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Thread: Problem 41

  1. #1
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    Problem 41

    1)Let $\displaystyle f(x)$ be a continous function on $\displaystyle \mathbb{R}$ solve the functional equation: $\displaystyle f(x+y)=f(x)f(y)$.
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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 $\displaystyle f(x)$ be a continous function on $\displaystyle \mathbb{R}$ solve the functional equation: $\displaystyle f(x+y)=f(x)f(y)$.
    $\displaystyle f(x)=c^{x}$ for some constant c.

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

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

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

    and
    $\displaystyle f(x)f(y)=c^{x}c^{y}=c^{x+y}$
    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 $\displaystyle n$ be a positive integer. Then
    $\displaystyle f(1)=f\left(\frac{1}{n}+\ldots+\frac{1}{n}\right)= f\left(\frac{1}{n}\right)^n$
    So
    $\displaystyle f\left(\frac{1}{n}\right)=f(1)^{\frac{1}{n}}$
    Now let $\displaystyle \frac{m}{n}$ be a positive rational number. Then
    $\displaystyle f\left(\frac{m}{n}\right)=f\left(\frac{1}{n}+\ldot s+\frac{1}{n}\right)=f\left(\frac{1}{n}\right)^m=f (1)^{\frac{m}{n}}$
    Now let $\displaystyle x$ be a positive real number. By continuity,
    $\displaystyle f(x)=\lim_{z\rightarrow x}f(z)=f(1)^x$
    where the last result is established by our definition of $\displaystyle f$ over the rational numbers.
    Assume $\displaystyle f(1)\neq0$, then it is trivial that $\displaystyle f(0)=1$. Thus, for any positive real number $\displaystyle y$,
    $\displaystyle f(-y)=\frac{f(-y)f(y)}{f(y)}=\frac{f(0)}{f(y)}=f(1)^{-y}$
    Therefore, $\displaystyle f(x)=f(1)^x$ for all $\displaystyle x$ or assume $\displaystyle f(1)=0$, then it is trivial that $\displaystyle f(x)=0$ for all $\displaystyle x$
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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,
    $\displaystyle f(1)=f\underbrace{\left(\frac{1}{n}+\ldots+\frac{1 }{n}\right)}_{n \textnormal{\small{ times}}}=\underbrace{f\left(\frac{1}{n}\right)\ldo ts f\left(\frac{1}{n}\right)}_{n\textnormal{\small{ times}}}=f\left(\frac{1}{n}\right)^n$
    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,
    $\displaystyle f(1)=f\underbrace{\left(\frac{1}{n}+\ldots+\frac{1 }{n}\right)}_{n \textnormal{\small{ times}}}=\underbrace{f\left(\frac{1}{n}\right)\ldo ts f\left(\frac{1}{n}\right)}_{n\textnormal{\small{ times}}}=f\left(\frac{1}{n}\right)^n$
    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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