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Is it possible to solve this without guessing the function?

  1. Jan 19, 2013 #1
    We know f(f(x))=2^x-1;
    We must calculate f(1)+f(0)
    Its posible to solve this without guessing the function??? If i derivate i get to f(0)=f(0) or f(1)=f(1).
     
  2. jcsd
  3. Jan 19, 2013 #2

    vela

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    Re: Its posible to solve this without guessing the function

    What does derivate mean?

    Yes, it's possible to solve without knowing what f(x) is.

    Hint: Consider f(f(f(x))) to show that ##f(2^x-1) = 2^{f(x)}-1##.
     
  4. Jan 19, 2013 #3
    Isnt f(2x-1)=2f(f(x))-1??? And in this case it becamse very easy.
    The hint leads me to f(0)=f(0)f(0)-1 , so f(0)=0 or 1.
    Same result for f(1).
    I guess the function is injective so the sum is 1???
    Actually i solved the function but with diferential equation and obviously i dont know the constant :D.
    Is this correct???
    f(f(f(x)))=2^(f(f(x)))-1
    f(f(f(x)))=f(2^x-1)=2^(f(f(x)))-1=f(2^x-1)=2^(2^x-1)-1
    x=0=>f(2^0-1)=2^(2^0-1) -1<=>f(0)=0;
    x=1=>f(2^1-1)=2^(2^1-1)-1<=>f(1)=1;
    f(1)+f(0)=1;
     
    Last edited: Jan 19, 2013
  5. Jan 19, 2013 #4

    vela

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    How'd you get that?
     
  6. Jan 19, 2013 #5
    No, sorry you are right. I messed up the argument of the function :D
    Thanks, i understood .
     
    Last edited: Jan 19, 2013
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