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Math Help - Compare A ,B

  1. #1
    Super Member dhiab's Avatar
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    Compare A ,B

    Hello :

    Compare A ,B:

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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    To potential helpers : please read this post : "http://www.mathhelpforum.com/math-help/number-theory/93501-facto-sum.html "
    before potentially losing your time answering this guy's questions.
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  3. #3
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    Hello, dhiab!

    Compare A\text{ and }B\!:\quad\begin{array}{ccc}<br />
A &=& \dfrac{1.0000004}{(1.0000006)^2} \\ \\[-3mm] B &=& \dfrac{(0.9999995)^2}{0.9999998} \end{array}

    \text{Let }d \:=\:\frac{a}{10^7},\:\text{where }a\text{ is a digit } \leq 6


    Then: . \begin{array}{ccc}A &\approx & \dfrac{1+d}{(1+d)^2} \\ \\ [-3mm]B & \approx& \dfrac{(1-d)^2}{1-d} \end{array}


    We have: . \overbrace{\frac{1+d}{(1+d)^2}}^{A} \quad^{>}_{<}\quad \overbrace{\frac{(1-d)^2}{1-d}}^{B}

    Then: . (1+d)(1-d)\quad ^{>}_{<}\quad (1-d)^2(1+d)^2

    . . . . . . . . . . 1 - d^2 \quad ^{>}_{<}\quad (1-d^2)^2

    . . . . . . . . . . 1 - d^2 \quad ^{>}_{<}\quad 1 - 2d^2 + d^4

    . . . . . . . . . . . . d^2\quad ^{>}_{<}\quad d^4


    Divide by positive d^2\!:\quad 1 \quad ^{>}_{<}\quad d^2 .[1]


    Since d < 1, then d^2 < 1

    Hence, [1] becomes: .  1 \;\;{\color{red}>}\;\;d^2


    Therefore: . A \;{\color{red}>}\;B

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  4. #4
    Super Member dhiab's Avatar
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    Quote Originally Posted by Bruno J. View Post
    To potential helpers : please read this post : "http://www.mathhelpforum.com/math-help/number-theory/93501-facto-sum.html "
    before potentially losing your time answering this guy's questions.
    Hello : I to aprecier your opinion, I see not or ' is the problem?,that is a olymiad question
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