As we’ll see in the next section, a differential equation looks like this: [late
As we’ll see in the next section, a differential equation looks like this: [latex]\frac{dP}{dt} = 0.03 \cdot P[/latex]. What I want to first talk about though are recurrence relations. Let me introduce these with a magic trick.
Pick a number between 1 and 100, and I’m going to guess it. But not before we mix it up a bit.
- Take your number and divide by five, and round to the nearest whole number.
- Then add [latex]36[/latex] to the result.
- Repeat steps (1) and (2) twice more, for a total of three iterations.
Done? I bet you ended with the number 45. Are you amazed?
This trick is based off the recurrence relation [latex]f_{t+1} = \frac{1}{5} f_t + 36[/latex]. Think of [latex]f_t[/latex] as the previous value, and [latex]f_{t+1}[/latex] as the new value. How do you get from one to another? Well, ignoring the rounding, you divide by [latex]5[/latex] and add [latex]36[/latex], and that’s exactly what [latex]f_{t+1} = \frac{1}{5} f_t + 36[/latex] is telling you to do. In order to use such a equation, we need an initial value or [latex]f_0[/latex]. In the trick, this was the original number you picked. Let’s create a graph with the initial value [latex]f_0 = 100[/latex]:
As you can see, this recurrence relation quickly converges to [latex]f_t = 45[/latex] by the time [latex]t = 3[/latex]. That’s why the trick works! If we started somewhere else, the graph looks much the same and it converges to 45 anyway.
However, recurrence relations are useful for more than just magic tricks.
Well, since this is a recurrence relation, we want to relate the quantity under consideration, [latex]h_t[/latex] to its value the next year, which is [latex]h_{t+1}[/latex]. So it will look something like
[latex]h_{t+1} = 1.5 h_t + 16[/latex]
but those aren’t the right values yet — just want to have some idea of where this is going.
The first thing we need to encode is the expansion by [latex]1\%[/latex]. We can take [latex]1\%[/latex], or [latex]0.01[/latex] and multiply by [latex]h_t[/latex] like so: [latex]0.01 h_t[/latex]. But the old forest is still there (except for the logging, which we’ll worry about in a second), so let’s add [latex]h_t[/latex] as well: [latex]0.01 h_t + h_t[/latex]. If we factor out [latex]h_t[/latex], we get \begin{align*}
0.01 h_t + h_t & = h_t(0.01 + 1) \\
& = h_t(1.01) \\
& = 1.01 h_t
\end{align*}
This is the growth by [latex]1\%[/latex]. What about that logging? Well, that’s not a percent change, so we’ll just subtract the [latex]2000[/latex] to represent the loss of biomass. So our final recurrence relation is–>
[latex]h_{t+1} = 1.01 h_t - 2000[/latex]
Well, let’s play with it a bit and see what happens. But before we can do that, we need an initial value [latex]h_0[/latex]. Let’s guess something. Since we are losing [latex]2000[/latex] a year, we’ll need a much bigger number than [latex]2000[/latex]. Let’s just guess that [latex]h_0[/latex] is [latex]50,\!000[/latex] metric tonnes.
Now we can compute several [latex]h_t[/latex] values:
\begin