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Math Help - Analysis inequality.

  1. #1
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    Analysis inequality.

    how do we get from here.

    x^2_x  > 2 - \frac{1}{10^n} (2x_n + \frac{1}{10^n})

    to here?

     > 2 - \frac{5}{10^n}

    It's done in one step in my textbook, so I'm assuming it's obvious, but I can't see it.
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  2. #2
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    Re: Analysis inequality.

    Quote Originally Posted by AlyoshaKaz View Post
    how do we get from here.
    x^2_x  > 2 - \frac{1}{10^n} (2x_n + \frac{1}{10^n}) to here?
     > 2 - \frac{5}{10^n}
    It's done in one step in my textbook, so I'm assuming it's obvious, but I can't see it.
    You seem to expect us to be mind readers, in that you have put us in the middle of a proof with no context whatsoever.

    It seems that  2 - \frac{1}{10^n} (2x_n + \frac{1}{10^n})>2 - \frac{5}{10^n}.

    That can only happen if 2x_n<5-\frac{1}{10^n}.

    You should post the complete problem and proof.
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  3. #3
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    Re: Analysis inequality.

    Ok, sorry. It's a proof for the Theorem:

    "For each positive real number a and each integer n > 1, there is a unique real number b such that  b^n = a

    In the special case b^2 = 2

    The proof begins with

    1^2 < 2 < 2^2
    (1.4)^2 < 2 < (1.5)^2
    (1.41)^2 < 2 < (1.42)^2

    etc.

    to give decimal b^2 = (1.414...)^2 = 2

    Then the method uses a table
    b b b^2
    1 1 1
    1.4 1.4 1.96
    1.41 1.41 1.9881
    1/414 1/414 1.999396

    etc.

    to prove so.

    Then to prove the least upper bound of the set of numbesr in the third column of the table is 2, we check that M=2 is an upper bound of E, which follows from the inequalities at the top. Then to show if M' < 2 there is a number in E that is greater then M'.

    We put   x_0 = 1, x_1 = 1.4, x_2 = 1.41, ...

    Then we have

    2 < (x_n + \frac{1}{10^n})^2 = x^2_n + \frac{2x_n}{10^n} + (\frac{1}{10^n})^2

    so

    x^2_n > 2 - \frac{1}{10^n}(2x_n + \frac{1}{10^n})

     > 2 - \frac{5}{10^n}
    Last edited by AlyoshaKaz; March 6th 2013 at 07:21 AM.
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  4. #4
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    Re: Analysis inequality.

    Also, on a side not, I'm not sure how the initial numbers are chosen for the inequalities at the beginning.
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  5. #5
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    Re: Analysis inequality.

    Is bumping allowed?
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