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Math Help - Approximation of 1/(2+x)

  1. #1
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    Approximation of 1/(2+x)

    I've just learned that 4) \frac{1}{1-x}=1+x+x^2+...+x^n where |x|<1. I am looking at an approximation of:

    \frac{1}{2+x} which makes use of a technique called scaling and also applies 4) to reach an approximation. The whole solution looks like:

    \frac{1}{2+x}=\frac{1/2}{1+x/2} \approx \frac{1}{2}(1-\frac{x}{2})

    The denominator of the equation goes from 2+x to 1+x/2?

    It is said that the approximation is developed by scaling and using 4). I am trying to determine what happened to the original equation to reach the approximation.

    Thanks for any responses...
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  2. #2
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    Re: Approximation of 1/(2+x)

    Quote Originally Posted by sepoto View Post
    I've just learned that 4) \frac{1}{1-x}=1+x+x^2+...+x^n where |x|<1. I am looking at an approximation of:

    \frac{1}{2+x} which makes use of a technique called scaling and also applies 4) to reach an approximation. The whole solution looks like:

    \frac{1}{2+x}=\frac{1/2}{1+x/2} \approx \frac{1}{2}(1-\frac{x}{2})

    The denominator of the equation goes from 2+x to 1+x/2?

    It is said that the approximation is developed by scaling and using 4). I am trying to determine what happened to the original equation to reach the approximation.

    Thanks for any responses...
    \frac{1}{2 \left(\frac{x}{2}+1\right)} = \frac{1}{2}\left(\frac{1}{1--\frac{x}{2}}\right)

    and now they take the first couple of terms of the series representation

    = \frac{1}{2} \left(1-\frac{x}{2}\right)
    Thanks from sepoto
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