$(1-x)^y ≈ e^{-xy}$

Question

Here is an approximation I often see in biology articles but don't really understand:

$$(1-x)^y ≈ e^{-xy}$$

I think this $e^{-xy}$ closely approximates $(1-x)^y$ whenever $x$ is small. Can you help me to understand the conditions for this approximation and why this approximation holds true.

Answer 1

When working with real exponents, it is useful to come back to the definition:

We have $(1-x)^y=e^{y\ln(1-x)}$

but, when $x$ is small, $\ln(1-x)\approx -x$ (since the tangent to the graph $y=\ln(1-x)$ has equation $y=-x$), so for any $y$ and for any $x$ small,

we obtain $y\ln(1-x)\approx -xy$.

Using the continuity of $\exp$, we obtain $(1-x)^y\approx e^{-xy}$.

Answer 2

We have by the Taylor series $$(1-x)^y\sim_01-yx$$ and $$e^{-xy}\sim_0 1-xy$$ so we have the given approximation.