2}+bx+c=0\ ,\ a\neq 0$, The roots $x_{1},x_{2}$ of the quadratic polynomial satisfy

$x_{1}+x_{2}=-{\frac {b}{a}},\quad x_{1}x_{2}={\frac {c}{a}}.$

Simplifying rational expressions

Consider the two polynomials

$p(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}$

and

$q(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+\cdots +b_{1}x+b_{0}$

When we take the quotient of the two we obtain

${\frac {p(x)}{q(x)}}={\frac {a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}}{b_{m}x^{m}+b_{m-1}x^{m-1}+\cdots +b_{1}x+b_{0}}}$

The ratio of two polynomials is called a rational expression. Many times we would like to simplify such a beast. For example, say we are given ${\frac {x^{2}-1}{x+1}}$ . We may simplify this in the following way:

${\frac {x^{2}-1}{x+1}}={\frac {(x+1)(x-1)}{x+1}}=x-1,\qquad x\neq -1$

This is nice because we have obtained something we understand quite well, $x-1$ , from something we didn't.

Formulas of multiplication of polynomials

Here are some formulas that can be quite useful for solving polynomial problems:

$(a\pm b)^{2}=a^{2}\pm 2ab+b^{2}$

$(a-b)(a+b)=a^{2}-b^{2}$

$(a\pm b)^{3}=a^{3}\pm 3a^{2}b+3ab^{2}\pm b^{3}$

$a^{3}\pm b^{3}=(a\pm b)(a^{2}\mp ab+b^{2})$

Polynomial Long Division

Suppose we would like to divide one polynomial by another. The procedure is similar to long division of numbers and is illustrated in the following example:

Example

Divide $x^{2}-2x-15$ (the dividend or numerator) by $x+3$ (the divisor or denominator) Similar to long division of numbers, we set up our problem as follows: ${\begin{array}{rl}\\x+3\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}-2x-15\end{array}}\end{array}}$ First we have to answer the question, how many times does $x+3$ go into $x^{2}$? To find out, divide the leading term of the dividend by leading term of the divisor. So it goes in $x$ times. We record this above the leading term of the dividend: ${\begin{array}{rl}&~~\,x\\x+3\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}-2x-15\end{array}}\\\end{array}}$ , and we multiply $x+3$ by $x$ and write this below the dividend as follows: ${\begin{array}{rl}&~~\,x\\x+3\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}-2x-15\end{array}}\\&\!\!\!\!-{\underline {(x^{2}+3x)~~~}}\\\end{array}}$ Now we perform the subtraction, bringing down any terms in the dividend that aren't matched in our subtrahend: ${\begin{array}{rl}&~~\,x\\x+3\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}-2x-15\end{array}}\\&\!\!\!\!-{\underline {(x^{2}+3x)~~~}}\\&\!\!\!\!~~~~~~-5x-15~~~\\\end{array}}$ Now we repeat, treating the bottom line as our new dividend: ${\begin{array}{rl}&~~\,x-5\\x+3\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x^{2}-2x-15\end{array}}\\&\!\!\!\!-{\underline {(x^{2}+3x)~~~}}\\&\!\!\!\!~~~~~~-5x-15~~~\\&\!\!\!\!~~~-{\underline {(-5x-15)~~~}}\\&\!\!\!\!~~~~~~~~~~~~~~~~~~~0~~~\\\end{array}}$ In this case we have no remainder.

Application: Factoring Polynomials

We can use polynomial long division to factor a polynomial if we know one of the factors in advance. For example, suppose we have a polynomial $P(x)$ and we know that $r$ is a root of $P$ . If we perform polynomial long division using P(x) as the dividend and $(x-r)$ as the divisor, we will obtain a polynomial $Q(x)$ such that $P(x)=(x-r)Q(x)$ , where the degree of $Q$ is one less than the degree of $P$.

Exercise

Use ^ to write exponents:

Factor $x-1$ out of $6x^{3}-4x^{2}+3x-5$.

Application: Breaking up a rational function

Similar to the way one can convert an improper fraction into an integer plus a proper fraction, one can convert a rational function $P(x)$ whose numerator $N(x)$ has degree $n$ and whose denominator $D(x)$ has degree $d$ with $n\geq d$ into a polynomial plus a rational function whose numerator has degree $\nu$ and denominator has degree $\delta$ with $\nu <\delta$ .

Suppose that $N(x)$ divided by $D(x)$ has quotient $Q(x)$ and remainder $R(x)$ . That is

$N(x)=D(x)Q(x)+R(x)$

Dividing both sides by $D(x)$ gives

${\frac {N(x)}{D(x)}}=Q(x)+{\frac {R(x)}{D(x)}}$

$R(x)$ will have degree less than $D(x)$ .

Example

Write ${\frac {x-1}{x-3}}$ as a polynomial plus a rational function with numerator having degree less than the denominator. ${\begin{array}{rl}&~~\,1\\x-3\!\!\!\!&{\big )}\!\!\!{\begin{array}{lll}\hline \,x-1\end{array}}\\&\!\!\!\!-{\underline {(x-3)~~~}}\\&\!\!\!\!~~~~~~~~~2~~~\\\end{array}}$ so ${\frac {x-1}{x-3}}=1+{\frac {2}{x-3}}$