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Thread: Functional Equation

  1. #1
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    Functional Equation

    Find all functions $\displaystyle f : \mathbb{R} - \left\{ -\frac{1}{3}, \frac{1}{3} \right\} \rightarrow \mathbb{R}$ that satisfy the identity

    $\displaystyle f \left(\frac{x+1}{1-3x} \right) = x-f(x)$
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  2. #2
    Super Member PaulRS's Avatar
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    Consider the change of variables: $\displaystyle
    x : \to {\textstyle{{x - 1} \over {1 + 3x}}}
    $

    Then we get: $\displaystyle
    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: $\displaystyle
    \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: $\displaystyle
    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: $\displaystyle
    x: \to {\textstyle{{x - 1} \over {1 + 3x}}}
    $

    Then: $\displaystyle
    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: $\displaystyle
    \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 $\displaystyle
    f\left( x \right)
    $
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  3. #3
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    Quote Originally Posted by PaulRS View Post
    and solve for $\displaystyle
    f\left( x \right)
    $
    How would I do this?
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  4. #4
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Winding Function View Post
    How would I do this?
    As you'd solve in a normal system of equations.
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  5. #5
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    Quote Originally Posted by PaulRS View Post
    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?
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  6. #6
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by Winding Function View Post

    I'm confused. - What are the "variables" of the equation, and how do I isolate them?
    Quote Originally Posted by PaulRS View Post

    and solve for $\displaystyle
    f\left( x \right)
    $
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  7. #7
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    Quote Originally Posted by Krizalid View Post
    But what do I do with terms like $\displaystyle f\left( \frac{x+1}{1-3x} \right)$?
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Winding Function View Post
    But what do I do with terms like $\displaystyle f\left( \frac{x+1}{1-3x} \right)$?
    What happens if you add them?
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