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Math Olympiads problem that I couldn't do.

  1. Dec 3, 2011 #1
    1. The problem statement, all variables and given/known data
    for every (x,y) in ℝ^2:

    f(x^2+f(y))= y-x^2

    2. Relevant equations

    Find all functions.

    3. The attempt at a solution
    I was wondering if rewriting it as x^2+f(y)=f^-1(y-x^2)?
  2. jcsd
  3. Dec 3, 2011 #2
    So, what is the question?? And what is f??
  4. Dec 3, 2011 #3
    I believe the idea is to find all functions satisfying that condition.
  5. Dec 3, 2011 #4
    Yes that's what it is. Find all functions satisfying that condition.
  6. Dec 3, 2011 #5
    Note that


    So if we define


    Then what can you say about [itex]g_x\circ g_x[/itex]?
  7. Dec 3, 2011 #6
    I don't get what you're trying to tell me with gx∘gx.

    I understand that its a category from gx to gx, so u can say: gx(g(x)). right?
  8. Dec 3, 2011 #7
    I'm sorry for my intrusion, but after having spent some time on this I would like to add my comment.
    From intuition and some trials I can say the [itex]f(x) = -x[/itex] works.
    So that:
    [itex]f(y) = -y[/itex]
    [itex]f(y) + x^2= -y+x^2[/itex]
    [itex]f(f(y) + x^2)= y-x^2[/itex]

    But is there an analytical or mechanical way to solve these problems ?

    I can't get this too. Maybe one should apply the derivative ???
  9. Dec 3, 2011 #8


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    I'd try just differentiating both sides with respect to x.
  10. Dec 3, 2011 #9
    Yes, but what if f is not differentiable?? :frown:
  11. Dec 3, 2011 #10
    But it's not a function of 2 variables. There's only one "input".
  12. Dec 3, 2011 #11
    f(f(y)+x^x)=y−x^2 can also be written as fof(y)+f(x^2)=y-x^2
  13. Dec 3, 2011 #12
    But the problem says "find all the solutions". So there are more than 1 ? What the others look like then ? :confused:
  14. Dec 3, 2011 #13


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    I'd say find all the differentiable ones first, then try thinking about if there could be more.
  15. Dec 3, 2011 #14


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    Treat y as a constant.
  16. Dec 3, 2011 #15
    I don't know that's what i'm trying to reach but i also got this one:

  17. Dec 3, 2011 #16
    Yes, but you got to watch out. f is not necessarily invertible. So the function [itex]f^{-1}[/itex] might not exist!!
  18. Dec 3, 2011 #17
    Yes but is it correct?
  19. Dec 3, 2011 #18
    math olympiads.jpg

    That's the question, its the second one if you understand some french micromass.
  20. Dec 3, 2011 #19
    Not until you show f to be invertible. What happens if x=0 ?
  21. Dec 3, 2011 #20
    for x=0 you get fof(y)=y and f(y)=f^-1(y)
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