Certainly! Let's consider an explicit function [tex]\( I(t) \)[/tex] where the function describes the value at time [tex]\( t \)[/tex] in terms of the value at time [tex]\( t-1 \)[/tex].
Given the function in an exponential growth form, it can be expressed recursively. Here's how you can express it:
1. Recursive form:
[tex]\[ I(t) = r \cdot I(t-1) \][/tex]
2. Initial condition:
[tex]\[ I(0) = I_0 \][/tex]
Where:
- [tex]\( I(t) \)[/tex] is the value at time [tex]\( t \)[/tex].
- [tex]\( r \)[/tex] is the growth rate.
- [tex]\( I_0 \)[/tex] is the initial value at time [tex]\( t=0 \)[/tex].
Let’s break down the recursive relationship:
- At [tex]\( t = 0 \)[/tex]:
[tex]\[ I(0) = I_0 \][/tex]
This is the initial value of the function.
- At [tex]\( t = 1 \)[/tex]:
[tex]\[ I(1) = r \cdot I(0) \][/tex]
- At [tex]\( t = 2 \)[/tex]:
[tex]\[ I(2) = r \cdot I(1) \][/tex]
Substituting [tex]\( I(1) \)[/tex] from the previous step:
[tex]\[ I(2) = r \cdot (r \cdot I(0)) = r^2 \cdot I_0 \][/tex]
- At [tex]\( t = 3 \)[/tex]:
[tex]\[ I(3) = r \cdot I(2) \][/tex]
Substituting [tex]\( I(2) \)[/tex] from the previous step:
[tex]\[ I(3) = r \cdot (r^2 \cdot I_0) = r^3 \cdot I_0 \][/tex]
Continuing this pattern, at time [tex]\( t \)[/tex], the value can be expressed as:
[tex]\[ I(t) = r^t \cdot I_0 \][/tex]
Thus, this recursive system, with its initial condition, completely describes the explicit form of the function [tex]\( I(t) \)[/tex].
