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I am wondering if there is a name and way to derive the following approximation:

$$\frac{1+x}{1+y} - 1 \approx x-y$$

I'm essentially interested in how to refer to this.

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    $\begingroup$ What's wrong with calling it a "Taylor approximation"? $\endgroup$ – kimchi lover Sep 29 '18 at 17:55
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I would call this a Taylor approximation. When $|y|\lt1$, $$ \begin{align} \frac{1+x}{1+y}-1 &=-1+(1+x)\left(1-y+y^2-y^3+\dots\right)\\ &=x-y-xy+y^2+xy^2-y^3-xy^3+\dots\\[6pt] &=x-y+O\!\left(\max(|x|,|y|)^2\right) \end{align} $$

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  • $\begingroup$ i think there is a typo in the denominator of the LHS - should be 1+y, not 1-y? $\endgroup$ – laszlopanaflex Sep 29 '18 at 19:51
  • $\begingroup$ Fixed it; thanks! $\endgroup$ – robjohn Sep 30 '18 at 0:43
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$$\frac{(1+x)}{(1+y)} - 1 = \frac{(1+x-1-y)}{(1+y)} =\frac{(x-y)}{(1+y)} \approx x-y$$

The approximation is valid when the value of $y$ can be neglected with respect to $1$

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