Difference between revisions of "Functional equation"

Revision as of 16:17, 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, $f(y) = -\frac{3}{7}y^2 - \frac{4}{7y^2}$

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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:
+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>

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Difference between revisions of "Functional equation"

Revision as of 16:17, 29 August 2024

Contents

Introductory Topics

The Inverse of a Function

Intermediate Topics

Cyclic Functions

Problem Examples

Advanced Topics

Injectivity and Surjectivity

See Also