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Math Help - Why does this function act like this for large z?

  1. #1
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    Why does this function act like this for large z?

    I have been told that, for large z,

    f(z) = [(z - 1) / (z + 1)]^(1/6)

    = [(1-(1/z)) / (1 + (1/z))] by dividing through by z

    = 1 - (1/3z) + (1/(18z^2)) + O(z^-3)

    so write

    f(z) = 1 - (1/3z) + g(z) where g(z) = O(z^-2) as z --> infinity

    I have no idea how why the 3rd and 4th lines are as they are. What is the working behind them?

    thanks
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  2. #2
    Senior Member Shanks's Avatar
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    Do you know the Talor fomulla in Caculas?
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  3. #3
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    Quote Originally Posted by Shanks View Post
    Do you know the Talor fomulla in Caculas?
    I can't see how to use it in this case
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  4. #4
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    He is suggesting you expand \frac{1- u}{1+ u} in a Taylor's series about u= 0.
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