# Real Analysis: The Real Numbers

## Polynomials and Rational Functions

### Monomials

A real monomial function, or simply a monomial, is a function $f : \mathbb{R} \rightarrow \mathbb{R}$ where $f$ is of the form $f(x) = ax^n$, where $a$ is a nonzero real number and $n \in \mathbb{N}.$ The exponent $n$ is the degree or power of the monomial, which is expressed mathematically as $\text{deg}(f) = n.$ The special case where $f(x) = 0$ is called the zero monomial, and its degree is undefined. Monomials are the building blocks of the more interesting polynomials.

### Polynomials

A real polynomial function, or simply a polynomial, is a function $f : \mathbb{R} \rightarrow \mathbb{R}$ where $f$ is of the form $f(x) = \sum\limits_{i=0}^n a_ix^i,$ where $a_i \in \mathbb{R}$ and $a_n \neq 0.$ In other words, a polynomial is a sum of monomials. The values $a_i$ are called the coefficients of the polynomial. As with monomials, the value of $n$ is the degree or power of the polynomial, which is again expressed mathematically as $\text{deg}(f) = n.$ The coefficient $a_n$ is called the leading coefficient. The polynomial $f(x) = 0$ is called the zero polynomial. It is identical to the zero monomial, but the term "zero polynomial" is used far more commonly, as polynomials are much more important objects of study. The leading coefficient of every other (i.e. nonzero) polynomial is always nonzero, as otherwise it would not be of degree $n$ but of some degree less than $n.$

For example, if $n = 0,$ then $f$ is just the constant function $f(x) = a_0.$ If instead $n = 3,$ then $f$ is a third degree polynomial. Note that $a_i$ can be zero for any of the non-leading terms. Thus if we consider the third degree polynomial $f(x) = 2x^3 - 5x,$ the coefficients are $a_0 = 0, a_1 = -5, a_2 = 0, a_3 = 2.$ Make sure not to confuse the degree of the polynomial for the number of terms it has, as the example here has two terms but is still of degree $3.$

### The Set of Polynomials

We denote the set of all real polynomials of degree at most $n$ as $\mathcal{P}_n(\mathbb{R}).$ The set of all real polynomials is denoted as simply $\mathcal{P}(\mathbb{R}).$ The reason for the usage of the $\mathbb{R}$ notation is that polynomials can be defined for coefficients of any field, such as the rationals or the complex numbers. In fact, polynomials can be defined over more abstract algebraic sets such as commutative rings, and some areas of math generalize them further. However, for the purposes of real analysis, we are mostly concerned with $\mathcal{P}(\mathbb{R}).$

Though the notation is not standard, for this section we will likewise refer to the set of all real monomials as $\mathcal{M}(\mathbb{R}),$ and the set of all real monomials of degree $n$ as $\mathcal{M}_n(\mathbb{R}).$

### Polynomial Names

The polynomials of the first few degrees have special names and special notation for their coefficients, which are listed here in a table:

 Degree Name Notation 0 Nonzero Constant $f(x) = c$ 1 Linear $f(x) = mx + b$ 2 Quadratic $f(x) = ax^2 + bx + c$ 3 Cubic $f(x) = ax^3 + bx^2 + cx + d$ 4 Quartic $f(x) = ax^4 + bx^3 + cx^2 + dx + e$ $\vdots$ $\vdots$ $\vdots$

The constant case is special, as the constant must be nonzero. If $f(x) = 0,$ then the degree is undefined. Polynomials of higher degrees follow the naming convention of the Latin ordinals, such as quintic, sextic, etc.

### Zeroes

Of particular importance to the study of polynomials are the values for which they are zero. Specifically, a value $x_0 \in \mathbb{R}$ of a polynomial $p$ such that $p(x_0) = 0$ is called a zero or root of $p$. Finding zeroes is a surprisingly difficult process. There is no general solution to finding the zeroes of polynomials of degree $5$ or higher, and often the easiest method is to use a computer program to find an approximate solution. However, proving that the formulas for finding zeroes of small polynomials are correct and that the computational methods for approximating zeroes for general polynomials are correct are both squarely in the domain of real analysis.

### Rational Functions

rational function is the quotient of two polynomial functions. Rational functions are undefined at the zeroes of the denominator polynomial for the simple reason that division by zero is undefined. The behavior of rational functions as they approach these zeroes in their denominator as well as their behavior as they tend toward positive and negative infinity is of particular to real analysis. However, we need to develop the concept of a limit to study these properties. That said, we can begin to study some other interesting algebraic properties of rational functions that do not require the use of limits.

## Problems

1. Consider the class $\mathcal{M}(\mathbb{R})$ of all real-valued monomials where $\mathcal{M}_n(\mathbb{R})$ is the set of all real-valued monomials of degree $n.$ Show the following:

1. $\mathcal{M}_n(\mathbb{R})$ is not closed under multiplication.

2. $\mathcal{M}(\mathbb{R})$ is closed under multiplication.

3. $\mathcal{M}_n(\mathbb{R})$ is not closed under addition.

4. $\mathcal{M}(\mathbb{R})$ is not closed under division.

1. Let $p, q \in \mathcal{M}_n(\mathbb{R})$ where $p(x) = a_px^n$ and $q(x) = a_qx^n.$ Their product is then

$(pq)(x) = p(x)q(x) \\ (pq)(x) = (a_px^n)(a_qx^n) \\ (pq)(x) = (a_pa_q)x^{2n}.$

Since $\text{deg}(pq) = 2n \neq n$, it follows that $pq \notin \mathcal{M}_n(\mathbb{R}).$

2. Let $p, q \in \mathcal{M}(\mathbb{R})$ where $p(x) = a_px^n$ and $q(x) = a_qx^m.$ Their product is then

$(pq)(x) = p(x)q(x) \\ (pq)(x) = (a_px^n)(a_qx^m) \\ (pq)(x) = (a_pa_q)x^{n+m}.$

Because $a_pa_q \neq 0$ and $n+m \in \mathbb{N},$ it follows that $pq \in \mathcal{M}(\mathbb{R}).$

3. Let $p, q \in \mathcal{M}_n(\mathbb{R})$ where $p(x) = a_px^n$ and $q(x) = a_qx^n.$ Their sum is then

$(p+q)(x) = p(x)+q(x) \\ (p+q)(x) = a_px^n + a_qx^m \\ (p+q)(x) = (a_p+ a_q)x^n.$

Though $\text{deg}(pq) = n$, if $a_p = -a_q,$ then $(p + q)(x) = 0$, and so $pq \notin \mathcal{M}_n(\mathbb{R}).$

4. Let $p, q \in \mathcal{M}(\mathbb{R})$ where $p(x) = a_px^n$ and $q(x) = a_qx^m.$ The product is then

$(p/q)(x) = \frac{p(x)}{q(x)} \\ (p/q)(x) = \frac{a_px^n}{a_qx^m} \\ (p/q)(x) = \frac{a_p}{a_q}x^{\frac{n}{m}}.$

Since the power of $x$ is a rational number and not a natural number, it follows that $p/q$ is not a monomial.

2. Show the following:

1. $\mathcal{P}(\mathbb{R})$ is closed under addition.

2. $\mathcal{P}_n(\mathbb{R})$ is not closed under multiplication for $n > 1.$

3. $\mathcal{P}(\mathbb{R})$ is closed under multiplication.

1. Let $p, q \in \mathcal{P}(\mathbb{R})$ where $\text{deg}(p) = n$ and $\text{deg}(q) = m.$ If $n = m$, then

$(p + q)(x) = p(x) + q(x) \\ (p + q)(x) = \sum\limits_{i=0}^n a_{0,i}x^i + \sum\limits_{i=0}^m a_{1,i}x^i \\ (p + q)(x) = \sum\limits_{i=0}^n (a_{0,i}+a_{1,i})x^i$

Since $\mathbb{R}$ is closed under addition, it follows that $(a_{0,i}+a_{1,i}) \in \mathbb{R}$, and so $(p + q) \in \mathcal{P}(\mathbb{R}).$

If instead $n < m.$ Then

$(p + q)(x) = p(x) + q(x) \\ (p + q)(x) = \sum\limits_{i=0}^n a_{0,i}x^i + \sum\limits_{i=0}^m a_{1,i}x^i \\ (p + q)(x) = \sum\limits_{i=0}^n a_{0,i}x^i + \sum\limits_{i=0}^n a_{1,i}x^i + \sum\limits_{i=n+1}^{m}a_{1,i}x^i \\ (p + q)(x) = \sum\limits_{i=0}^n (a_{0,i}+a_{1,i})x^i + \sum\limits_{i=n+1}^{m}a_{1,i}x^i$

Let $c_i = (a_{0,i}+a_{q,i})$ for $0 \leq i \leq n$ and let $c_i = a_{1,i}$ for $n + 1 \leq i \leq m.$ We can now substitute these values and simplify:

$(p + q)(x) = \sum\limits_{i=0}^n c_ix^i + \sum\limits_{i=n+1}^{m}c_ix^i \\ (p + q)(x) = \sum\limits_{i=0}^m c_ix^i$

Again we see that because $\mathbb{R}$ is closed under addition, $c_i \in \mathbb{R},$ and thus $(p + q) \in \mathcal{P}(\mathbb{R}).$

2. Consider $p(x) = ax^n$ where $n > 0$. Then $(pp)(x) = p(x)p(x) = (ax^n)(ax^n) = a^2x^{2n}.$ Since $\text{deg}(pp) = 2n \neq n,$ it follows that $\mathcal{P}_n(\mathbb{R})$ is not closed under multiplication.

3. First we show that polynomials are closed under multiplication by monomials. Let $p \in \mathcal{P}_n(\mathbb{R})$ where $p = \sum\limits_{i=0}^n a_ix^i$ and $q \in \mathcal{M}(\mathbb{R})$ where $q(x)=bx^m.$ Then $(pq)(x) = p(x)q(x) = \left(\sum\limits_{i=0}^n a_ix^i\right)(bx^m) = \sum\limits_{i=m}^{n+m} (a_ib)x^i.$ Therefore $pq \in \mathcal{P}_{n+m}(\mathbb{R}).$

Next we note that a polynomial is a sum of monomials. Thus for some $q \in \mathcal{P}_m(\mathbb{R})$ where $q = \sum\limits_{i=0}^m b_ix^i,$ we see that $\left(\sum\limits_{i=0}^n a_i x^i\right)\left(\sum\limits_{i=0}^m b_i x^i\right) = \sum\limits_{i=0}^m \left(\sum\limits_{j=0}^n (a_ib_j) x^{i+j}\right).$ The internal sum produces a polynomial, and the outside sum sums polynomials. Since $\mathcal{P}(\mathbb{R})$ is closed under addition, it follows that $pq \in \mathcal{P}(\mathbb{R}).$

1. Show that the set of rational functions is closed under division.

2. What is the domain of the quotient of two rational functions in terms of its underlying polynomials?

1. Let $p, q, s, t \in \mathcal{P}(\mathbb{R})$, and define the two rational functions $a(x) = \frac{p(x)}{q(x)}$ and $b(x) = \frac{s(x)}{t(x)}.$ Define $h(x) = \frac{a(x)}{b(x)}.$ Some algebra reveals

$h(x) = \dfrac{a(x)}{b(x)} \\ h(x) = \dfrac{\frac{p(x)}{q(x)}}{\frac{s(x)}{t(x)}} \\ h(x) = \dfrac{p(x)t(x)}{q(x)s(x)}$

Because $\mathcal{P}(\mathbb{R})$ is closed under multiplication, it follows that $h$ is also a rational function.

2. The domain of the original rational functions is $\text{Dom}(a) = \{ x : q(x) \neq 0 \}$ and $\text{Dom}(b) = \{ x : t(x) \neq 0 \}.$ The domain of $\frac{a}{b}$ is $\text{Dom}\left(\frac{a}{b}\right) = \{ x : x \in \text{Dom}(a) \text{ and } x \in \text{Dom}(b)\}.$ This expands to $\text{Dom}\left(\frac{a}{b}\right) = \{ x : q(x) \neq 0 \text{ and } s(x) \neq 0 \text{ and } t(x) \neq 0 \}.$

3. Zeroes of the Linear Function: Find all zeroes for linear functions, i.e. functions of the form $f(x) = mx + b.$

$f(x) = 0\\ mx + b = 0 \\ mx = -b \\ x = \dfrac{-b}{m}$

4. Quadratic Formula: Find all zeroes of functions of the form $f(x) = ax^2 + bx + c.$

First, set the function equal to zero:

$ax^2 + bx + c = 0$

Next, divide each side by $a.$ This is possible, since $a \neq 0,$ as otherwise $f$ is not a quadratic function:

$x^2 + \dfrac{b}{a}x + \dfrac{c}{a} = 0$

Isolate the $x$ terms by moving the constant to the other side:

$x^2 + \dfrac{b}{a}x = -\dfrac{c}{a}$

The next step is called completing the square. An equation of the form $(x + k)^2$ expands to $x^2 + 2kx + k^2.$ If we add $\left(\dfrac{b}{2a}\right)^2,$ we can work backwards to the factored form:

$x^2 + \dfrac{b}{a}x + \left(\dfrac{b}{2a}\right)^2 = -\dfrac{c}{a} + \left(\dfrac{b}{2a}\right)^2 \\ \left(x + \dfrac{b}{2a}\right)^2 = -\dfrac{c}{a} + \left(\dfrac{b}{2a}\right)^2$

From here, we take the square root of each side. However, we are interested in both the positive and the negative square roots, as both will work in this equation:

$x + \dfrac{b}{2a} = \pm \sqrt{-\dfrac{c}{a} + \left(\dfrac{b}{2a}\right)^2}$

From here, we simplify to get a tidy expression:

$x = \dfrac{-b}{2a} \pm \sqrt{-\dfrac{c}{a} + \left(\dfrac{b}{2a}\right)^2} \\ x = \dfrac{-b}{2a} \pm \sqrt{-\dfrac{c}{a} + \dfrac{b^2}{4a^2}} \\ x = \dfrac{-b}{2a} \pm \sqrt{-\dfrac{4ac}{4a^2} + \dfrac{b^2}{4a^2}} \\ x = \dfrac{-b}{2a} \pm \sqrt{\dfrac{b^2 - 4ac}{4a^2}}\\ x = \dfrac{b \pm \sqrt{b^2 - 4ac}}{2a}$

The term under the square root is called the discriminant. A quadratic equation only has real zeroes if its discriminant is non-negative, as square roots of negative numbers are not defined.