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Thread: Help with 2 Other Questions ??

  1. #1
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    Help with 2 Other Questions ??

    i If someone sends an answer
    please write that in mathematical form

    i have problems with my english
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  2. #2
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    For any $\displaystyle n>0$

    Consider the vector space $\displaystyle \mathbb{R}^n$ over $\displaystyle \mathbb{R}$.

    Now, let
    $\displaystyle \bold{u}=\left<1,\frac{1}{2},\frac{1}{3},...,\frac {1}{n}\right>$

    $\displaystyle \bold{v}=\left<1,\frac{1}{2^3},\frac{1}{3^3},...,\ frac{1}{n^3}\right>$

    Then by Cauchy-Schwartz Inequality we have,
    $\displaystyle \bold{ab}\leq ||\bold{a}||||\bold{b}||$

    Now, (the dot product)
    $\displaystyle \bold{ab}=1+\frac{1}{2^4}+...\frac{1}{n^4}$

    And the norms,
    $\displaystyle ||\bold{a}||=\sqrt{1+\frac{1}{2^2}+...+\frac{1}{n^ 2}}$
    $\displaystyle ||\bold{b}||=\sqrt{1+\frac{1}{2^6}+...+\frac{1}{n^ 6}}$
    Thus,

    $\displaystyle 1+\frac{1}{2^4}+...+\frac{1}{n^4}\leq \sqrt{1+\frac{1}{2^2}+...+\frac{1}{n^6}}\cdot \sqrt{1+\frac{1}{2^6}+...+\frac{1}{n^6}}$

    Square both sides (non-negative terms),
    $\displaystyle \left(1+\frac{1}{2^4}+...+\frac{1}{n^4}\right)^2\l eq$$\displaystyle \left(1+\frac{1}{2^2}+...+\frac{1}{n^2}\right)\cdo t$$\displaystyle \left(1+\frac{1}{2^6}+...+\frac{1}{n^6}\right)$
    Last edited by ThePerfectHacker; Nov 2nd 2006 at 07:28 AM.
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  3. #3
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    Let,
    $\displaystyle x\geq 1$
    Then,
    $\displaystyle \frac{1}{x^n}-\frac{1}{x^{n+1}}=\frac{x^{n+1}-x^n}{x^{2n+1}}\leq x^{n+1}-x^n$
    Thus,
    $\displaystyle x^{n+1}-x^n\geq \frac{1}{x^n}-\frac{1}{x^{n+1}}$
    Thus,
    $\displaystyle x^{n+1}+\frac{1}{x^{n+1}}\geq x^n+\frac{1}{x^n}$

    Now, if $\displaystyle 0<x<1$
    Then,
    $\displaystyle \frac{1}{x}>1$
    Thus, (by the previous discussion),
    $\displaystyle (1/x)^{n+1}+\frac{1}{(1/x)^{n+1}}\geq (1/x)^n+\frac{1}{(1/x)^n}$
    Thus,
    $\displaystyle x^{n+1}+\frac{1}{x^{n+1}}\geq x^n+\frac{1}{x^n}$
    Last edited by ThePerfectHacker; Nov 2nd 2006 at 07:30 AM.
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