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The first equation from Barton 1979 (a paper in population genetics) is

$$\Delta p = \left(\frac{p^2+pq(1-s)}{1-2pqs}\right)-p$$

, where $1>s>0$, $1>p>0$ and $p+q=1$. The first expression on the RHS can be understood as the state of $p$ at the following time step. He directly indicates that the equivalent equation in a continuous time model is

$$\frac{dp}{dt}≈spq(p-q)$$

Can you help me to make this "conversion" from discrete time to continuous time?

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Starting out with the discrete form:

$\frac{\Delta p}{\Delta t}=\frac{p^2+pq(1-s)}{1-2pqs}-p$

bring the solitary $p$ back into the fraction:

$\frac{\Delta p}{\Delta t}=\frac{p^2+pq(1-s)-p(1-2pqs)}{1-2pqs}$

Multiplying this out gives:

$\frac{\Delta p}{\Delta t}=\frac{p^2+pq-pqs-p+2p^2qs}{1-2pqs}$

make the substitution $q = (1-p)$. Then,

$\frac{\Delta p}{\Delta t}=\frac{p^2+p(1-p)-p(1-p)s-p+2p^2(1-p)s}{1-2p(1-p)s}$

Multiplying out gives:

$\frac{\Delta p}{\Delta t}=\frac{p^2+p-p^2-ps+p^2s-p+2p^2s-2p^3s}{1-2ps+2p^2s}=\frac{3p^2s-ps-2p^3s}{1-2ps+2p^2s}$

We will call this last quantity $LHS$

Now to find out how to get this quantity from the final result:

$\frac{dp}{dt}=pqs(p-q)$

Multiply this out:

$\frac{dp}{dt}=p^2qs-pq^2s$

Substitute $q=(1-p)$:

$\frac{dp}{dt}=p^2(1-p)s-p(1-p)^2s=p^2(1-p)s-p(1-2p+p^2)s$,

$\frac{dp}{dt}=p^2s-p^3s-ps+2p^2s-p^3s$

Collecting terms:

$\frac{dp}{dt}=3p^2s-ps-2p^3s=RHS$

Now compare $LHS$ and $RHS$ - they appear identical in 'numerator', but what has happened to the divisor $1-2ps+2p^2s$ ? I can only assume that they have decided to neglect the terms other than $1$ in the denominator, since they have stated $s<<1$ earlier in the paper.

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