[High School - Precalc] Mouse Problem - easy shortcut I am missing?

[aTzUeLo1191]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

@Meaculpa: Your method seems right too...perhaps I just made some addition mistake. I'll have to see when I get the problem back.

@Trumaestro: I'm not really sure how to use the equation because B isn't 3. I don't any idea what to put in for that variable

@emerald: You lost me at "recursion mathematics".

I should get the answer w/ a method on Monday; perhaps someone here is right after all.

[Commandersa1]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

I got 1550. :/

This was my method:

http://img367.imageshack.us/img367/5570/micekb2.jpg

I cannot see how I could mess up, and igotrhythm got close to that answer too.

[QED Stepfiles]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

OK Let’s try this using matrices… since everybody seems to be getting different answers.

First we have to come up with a recursion. We have that

M_n = m_{n-1} + 3m_{n-3}

In other words, the number of mice at time n is equal to the number of mice from time n-1 added to the number of mature mice, which is m_{n-3}, multiplied by 6 (how many mice are produced by each pair), divided by 2 (because we need a pair of mice to reproduce).

So, the matrix for this equation is

1 0 3
1 0 0
0 1 0

Call this matrix A.

So, generally from here we would find eigenvalues to decouple the equation… but just from inspection I’m pretty sure there are complex eigenvalues and I don’t want to chug through the math…

Basically, this matrix when multiplied by a vector that describes how many mice there are at three subsequent times (say, 1,2,3), will give you how many mice there are at 2,3,4. You can check this if you want. Multiply the matrix on the right by a column vector (a3, a2, a1), and you'll get out the result

(a3+3a1, a3, a2) which is precisely the description of how many mice there are at (a4, a3, a2)

The number of mice there are at times 1,2,3 is 8, 14, 20.

We want to find the mice at time 9 (since this is precisely one year later). Therefore, we just multiply A^6 * (8,14,20), since we would like to go from time 3 to time 9, which requires 6 applications of our matrix.

We can use a calculator to do this, and we get:

974, 536, 278

Meaning, after a year there are 974, after 320 days there are 536, and after 280 days there are 278.

We could continue this if you like... if you would like to figure out how many mice there are after 40*n days, just multiply the vector (8,14,20) by A^(n-3) on the left, and the first number is your answer.

Basically, matrix multiplication makes everything easier.

=D

So yes, 974 is the correct answer. I’m 100% sure.

Hope this helps.

[emerald000]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

You should have kept on reading after that, it isn't as hard as it seem. Suppose we want to find the number of mice at generation 5. We would need to find m_4. (m_0 being the first one)

In recursive mathematics, all you need to do is to apply the formula(s) a lot of times in order to simplify to data you know. Here we will start with m_4, and try to end with m_0, f_0, f_1 and f_2. It can get quite messy once you start to try higher numbers, so the usage of a computer is often useful to speedup the process.

Given that m_n = m_(n-1) + 6 f_(n-1) and f_n = f_(n-1) + 3 f_(n-3), and that m_0=2 and f_0=f_1=f_2=1

m_4 = m_3 + 6 f_3
m_4 = (m_2 + 6 f_2) + 6 (f_2 + 3 f_0)
m_4 = ((m_1 + 6 f_1) + 6 f_2) + 6 (f_2 + 3 f_0)
m_4 = (((m_0 + 6 f_0) + 6 f_1) + 6 f_2) + 6 (f_2 + 3 f_0)

We know the value of m_0, f_0, f_0 and f_0. All we have to do now is to put the values back into that equation.

m_4 = (((2 + 6 (1)) + 6 (1)) + 6 (1)) + 6 ((1) + 3 (1))
m_4 = 44

---

I also doubt that linear algebra is going to help him much in his problem...

[QED Stepfiles]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

Emerald's method is completely correct, but the problem is that it's just really... really... really messy. Even with a computer, it's much easier to just reorganize the problem in terms of a matrix multiplication, and just to go from there. After all, telling a computer to raise a square matrix to some integral power is much easier than asking it to solve a recursion. Both methods definitely work, though.

In terms of matrix multiplication, however, if you want a convenient online program to do it for you, try here:

http://wims.unice.fr/wims/en_tool~linear~matmult.html

Good luck!

And by the way, the reason why linear algebra would help is that if we can find a basis of eigenvectors that correspond to the matrix that describes the recursion, we can diagonalize the matrix, and diagonal matrices are really easy to work with (if you raise it to a power "n", then just raise each element down the diagonal to a power "n"). As far as I know, it's probably the most elegant solution to this problem, and it doesn't require a computer to solve at all (you just need to find the roots to some polynomials, and in a 3x3 case that's easy enough to do by hand). This is precisely how one would try to find an explicit expression for the Fibonacci sequence (that's actually an easier problem, since that's only a 2x2 matrix and 2nd degree polynomials are easily solved by the quadratic formula).

[MeaCulpa]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

Hmm...I agree with emerald's f_9 value:

1 female initially (reproduces 10 times)
3 born on 0 (each reproduce 7 times)
3 born on 40 (each reproduce 6 times)
3 born on 80 (each reproduce 5 times)
12 born on 120 (each reproduce 4 times)
21 born on 160 (each reproduce 3 times)
30 born 200 (each reproduce 2 times)
66 born on 240 (each reproduce once)
___
139 reproductive females by day 360.

But I don't think 974 is high enough of a value, considering the number of times each can reproduce:

(1 x (10 x 6)) + (3 x (7 x 6)) + (3 x (6 x 6)) + (3 x (5 x 6)) + (12 x (4 x 6)) + (21 x (3 x 6)) + (30 x (2 x 6)) + (66 x (1 x 6)) = 1806.

[igotrhythm]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

Haha, this is good stuff...all the different solutions we've come up with. XD

You should show your teacher this on Monday and give out credits to anyone that ends up being right.

[aTzUeLo1191]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

Bumping w/ answer.

Teacher finally gave it back today, one or two people in our class got it right - 1808. Sadly, no one here did, and neither did I.

I didn't get a method on how he or anyone else found it, but if you all want, I could ask him.

Edit: Damn I just noticed Mea was pretty close...did you forget to add the two parents?

[MeaCulpa]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

Oh, yeah I guess I did. Just looking at my calculation in my last post, I calculated the total progeny without Adam and Eve. I guess it wouldn't hurt to find out how exactly he solved it, and if there is a proper shortcut after all.

[Commandersa1]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

I would like to know the actual process in solving this. :/

[Patashu]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

I'll try and sum up the problem so I don't have to type up a long paragraph:

2 mice are on an island with unlimited food and no predators. They have 6 children on the first day, 3 M, 3 F, and reproduce the same every 40 days after that. Children reproduce after 120 days, every 40 days. How many mice will there be after 1 year?

Is there a really simple shortcut I am missing? A problem like this is kinda tough to write out every stage at a time.

Looks like an iterative function to me. A pair of mice produce three reproducing pairs that kick in three steps later, so let's have f_n counting the number of reproducing mice pairs and g_n counting the total number of mice at any given 40 day interval.
f_0, f_1, f_2 = 1
g_0 = 8

f_n = f_(n-1) + 3*f_(n-3) (the mice pairs from exactly 120 days ago are considered breeders from this term on)
g_n = g_(n-1) + 6*f_n (calculate this after calculating f_n)

f_3 = 4
f_4 = 7
f_5 = 10
f_6 = 22
f_7 = 43
f_8 = 73
f_9 = 139

g_1 = 14
g_2 = 20
g_3 = 44
g_4 = 86
g_5 = 146
g_6 = 278
g_7 = 536
g_8 = 974
g_9 = 1808

So after 360 days there are 1808 mice. After 365.251 days there will still be 1808 mice

[QED Stepfiles]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

Ah ****... I forgot that we start at time zero, not at time one...

If you take the matrix that I have in my post and take it to the 7th power rather than the 6th, you get the correct answer.

Whoops...

[emerald000]

Re: [High School - Precalc] Mouse Problem - easy shortcut I am missing?

And using m_10 with my recursion formula. At least I know my technique works (and the one Patashu posted since it is the exact same one).