Approximating $\ln{\frac{x-y}{x+y}}$

Question

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?

Answer 1

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

$$ \frac{x-y}{x+y} = \frac{x-y}{x}\left(1-\frac{y}{x} + \ldots\right) \approx 1 -2\frac{y}{x}$$

Answer 2

Since you admit that $x>>y$, let is define $z=\frac{y}{x}$ and rewrite
$$\log \left(\frac{x-y}{x+y}\right)=\log \left(\frac{1-z}{1+z}\right)$$ and use Taylor expansion for both numerator and denominator. So, as the beginning of the expansion, you have $$\log \left(\frac{1-z}{1+z}\right)=-2 z-\frac{2 z^3}{3}-\frac{2 z^5}{5}-\frac{2 z^7}{7}-\frac{2 z^9}{9}+O\left(z^{10}\right)$$ which allows pretty good estimations even if $x$ is not very large when compared to $y$.

For illustration purposes, let me take $x=2y$, that is to say $z=\frac{1}{2}$. Limited to the only terms given above, the result is $-1.09850$ while the exact value is $-1.09861$. Limited to only the first two terms, the approxiamtion gives $-1.08333$.

Answer 3

When you approximate $x+y$ by $x$, the value changed by a relative amount $y/x$. That looked negligible, but since the final answer is also that size, the small changed halved the answer.

Answer 4

$${x-y\over x+y}={x+y-2y\over x+y}=1-{2y\over x+y}\approx 1-{2y\over x}$$