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Very important

  1. Apr 30, 2008 #1
    Very important....

    Let R denote the set of real numbers. Find all nondecreasing continuous
    functions g : R R such that

    g(x+g(y))=g(g(x))+g(y) for all x,y∈R
     
  2. jcsd
  3. Apr 30, 2008 #2

    HallsofIvy

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    And why is that "very important"?
     
  4. Apr 30, 2008 #3

    tiny-tim

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    Looks monotonous to me … :rolleyes:
     
  5. May 2, 2008 #4
    I think u see it easy.....

    I read this problem in abook,and I liked it
    so can you solve it???
     
  6. May 2, 2008 #5

    I read this problem in abook,and I liked it
    so can you solve it???
     
  7. May 2, 2008 #6

    tiny-tim

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    Hi vip89! :smile:

    Hint: try differentiating with respect to x, and y, separately. :smile:
     
  8. May 2, 2008 #7

    HallsofIvy

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    Oh, I see. That hint didn't seem to make much sense until I did!

    (Funny how often that happens.)
     
  9. May 2, 2008 #8

    tiny-tim

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    Oh my goodness! Does it work? :smile:

    I didn't actually try it! :rolleyes:
     
  10. May 3, 2008 #9
    ......

    oh.......I cant solve it.!!
    can u send me the answer??I need it quickly and there is no time to solve it because I have final exams


    Thankx very much
     
  11. May 3, 2008 #10

    tiny-tim

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    Hi vip89! :smile:

    It's because you have final exams that you need to practise the technique …in case it comes up!

    (and if it doesn't come up … why are you bothered? :rolleyes:)

    So … differentiating with respect to x … what do you get? :smile:
     
  12. May 3, 2008 #11
    I have acomption with my friend ,who find the answer first
     
  13. May 3, 2008 #12

    Redbelly98

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    I am going to stay with the topic as given in the subject line here.

    Why is the competition with your friend important?
     
  14. May 4, 2008 #13
    ok , it is not important
    can u solve it ,pls??
     
  15. May 4, 2008 #14

    HallsofIvy

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    Wouldn't that be cheating?
     
  16. May 4, 2008 #15

    Redbelly98

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    Actually no, I can't. But I do have some advice for you.

    In this forum, you'll pretty much never get other people to solve entire problems for you. People like to see some attempt that you tried to solve it first. So far you haven't done that.

    Did you try the suggestion from tiny-tim in message #6 ? If not, then try it. If you did try it then explain how far you got with it, or why you were unable to get very far with it, or just what you don't understand.

    And lastly, welcome to physicsforums.com :smile:
     
  17. May 6, 2008 #16
    ok
    I will try
    and I want to say that is not cheating because It is not homework
     
  18. May 9, 2008 #17
    That is wt I did:
     

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