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Math Help - functional equation

  1. #1
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    functional equation

    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 x, f(x)=1/x, then f(f(x))=x again. Similarly, if f(x)=-x, then f(f(x))=x again. This last one is what we want. So

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

    Q.E.D.
    Last edited by AlephZero; July 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, 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, 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: f^k(x)=f(f(...f(x))) ( where f appears k times )

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

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

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

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

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

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

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