Functional equation

A functional equation, roughly speaking, is an equation in which some of the unknowns to be solved for are functions. For example, the following are functional equations:

$f(x) + 2f\left(\frac1x\right) = 2x$
$g(x)^2 + 4g(x) + 4 = 8\sin{x}$

Introductory Topics

The Inverse of a Function

The inverse of a function is a function that "undoes" a function. For an example, consider the function: $f(x) = x^2 + 6$ . The function $g(x) = \sqrt{x-6}$ has the property that $f(g(x)) = x$ . In this case, $g$ is called the (right) inverse function. (Similarly, a function $g$ so that $g(f(x))=x$ is called the left inverse function. Typically the right and left inverses coincide on a suitable domain, and in this case we simply call the right and left inverse function the inverse function.) Often the inverse of a function $f$ is denoted by $f^{-1}$ .

Intermediate Topics

Cyclic Functions

A cyclic function is a function $f(x)$ that has the property that:

$f(f(\cdots f(x) \cdots)) = x$

A classic example of such a function is $f(x) = 1/x$ because $f(f(x)) = f(1/x) = x$ . Cyclic functions can significantly help in solving functional identities. Consider this problem:

Find $f(x)$ such that $3f(x) - 4f(1/x) = x^2$ . Let $x=y$ and $x = 1/y$ in this functional equation. This yields two new equations:

$3f(y) - 4f\left(\frac1y\right) = y^2$

$3f\left(\frac1y\right)- 4f(y) = \frac1{y^2}$

Now, if we multiply the first equation by 3 and the second equation by 4, and add the two equations, we have:

$-7f(y) = 3y^2 + \frac{4}{y^2}$

So, clearly, $f(y) = -\frac{3}{7}y^2 - \frac{4}{7y^2}$

Problem Examples

Advanced Topics

Functions and Relations

Given a set $\mathcal{X}$ and $\mathcal{Y}$ , the Cartesian Product of these sets (denoted $\mathcal{X}\times \mathcal{Y}$ ) gives all ordered pairs $(x,y)$ with $x \in \mathcal{X}$ and $y \in \mathcal{Y}$ . Symbolically, $\mathcal{X}\times \mathcal{Y} = \{ (x,y) \ | \ x \in \mathcal{X} \ \text{and} \ y \in \mathcal{Y}\}$

A relation $R$ is a subset of $\mathcal{X}\times \mathcal{Y}$ . A function is a special time of relation where for every $y \in \mathcal{Y}$ in the ordered pair $(x,y)$ , there exists a unique $x \in \mathcal{X}$ .

Injectivity and Surjectivity

Consider a function $f: \mathcal{X} \rightarrow \mathcal{Y}$ be a function $f$ from the set $\mathcal{X}$ to the set $\mathcal{Y}$ , i.e., $\mathcal{X}$ is the domain of $f(x)$ and $\mathcal{Y}$ is the codomain of $f(x)$ .

The function $f(x)$ is injective (or one-to-one) if for all $a, b$ in the domain $\mathcal{X}$ , $f(a)=f(b)$ if and only if $a=b$ . Symbolically, \begin{equation} f(x) \ \text{is injective} \iff (\forall a,b \in \mathcal{X}, f(a)=f(b)\implies a=b). \end{equation}

The function $f(x)$ is surjective (or onto) if for all $a$ in the codomain $\mathcal{Y}$ there exists a $b$ in the domain $X$ such that $f(b)=a$ . Symbolically, \begin{equation} f(x) \ \text{is surjective} \iff \forall a \in \mathcal{Y},\exists b \in \mathcal{X}: f(b)=a. \end{equation}

The function $f(x)$ is bijective (or one-to-one and onto) if it is both injective and subjective. Symbolically, \begin{equation} f(x) \ \text{is bijective} \iff \forall a \in \mathcal{Y},\exists! b \in \mathcal{X}: f(b)=a. \end{equation}

The function $f(x)$ has an inverse function $f^{-1}(x)$ , where $f^{-1}(f(x)) = x$ , if and only if it is a bijective function.

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