<s>[INST] Approximating $\ln{\frac{x-y}{x+y}}$ For $x>>y$,
$$\ln{\frac{x-y}{x+y}} = \ln{\left[ x\left( 1-y/x \right) \right]} -\ln{\left[ x\left( 1+y/x \right) \right]} \approx -2\frac{y}{x}$$
However, the following does not work:
$$\ln{\frac{x-y}{x+y}} \approx \ln{\frac{x-y}{x}} =  \ln{\left( 1-\frac{y}{x} \right)} \approx -\frac{y}{x}$$
Can anyone explain why the above approximation is not valid?
 [/INST] The correct way to expand $\dfrac{x-y}{x+y}$ is using the binomial theorem.
$$ \frac{1}{x+y} = \frac1x\left(1+\frac{y}{x}\right)^{-1} = \frac1x\left(1-\frac{y}{x} + \ldots\right)$$
Hence
$$ \frac{x-y}{x+y} = \frac{x-y}{x}\left(1-\frac{y}{x} + \ldots\right) \approx 1 -2\frac{y}{x}$$
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