I would like to better understand, when the function $e^{pt}$ is the "better" choice and when (if at all) $(1+p)^t$ should be used.
To give a classic example: Say we want to model radioactive decay of some element $A$. Let $t$ be in units of one half-life of $A$. Then $p=-\frac{1}{2}$.
Now I'd say the standard approach to modeling this is via the function $$A_1(t)=A_0\cdot \left(1-\frac{1}{2}\right)^t.$$
On the other hand, if we approach this problem as an ODE, we can say that at any point $t$ we want $A(t)$ to decrease at a rate of half of its momentary amount: $$\frac{d}{dt}A_2(t)=-\frac{1}{2}A_2(t)~,$$
which leads to the function $$A_2(t)=A_0\cdot e^{-\frac{1}{2}t}.$$
But which approach would be "better" here? I think $A_1$ is much more commonly (if not exceptionally) used when it comes to modeling atomic decay. On the other hand, I know that $$e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$$ which essentially means that the rate of change of $A_2$ is continously updated, while $A_1$ is updated discretely, right? That's the best way I can phrase it at the moment.
So to conclude: Does this mean that $A_2$ is always the "better", more accurate choice or are there situations where $A_1$ is actually "correct"? I hope I could make myself clear and thank you!
It depends how large the decay period is. Let´s say the length of the decay period is $\frac1n$. Then in $n$ periods the amount of units of the atom after t periods is
$A_t=A_0\cdot \left(1-\frac{\frac12}{n}\right)^{n\cdot t}$, where $A_0$ is the initial amount. Important: $\frac1{2}$ is the decay rate in a period of the legnth $\frac1n$.
The period can get smaller and smaller, which means that $n\to \infty$.
$$A_t=A_0\cdot \lim_{n\to \infty} \left(1-\frac{\frac12}{n}\right)^{n\cdot t}=A_0\cdot e^{-\frac12\cdot t}$$
Conclusion: The smaller the decay/growth period gets the more it is reasonable to apply the continuous growth/decay model.