ALL functions such that:

  1. 1. The problem statement, all variables and given/known data

    With proof, Find all functions f: R->R such that f(x+y)=f(x)+f(y)+1 and f(1)=0

    2. Relevant equations



    3. The attempt at a solution

    First I cried...then...

    Since f(1)=0, f(2)=1, f(3)=2.....= x-1
    I'm having a really hard time conceptualizing this problem and I'm not sure what the answer is suppose to look like. Is it a finite or infinite set of functions? I'm not sure. The proof part is kinda throwing me off too. Any help would be great..
     
    Last edited: Sep 20, 2011
  2. jcsd
  3. f(1) = 0 and f(x + y) = ?
     
  4. Dick

    Dick 25,893
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    You can find the values of more than just the integers. E.g. what's f(1/2)? Can you find the value of f(1/n) where n is an integer? Can you find the value of f(m/n) where m and n are integers? Are you given that f is continuous or anything like that?
     
  5. yes f is continuous, and yes i can find f(1/2), but what i dont understand is why this problem has more than one solution (y=x-1)
     
  6. Dick

    Dick 25,893
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    Why do you think it has more than one solution?
     
  7. LCKurtz

    LCKurtz 8,404
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    And if you assume it is differentiable you can think about its derivative...
     
  8. @flyingpig: f(x+y)=f(x)+f(y)+1..it isn't given as anything else
     
  9. Looking at all the post, this question is probably out of my league. So I will just erase my answer.

    Sorry about that
     
  10. Dick

    Dick 25,893
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    You don't need to assume it's differentiable. I think continuous is enough, isn't it?
     
  11. LCKurtz

    LCKurtz 8,404
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    It might be, I don't know. I haven't worked through it. But if it is differentiable that gives an easy way to find a family of functions that work, which must be included in the "all" functions.
     
  12. Dick

    Dick 25,893
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    I did work through it. To use differentiability, which I DON'T think is given, you need to find the value of something like f(1/n). Then it's straightforward. But once you've found f(1/n) why not find f(m/n) (m and n integers) and use continuity and skip the differentiability assumption? That's what I'm trying to push iceblits to do.
     
  13. oh I see...I'll try. I'm just having a hard time conceptualizing this because I feel like y=x-1 is the only functions that can satisfy all the points. Thank-you for all the responses though..I will keep trying this problem until I solve it
     
  14. Dick

    Dick 25,893
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    y=x-1 IS the only function that works. You just have to prove it.
     
  15. oh! i see!!!!!!
     
  16. ok now i have some direction lol...I was thinking there was an infinite set of functions or something (like some wierd sine wave going diagonally up)..or like the line y=x-1 except if you imagine it like a string, push the string together to get various functions..or something like that
     
  17. LCKurtz

    LCKurtz 8,404
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    No, it isn't the only one.

    [Edit]Woops, ignore that. I agree. I forgot f(1) = 0. :frown:
     
  18. Dick

    Dick 25,893
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    That makes me feel better.
     
  19. gahhh..ok so if I let x and y both equal n, then, f(2n)=2f(n)+1. if I let x and y equal 1/n I get f(2/n)=2f(1/n)+1 and if I let x and y equal m/n, I get f(2m/n)=2f(m/n)+1....I dont think that shows anything (sorry, I haven't taken a proof class or anything yet :( ). Should I be trying to manipulate f(x+y)=f(x)+f(y)+1 to achieve y=x-1 or should I assume y=x-1 is the only solution and then show that any if any other function exists to satisfy the condition, it would have to be y=x-1? I guess I'm still lost haha..I thought I would be able to prove that y=x-1 is the only solution really easily but i guess i cant...
     
  20. Dick

    Dick 25,893
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    You found f(1/2), right? How would you find f(1/3)? Does that help you to find a general way to find f(1/n)? Here's a hint. Can you figure out a formula for f(nx) where n is an integer and x is any real? Formally, it's an induction proof, but if you haven't done induction you should still be able to figure out the formula just by thinking about it.
     
  21. ye f(1/2+1/2)=f(1/2)+f(1/2)+1...f(1)=2f(1/2)+1, f(1/2)=-1/2..wouldn't the formula just be y=x-1?
     
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