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Thread: Functions

  1. #1
    Junior Member
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    Functions

    Hi everyone,

    I am asked to determine all functions f continious in $\displaystyle \mathbb{R}$ such as:

    $\displaystyle {f(2009)=2009^{2008}\atop (\forall(x;t)\in \mathbb{R}^2) f(x+t)=f(x)+f(t)}$

    I don't know how to start?

    Can you help me please?

    And thank you anyway.
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let $\displaystyle x=t=0\Rightarrow f(0)=2f(0)\Rightarrow f(0)=0$

    $\displaystyle f(1)=f\left(\underbrace{\frac{1}{n}+\frac{1}{n}+\l dots+\frac{1}{n}}_{n}\right)=nf\left(\frac{1}{n}\r ight)\Rightarrow f\left(\frac{1}{n}\right)=\frac{1}{n}\cdot f(1)$

    $\displaystyle f\left(\frac{m}{n}\right)=\frac{m}{n}\cdot f(1)$

    Then $\displaystyle f(x)=f(1)\cdot x, \ \forall x\in\mathbb{Q}$

    Let $\displaystyle x\in\mathbb{R}-\mathbb{Q}$. Then exists a sequence $\displaystyle (x_n)_{n\in\mathbb{N}}, \ x_n\in\mathbb{Q}$ such as $\displaystyle \lim_{x\to\infty}x_n=x$

    $\displaystyle f(x_n)=f(1)\cdot x_n$

    Apply the limit to both sides:

    $\displaystyle \lim_{n\to\infty}f(x_n)=f(1)\lim_{x\to\infty}x_n\R ightarrow f(x)=f(1)\cdot x, \ \forall x\in\mathbb{R}$

    $\displaystyle f(2009)=2009^{2008}\Rightarrow f(1)\cdot 2009=2009^{2008}\Rightarrow f(1)=2009^{2007}\Rightarrow f(x)=2009^{2007}x$
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  3. #3
    Junior Member
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    Thank you very much
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