Julia's question at Yahoo Answers regarding a linear recurrence

[MarkFL]

Here is the question:

Pre Calculus homework help please!?

A tree farm in the year 2010 has 10,000 Douglas fir trees on its property. Each year thereafter 10% of the fir trees are harvested and 750 new fir saplings are then planted in their place.

a) Write a recursive sequence that gives the current number t"sub"n of fir trees on the farm in the year n, with n=0 corresponding to 2010.

b) Use the recursive formula from part a to find the numbers of fir trees for n=1, 2, 3, and 4. Interpret the values in context.

c) Use a graphing utility to find the number of fir trees as time passes infinitely. Explain your result.***Please show work and explain how you got the answers because I have no idea what to do! Thanks so much!

Here is a link to the question:

Pre Calculus homework help please!? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello Julia,

a) Let $T_n$ denote the number of trees present in year $n$. Each year, 10% of the trees are removed (leaving 90%) and 750 trees are added, and at time $n=0$ we have 10000 trees, hence we may state the linear inhomogeneous recurrence:

\(\displaystyle T_{n+1}=0.9T_{n}+750\) where \(\displaystyle T_0=10000\)

b) Using the result from part a), we may state:

\(\displaystyle T_{1}=0.9T_{0}+750=0.9\cdot10000+750=9750\) (This is the number of trees in 2011).

\(\displaystyle T_{2}=0.9T_{1}+750=0.9\cdot9750+750=9525\) (This is the number of trees in 2012).

\(\displaystyle T_{3}=0.9T_{2}+750=0.9\cdot9525+750=9322.5 \approx9323\) (This is the number of trees in 2013).

\(\displaystyle T_{4}=0.9T_{3}+750=0.9\cdot9322.5+750=9140.25 \approx9140\) (This is the number of trees in 2014).

c) To find the value of \(\displaystyle \lim_{n\to\infty}T_{n}\), let's find the closed form.

(1) \(\displaystyle T_{n+1}=0.9T_{n}+750\)

(2) \(\displaystyle T_{n+2}=0.9T_{n+1}+750\)

Subtracting (1) from (2), using symbolic differencing, we obtain the linear homogeneous recurrence:

\(\displaystyle T_{n+2}=1.9T_{n+1}-0.9T_{n}\)

The characteristic roots are:

\(\displaystyle \lambda=0.9,1\) hence:

\(\displaystyle T_n=k_1+k_2(0.9)^n\)

Using the initial values, we may determine the parameters $k_i$:

\(\displaystyle T_0=k_1+k_2(0.9)^0=k_1+k_2=10000\)

\(\displaystyle T_1=k_1+k_2(0.9)^1=k_1+0.9k_2=9750\)

Solving this system, we find \(\displaystyle k_1=7500,\,k_2=2500\) and so we have:

\(\displaystyle T_n=7500+2500(0.9)^n\) and low it is easy to see that:

\(\displaystyle \lim_{n\to\infty}T_{n}=7500\)

Here is a plot of the recurrence:

https://www.physicsforums.com/attachments/823._xfImport

To Julia and any other guests viewing this topic, I invite and encourage you to post other linear recurrence questions here in our http://www.mathhelpboards.com/f15/ forum.

Best Regards,

Mark.