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    Can somebody help me with the following question:

    An approximate value for 0.8^1/2
    is found by substituting
    x = 0.1 into the first three terms of the binomial expansion for (1-2x)^1/2.The percentage error in this approximate value is:

    a) 0.064%
    b) 6.4%
    c) 0.057%
    d) 1.18%


    In the previous question I expanded (4-x)/(1-2x)^1/2 up to x^2, not sure if this has anything to do with the question. I'm guessing the percntage error comes from the expansion only covering up to x^3 and not infinite, but I have no idea about how to get the answer.

    I'm on AQA, and the questions are from MEI, so that's probably why I haven't been taught it, still would like to know the answer. Cheers.
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    (Original post by FishyTim)
    Can somebody help me with the following question:

    An approximate value for 0.8^1/2
    is found by substituting
    x = 0.1 into the first three terms of the binomial expansion for (1-2x)^1/2.The percentage error in this approximate value is:

    a) 0.064%
    b) 6.4%
    c) 0.057%
    d) 1.18%


    In the previous question I expanded (4-x)/(1-2x)^1/2 up to x^2, not sure if this has anything to do with the question. I'm guessing the percntage error comes from the expansion only covering up to x^3 and not infinite, but I have no idea about how to get the answer.

    I'm on AQA, and the questions are from MEI, so that's probably why I haven't been taught it, still would like to know the answer. Cheers.
    When calculating percentage errors, you're comparing the result of substituting into the bracket against the result attained by substituting into your expansion.

    Percentage error = 100 x (est - original)/(est)
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    (Original post by Indeterminate)
    When calculating percentage errors, you're comparing the result of substituting into the bracket against the result attained by substituting into your expansion.

    Percentage error = 100 x (est - original)/(est)
    Seems so obvious now you say it haha, was racking my brains for ages. Thanks.
 
 
 
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