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Math Help - alternating series remainder estimate

  1. #1
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    alternating series remainder estimate

    alternating series remainder estimate-imag0975.jpg

    Number 46

    I found the first 7 terms to be .78343057
    Since I want to 7 places I added the value when n=8 (5.96x10^-8) to my initial value
    So .78343057<f(x)<.78343051

    So I claim that the sum of the series should be .7834305

    My teacher told me this is wrong but could not explain why. His value was .783431

    Where is my mistake?

    Listing from phone so hard to type out step by step at the moment.
    Last edited by Bowlbase; May 9th 2012 at 04:27 PM. Reason: incorrect value added. should be 5.96x10^-8
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  2. #2
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    Re: alternating series remainder estimate

    You have rounded incorrectly. you have .78343057. Because 7 is closer to 10 than it is to 0, that is closer to .783431 than to .783430.
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  3. #3
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    Re: alternating series remainder estimate

    That doesn't make a lot of sense to me. I used my calculator to do the summation to n=7. I left out the other digits past 7 because that is the point where the two values differ. Even if it is a rounding error at 7, wouldn't it mean that instead of ...3057 I would have ...3060?

    And besides, I thought the point of the exercise is to find the accuracy of the value to a certain decimal number. Why would I round it in the end? If he was just rounding I can see the mistake but it seems counter-productive if that is the case.
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  4. #4
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    Re: alternating series remainder estimate

    |S - S_7| < \frac{1}{8^8}

    -\frac{1}{8^8} < S - S_7 < \frac{1}{8^8}

    S_7 - \frac{1}{8^8} <  S < S_7 + \frac{1}{8^8}

    0.7834305082 < S < 0.7834306274
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