Understanding Engineering Mathematics

Fory=f(x)=x^2 obtain a linear approximation tof(x), and hence evaluate( 1. 01 )^2

approximately.

3.You may have heard talk of ‘chaos’ in recent years. This question, using an iterative
process, gives you a simple introduction to this notion. It is based on what is called the
‘logistic map’, which is an iteration scheme given by

xn+ 1 =Axn( 1 −xn)

whereAis some constant. Don’t worry where this comes from, just accept that it has

a wide applicability in science and engineering and it is important to consider when

it yields a convergent solution forxn– i.e. when the sequence it generates converges

to a definite value as we perform the iterations, as we did in Section 14.8. It is found

that this depends crucially on the value ofA. Experiment with the iterations (15 – 20

should do, on calculator or computer) for the cases ofA= 2 , 1 +

√

5 and 4, with a

starting value ofx 0 = 0 .4. The caseA=4 gives a ‘chaotic’ result as you will soon

discover.

14.14 Answers to reinforcement exercises

1.(i) 3.1, 3.01, 3.001, 3.000001 (ii) 3

2. 2
   3.4/3

4.δs= 32 tδt+ 16 (δt)^2

δs

δt

= 32 t+ 16 (δt)

(i) 67.2 (ii) 65.6 (iii) 64.16 (iv) 64.016 64 m/sec.

5.(i) 16 (ii) 3 (iii) 0 (iv) limit does not exist

(v)− 4 (vi) 1 (vii) 32 (viii) limit does not exist

6.(i) 1 (ii) 3 (iii) α/β (iv) 1

(v) (α/β)γ (vi) 0 (vii) 0

7. (i) continuous for allx (ii) discontinuous atx= 1
   (iii) discontinuous atx=− 4 (iv) discontinuous atx= 0
   (v) discontinuous atx=− 2 (vi) continuous for allx
   (vii) discontinuous atx= 2 (viii) discontinuous atx= 1


8. 3 x^2 ,3


9. (i) 1,2,3,4,(∞) (ii)

4

5

, 1 ,

6

5

,

7

5

,(∞) (iii) 0,− 1 , 0 ,1 (divergent)

(iv) 0, 1, 0,

1

2

,(0) (v) 1,

1

√

2

,

1

√

3

,

1

2

,( 0 ) (vi)

2

3

,

2

5

,

2

7

,

4

9

,

(

1

2

)