Find the micro-organism growth after a certain amount of minutes

Question

For an experiment,

I put in a cell growing medium, 1 bacterium of type A and 3 bacteria of type B and every minute I add 1 bacterium of type A and 3 bacteria of type B. So, after 1 minute it means that my cell growing medium will contains 6 bacteria of type A and 9 bacteria of type B. Both bacteria grows at different speed.

The first bacterium of type A is growing at $A(t) = 4t+1$
The second bacterium of type B is growing at $B(t) = 2^t$

I am trying to figure the total amount of bacteria I will have after 5 minutes.

Just to sum up what I have so far:

• For A:
  • t = 0, $[A]$
  • t = 1, $4*1+1=5$ then I add $1$, $6$
  • t = 2, $4*6+1=25$ then I add $1$, $26$
  • t = 3, $4*26+1=105$ then I add $1$, $126$
• For B:
  • t = 0, $[BBB]$
  • t = 1, $2^3=8$ -- not valid
  • t = 2, can't figure that one.

I'm really curious to see the answer and how to solve this kind of problem.

Thanks

Answer 1

Based on your numbers at $1$ and $2$ minutes, you really have $A(n+1)=4A(n)+1$ and $B(n+1)=2^{B(n)}$ with $A(0)=1, B(0)=3$. You should follow each chain separately as they don't interact. The number of $B$'s is then $B(t)=(2\uparrow \uparrow(t-1))^3$. The $A$'s are a linear recurrence The up-arrow is a chain of exponentials, one taller each stage.

Answer 2

Solution with new data values in mind:

You mean $A(t) = 4 A(t-1) + 1$ is how they grow, and then you add $1$, so in fact $$A(t) = 4A(t-1) + 2$$

Similarly, $$B(t) = 2^t B(t-1) + 3$$

This gives $B(1) = 9$ and $B(2) = 39$, not $2051$.

Solving the recurrence relationships gives:

$$ A(t) = \frac{5}{3} 4^t - \frac{2}{3} = \frac{1}{3} \left(5 \cdot 4^t - 2\right)$$

I have seen the explicit solution for $B(t)$ using Mathematica and it isn't pretty. You'd be better off just looking at the recurrence equation but anyway...

$$ B(t) = 2^{\frac{1}{2} \left(t^2+t-2\right)} \left(\sum _{k=0}^{t-1} 3\ 2^{-\frac{1}{2}k(k+3)}+6\right) $$