Name and derivation of the approximation $\frac{1+x}{1+y} - 1 \approx x-y$?

514 Views Asked by At

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.

2

There are 2 best solutions below

2
On BEST ANSWER

I would call this a Taylor approximation. When $|y|\lt1$, $$ \begin{align} \frac{1+x}{1+y}-1 &=-1+(1+x)\left(1-y+y^2-y^3+\dots\right)\\ &=x-y-xy+y^2+xy^2-y^3-xy^3+\dots\\[6pt] &=x-y+O\!\left(\max(|x|,|y|)^2\right) \end{align} $$

0
On

$$\frac{(1+x)}{(1+y)} - 1 = \frac{(1+x-1-y)}{(1+y)} =\frac{(x-y)}{(1+y)} \approx x-y$$

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