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Evaluating, after obtaining a bionomial expansion equation watch

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    Find the first four terms of (1-\frac{1}{x})^\frac{1}{2}

    By using Bionomial expansion I got 1-\frac{1}{2x}-\frac{1}{8x^2}-\frac{1}{16x^3} - this is correct

    Part b. By substituting x=100 find a value for \sqrt{99}
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    (Original post by Dexter212)
    Example: when x=100 how could I find a value of \sqrt{99}

    The equation I have is :

    1-\frac{1}{2x}-\frac{1}{8x^2}-\frac{1}{16x^3}
    Please post the whole question.
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    (Original post by Notnek)
    Please post the whole question.
    There.
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    (Original post by Dexter212)
    Find the first four terms of (1-\frac{1}{x})^\frac{1}{2}

    By using Bionomial expansion I got 1-\frac{1}{2x}-\frac{1}{8x^2}-\frac{1}{16x^3} - this is correct

    Part b. By substituting x=100 find a value for \sqrt{99}
    If you plug 100 into (1-\frac{1}{x})^\frac{1}{2} then that gives you \sqrt{0.99}. So plugging 100 into

    1-\frac{1}{2x}-\frac{1}{8x^2}-\frac{1}{16x^3}

    will give you an approximation for \sqrt{0.99}.

    But you actually require an approximation for \sqrt{99}. What can you do to an approximation to \sqrt{0.99} to turn it into an approximation to \sqrt{99} ?
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    (Original post by Notnek)
    If you plug 100 into (1-\frac{1}{x})^\frac{1}{2} then that gives you \sqrt{0.99}. So plugging 100 into

    1-\frac{1}{2x}-\frac{1}{8x^2}-\frac{1}{16x^3}

    will give you an approximation for \sqrt{0.99}.

    But you actually require an approximation for \sqrt{99}. What can you do to an approximation to \sqrt{0.99} to turn it into an approximation to \sqrt{99} ?
    Multiply by 10?
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    (Original post by Dexter212)
    Multiply by 10?
    Correct
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    (Original post by Notnek)
    Correct
    Thank you.
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    (Original post by Notnek)
    Correct
    when asked to find the validity of (2+x)^-1 I know this is the same as
    2^{-1}(1+\frac{x}{2})^{-1}

    there fore I thought the valitidy would be

    -2<x<2

    but it is |x|<2 why?
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    (Original post by Dexter212)
    when asked to find the validity of (2+x)^-1 I know this is the same as
    2^{-1}(1+\frac{x}{2})^{-1}

    there fore I thought the valitidy would be

    -2<x<2

    but it is |x|<2 why?
    |x| < 2 is the same as -2<x<2. Think about this if you're unsure and let us know if you need an explanation.
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    (Original post by Notnek)
    |x| < 2 is the same as -2<x<2. Think about this if you're unsure and let us know if you need an explanation.
    Im going to need an Explanation.
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    (Original post by Dexter212)
    Im going to need an Explanation.
    Well what does |x| mean? Are you happy that e.g. |-3| = 3 which is >2. Think about -2<x<2 and check that any number in this interval will satisfy |x|<2.
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    (Original post by Notnek)
    Well what does |x| mean? Are you happy that e.g. |-3| = 3 which is >2. Think about -2<x<2 and check that any number in this interval will satisfy |x|<2.
    Oh SNAP!

    damn. thank you very much





 
 
 
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