Exponential growth with different time steps

Consider again our example forest with trees:

trees:500→⋅21000→⋅22000→⋅2...→⋅2yyear:2010→+102020→+102030→+10...→+10x\begin{array}{rllll} \text{trees}:& 500 &\xrightarrow[]{\cdot 2} & 1000 &\xrightarrow[]{\cdot 2}& 2000 &\xrightarrow[]{\cdot 2} & ... & \xrightarrow[]{\cdot 2} & y\\ \text{year}:& 2010 & \xrightarrow[]{+10} & 2020 &\xrightarrow[]{+10} & 2030 &\xrightarrow[]{+10} & ... &\xrightarrow[]{+10} & x\\ \end{array}

For the time step τ=10\tau=10 years the growth factor is u=2u=2. What is the growth factor uu for the step size τ=5\tau=5, that is, by what factor increases the number of trees every 55 years? Clearly uu will be smaller than 22, but what is the exact number? To find out, we draw again the diagram, but with step size τ=5\tau=5:

trees:500→⋅u?→⋅u1000year:2010→+52015→+52020\begin{array}{rllll} \text{trees}:& 500 &\xrightarrow[]{\cdot u} & ? &\xrightarrow[]{\cdot u}& 1000\\ \text{year}:& 2010 & \xrightarrow[]{+5} & 2015 &\xrightarrow[]{+5} & 2020\\ \end{array}

We see that

500⋅u⋅u=1000500u2=2∣:500u2=2∣u=2\begin{array}{rll} 500\cdot u\cdot u &=& 1000\\ 500 u^2&=& 2\quad \vert :500\\ u^2 &=& 2\quad \vert \sqrt{}\\ u&=&\sqrt{2} \end{array}

What about a time step of τ=20\tau=20? What is the growth factor then? Again, we can draw the diagram:

trees:500→⋅21000→⋅22000year:2010→+102020→+102030\begin{array}{rr} \text{trees}:& 500 & \xrightarrow[]{\cdot 2} & 1000 & \xrightarrow[]{\cdot 2} & 2000\\ \text{year}:& 2010 & \xrightarrow[]{+10} & 2020 & \xrightarrow[]{+10} & 2030\\ \end{array}

And we see that to get from 20102010 to 20302030 we have to multiply by

u=2⋅2=4u=2\cdot 2=4

Let's generalise. The following is straightforward to see:

Theorem 1: Change of time step size in exponential growth

Consider a quantity that grows exponentially with growth factor uu for every step in time τ\tau:

trees:y0→⋅u...year:x0→+τ...\begin{array}{rlll} \text{trees}:& y_0 &\xrightarrow[]{\cdot u} & ...\\ \text{year}:& x_0 & \xrightarrow[]{+\tau} & ...\\ \end{array}

If we multiply the time step by some factor a>0a>0, so that the new time step is aτa\tau:

trees:y0→⋅v...year:x0→+aτ...\begin{array}{rllll} \text{trees}:& y_0 &\xrightarrow[]{\cdot v} & ...\\ \text{year}:& x_0 & \xrightarrow[]{+a\tau} & ...\\ \end{array}

then the new growth factor vv is given by v=uav=u^a.

Proof

Let's prove this. Assume the quantity at time x0x_0 is y0y_0. For time step τ\tau the growth factor is uu, thus we have

f(x)=y0⋅u(x−x0)/τf(x)=y_0\cdot u^{(x-x_0)/\tau}

Now, what is the growth factor for time step aτa\tau, where aa is a real number? Lets figure out the quantity at time x0+aτx_0+a\tau:

f(x0+τ)=y0⋅u(x0+aτ−x0)/τ=y0⋅ua\begin{array}{lll} f(x_0+\tau)&=&y_0\cdot u^{(x_0+a\tau-x_0)/\tau}\\ &=&y_0\cdot u^a\\ \end{array}

So we see that to get from x0x_0 to x0+aτx_0+a\tau we need to multiply y0y_0 by the factor uau^a.

In the example above we had had a step size τ=10\tau=10 and the growth factor was u=2u=2. We then have seen that for step size 55 (that is, a=0.5a=0.5) the growth factor is 20.5=22^{0.5}=\sqrt{2}, and for the step size 2020 (that is, a=2a=2) we have seen that the growth factor is 22=42^2=4.

Exercise 1

The number of cells is 55, and the cell number doubles every quarter of an hour. Determine the growth factor for

  1. every hour

  2. every 1010 minutes

Solution

Growth factor u=2u=2 for τ=14\tau=\frac{1}{4} hour

  1. a=4a=4, thus the cell number is multiplied by 24=16‾2^4=\underline{16} every hour.
  2. 1010 minutes are 16\frac{1}{6} hour. So find aa with 16=a14\frac{1}{6}=a\frac{1}{4}, thus a=46=23a=\frac{4}{6}=\frac{2}{3}. Thus, the cell number multiplies by the factor 223=1.587‾2^\frac{2}{3}=\underline{1.587} every 1010 minutes.