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The formula for the next generation $p$ is as following:

$$p_{n}=\frac{p_{n-1}(1-p_{n-1})}{1-{p_{n-1}}^2}$$

  • $p_n$ is the percentage of the new allele in the next generation
  • Also $0 <p_{n-1} < 1$

I was wondering if there is just a standard formula for $p_n$, because I was asked to calculate the percentage of the $50$th generation. My only conclusion is that I have to enter the value 50 times into the formula, but I don't know if there is a more efficient way to calculate this.

Adnan
  • 1,021

2 Answers2

1

Assuming $p_{n-1}<1$, a factor of $1-p_{n-1}$ can be pulled out of both the numerator and denominator to get

$$p_n=\frac{p_{n-1}}{1+p_{n-1}} \tag{1}$$

Defining a reciprocal series $q_n=\frac{1}{p_n}$ we get by inverting (1)

$$q_n=q_{n-1}+1 \tag{2}$$

Hence for $k\ge1$ we have by repeated application of (2):

$$q_n=q_{n-k}+k \tag{3}$$

so, for example, $q_{50}=q_0+50$ whence $p_{50}=\left(\frac{1}{p_0}+50\right)^{-1}$.

Marconius
  • 5,635
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Let $p_n = \frac{ p_{n-1}}{1+ mp_{n-1}}$.
Then for $m=1$ we have the correct formula. However, this gives us \begin{eqnarray} p_{n+1} &=& \frac{ p_{n}}{1+ m p_n} \\ &=& \frac{ \frac{p_{n-1}}{1+m p_{n-1}}}{1+ \frac{p_{n-1}}{1+ m p_{n-1}}}\\ &=&\frac{ \frac{p_{n-1}}{1+m p_{n-1}}}{\frac{1 + mp_{n-1}+p_{n-1}}{1+ m p_{n-1}}}\\ &=& \frac{ p_{n-1}}{ 1 + (m+1)p_{n-1}} \end{eqnarray} So if we start with $p_2 = \frac{p_1}{1+ p_1}$ we find $p_{50} = \frac{ p_1}{1+49p_1}$.

Hetebrij
  • 3,981