- 17 Jun '05 14:40A 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. - 17 Jun '05 16:45this 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. - 19 Jun '05 10:05 / 1 edit

Ok here's a hint:*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.

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. - 19 Jun '05 10:13

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.

- 20 Jun '05 03:40

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

Or you can save me the trouble... - 21 Jun '05 11:14 / 1 editLet 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. - 21 Jun '05 14:17

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