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Math Help - Problem 41

  1. #1
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    Problem 41

    1)Let f(x) be a continous function on \mathbb{R} solve the functional equation: f(x+y)=f(x)f(y).
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  2. #2
    Super Member angel.white's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    1)Let f(x) be a continous function on \mathbb{R} solve the functional equation: f(x+y)=f(x)f(y).
    f(x)=c^{x} for some constant c.

    then
    f(x+y)=c^{x+y}

    and
    f(x)f(y)=c^{x}c^{y}=c^{x+y}
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  3. #3
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    Quote Originally Posted by angel.white View Post
    f(x)=c^{x} for some constant c.

    then
    f(x+y)=c^{x+y}

    and
    f(x)f(y)=c^{x}c^{y}=c^{x+y}
    How do you know there are no other solutions?
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  4. #4
    Super Member angel.white's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    How do you know there are no other solutions?
    I suppose It's not.

    I'll add another function while I'm at it, though

    f(x)=0

    then f(x+y)=0

    and f(x)f(y)=0*0 = 0

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  5. #5
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    Let n be a positive integer. Then
    f(1)=f\left(\frac{1}{n}+\ldots+\frac{1}{n}\right)=  f\left(\frac{1}{n}\right)^n
    So
    f\left(\frac{1}{n}\right)=f(1)^{\frac{1}{n}}
    Now let \frac{m}{n} be a positive rational number. Then
    f\left(\frac{m}{n}\right)=f\left(\frac{1}{n}+\ldot  s+\frac{1}{n}\right)=f\left(\frac{1}{n}\right)^m=f  (1)^{\frac{m}{n}}
    Now let x be a positive real number. By continuity,
    f(x)=\lim_{z\rightarrow x}f(z)=f(1)^x
    where the last result is established by our definition of f over the rational numbers.
    Assume f(1)\neq0, then it is trivial that f(0)=1. Thus, for any positive real number y,
    f(-y)=\frac{f(-y)f(y)}{f(y)}=\frac{f(0)}{f(y)}=f(1)^{-y}
    Therefore, f(x)=f(1)^x for all x or assume f(1)=0, then it is trivial that f(x)=0 for all x
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  6. #6
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    Quote Originally Posted by topsquark View Post
    The notation "f(x)" does not imply that it is f times x, as you appear to be using.

    -Dan
    ? I don't understand. How did you think that?
    Maybe this will clarify,
    f(1)=f\underbrace{\left(\frac{1}{n}+\ldots+\frac{1  }{n}\right)}_{n \textnormal{\small{ times}}}=\underbrace{f\left(\frac{1}{n}\right)\ldo  ts f\left(\frac{1}{n}\right)}_{n\textnormal{\small{ times}}}=f\left(\frac{1}{n}\right)^n
    The second equality is straight from the problem.
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by math sucks View Post
    ? I don't understand. How did you think that?
    Maybe this will clarify,
    f(1)=f\underbrace{\left(\frac{1}{n}+\ldots+\frac{1  }{n}\right)}_{n \textnormal{\small{ times}}}=\underbrace{f\left(\frac{1}{n}\right)\ldo  ts f\left(\frac{1}{n}\right)}_{n\textnormal{\small{ times}}}=f\left(\frac{1}{n}\right)^n
    The second equality is straight from the problem.
    Sorry. Now I see what you are doing. My bad!

    -Dan
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  8. #8
    Super Member Rebesques's Avatar
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    [cheater] What of discontinuous f?

    [/cheater]
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