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Thread: sequence

  1. #1
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    sequence

    Hi everybody

    $\displaystyle f(x)=Arctan\frac{1}{x}; x \in [ \frac{\sqrt{3}}{3};1] $
    $\displaystyle \{{ U_0=\frac{\sqrt{3}}{3} \atop U_{n+1}=f(U_n)}, \forall n \in \mathbb {N}$

    I must show that $\displaystyle (\forall n \in \mathbb{N}): \frac{\sqrt{3}}{3} \le U_n \le 1$ (i used induction but i don't know how to show that: $\displaystyle \frac{\sqrt{3}}{3} \le U_{n+1} \le 1)$

    Can you help me please?
    Last edited by lehder; Nov 11th 2009 at 02:24 PM.
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  2. #2
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    Quote Originally Posted by lehder View Post
    Hi everybody

    $\displaystyle f(x)=Arctan\frac{1}{x}; x \in [ \frac{\sqrt{3}}{3};1] $
    $\displaystyle \{{ U_0=\frac{\sqrt{3}}{3} \atop U_{n+1}=f(U_n)}, \forall n \in \mathbb {N}$

    I must show that $\displaystyle (\forall n \in \mathbb{N}): \frac{\sqrt{3}}{3} \le U_n \le 1$ (i used induction but i don't know how to show that: $\displaystyle \frac{\sqrt{3}}{3} \le U_{n+1} \le 1)$

    Can you help me please?
    No wonder you didn't succeed: it is false, and already from the beginning!:

    $\displaystyle U_1=f(U_0)=f\left(\frac{1}{\sqrt{3}}\right)=Arctan \sqrt{3} =\frac{\pi}{3}>1$...

    Tonio
    Last edited by tonio; Nov 11th 2009 at 05:53 PM.
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