MHB Julia's question at Yahoo Answers regarding a linear recurrence

  • Thread starter Thread starter MarkFL
  • Start date Start date
  • Tags Tags
    Linear Recurrence
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
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.
 
Mathematics news on Phys.org
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:

$$T_{n+1}=0.9T_{n}+750$$ where $$T_0=10000$$

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

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

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

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

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

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

(1) $$T_{n+1}=0.9T_{n}+750$$

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

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

$$T_{n+2}=1.9T_{n+1}-0.9T_{n}$$

The characteristic roots are:

$$\lambda=0.9,1$$ hence:

$$T_n=k_1+k_2(0.9)^n$$

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

$$T_0=k_1+k_2(0.9)^0=k_1+k_2=10000$$

$$T_1=k_1+k_2(0.9)^1=k_1+0.9k_2=9750$$

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

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

$$\lim_{n\to\infty}T_{n}=7500$$

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.
 

Attachments

  • julia2.jpg
    julia2.jpg
    23.9 KB · Views: 70
Last edited:
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top