MHB Recursive Function for Virus Spread: Finding the General Solution

[Tekilou]

Could anyone please help me I am especially stuck on the second part of the question. Thanks very much I really appreciate it.

 

[Prove It]

What did you have for part (i)?

 

[Kansas Boy]

Did you try writing a few numbers to see if you can see a pattern?

On the first day 1 person has the virus and he gives it to 2 other people.

On the second day 1+ 2= 3 people have the virus.
Each of the 3 give the virus to 2 people, a total of 6 new people.

On the third day, 3+ 6= 9 people have the virus.
Each of the 9 people give the virus to 2 people, a total of 18 new people.

On the fourth day a total of 9+ 18= 27 people have the virus.
Each of the 27 people give the virus to 2 people, a total of 54 new people.

On the fifth day a total of 27+ 54= 81 people have the virus.
Each of the 81 people give the virus to 2 people, a total of 162 new people.

On the sixth day a total of 81+ 162= 243 people have the virus.So far we have 1, 3, 9, 27, 81, 243. I see that those are all powers of 3!
$1= 3^0$, $3= 3^1$, $9= 3^2$, $27= 3^3$, $81= 3^4$, $243=3^5$.

I would conjecture that the number of sick people on the nth day is $3^{n-1}$.

Let $x_n$ be the number of sick people on the nth day. Those $x_n$ people each give the virus to 2 people for a total of $2x_n$ new people so on the n+1 day, $x_{n+1}= x_n+ 2x_n= 3x_n$. Does $x_n= 3^{n-1}$ satisfy that? $x_{n+1}= 3^{n+ 1-1}= 3^n= 3^n+ 2(3^n)= 3(3^n)= 3^{n+1}$. Yes, $x_n= 3^{n-1}$ satisfies this "recurrance relation". Further, $x_1= 3^{1- 1}= 3^0= 1$ as required.

 

[Kansas Boy]

IF we were a little more "mathematically sophisticated" we could have started from the "recurrence relation". If, on the nth day, $x_n$ people have the virus, each person gives the virus to two new people so a total of $2x_n$ new people. On the n+1 day, $x_{n+ 1}= x_n+ 2x_n= 3x_n$. Since we are repeatedly multiplying by 3, it is easy to see that the general solution to $x_{n+ 1}= 3x_n$ is $x_n= C3^n$ for some constant, C. We have $x_1= 1$ so must have $1= C(3^1)= 3C$ so $C=\frac{1}{3}$. $x_n= \left(\frac{1}{3}\right)3^{n}= 3^{n- 1}$.