# Understanding a derivative in a biology article

From this biology article (5th page, right column)

They have...

$$1-P_t ≈ 1-(1+s)P_{t+1}+\frac{p_{t+1}^2}{2}$$

and they conclude that...

$$\frac{dP}{dt} ≈ -sP + \frac{P^2}{2}$$

I don't really understand this link. How did they do that?

• What page? ${}{}$ – apnorton Feb 8 '14 at 20:08
• @anorton see update at the first line of my post. – Remi.b Feb 8 '14 at 20:10
• $P_{t+1}-P_t = -sP_{t+1}+P^2_{t+1}/2$. The left part is "something like" derivative (see numerical approximations of derivative). – TZakrevskiy Feb 8 '14 at 20:12

I think that $1$ is small when compared to $t$, so they read $P_{t+1}$ as $P_t + dP/dt$. The first equation then becomes $$1 - P \approx 1 - (1 + s)(P + dP/dt) + \tfrac{1}{2}(P + dP/dt)^2.$$ Solving this for $dP/dt$ and ignoring the term $(dP/dt)^2$ gives their result.
• Actually, the approximation $1\ll t$ is quite irrelevant here and there is no convincing reason to "ignore" $(dP/dt)^2$. – Did May 29 '14 at 9:06