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Thread: Functional equation

  1. #1
    MHF Contributor alexmahone's Avatar
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    Functional equation

    Find all functions of $\displaystyle R\to R$ with the property
    $\displaystyle f(xf(x) + f(y)) = (f(x))^2 + y$
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  2. #2
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    Quote Originally Posted by alexmahone View Post
    Find all functions of $\displaystyle R\to R$ with the property
    $\displaystyle f(xf(x) + f(y)) = (f(x))^2 + y$
    I might be barking up the wrong tree here but can't you apply $\displaystyle f^{-1}$ to both sides and get $\displaystyle x f(x) + f(y) = x f(x) + f^{-1}(y)$ .....?

    In which case the problem boils down to finding all functions such that $\displaystyle f(y) = f^{-1}(y)$, that is, functions which are their own inverse ......

    Such functions are called involutions.
    Last edited by mr fantastic; Oct 18th 2008 at 04:36 AM. Reason: Added the involution business
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by mr fantastic View Post
    $\displaystyle x f(x) + f(y) = x f(x) + f^{-1}(y)$
    No, the RHS of this equation is incorrect.
    We do not know that $\displaystyle f^{-1}(a+b)=f^{-1}(a)+f^{-1}(b)$
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  4. #4
    Super Member Showcase_22's Avatar
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    $\displaystyle
    f(xf(x) + f(y)) = (f(x))^2 + y
    $

    $\displaystyle f(xf(x)+f(y))-(f(x))^2=y$

    $\displaystyle (\sqrt{f(xf(x)+f(y)}+f(x))(\sqrt{f(xf(x)+f(y))}-f(x)=y$

    Does this help in any way? I was thinking that somehow I could get the $\displaystyle f(xf(x)+f(y))$ to cancel out but I couldn't get it to work.

    The RHS can be multiplied by$\displaystyle \frac{\sqrt{f(xf(x)+f(y))}-f(x)}{\sqrt{f(xf(x)+f(y)}-f(x)}$ to give:

    $\displaystyle \sqrt{f(xf(x)+f(y)}+f(x)=\frac{y(\sqrt{f(xf(x)+f(y ))}-f(x))}{f(xf(x)+f(y))-(f(x))^2}$

    but I don't think that's helping.

    It's an interesting question. I hope someone here can solve it!

    EDIT: Perhaps you could try case analysis. When x=0:

    [MATh]f(f(y))=(f(0))^2+y[/tex]

    f(0) has an unknown value so it might not help. However, we have successfully eliminated a term which is beneficial!
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  5. #5
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    Quote Originally Posted by alexmahone View Post
    No, the RHS of this equation is incorrect.
    We do not know that $\displaystyle f^{-1}(a+b)=f^{-1}(a)+f^{-1}(b)$
    Hmmm ..... you're right.

    I know f(x) = x works (an involution, by the way). I haven't checked involutions of the form f(x) = (ax + b)/(cx + d), where there's a relationship between a, b, c, and d that I can't be bothered remembering .... might be worth a check.
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  6. #6
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    I believe that in the case of f(0)=0 a function being an involution is a necessary but not sufficient criteria..
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  7. #7
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    Quote Originally Posted by alexmahone View Post
    Find all functions of $\displaystyle R\to R$ with the property
    $\displaystyle f(xf(x) + f(y)) = (f(x))^2 + y$
    1) $\displaystyle f$ is onto. (it's actually one-to-one as well, but we don't need that!)

    Proof: let $\displaystyle z \in \mathbb{R}.$ choose $\displaystyle y=z-(f(0))^2.$ then: $\displaystyle f(f(y))=(f(0))^2+y=z. \ \ \Box$


    2) $\displaystyle f(0)=0.$

    Proof: by 1) there exists $\displaystyle x \in \mathbb{R}: \ f(x)=0.$ thus: $\displaystyle x=x+(f(x))^2=f(xf(x)+f(x))=f(0)=f(f(x))=(f(0))^2+x .$ thus: $\displaystyle f(0)=0. \ \ \Box$


    3) $\displaystyle f(x)=\pm x.$

    Proof: we have: $\displaystyle f(f(x))=(f(0))^2+x=x.$ thus: $\displaystyle x^2=(f(f(x)))^2=f(f(x)f(f(x))+f(0))=f(xf(x))=(f(x) )^2.$ thus: $\displaystyle f(x)=\pm x. \ \ \Box$
    Last edited by NonCommAlg; Oct 20th 2008 at 02:05 AM.
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