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A model for the number of lobsters caught per year is based on the assumption that the number of lobsters caught in a year is the average of the number caught in the two previous years.
(a) Find a recurrence relation for {Pn}, where Pn is the number of lobsters caught in year n, under the assumption for this model.
(b) Find Pn if 100000 lobsters were caught in year 1 and 300000 were caught in year 2.
$\displaystyle P_{n+2}=\frac{P_n + P_{n+1}}{2}$
Hello, bhuvan!
I know some techniques for this problem.
I hope they're appropriate for your course.
Quote:
A model for the number of lobsters caught per year is based on the assumption
that the number of lobsters caught in a year is the average of the number
caught in the two previous years.
(a) Find a recurrence relation for $\displaystyle P(n)$, the number of lobsters caught in year $\displaystyle n.$
. . $\displaystyle P(n) \;=\;\frac{P(n-1) + P(n-2)}{2}$
Quote:
(b) Find $\displaystyle P(n)$ if $\displaystyle P(1) = 100,\!000\text{ and }P(2) = 300,\!000.$
From (a), we have: .$\displaystyle 2P(n) \;=\;P(n-1) + P(n-2)$ .[1]
We conjecture that $\displaystyle P(n)$ is an exponential function: .$\displaystyle P(n) \,=\,X^n$
Then [1] becomes: .$\displaystyle 2X^n \:=\:X^{n-1} + X^{n-2} \quad\Rightarrow\quad 2X^n -X^{n-1} - X^{n-2} \:=\:0 $
. . Divide by $\displaystyle X^{n-2}\!:\quad 2X^2 - X - 1 \:=\:0 \quad\Rightarrow\quad (X - 1)(2X + 1) \:=\:0$
. . And we have two roots: .$\displaystyle X \;=\;1,\;\text{-}\tfrac{1}{2}$
Then: .$\displaystyle f(n) = 1^n\:\text{ or }\:f(n) = \left(\text{-}\tfrac{1}{2}\right)^n $
Form a linear combination of these roots:
. . $\displaystyle f(n) \;=\;A\left(1^n\right) + B\left(\text{-}\tfrac{1}{2}\right)^n \quad\Rightarrow\quad f(n) \;=\;A + B\left(\text{-}\tfrac{1}{2}\right)n$
We know the first two values of this sequence:
. . $\displaystyle \begin{array}{ccccc}
f(1) = 100,\!000\!: & A - \frac{1}{2}B &=& 100,\!000 & {\color{blue}[2]} \\ \\[-3mm]
f(2) = 300,\!000\!: & A + \frac{1}{4}B &=& 300,\!000 & {\color{blue}[3]} \end{array}$
Subtract [2] from [3]: .$\displaystyle \tfrac{3}{4}B \:=\:200,\!000 \quad\Rightarrow\quad B \:=\:\frac{800,\!000}{3}$
Substitute into [3]: .$\displaystyle A + \tfrac{200,\!000}{3} \:=\:300,000 \quad\Rightarrow\quad A \:=\:\frac{700,\!000}{3} $
Therefore: .$\displaystyle f(n) \;=\;\frac{700,\!000}{3} + \frac{800,\!000}{3}\left(\text{-}\frac{1}{2}\right)^n \quad\Rightarrow\quad f(n) \;=\;\frac{100,\!000}{3}\bigg[7 + \frac{8}{(\text{-}2)^n} \bigg] $
Thanks you very much !!
Do you know how to solve below problem i am trying to solve this but i am confuse..
Consider the non homogeneous linear recurrence relation an=2an-1+2^n
(a) show that an=n2^n is a solution of this relation.
i tried to solve this by putting an value in
2an-1+2^n=2(n2^n)+2^n)=2^n(2n+1) but i am not getting right answer as far as i think.
It works !
$\displaystyle 2a_{n-1}+2^n=2(n-1)2^{n-1}+2^n=2n2^{n-1}-2.2^{n-1}+2^n=n2^n=a_n$
can you please explain it ? i am confuse how you got 2(n-1)2^n-1
in 2(n-1)2^n-1+2^n since we have an=n2^n ??
Never mind i got it.. Thank You for your effort.
