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Functional Equation

  1. Apr 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Suppose a function satisfies the conditions
    1. f(x+y) = (f(x)+f(y))/(1+f(x)f(y)) for all real x & y
    2. f '(0)=1.
    3. -1<f(x)<1 for all real x
    Show that the function is increasing throughout its domain. Then find the value:
    Limitx -> Infinity f(x)x


    3. The attempt at a solution
    I proceed by putting x,y=0 in eq 1.
    I get the following roots for f(0)={-1,0,1}
    But if I take f(0)={-1,1}, f(x) will become a constant function and will be equal to +1 when f(0)=1 and -1 when f(0)=-1, thereby violating condition 3
    So f(0)=0


    From equation 1: I assume 'y' as a constant and differentiate wrt x
    f ' (x+y)=(f ' (x)(1-f2(y))) / (1+f(x)f(y))2
    I put x=0;
    I get f ' (y)=1-f2(y) Using condition 3; I prove that the derivative is always positive.
    I have been able able to solve the first part of the question. But I couldn't evaluate the limit
    Limitx -> Infinity f(x)x. Please help me on the limit part.
     
    Last edited: Apr 23, 2009
  2. jcsd
  3. Apr 23, 2009 #2
    It may be helpful to "cheat" and use the fact that f(x) is really tanh x, to figure out what to do. Then go back and do it without using that fact.

    First show lim f(y)=1 as y approaches infinity.

    After that, then you find your limit, which has indeterminate form 1^infty, by using natural log and l'Hopital, just like you would do if you knew f was tanh. Unfortunately with f, you don't have all the trig identities at your disposal. Take a stab at it and ask again if you get stuck.
     
  4. Apr 23, 2009 #3

    Dick

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    Homework Helper

    You could take Billy Bob's hint farther and prove that f(x) really is tanh(x). Take f(x+e) (e is epsilon). Put that into your formula for f and rearrange it into a difference quotient and take the limit as e->0. Notice since f'(0)=1, lim f(e)/e ->1. That will give you a differential equation to solve for f.
     
  5. Apr 29, 2009 #4
    Thanks a lot for helping me solve my problem :D
     
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