Approximation of a negative exponential model?

Question

I am trying to get an approximation of this model, it is a negative exponential model introduced by Olson in 1963 "Energy storage and the balance of producers and decomposers in ecological systems". This model basically models the amount of fuel accumulation over t years in a landscape/forest.

$W_t = A(1-e^{-kt})$

$W_{t} = $ Fuel load at time $t$

$A =$ Constant

$k =$ Constant

$t =$ time

So basically I am trying to solve for

$W_{t+1} - W_{t} = A(1-e^{-k(t+1)}) - A(1-e^{-kt})$

I tried using logs and I still cant get a relationship. Is it even possible?

Thanks!

Answer 1

You can simplify further:

\begin{align} W_{t+1} - W_{t} &= A(1-e^{-k(t+1)}) - A(1-e^{-kt}) \\ &= A-Ae^{-kt}e^{-k}-A+Ae^{-kt} \\ &=Ae^{-kt}(1-e^{-k}) \\ &= const \cdot e^{-kt} \end{align}

Answer 2

Assume $k \ll 1$ then $W_{t+1}-W_t\approx \frac{dW_t}{dt}=kAe^{-kt}$. This is an over estimate of the difference, with absolute value of the error bounded by $\varepsilon(t)=k^2Ae^{-kt}/2$

If you want it exactly though:

$$ W_{t+1}-W_t=Ae^{-kt}-Ae^{-k(t+1)}=Ae^{-kt}(1-e^{-k}) $$

An Octave session illustrating this is shown:

>> k=0.01;A=1;
>>
>> # Define W
>> W = @(t,A,k) A.*(1-exp(-k.*t));
>>
>> #calculate W_6-W_5
>> W5=W(5,A,k)
W5 =  0.048771
>> W6=W(6,A,k)
W6 =  0.058235
>> W6-W5
ans =  0.0094649
>>
>> #now calc approx difference
>> k*exp(-k*5)
ans =  0.0095123
>>
>> #now calc exact difference
>> A.*exp(-k.*t).*(1-exp(-k))
ans =  0.0094649
>>