Approximation of $\frac{1+a}{1+b}$

Question

I've found the following assertion on an economics book:

For $r$ and $g$ small enough, $\frac{1+r}{1+g}\approx 1+r-g$

(where $r$ is the interest rate and $g$ is the growth rate of the economy)

I would like to know why this is true. I've tried to find the solution by myself, but I really don't know where to start from. What kind of approximation is this?

Answer 1

This is just a Taylor expansion of the function $f(r,g) = \frac{1+r}{1+g}$ at the point $(r,g) = (0,0)$.

The partial derivatives of $f$ are $$\frac{\partial f}{\partial r} = \frac{1}{1+g}$$ and $$\frac{\partial f}{\partial g} = -\frac{1+r}{(1+g)^2}.$$

Hence, the Taylor expansion of first order at $(0,0)$ is $1 + r - g$.

Answer 2

Provided $|b| \ll 1$ we have $$\dfrac{1+a}{1+b} = (1+a)(1+b)^{-1} = (1+a) \sum_{n=0}^{\infty} (-1)^nb^n$$ $$\approx (1+a)(1-b) = 1+a-b-ab \approx 1+a-b$$

Answer 3

Let $f(r,g) = { 1+r \over 1+g }$, then ${\partial f(0,0) \over \partial r} = 1$, ${\partial f(0,0) \over \partial g} = -1$.

Hence for small $r,g$ we have $f(r,g) \approx f0,0) +{\partial f(0,0) \over \partial r} r + {\partial f(0,0) \over \partial g} g = f(0,0)+r - g $.

Answer 4

$$\frac{1+a}{1+b}=(1+a)(1+b)^{-1}$$ as $x\to 0$, $(1+x)^n\approx 1+nx$ $$\to (1+a)(1-b) \approx 1-ab+a-b$$ further $ab$ is negligible as it is very small. $$\frac{1+a}{1+b}\approx 1+a-b$$