2. Functions

$f: A \to B$ is a function $\iff \forall a \in A$ is assigned to a unique $b \in B$. We say that:

\(f: A \to B\) where $A$ is the domain, and $B$ is to codomain.

The range of $f(A)={b \in B: b=f(a) \exists a \in A} \subseteq B$

Definitions

A function $f: \mathbb{R \to R}$ us considered:

\(\begin{aligned} \text{i)}\;& \underline{\text{Even}}, && \text{if } f(-x) = f(x), \quad \forall x \in \mathbb{R}. \\ \text{ii)}\;& \underline{\text{Odd}}, && \text{if } f(-x) = -f(x), \quad \forall x \in \mathbb{R}. \\ \text{iii)}\;& \underline{\text{Periodic}}, && \text{if } \exists T > 0 \text{ such that } f(x) = f(x + T), \quad \forall x \in \mathbb{R}. \\ \text{iv)}\;& \underline{\text{Increasing}}, && \text{if } f(x) \ge f(y), \quad \forall x > y,\; x, y \in \mathbb{R}. \\ \text{v)}\;& \underline{\text{Decreasing}}, && \text{if } f(x) \le f(y), \quad \forall x > y,\; x, y \in \mathbb{R}. \end{aligned}\)

Polynomials

A polynomial $p: \mathbb{R \to R}: p(x) = \sum_{n=0}^{d}a_nx^n$, where $d$ is the degree of the polynomial, $a_n$ is the $n^\text{th}$ coefficient.

Note: let $deg(p(x))$ be the degree of $p$ (i.e. $d$ from the aforementioned equation).

Polynomial Division

let $p, d$ be polynomials, $deg(p), deg(q) \not\equiv 0$, then $\exists$ polynomials $q$ (quotient) and $r$ (remainder), such that:

\[\boxed{ \begin{aligned} \text{i)}\;& p(x) = q(x)d(x) + r(x), \quad \forall x \in \mathbb{R}. \\ \text{ii)}\;& \text{either } r \equiv 0, \text{ or } deg(r) < deg(d). \end{aligned} }\]

if $r(x)=0, \forall x \in \mathbb{R}$ , we say that $d$ is a factor of $p$.

Polynomial Division Algorithm (Polynomial Long Division)

We write polynomial divisions using this method:

\[\boxed{ \begin{array} \quad\quad\quad q(x)+r(x) \\ d(x)\; |\overline{\hspace{2em} f(x) \hspace{2em}} \end{array} }\]

$\underline{\text{Example:}} \; p(x)=2x^3-3x^2+4x+5, d(x)=x+2$

\[\begin{aligned} &\hspace{5em} q(x) + r(x) \\[3pt] &(x + 2)\; \big|\,\overline{ 2x^3 - 3x^2 + 4x + 5 } \\[4pt] &\hspace{2.7em} \underline{-(2x^3 + 4x^2)} \\[3pt] &\hspace{5.3em} -7x^2 + 4x + 5 \\[2pt] &\hspace{4.6em} \underline{-(-7x^2 -14x)} \\[3pt] &\hspace{9.3em} 18x + 5 \\[2pt] &\hspace{8.1em} \underline{-(18x + 36)} \\[3pt] &\hspace{10.9em} -31 \rightarrow \text{remainder} \\[2pt] \end{aligned}\] \[\therefore q(x) = 2x^2 -7x + 18, \quad r(x) = -31\]

So

\[\boxed{p(x) = (x + 2)(2x^2 - 7x + 18) - 31}\]

$\underline{\text{Theorem 1.16:}}$ $p$ is a real polynomial, let $a \in \mathbb{R}$, $\therefore a$ is a root of $p \iff (x-a)$ is a factor of $p$.

$\underline{\text{Proof:}}$ By the division algorithm, $\exists$ polynomials $q, r$: $p(x)=q(x)(x-a)+r(x), \forall x \in \mathbb{R}$, where $r \equiv 0$ or $deg(r) < deg(x-a) = 1 \Rightarrow deg(r) = 0$, so $r(x)$ is a constant (say, $r(x)=c$), so in ($\star$) we have $p(x) = q(x)(x-a) + c, \forall x \in \mathbb{R} \Rightarrow$ if $a$ is a root, $p(a)=0, \therefore q(a)(a-a)+c = 0 = q(a)(0) + c, c = 0$.

$\therefore p(x) = q(x)(x-a), x-a$ is a factor, $a$ is a root.

Quadratics

\(\boxed{\Delta \text{ (discriminant)} = b^2 - 4ac}\)

The roots of a quadratic can be found with the quadratic formula:

\[\boxed{x=\dfrac{-b \pm \sqrt{\Delta}}{2a}}\]

Conditions on $\Delta$: \(\begin{aligned} & \underline{\Delta < 0}: & \text{if xyz,} \dots \\ & \underline{\Delta = 0}: & \text{if xyz,} \dots \\ & \underline{\Delta > 0}: & \text{if xyz,} \dots \end{aligned}\)

Rational functions

Let $p, q$ be polynomials, $f(x) = \dfrac{p(x)}{q(x)}$. If $deg(q) < deg(p), f$ is considered a proper rational function.

Partial functions

Let $h(x) = \dfrac{p(x)}{q(x)}$, and be a proper rational function.

$\underline{\text{Goal:}}$ Express $h$ as a sum of several factors with simpler denominators.

$\underline{\text{Method:}}$

• Factor $g(x)$
• …
• See lecture notes of 06.10.2025 on Moodle page