Functional Equation

• December 24th 2008, 06:59 AM
Winding Function
Functional Equation
Find all functions $f : \mathbb{R} - \left\{ -\frac{1}{3}, \frac{1}{3} \right\} \rightarrow \mathbb{R}$ that satisfy the identity

$f \left(\frac{x+1}{1-3x} \right) = x-f(x)$
• December 24th 2008, 09:18 AM
PaulRS
Consider the change of variables: $
x : \to {\textstyle{{x - 1} \over {1 + 3x}}}
$

Then we get: $
f\left( x \right) = {\textstyle{{x - 1} \over {1 + 3x}}} - f\left( {{\textstyle{{x - 1} \over {1 + 3x}}}} \right)
$
( check that the denominator is never 0)

So we've got these 2 equations: $
\left[ \begin{array}{l}
f\left( x \right) = {\textstyle{{x - 1} \over {1 + 3x}}} - f\left( {{\textstyle{{x - 1} \over {1 + 3x}}}} \right) \\
f\left( {{\textstyle{{x + 1} \over {1 - 3x}}}} \right) = x - f\left( x \right) \\
\end{array} \right.
$

Thus: $
f\left( {{\textstyle{{x + 1} \over {1 - 3x}}}} \right) - f\left( {{\textstyle{{x - 1} \over {1 + 3x}}}} \right) = x - {\textstyle{{x - 1} \over {1 + 3x}}} = {\textstyle{{3x^2 + 1} \over {3x + 1}}}

$

Again take: $
x: \to {\textstyle{{x - 1} \over {1 + 3x}}}
$

Then: $
f\left( x \right) - f\left( {{\textstyle{{x + 1} \over {1 - 3x}}}} \right) = {\textstyle{{3 \cdot \left( {{\textstyle{{x - 1} \over {1 + 3x}}}} \right)^2 + 1} \over {3 \cdot \left( {{\textstyle{{x - 1} \over {1 + 3x}}}} \right) + 1}}}
$

Consider then the system of equations: $
\left\{ \begin{array}{l}
f\left( x \right) - f\left( {{\textstyle{{x + 1} \over {1 - 3x}}}} \right) = {\textstyle{{3 \cdot \left( {{\textstyle{{x - 1} \over {1 + 3x}}}} \right)^2 + 1} \over {3 \cdot \left( {{\textstyle{{x - 1} \over {1 + 3x}}}} \right) + 1}}} \\
f\left( {{\textstyle{{x + 1} \over {1 - 3x}}}} \right) = x - f\left( x \right) \\
\end{array} \right.
$

and solve for $
f\left( x \right)
$
• December 24th 2008, 10:11 AM
Winding Function
Quote:

Originally Posted by PaulRS
and solve for $
f\left( x \right)
$

How would I do this?
• December 24th 2008, 10:12 AM
PaulRS
Quote:

Originally Posted by Winding Function
How would I do this?

As you'd solve in a normal system of equations.
• December 24th 2008, 05:41 PM
Winding Function
Quote:

Originally Posted by PaulRS
As you'd solve in a normal system of equations.

I'm confused. - What are the "variables" of the equation, and how do I isolate them?
• December 24th 2008, 06:23 PM
Krizalid
Quote:

Originally Posted by Winding Function

I'm confused. - What are the "variables" of the equation, and how do I isolate them?

Quote:

Originally Posted by PaulRS

and solve for $
f\left( x \right)
$

:)
• December 24th 2008, 07:56 PM
Winding Function
Quote:

Originally Posted by Krizalid
:)

But what do I do with terms like $f\left( \frac{x+1}{1-3x} \right)$?
• December 24th 2008, 10:31 PM
Mathstud28
Quote:

Originally Posted by Winding Function
But what do I do with terms like $f\left( \frac{x+1}{1-3x} \right)$?

What happens if you add them?