describing a decay process with exponentials and differential equations

Question

I have a process of degradation of some material that proceeds like this across time $t$:

$C_t = C_{t-1} + RC_{t-1}$

where $C_t$ is the amount of material at time $t$ and $R$ is a (negative) rate of degradation. Starting with initial amount $C_0 = 1$ and $R = -0.5$, we have:

$C_0 = 1, C_1 = 0.5, C_2 = 0.25, C_3 = 0.125, C_4 = 0.0625, C_5 = 0.03125, \dots$

what is the correct way to describe $C_t$ in terms of $\exp()$ decay? This form doesn't match the results:

$C_t = C_0\exp(-0.5t)$

since $C_1 = \exp(-0.5) = 0.60653 \neq 0.5$.

Answer 1

This is a discrete case.

$$C_t =(1+R)C_{t-1}=(1+R)^2 C_{t-2}=\dots =(1+R)^t C_0$$

So the equation is $C_t = (1+R)^t C_0$.

Answer 2

First of all, rather than a "negative rate of degradation", it's better to think about the ratio of the quantities of material before and after a timestep, like this:

$$C_t = (1 + R)C_{t-1} = \rho C_{t-1}.$$

So the ratio is $\rho = 1 + R$, where $R$ is your "negative rate of degradation". For $R = -0.5$, we find that $\rho = 1 + (-0.5) = 0.5$.

By induction, one can show that

$$C_t = \rho^t C_0.$$

If you want to use the $\exp()$ function, follow these steps:

\begin{align} \rho &= e^{\ln \rho} \\ \rho^t &= \left(e^{\ln \rho}\right)^t = e^{(\ln\rho)t} = \exp((\ln\rho) t) \\ C_t &= \rho^t C_0 = \exp((\ln\rho) t) C_0 \end{align}

For the case where $R = -0.5$, we have $\ln \rho = \ln(0.5) \approx -0.69314718056$, so

$$ C_t \approx \exp(-0.69314718056 \, t) C_0.$$