Difference between revisions of "Functional equation"

Latest revision as of 16:59, 29 August 2024

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.

@@ Line 21: / Line 21: @@
 A classic example of such a function is <math>f(x) = 1/x</math> because <math>f(f(x)) = f(1/x) = x</math>.  Cyclic functions can significantly help in solving functional identities.  Consider this problem:
-Find <math>f(x)</math> such that <math>3f(x) - 4f(1/x) = x^2</math>.  In this functional equation, let <math>x=y</math> and let <math>x = 1/y</math>.  This yields two new equations:
+Find <math>f(x)</math> such that <math>3f(x) - 4f(1/x) = x^2</math>.  Let <math>x=y</math> and <math>x = 1/y</math> in this functional equation.  This yields two new equations:
 <math>3f(y) - 4f\left(\frac1y\right) = y^2</math>
@@ Line 36: / Line 36: @@
 * [[2006_AMC_12A_Problems/Problem_18 | 2006 AMC 12A Problem 18]]
 * [[2007_AIME_II_Problems/Problem_14 | 2007 AIME II Problem 14]]
+==Advanced Topics==
+===Functions and Relations===
+Given a set <math>\mathcal{X}</math> and <math>\mathcal{Y}</math>, the Cartesian Product of these sets (denoted <math>\mathcal{X}\times \mathcal{Y}</math>) gives all ordered pairs <math>(x,y)</math> with <math>x \in \mathcal{X}</math> and <math>y \in \mathcal{Y}</math>. Symbolically,
+<cmath>
+\mathcal{X}\times \mathcal{Y} = \{ (x,y) \  | \  x \in \mathcal{X} \ \text{and} \  y \in \mathcal{Y}\}
+</cmath>
+A relation <math>R</math> is a subset of <math>\mathcal{X}\times \mathcal{Y}</math>. A function is a special time of relation where for every <math>y \in \mathcal{Y}</math> in the ordered pair <math>(x,y)</math>, there exists a unique <math>x \in \mathcal{X}</math>.
+===Injectivity and Surjectivity===
+Consider a function <math>f: \mathcal{X} \rightarrow \mathcal{Y}</math> be a function <math>f</math> from the set <math>\mathcal{X}</math> to the set <math>\mathcal{Y}</math>, i.e., <math>\mathcal{X}</math> is the domain of <math>f(x)</math> and <math>\mathcal{Y}</math> is the codomain of <math>f(x)</math>.
+The function <math>f(x)</math> is injective (or one-to-one) if for all <math>a, b</math> in the domain <math>\mathcal{X}</math>, <math>f(a)=f(b)</math> if and only if <math>a=b</math>. 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 <math>f(x)</math> is surjective (or onto) if for all <math>a</math> in the codomain <math>\mathcal{Y}</math> there exists a <math>b</math> in the domain <math>X</math> such that <math>f(b)=a</math>. 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 <math>f(x)</math> 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 <math>f(x)</math> has an inverse function <math>f^{-1}(x)</math>, where <math>f^{-1}(f(x)) = x</math>, if and only if it is a bijective function.
 ==See Also==
 *[[Functions]]
-*[[Polynomials]]
 *[[Cauchy Functional Equation]]
-[[Category:Definition]]
+[[Category:Algebra]]
-[[Category:Elementary algebra]]
+[[Category:Definition]]]

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