A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Empty

  • one year ago

What are all possible functions on the reals that satisfy this functional equation? $$f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x)$$ source: http://www.imo2015.org/solution.php?lang=en

  • This Question is Closed
  1. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Well... the identity function works for starters... :)

  2. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Haha yeah, the trouble is showing this is the only possible function.

  3. danica518
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    really thats the only one?

  4. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    In fact, f(x) = c does not appear to work for any constant....

  5. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    No, that's just wishful thinking coming in, that or find all the other solutions whatever they may be.

  6. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I have no idea what the general solution is!

  7. danica518
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can a function be both multiplicative and also obey lienar Super position f(xy)=f(x)+f(y) and f(x)*F(y)?

  8. danica518
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok i guess not

  9. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @danica518 the obvious solution \(f(x)=x\) is both

  10. danica518
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    other than identity

  11. danica518
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but the identity works? lemme see

  12. danica518
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1437170220728:dw|

  13. danica518
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    sorry i shouldnt have said Linear SSP i just mean a function that can do this f(xy)=f(x)+f(y) and f(x)*F(y)? which i dont think is possible

  14. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    It looks like to what you're saying dan, f(x)=0 is the only solution, since you have: $f(x)f(y)=f(xy)=f(x)+f(y)$ which is basically just saying $ab=a+b$ so unless you put more conditions on it idk...?

  15. danica518
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i nothing something similar going there

  16. danica518
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    like for the addition the x is added for multiplication the y is multiplied f(x+f(x+y))= x+f(x+y) and f(xy)=y*f(x)

  17. danica518
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    suppose we take them separetly can we find a function that would satisfy these 2 conditions separetely

  18. danica518
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    f(x+f(x+y))= x+f(x+y) f(xy)=y*f(x) is there a solution to both of these equations, other than f(x)=0 and f(x)=x

  19. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    consider \(x=y=0\) so $$f(0+f(0))+f(0)=f(0)\\f(f(0))=0$$ now consider \(x=f(0),y=0\) so $$f(f(0)+f(f(0)+0))+f(0)=f(0)+f(f(0))\\f(f(0)+f(f(0)))=f(0)+f(f(0))\\f(f(0))=f(0)$$ implying \(f\) has a fixed point, \(f(0)\to f(0)\) now consider \(x=0,y=f(0)\) $$f(0+f(f(0)))+f(0)=f(f(0))+f(0)f(0)\\f(0)+f(0)=f(0)^2\\f(0)^2-2f(0)=0\\f(0)\left(f(0)-2\right)=0$$ so either \(f(0)=0\) or \(f(0)=2\)

  20. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but if \(f(0)=2\) then \(f(2)=0\) which contradicts that \(f(0)\) is a fixed point (\(f(2)=2\)), so it must be that \(f(0)=0\) i think

  21. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Potential typo in lines 6 and 7 of original calculation @oldrin.bataku

  22. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    The f(0) on the RHS should not be there since it gets cancelled by the f(0) on the LHS.

  23. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    now consider the general case \(x=0\) so $$f(f(y))+f(0)=f(y)+yf(0)$$using the fact \(f(0)=0\) this reduces to teh functional equation $$f(f(y))=f(y)$$ so all \(f(y)\) are fixed points, which suggests that the only functions that work are restrictions of \(f(x)=x\)

  24. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @jtvatsim ? what are you talking about

  25. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    The part where you plug in x = f(0) and y = 0 does not yield f(f(0)) = f(0) rather, it duplicates the previous substitution giving f(f(0)) = 0 again.

  26. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    However, I don't think it changes the main thrust of the proof. :)

  27. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    which line is this? can you type it out?

  28. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Sure.

  29. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    Let x = f(0) and y = 0. Then, \[f(f(0) + f(f(0)+0)) + f(0) = f(0) + f(f(0))\] \[f(f(0) + f(f(0))) + f(0) = f(0) + f(f(0))\] \[f(f(0)) = 0\] rather than \[f(f(0)) = f(0) \]

  30. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh, oops, so then \(f(0)\) need not be a fixed point, in which case \(f(0)=2\) might still work

  31. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but $$f(f(x))=x$$ still holds so \(f^2\) is still the identity, which means we want a function $f$ that is an involution but yields \(f(0)=2\). the most obvious one is \(f(x)=2-x\), which seems to be a valid solution, but there might be more

  32. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Is there some thing we can do with this \(x+f(x+y)\) term that appears twice inside and outside the function and do some sort of infinite recursive plugging in?

  33. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    thought about it, but i'm not sure how to formalize it

  34. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    this is hard

  35. jtvatsim
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 3

    This is International Math Olympiad...

  36. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Here's my solution for integer n I made by combining both into one with a cute little discrete fourier transform... :P \[f(x)=1+(-1)^n(x-1)\]

  37. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.