Bacteria

A new species of bacteria called blue and red behave in the following way: after one minute every blue bacterium turns into 6 red ones, and every red bacterium turns into 1 blue and 1 red bacterium. During an experiment you put 1 blue and 1 red bacterium into a test-tube, so that after one minute there were 1 blue bacterium and 7 red ones; and after two minutes there were 7 blue ones and 13 red ones. Prove that after n minutes there will be:

3/5 * 3^n + 2/5 * (-2)^n
Blue bacteria in the test-tube.

I wont provide any hints at the moment but if later no one makes any headway, which i doubt, then i'll post some hints.

this is a version of the game of "life". Anyone remember that one? It
was in a Scientific American article many years ago. Used a board
something like a Go board, but with a theoretically infinite number
of intersections but a Go board works for most iterations.
You have black and white stones, put down some initial pattern
then some stones die, some reproduce and the pattern changes
in response to a simple set of rules. Each intersection has 8 surrounds
and depending on the count of black and white stones in those
surrounds, the pieces live or die. Like if there is only one of its color
in the 8 surrounds, it dies of loneliness, if there are two and only two
it reproduces, etc. Each generation produces a new pattern that,
depending on the original configuration, can make a self reproducing
shape that eventually runs off the board. Some shapes sit there
and go back and forth between two configurations, a stalemate
and others dribble into a static state or dies off completely.

Originally posted by sonhouse
this is a version of the game of "life". Anyone remember that one? It
was in a Scientific American article many years ago. Used a board
something like a Go board, but with a theoretically infinite number
of intersections but a Go boa ...[text shortened]... e
and others dribble into a static state or dies off completely.

Ok here's a hint:

This is a proof by Induction.

Also note that,

b1 = 1, r1 = 7
b2 = 7, r2 = 13
b3 = 13, r3 = 55

See the pattern? Now you just need to prove it.

Originally posted by elopawn
Ok here's a hint:

This is a proof by Induction.

Also note that,

b1 = 1, r1 = 7
b2 = 7, r2 = 13
b3 = 13, r3 = 55

See the pattern? Now you just need to prove it.

Well, induction means showing that it is true in one case, then assuming it is true for a particular case (n=k), then showing it is true for the next case (k + 1).

Originally posted by elopawn
Ok here's a hint:

This is a proof by Induction.

Also note that,

b1 = 1, r1 = 7
b2 = 7, r2 = 13
b3 = 13, r3 = 55

See the pattern? Now you just need to prove it.

I can see more than one possible pattern, I will have to do the next step to decide which pattern it actually is.

Or you can save me the trouble... 😉

This is a toughie. The induction isn't that hard, but how did you come up with the formula? The neatest thing I found was that the ratio of red to blue approaches 3. 😕 Not too helpful here.

Let b_t = number of blues after t minutes
Let r_t = number of reds after t minutes

What is b_(t+1) in terms of b_t and r_t?
What is r_(t+1) in terms of b_t and r_t?

Ok, lets see,

b_t = r_(t-1)
r_t = 6b_(t-1) + r_(t-1)

b_(t-1) = r_(t-2)
r_(t-1) = 6b_(t-2) + r_(t-2) = 6b_(t-2) + b_(t-1)

Thus: b_t = b_(t-1) + 6b(t-2)

Now, it should be easier.

Originally posted by elopawn
Let b_t = number of blues after t minutes
Let r_t = number of reds after t minutes

What is b_(t+1) in terms of b_t and r_t?
What is r_(t+1) in terms of b_t and r_t?

Ok, lets see,

b_t = r_(t-1)
r_t = 6b_(t-1) + r_(t-1)

b_(t-1) = r_(t-2)
r_(t-1) = 6b_(t-2) + r_(t-2) = 6b_(t-2) + b_(t-1)

Thus: b_t = b_(t-1) + 6b(t-2)

Now, it should be easier.

I got that far already, but converting that into the exponential formula is giving me the runs. Help!

Originally posted by PBE6
I got that far already, but converting that into the exponential formula is giving me the runs. Help!

hmm, i'm still working on it as well, i have a feeling it has got to do with the recurrance relation.