IMP 3 POW 1 help

[Meow_Mix]

Hello, I am stumped. The problem reads something like this:

A female rat and a male rat live on an island, they produce a litter on January 1st, and every 40 days after that. The number of young produced in every litter is six, and three of those six are females. The original female gives birth to six young on January first, and produces another litter of six every fourty days after that. Each female born on the island will produce her first litter 120 days after her birth and then produce a new litter every fourty days after. What will be the total number of rats the following january 1st, including the original two?

Thanks a bunch, I am stumped, and have to write a 7 paragraph essay on it. Please explain how you got the answer, I have been messing around with graphs and such, but no luck yet. Plz answer asap.

 

[stapel]

For how many generations have you computed their progeny? What formula have you arrived at? How far have you gotten?

Eliz.

 

[pka]

Notation: S<SUB>n</SUB> will be the TOTAL number of females alive n*40 days after January 1.

S<SUB>0</SUB> (on January 1) will be 4, the producing female and three baby females.
S<SUB>1</SUB>will be 7, the three new babies from the one producing female plus the 4 already there from S<SUB>0</SUB>
S<SUB>2</SUB>will be 10, still just one producing female, why?
When we get to S<SUB>3</SUB> we have 4 producing females, it is 120 days after January 1.
S<SUB>3</SUB>=22 = S<SUB>2</SUB>+ 3*S<SUB>0</SUB>
In general, S<SUB>n</SUB>= S<SUB>n-1</SUB>+ 3* S<SUB>n-3</SUB>.
Now don’t forget the boys. There are as many boys as girls.

 

[beevaj]

Okay? I didn't get that. Please explain. And what is the answer? I might get it wrong.

 

[stapel]

Okay? I didn't get that.

You don't get which? Please be specific.

And what is the answer? I might get it wrong.

If you aren't sure of your answer, please feel free to post all of your steps and reasoning, and we'll be glad to take a look at it.

Thank you.

Eliz.

 

[beevaj]

okay.. i did a chart... my teacher recommend me to write 40 nine times down the page... you knoe..
Female & Male
40 -6

40 -6

40 -6

you know.. and then continue the assumptions the book has given me... at the end I got 594 rats at the end of the year.. but.. Im confused... One of my friend told me that its over 1000.. but.. I dont get it..

 

[beevaj]

My WORK!

6*9=54
18*18=324
36*6=216

All of them add together is 594.

 

[pka]

Thanks Dennis!

 

[beevaj]

Thanks ALOT! I really appreciate the help!

 

[galactus]

Here's my way. It differs from pka's though, which makes me apprehensive about posting it.

We start out with 8 rats because the first pair have a litter on Jan. 1

40 days later, another 6 are born, now we have 14.

40 days after that, another 6, now we have 20.

40 days after that, the first litter will start having little ones and for each

pair of grown rates we get 6 more. So, we add 3 rats for every rat.

This gives us the recursion, $\displaystyle R_{n}=R_{n-1}+3R_{n-3},\;\ for\;\ n\geq{4}$

$\displaystyle R_{0}=8$
$\displaystyle R_{1}=14$
$\displaystyle R_{2}=20$
$\displaystyle R_{3}=20+3(8)=44$
$\displaystyle R_{4}=44+3(14)=86$
$\displaystyle R_{5}=86+3(20)=146$
$\displaystyle R_{6}=146+3(44)=278$
$\displaystyle R_{7}=278+3(86)=536$
$\displaystyle R_{8}=536+3(146)=974$
$\displaystyle R_{9}=974+3(278)=1808$

There is probably a closed form which goes with this, similar to the Fibonacci thing I would suppose.

 

[Denis]

pka, how do you get 38 at S2?
looks like you got the 3 females born Jan 1st getting pregnant 40 days too soon :?

I make it 20 at S2, then 44 at S3; no?

Edit: galactus, methinks you're short one 40 day period;
you need numbering from 0 to 9 ; it's at period 0 (Jan 1) that there is 8.

 

[galactus]

In just seen this famous problem posted again on another site.

I got to wondering if anyone knows of or has derived the closed form solution for the nth term in this sequence.

Anyone familiar with Fibonacci knows that Binet's formula is used for the nth term of that sequence. I wonder what this one is.

The solutions of $\displaystyle x^{2}-x-1=0$ is the ratio in the Fibonacci series.

The rat series has the solution to $\displaystyle x^{3}-x^{2}-3$ as the ratio.

That is, $\displaystyle \lim_{x\to\infty}\frac{R_{n}}{R_{n-1}}=1.86370652782$

 

[galactus]

Edit: galactus, methinks you're short one 40 day period;

You're right as rain, Denis. Now, get to working on the closed form. :lol:

 

[galactus]

Not that anyone gives a rat's ***, but I found a closed-form solution to the problem.

\(\displaystyle \L\\\sum_{roots\;\ of\;\ 3x^{3}+x-1}\frac{2(3R^{2}+3R+4)(\frac{1}{R})^{n}}{R(9R^{2}+1)}\)


I posted it here, originally: http://math2.org/mmb/thread/37302



It works, though it's complicated.