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Difference between revisions of "Functional equation"

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<math>f(f(\cdots f(x) \cdots)) = x</math>
 
<math>f(f(\cdots f(x) \cdots)) = x</math>
  
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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) = 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:
 
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:

Revision as of 11:51, 26 May 2008

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 $3f(x) - 4f(1/x) = x^2$. In this functional equation, let $x=y$ and let $x = 1/y$. 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 subtract the second equation from the first, we have:

$25f(y) = 3y^2 - \frac{4}{y^2}$

So, clearly, $f(y) = \frac{3}{25}y^2 - \frac{4}{25y^2}$

Problem Examples

See Also

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