# Recursions, Recurrence Relations, Difference Equations, Equations for Population Growth, Fibonacci Sequences and Binet's Formula

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1. Solve xn=axn-1+b when a=1.

2. An isotope of carbon called carbon-14 (14C) is used to establish the age of artifacts and fossils. It decays so that every 5000 years an amount of 14C is reduced to 54.44256% of its initial valued A archaeologist finds a fossil that contains 16% of the amount of 14C it contained when it was alive.

a) Find a difference equation to describe the amount of 14C at any time. Be careful how you define the periods.

b) How old is the fossil?

3. The Fibonacci sequence can also be written 0, 1, 1, 2, 3, 5,.... Find the closed form solution of the Fibonacci recurrence relation when fo = 0 and f1 = 1. This form of the solution is called Binet's Formula.

4. A forest ranger wants to repopulate a forest. He writes a letter to two of his friends (the first generation of friends) asking each of them to buy a tree and to write letters to two of their friends (the second generation of friends). Let t1 = 2 be the total number of trees purchased by the first generation and let t2=6 be the total number of trees purchased by the first and second generations together. Then t is the total number of trees purchased by generations 1, 2, . . ., n.

a) Find a difference equation for t,

b) Use iteration to find a (closed form) solution for the difference equation.

c) How many frees will be donated in all by the first 10 generations of friends?

5. Let r be the growth rate of an animal population between year 0 and year 1.

a) If the growth rate doubles each year, what is the growth rate between year k and year k + 1

b) Write down a difference equation describing th? size of the population in year k in terms of the population in year k?I.

c) If the size of the population at year 0 is 1000 and the size at year 1 is 1100, what is the size in year k?

6. An animal population increases by 25% each year. Its size in the year k = 0 is 1000. If 400 of the animals are trapped each year, find the size of the population after k years. After how many years is the population size reduced to below 350?

7. A certain population of birds is decreasing as a result of competition for limited resources. The decrease in population size between the (k ? 1)th year and the kth year is equal to one-third of the size atthe(k?2)thyear.

a) Find a difference equation for the size of the population.

b) Solve the equation if the original population sizes were 15,000 in year 0 and 14,500 in year 1. c) What is the size of the population in year 10?

8. Find the general solutions of each of the following:

a)xk+2?2xk+1 +xk = 0

b) xk+3 ? 7Xk+2 + 35x+1 ? 50xk = 0 Hint: one of the eigenvalues is 2.

9. The number of new cases of an epidemic occurring daring the kth week is equal to four times the number of cases Xk-2 that existed at the end of the (k ? 2)th week,

a) What is the difference equation satisfied by Xk?

b) Solve the difference equation with the initial conditions x0 = I and x1=4.

c) Solve a) and b) under the additional assumption that during the kth week half of the cases that existed at the end of the previous week are cured.

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2. a. The difference equation is: xn = (0.5444)nx0, where x0 is the initial amount, where n is the time interval counted at every 5000 years

b. Now, xn = 0.16x0

Plugging in a, 0.16x0 = (0.5444)nx0

Therefore, 0.16 = (0.5444)n

Or log 0.16 = n log 0.5444

Therefore, n = log (0.16)/ log(0.5444) = 3.01376

But n is counted for every 5000 years, therefore, the age of the fossil = 3.01376*5000

= 15,068.81 years

3. To find the closed form of the Fibonacci relation:

The general solution to equation Fn = rn will be all possible combinations of roots r1 and r2:

Constants A and B can be found from the boundary conditions F0=0 and F1=1:

Therefore, Fk

4. Clearly, the first few terms are as follows:

In each generation, the number of trees is twice the price generation.

Therefore, the number of trees in successive generations is: 2, 4, 8, 16, ....

Correspondingly,

t1 = 2

t2 = 2 + 4 = 6

t3 = 6 + 8 = 14

t4 = 14 + 16 = 30

......

a. The general term is: tn = tn-1 + 2n; n = 2, 3, 4, .... ; t1 = 2

b. Clearly, tn-1 is the sum of the first n-1 terms of a GP whose first term is 2 and whose common ration is 2.

tn-1 =

Now, tn = tn-1 + 2n = 2n - 2 + 2n = 2n+1 - 2

c. For n = 10,

t10 = 211 - 2 = 2046

5.

(a) ...

#### Solution Summary

Recursions, Recurrence Relations, Difference Equations, Equations for Population Growth, Fibonacci Sequences and Binet's Formula are investigated. The solution is detailed and well presented.