FACULTY OF SCIENCE AND TECHNOLOGY

SEMESTER MAY / 2012

SBMA4603

NUMERICAL METHODS

Name :

Matric num. :

PPK :

QUESTION 1

f(x) has a root of order 9 at x=p.

From Lemma 2.1 (page54)

f(x) can be expressed as f(x) = (x-p)m. h(x) ; h(x)≠0

When m=9;

So that, f(x) = (x-p)9. h(x)

Compute f’(p) :

From f(x)= (x-p)9. H(x)

We use a product rule of derivatives

f’(x) = u’. v + v’. u

= 9(x-p)8(1). h(x)+ h’(x).(x-p)9

= 9(x-p)8. h(x) + h’(x).(x-p)9

f’(p) = 9(p-p)8. h(x) + h’(x).(p-p)9

= 9(0)8. h(x) + h’(x).(0)9

= 0.

Compute f”(p) :

From f’(x) = 9(x-p)8. h(x) + h’(x).(x-p)9

f”(x) = 9(8)(x-p)7(1). h(x) + h’(x).9(x-p)8 + h”(x).(x-p)9+9(x-p)8(1).h(x)

= 9(8)(x-p)7. h(x) + 2[9(x-p)8 . h’(x)] + h”(x).(x-p)9

f”(p) = 9(8)(p-p)7. h(x) + 2[9(p-p)8 . h’(x)] + h”(x).(p-p)9

= 9(8)(0)7. h(x) + 2[9(0)8 . h’(x)] + h”(x).(0)9

= 0.

Compute f (9) (p) :

From f”(x) = 9(8)(x-p)7. h(x) + 2[9(x-p)8 . h’(x)] + h”(x).(x-p)9

f(9)(x) = 9(8)(7)(6)(5)(4)(3)(2)(1) (x-p)0. h(x) + 9[9(8)(7)(6)(5)(4)(3)(2) (x-p)1. h8(x)] +

h9(x).(x-p)9

f(9)(p) = 362880h(x) + 3265920(x-p)1. h8(x) + h9(x).(x-p)9

= 362880h(x) + 3265920(p-p)1. h8(x) + h9(x).(p-p)9

= 362880h(x) + 3265920(0)1. h8(x) + h9(x).(0)9

= 362880h(x) + 0 + 0

= 362880h(x)

QUESTION 1 (a)

We know that g(x) = ax3 + bx2 + cx + d.

The order of roots for g(x) = x3 + 3x2 - 4;

We found that a=1, b=3, c=0, d= - 4.

From calculation by calculator scientific, the roots for equation g(x) = x3 + 3x2 – 4 are:

x1=1 , x2= -2

when g(x)=0; g(x) = x3 + 3x2 – 4 = 0

g’(x) = 3x2 + 6x = 0

g”(x) = 6x + 6 = 0

when root x=1; g(1) = (1)3 + 3(1)2 – 4 = 0

g’(1) = 3(1)2 + 6(1) = 9 ; g’(1) ≠0

when g’(1) ≠0 ;

thus, the order = 1

The root x=1 , the order = 1 (x-1)1 = (x-1)

When root x=-2; g(-2) = (-2)3 + 3(-2)2 – 4 = 0

g’(-2) = 3(-2)2 + 6(-2) = 0

g”(-2) = 6(-2) + 6 = -6 ; g”(-2) ≠0

when g”(-2) ≠0 ;

thus, the order = 2

The root x = -2 , the order = 2 (x+2)2

Thus, x3 + 3x2 – 4 = (x+2)2 (x-1)

QUESTION 1 (b)

The order of the roots for h(x) = x3 - 9x2 + 27x - 27

We found that a=1, b= -9, c=27, d= -27

From calculation by calculator scientific, the roots for equation h(x) = x3 - 9x2 + 27x – 27 is:

x=3.

When h(x)=0; h(x) = x3 - 9x2 + 27x – 27 = 0

h’(x) = 3x2 - 18x + 27 = 0

h”(x) = 6x – 18 = 0

when root x=3; h(3) = (3)3 – 9(3)2 + 27(3)– 27 = 0

h’(3) = 3(3)2 – 18(3) + 27 = 0

h”(3) = 6(3) – 18 = 0

thus, the order = 3

The root x=3, the order = 3 (x-3)3

Thus, x3 - 9x2 + 27x – 27 = (x-3)3

QUESTION 2 (a)

The given approximations; and

≈ 0.01101 × 2-1

+ ≈ 0.01010 × 2-1

≈ 0.10111 × 2-1

≈ 0.01010 × 2-1

- ≈ 0.001101 × 2-1

≈ 0.000101 × 2-1

≈ 0.10111 × 2-1

+ ≈ 0.000101 × 2-1

≈ 0.110011 × 2-1

≈ 0.110011 × 2-1

- ≈ 0.001101 × 2-1

≈ 0.001110 × 2-1

Thus, the approximation for fraction 1/3 is 0.001110 × 2-1

QUESTION 2 (b)

First calculate :

And we are already found = 0.10111 × 2-1

≈ 0.10111 × 2-1

- ≈ 0.001101 × 2-1

≈ 0.100001 × 2-1

Then, calculate

≈ 0.100001 × 2-1

+ ≈ 0.100001 × 2-1

≈ 1.000010 × 2-1

Thus, the approximation for fraction 8/15 is 1.000010 × 2-1

QUESTION 2 (c)

≈ 0.01101 × 2-2 ≈ 0.001101× 2-1

+ ≈ 0.01101 × 2-1 ≈ 0.01101 × 2-1

≈ 0.100111 × 2-1

≈ 0.100111 × 2-1

- ≈ 1.010× 2-3 ≈ 0.01010 × 2-1

≈ 0.010011 × 2-1

Thus, the approximation for fraction 1 /10 + 1/5 - 1/6 is 0.010011 × 2-1

QUESTION 3 (a)

An infinite geometric series can be written as:

cr0 + cr1 + cr2 + …

cr0 = c =

Hence, we found c = . So that, we can find the value for r1.

So, by replacement c = into cr1 :

cr1 =

(1/4) r1 =

r1 = ÷

= × 4

Thus, r1 =

When |r| <1; observe that c= and r1 = into Equation (1.8) (on page 9) :

Thus, is proven.

QUESTION 3 (b)

(www.mathsisfun.com/binary-decimal-hexadecimal-converter.html)

The decimal fraction 8/7 is = 1.14285714285714

By converting the decimal fraction to the binary;

1.14285714285714 = 1.001 001 001 001 001 001 001 001 001 001 01

The binary representation of the decimal fraction 8/7 is

= 1.001 001 001 001 001 001 001 001 001 001 01

QUESTION 4 (a)

Thus, linear equation in matrix form is

QUESTION 4 (b)

Express the coefficient matrix of above system as a product of an upper and lower triangular

matrices.

Let the diagonal ;

By comparing RHS with LHS, we have

Hence, by replacement all the values into matrix.

Thus,

References

1. Assoc Prof Dr Ishak Hashim(2011). SBMA4603 Numerical Methods. Open University

Malaysia. Meteor Doc. Sdn. Bhd. Selangor Darul Ehsan.

2. www.mathsisfun.com/binary-decimal-hexadecimal-converter.html

3. www.easycalculation.com

4. www.easysurf/fracton.htm