Expand \(\sqrt{\left( \frac{1-x}{1+x} \right)}\) in ascending powers of \(x\), up to and including the term in \(x^2\), simplifying the coefficients.
Solution
To expand \(\sqrt{\left( \frac{1-x}{1+x} \right)}\), we can write it as \((1-x)^{\frac{1}{2}}(1+x)^{-\frac{1}{2}}\).
First, expand \((1-x)^{\frac{1}{2}}\) using the binomial series:
\((1-x)^{\frac{1}{2}} = 1 - \frac{1}{2}x + \frac{1}{8}x^2 + \cdots\)
Next, expand \((1+x)^{-\frac{1}{2}}\) using the binomial series:
\((1+x)^{-\frac{1}{2}} = 1 - \frac{1}{2}x + \frac{3}{8}x^2 + \cdots\)
Now, multiply these expansions together and collect terms up to \(x^2\):
\((1-x)^{\frac{1}{2}}(1+x)^{-\frac{1}{2}} = (1 - \frac{1}{2}x + \frac{1}{8}x^2)(1 - \frac{1}{2}x + \frac{3}{8}x^2)\)
\(= 1 - \frac{1}{2}x - \frac{1}{2}x + \frac{1}{4}x^2 + \frac{1}{8}x^2 - \frac{3}{16}x^2 + \cdots\)
\(= 1 - x + \frac{1}{2}x^2\)
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