# A new flu virus is introduced when a stranger visits an isolated village of 8000 people. Every infected person infects two more each day. How do you write an exponential function to model the number of UNINFECTED people?

Jul 26, 2018

$u \left(n\right) = 8000 - {3}^{n} \cdot 2$

#### Explanation:

Let i(n) be the number of infected people and u(n) the number of uninfected people, both on the n-th day.

Let x(n) be the number of people who got infected on the n-th day. The relation between i(n) and x(n) is given by:

$i \left(n\right) = i \left(n - 1\right) + x \left(n\right)$

Let's assume the stranger leaves before the next day.

Taking the day the stranger arives as the 0-th day and assuming he infects two people on that day, then

$i \left(0\right) = x \left(0\right) = 2$

$u \left(0\right) = 8000 - 2 = 7998$

The two people infected now will also infect two other people, meaning that:

$x \left(1\right) = 2 i \left(0\right) = 4$
$i \left(1\right) = i \left(0\right) + x \left(1\right) = 6$
$u \left(1\right) = 7998 - x \left(1\right) = 7994 = 8000 - i \left(1\right)$

Some more relations become clear:

$u \left(n\right) = u \left(n - 1\right) - x \left(n\right)$
$u \left(n\right) = 8000 - i \left(n\right)$

As every person infects two people per day, then

$x \left(n\right) = 2 i \left(n - 1\right)$

$\therefore i \left(n\right) = i \left(n - 1\right) + 2 i \left(n - 1\right) = 3 i \left(n - 1\right)$

This a recurrence relation for the number of infected people at a given day:

$i \left(n\right) = \left\{\begin{matrix}3 i \left(n - 1\right) \text{ if " n!= 0 \\ 2 " if } n = 0\end{matrix}\right.$

$\implies i \left(1\right) = 3 i \left(0\right) = 6 = {3}^{1} \cdot 2$

$i \left(2\right) = 3 i \left(1\right) = 18 = {3}^{2} \cdot 2$

$i \left(3\right) = 3 i \left(2\right) = 54 = {3}^{3} \cdot 2$

Analogously, we can figure out that

$i \left(n\right) = {3}^{n} \cdot 2$

Hence the number of uninfected people on day n is

$u \left(n\right) = 8000 - {3}^{n} \cdot 2$