Difference between revisions of "Functional equation"
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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) + | + | *<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> | ||
+ | |||
==Introductory Topics== | ==Introductory Topics== | ||
Line 8: | Line 9: | ||
===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) | + | 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) = | + | 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>. | + | 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 | + | Now, if we multiply the first equation by 3 and the second equation by 4, and add the two equations, we have: |
+ | |||
+ | <math>-7f(y) = 3y^2 + \frac{4}{y^2}</math> | ||
+ | |||
+ | So, clearly, <math>f(y) = -\frac{3}{7}y^2 - \frac{4}{7y^2}</math> | ||
+ | |||
+ | === Problem Examples === | ||
+ | * [[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>. | ||
− | <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== | ==See Also== | ||
*[[Functions]] | *[[Functions]] | ||
− | *[[ | + | *[[Cauchy Functional Equation]] |
+ | |||
+ | [[Category:Algebra]] | ||
+ | [[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:
Contents
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: . The function has the property that . In this case, is called the (right) inverse function. (Similarly, a function so that 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 is denoted by .
Intermediate Topics
Cyclic Functions
A cyclic function is a function that has the property that:
A classic example of such a function is because . Cyclic functions can significantly help in solving functional identities. Consider this problem:
Find such that . Let and in this functional equation. This yields two new equations:
Now, if we multiply the first equation by 3 and the second equation by 4, and add the two equations, we have:
So, clearly,
Problem Examples
Advanced Topics
Functions and Relations
Given a set and , the Cartesian Product of these sets (denoted ) gives all ordered pairs with and . Symbolically,
A relation is a subset of . A function is a special time of relation where for every in the ordered pair , there exists a unique .
Injectivity and Surjectivity
Consider a function be a function from the set to the set , i.e., is the domain of and is the codomain of .
The function is injective (or one-to-one) if for all in the domain , if and only if . 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 is surjective (or onto) if for all in the codomain there exists a in the domain such that . 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 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 has an inverse function , where , if and only if it is a bijective function.