“A model should be as simple as possible but no simpler…”
Albert Einstein
Define the value of the population at time \(t=0\)
\[\begin{eqnarray} N(t = 0) &=& N_0 \\ &=& 7\ 000\ 000\ 000 \end{eqnarray}\]
Incrementing the human population in 1 year time steps
\[N(t + 1) = \lambda \times N(t) + N(t)\]
where \(\lambda = 0.015\)
Incrementing the human population in 1 month time steps
\[N(t + 1) = (\lambda / 12) \times N(t) + N(t)\]
where \(\lambda = 0.015\)
Incrementing the human population in 1 year time steps
\[N(t + 1) = \lambda \times N(t) + N(t)\]
where \(\lambda = 0.015\)
\[N(t + 1) = \lambda \times N(t) + N(t)\]
\[ \mathit{i.e.\quad} N(t + 1) = (\lambda + 1) \times N(t)\]
\[\begin{eqnarray} N(t + 2) &=& (\lambda + 1) \times N(t + 1) \\ &=& (\lambda + 1)\times(\lambda + 1) \times N(t) \\ &=& (\lambda + 1)^2 \times N(t) \end{eqnarray}\]
\[N(t + n) = (\lambda + 1)^n \times N(t)\]
So, the number of years (\(n\)) taken to double is the \(n\) that satisfies
\[(\lambda + 1)^n = 2\]
\[\begin{eqnarray} (\lambda + 1)^n &=& 2 \\ \log((\lambda + 1)^n) &=& \log(2) \\ n \times \log(\lambda + 1) &=& \log(2) \\ n &=& \log(2) / \log(\lambda + 1) \\ n &=& 46.6 \end{eqnarray}\]