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Thread: functional equation

  1. #1
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    functional equation

    $\displaystyle solve f(x)+2x=f(f(x))$
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  2. #2
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    Solve using some knowledge of "involuntary functions." For example, if for nonzero $\displaystyle x,$ $\displaystyle f(x)=1/x,$ then $\displaystyle f(f(x))=x$ again. Similarly, if $\displaystyle f(x)=-x,$ then $\displaystyle f(f(x))=x$ again. This last one is what we want. So

    $\displaystyle f(x)=-x \implies f(x)+2x=x=f(f(x)),$

    Q.E.D.
    Last edited by AlephZero; Jul 8th 2009 at 07:39 AM. Reason: typo
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  3. #3
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    Unless you were forced to come up with an answer, I think the word should be "involutary".

    By the way, $\displaystyle f(x)=2x$ solves the equation as well. Maybe there are other solutions?
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  4. #4
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    Quote Originally Posted by halbard View Post
    Unless you were forced to come up with an answer, I think the word should be "involutary".

    By the way, $\displaystyle f(x)=2x$ solves the equation as well. Maybe there are other solutions?
    Freudian slip, I guess...

    I'd be quite curious to know if anyone can come up with other solutions. I would conjecture that they would be significantly more complicated.
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  5. #5
    Super Member PaulRS's Avatar
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    Just some stuff I've found out about the equation...

    Observation 1 : We write: $\displaystyle f^k(x)=f(f(...f(x)))$ ( where f appears k times )

    Your equation then turns into: $\displaystyle f^1(x)+2f^{0}(x)=f^2(x)$ very suggestive, isn't it?

    This clearly generalises to: $\displaystyle f^{n+1}(x)+2f^{n}(x)=f^{n+2}(x)$ now fix $\displaystyle x$ and set: $\displaystyle a_n=f^n(x)$ solving the recurrence relation: $\displaystyle f^k(x)=A(x)\cdot{(-1)^k}+B(x)\cdot{2^k}$

    Setting: $\displaystyle k=0$ we have the relation: $\displaystyle x=A(x)+B(x)$ ( in particular your solutions correspond to the cases $\displaystyle A(x)=0$ and the other $\displaystyle B(x)=0$ )

    Now, this implies: $\displaystyle f(x)=-A(x)+2B(x)=-x+3B(x)$ and going to the original equation we get the following restriction $\displaystyle 2B(x)=B(3B(x)-x)=B(f(x))$

    So it seems it all comes down to solving that functional equation.

    Observation 2: $\displaystyle f$ is injective. To see this, suppose $\displaystyle f(x)=f(y)$ then $\displaystyle f(f(x))=f(f(y))$ thus: $\displaystyle f(x)+2x=f(y)+2y$ but then $\displaystyle 2x=2y$ and so $\displaystyle x=y$

    Observation 3: It follows from 2 that since $\displaystyle f(f(0))=f(0)$ we must have $\displaystyle f(0)=0$
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