Difference between revisions of "Functional equation"

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Functional Equations are equations that involve [[functions]].  For an example, these are some examples of functional equations:
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A '''functional equation''', roughly speaking, is an equation in which some of the unknowns to be solved for are [[function]]s.  For example, the following are functional equations:
  
*<math>f(x) + f\left(\frac1x\right) = 2x</math>
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*<math>f(x) + 2f\left(\frac1x\right) = 2x</math>
 
*<math>g(x)^2 + 4g(x) + 4 = 8\sin{x}</math>
 
*<math>g(x)^2 + 4g(x) + 4 = 8\sin{x}</math>
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==Introductory Topics==
 
==Introductory Topics==
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===The Inverse of a Function===
 
===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)<math> = x^2 + 6</math>.  The function <math>g(x) = \sqrt{x-6}</math> has the property that <math>f(g(x)) = x</math>.  In this case, <math>g</math> is called the inverse function.  Often the inverse of a function <math>f</math> is denoted by <math>f^{-1}</math>.
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The inverse of a function is a function that "undoes" a function.  For an example, consider the function: <math>f(x) = x^2 + 6</math>.  The function <math>g(x) = \sqrt{x-6}</math> has the property that <math>f(g(x)) = x</math>.  In this case, <math>g</math> is called the '''(right) inverse function'''(Similarly, a function <math>g</math> so that <math>g(f(x))=x</math> 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 <math>f</math> is denoted by <math>f^{-1}</math>.
  
 
==Intermediate Topics==
 
==Intermediate Topics==
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<math>f(f(\cdots f(x) \cdots)) = x</math>
 
<math>f(f(\cdots f(x) \cdots)) = x</math>
  
A classic example of such a function is <math>f(x) = 1/x</math> because <math>f(f(x)) = f(1/x) = f(x)</math>.  Cyclic functions can significantly help in solving functional identities.  Consider this problem:
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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:
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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>
 
<math>3f(y) - 4f\left(\frac1y\right) = y^2</math>
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<math>3f\left(\frac1y\right)- 4f(y) = \frac1{y^2}</math>
 
<math>3f\left(\frac1y\right)- 4f(y) = \frac1{y^2}</math>
  
Now, if we multiply the first equation by 3 and the second equation by 4, and substract the second equation from the first, we have:
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Now, if we multiply the first equation by 3 and the second equation by 4, and add the two equations, we have:
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<math>-7f(y) = 3y^2 + \frac{4}{y^2}</math>
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So, clearly, <math>f(y) = -\frac{3}{7}y^2 - \frac{4}{7y^2}</math>
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=== Problem Examples ===
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* [[2006_AMC_12A_Problems/Problem_18 | 2006 AMC 12A Problem 18]]
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* [[2007_AIME_II_Problems/Problem_14 | 2007 AIME II Problem 14]]
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==Advanced Topics==
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===Functions and Relations===
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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,
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<cmath>
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\mathcal{X}\times \mathcal{Y} = \{ (x,y) \  | \  x \in \mathcal{X} \ \text{and} \  y \in \mathcal{Y}\}
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</cmath>
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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>.
  
<math>25f(y) = 3y^2 - \frac{4}{y^2}</math>
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===Injectivity and Surjectivity===
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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>.
  
So clearly, <math>f(y) = \frac{3}{25}y^2 - \frac{4}{25y^2}</math>
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 +
 
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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,
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\begin{equation}
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f(x) \ \text{is injective} \iff (\forall a,b \in \mathcal{X}, f(a)=f(b)\implies a=b).
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\end{equation}
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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,
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\begin{equation}
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f(x) \ \text{is surjective} \iff \forall a \in \mathcal{Y},\exists b \in \mathcal{X}: f(b)=a.
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\end{equation}
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 +
 
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The function <math>f(x)</math> is bijective (or one-to-one and onto) if it is both injective and subjective. Symbolically,
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\begin{equation}
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f(x) \ \text{is bijective} \iff \forall a \in \mathcal{Y},\exists! b \in \mathcal{X}: f(b)=a.
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\end{equation}
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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==
 
==See Also==
 
*[[Functions]]
 
*[[Functions]]
*[[Polynomials]]
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*[[Cauchy Functional Equation]]
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[[Category:Algebra]]
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[[Category:Definition]]]

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.

See Also