I have a geometric decay sequence given by $a_n=a_0*(1-d)^n$. It seems intuitively obvious that as the difference between successive values of n tends to zero, the sequence increasingly closely approximates an exponential relationship of the form $\frac{dy}{dx}=-k \Rightarrow y=y_0e^{(-kx)}$ (Using the boundary condition $y(x=0)=y_0$).
How do I prove this? So far, what I've tried has been to try to rearrange the formula to the sequence so I can take some kind of limit, but I haven't got much further than that.
The $a_n$ notation is more common for integer $n$, so the relation could be better written: $$a(x) = a(0) \cdot (1-d)^x$$
Using the identity $z = e^{\ln z}$ the above can be written as: $$a(x) = a(0) \cdot e^{\ln(1-d) \;\cdot\; x}$$
So $a(x)$ is in fact an exponential of the form $y_0 \cdot e^{-k \, x}$ for $y_0=a(0)$ and $k = -\ln(1-d) \gt 0$.