APM Solutions for Assignment 1 Semester 2

APM1514/201/2/2018




Tutorial letter 201/2/2018

Mathematical Modelling
APM1514

Semester 2


Department of Mathematical Sciences


Ths tutorial letter contains solutions for assignment 01.




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Define tomorrow. of south africa

, Solutions to assignment 01

1. (a) From an+1 = a2n , and a0 = 2

for n = 0; a1 = a20 = 22
2 2
n = 1; a2 = a21 = (22 ) = 22
 2
2 3
n = 2; a3 = a22 = (22 ) = 22
 2 2
2 2 2 4
n = 3; a4 = a3 = (2 ) = 22
 ! 2
 2  2 2
2  = 225
n = 4; a5 = a24 =  (22 )


(b) Equilibruim points
from: an+1 = a2n , we have
a = a2
0 = (a2 − a)
0 = a (a − 1)
∴ a = 0 or a = 1
(c)
2
a1 = a20 , a2 = a20 , a3 = a2
4 5
a4 = a20 , a5 = a20

n
(d) an = a20
An


2. (a) A(n+1) = An + 100 − 3000
A(n+1) = 1.01An − 3000
(b) equilibrium point

A = 1.01A − 3000

A
= 3000 ∴ A∗ = 300 000
100

(c) Since A0 = 200 000 < A∗ = 3000 00, it follows that the loan will be paid out completely.
(d) Yes, it will, and will not, for example
(i) will not change,
(ii) yes it will change to
A(n+1) = 1.01An − 1600
(iii) yes it will change to  
2
A(n+1) = An + An − 3000
100



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