NUMERICAL METHOD PDF

TOPIC 2:
NUMERICAL METHOD

BFMI / DEC 2020 / DBM30043

Solution of • Gaussian Elimination Method
Linear • LU Decomposition Method
• ( Crout & Doolittle)
Equation

Solution of • Newton Raphson Method
Polynomial • Fixed Point Iteration Method

Equation

BFMI / JUN 2020 / DBM30043

GAUSSIAN ELIMINATION METHOD

BBFMFMI /I J/ UJUNN20220020/ D/ DBBMM3030040343

Gaussian Elimination Method
STEPS OF USING GAUSSIAN
ELIMINATION METHOD

STEP 1: Change Into Matrix Form

STEP 2: Transform 1

STEP 3: Transform 2

STEP 4: Find Value Of Variables Of Matrix X

BFMI / JUN 2020 / DBM30043

Gaussian Elimination Method 1

a11x1  a12 x2  a13x3  b1
a21x1  a22 x2  a23x3  b2
a31x1  a32 x2  a33x3  b3

STEP 1: CHANGE INTO MATRIX FORM

a11 a12 a13   x1  b1 
a21    b2 
a22 a23   x2   

a31 a32 a33   x3  b3 

BFMI / JUN 2020 / DBM30043

Gaussian Elimination Method

STEP 2: TRANSFORM 1(involving rows 2 and 3)

Make a21 = 0 Make a31 = 0

a '21  a21  a11  a21   0 a ' 31  a31  a11 a31   0
a11 a11

a '22  a22  a12  a21   ? a '32  a32  a12  a31   ?
a11 a11

a '23  a23  a13  a21   ? a ' 33  a33  a13 a31   ?
a11 a11

b'2  b2  b1  a21   ? b'3  b3  b1  a31   ?
a11 a11

BFMI / JUN 2020 / DBM30043

Gaussian Elimination Method

The 1st level transform:

BFMI / JUN 2020 / DBM30043

Gaussian Elimination Method

Make a21 = 0

The 2nd level transform:

BFMI / JUN 2020 / DBM30043

Gaussian Elimination Method

STEP 4: FIND THE VALUE x1 , x2 ,x3

BFMI / JUN 2020 / DBM30043

Gaussian Elimination Method (Method 1)

Example 1 : Solve this 3 simultaneous linear equation using Gaussian Elimination Method

STEP 1: CHANGE INTO MATRIX FORM

 3 2 1  x  10
7 1 6  y = 8 
 3 0 2   z   5 

BFMI / JUN 2020 / DBM30043

Gaussian Elimination Method

STEP 2: TRANSFORM 1(involving rows 2 and 3)

Make a21 = 0 Make a31 = 0

BFMI / JUN 2020 / DBM30043

Gaussian Elimination Method

The 1st level transform:

BFMI / JUN 2020 / DBM30043

Gaussian Elimination Method

STEP 3: TRANSFORM 2 (involving row 3)
Make a32 = 0

BFMI / JUN 2020 / DBM30043

The 2nd level transform:
BFMI / JUN 2020 / DBM30043

Gaussian Elimination Method

STEP 4: FIND THE VALUE x , y , z

BFMI / JUN 2020 / DBM30043

GUaSuINsGsiaGnAEUlSimSiInAaNtioEnLIMeINthAoTdION (METHOD 2)
USING GAUSSIAN ELIMINATION (METHOD 2)

a11x1  a12 x2  a13x3  b1
a21x1  a22 x2  a23x3  b2
a31x1  a32 x2  a33x3  b3

STEP 1: CHANGE INTO MATRIX FORM

a11 a12 a13   x1  b1 
a21    b2 
a22 a23   x2   

a31 a32 a33   x3  b3 

BFMI / JUN 2020 / DBM30043

STEP 2: CREATE THE TABLE

A B Elimination Operation
Coefficient

 a11 a12 a13   b1  a21 a '21  a21  a11  a21   0
 a21 a22 a23  b2  a11 a11
 a31 a32 a33   b3 
a '22  a22  a12  a21   ?
a11

a '23  a23  a13  a21   ?
a11

b'2  b2  b1  a21  ?
a11

 a11 a12 a13   b1  a ' 31  a31  a11 a31   0
0 a'22 a'23  b2  a11
 a31 a33  b3  a31
a32 a11 a '32  a32  a12  a31   ?
a11

a ' 33  a33  a13  a31   ?
a11

b'3  b3  b1  a31   ?
a11

 a11 a12 a13   b1  , a"33  a '33  a '23  a '32   ?
0 a ' 22 a'23  b'2  a '22
a 32
a2, 2

a "  a ' 32  a ' 22  a ' 32   0
32 a ' 22
 0 a '32 a '33   b'3 
 a '32 
b"3  b'3  b'2 a '22 ?

 a11 a12 a13   b1 ' 
0 a ' 22 a ' 23  b 2 

 0 0 a"33  b"33 
BFMI / JUN 2020 / DBM30043

Gaussian Elimination Method

STEP 3: CHANGE INTO NEW MATRIX FORM TO FIND THE
VALUE OF X1, X2, X3

STEP 4: FIND THE VALUE x1 , x2 ,x3

BFMI / JUN 2020 / DBM30043

Gaussian Elimination (Method 2)

Example 2: Solve this 3 simultaneous linear equation using Gaussian Elimination Method

x1  2x2  5x3  6
4x1  4x2  3x3  7
5x1  7x2 13x3  9

STEP 1: CHANGE INTO MATRIX FORM

 1 2 5   x1  =  6
4 4 3   x2  7

 5 7  13  x3   9

BFMI / JUN 2020 / DBM30043

STEP 2: CREATE THE TABLE

A B Elimination Operation
Coefficient

 1 2  5   6 4  4  4(1)  0
4 4 3   7 
 5 7 13  9 4 4  4  4(2)  4
1  3  3  4(5)  17

7  7  4(6)  31

1 2  5   6 5  5  5(1)  0
0  4 17   31 
5 7 13  9 5 7  7  5(2)  3
1 2  5  6 1 13  13  5(5)  12
0  4 17   31 
0  3 12   21   9  9  5(6)  21

 3  3  3 (4)  0
−3 4
12  12  3 (17)   3
−4 44

21  21  3 (31)   9
44

 1 2  5    6 
 
 0 4 17   31 
0 3   9 
0  
 4  4 

BFMI / JUN 2020 / DBM30043

Gaussian Elimination Method

STEP 3: CHANGE INTO NEW MATRIX FORM TO FIND
THE VALUE OF X1, X2, X3

 1 2  5   x1    6 
 4   x2 
 0 0 17   x3    31 
3   9 
0  
 4  4

STEP 4: FIND THE VALUE x1 , x2 ,x3

x1  2x2  5x3  6  4x2  17 x3  31  3 x3  9
x1  (2x5)  (5x3)  6  4x2  17(3)  31 4 4
x1  1 x2  5
x3  3

BFMI / JUN 2020 / DBM30043

LU DECOMPOSITION METHOD

Method Lower (L) Upper (U) Multiplication
Crout Method
 l11 0 0   1 u12 u13   l11 l11u12 l11u13 
Doolittle Method  l21 l22 0  0 1 u23 
 l31 l32 l33   0 0 1   l21 l u21 12  l22 l u21 13  l22u23 

 1 0 0 u11 u12 u13   l31 l u31 12  l32 l31u13  l32u23  l33 
 l21 1 0  0 u22 u23 
 l31 l32 1   0 0 u33   u11 u12 u13 
 l21u11 l21u12  U 22 l21u13  u23 
 l31u11 l31u12  l32u22 l31u13  l32u23  U 33 

BFMI / JUN 2020 / DBM30043

Apply LU Decomposition using Crout Method

a11x1  a12 x2  a13x3  b1
a21x1  a22 x2  a23x3  b2
a31x1  a32 x2  a33x3  b3

STEP 1: CHANGE INTO MATRIX FORM

a11 a12 a13   x1  b1 
a21    b2 
a22 a 23   x 2   

a31 a32 a33   x3  b3 

BFMI / JUN 2020 / DBM30043

Apply LU Decomposition using Crout Method

STEP 2: Find LU

A= L U

 a 11 a12 a 13   ll1211 0 0   1 u12 u13 
 a 21 a 22 a 23  l 22 0   0 1 u23 

 a31 a32 a33  l31 l32 l33   0 0 1 

 l11 l11u12 l11u13 

 l21 l21u12  l22 l u21 13  l22u23 

 l31 l31u12  l32 l31u13  l32u23  l33 

a11 = l11 a12 = l11u12 a13 = l11u13

a21 = l21 a22 = l21u12 + l22 a23 = l21u13 + l22u23

a31 = l31 a32 = l31u12 + l32 a33 = l31u13 + l22u23 + l33

BFMI / JUN 2020 / DBM30043

Apply LU Decomposition using Crout Method

STEP 3: Find Ly=B

L =y B

 1 0 0    y1   b1 
 l 21 1 0   y2   b2 

l31 l32 1  y3  b3 

STEP 4: Ux=y

U x= B

 1 u12 u13  x1    y1 
 0 1 u23  x2   y2 

 0 0 1  x3   y3 

BFMI / JUN 2020 / DBM30043

Apply LU Decomposition using Crout Method

Example 3: Solve this 3 simultaneous equation below using Crout method
3 p  6q  5r  6
 4q  3r  4  0
4 p  8q  8r  10

STEP 1: CHANGE INTO MATRIX FORM

3  6 5  p  6 

0  4 3   q    4 
    

4 8  8  r  10

BFMI / JUN 2020 / DBM30043

Apply LU Decomposition using Crout Method

STEP 2: Find LU

A = L U

 a 11 a12 a13    l 11 0 0   1 u12 u13 
 a 21 a 22 a 23   l 21 l 22 0   0 1 u23 
a 32
 a31 a33  l31 l32 l33  0 0 1 

 3  6 5   l11 l11u12 l11u13 

 0  4 3   l21 l21u12  l22 l u21 13  l22u23 
 4 8  8 l31 l31u12  l32 l31u13  l32u23  l33 

a11 = l11 a12 = l11u12 a13 = l11u13
-6= (3)u12
3 = l11 12 = −2 5 = (3)u13
13 = 5/3
a21 = l21 a22 = l21u12 + l22 3 = l21u13 + l22u23
3 = 0 5/3 + −4 23
0 = l21 −4 = 0(−2) + 22 23 = −3/4
22 = −4
a31 = l31 a33 = l31u13 + l32u23 + l3
a32 = l31u12 + l32
1 = l31 −8 = 1 5/3 + 10 −3/4 + 33
8=(1)(-2)+ l32 l33= -13/6

10= l32

BFMI / JUN 2020 / DBM30043

Apply LU Decomposition using Crout Method

STEP 3: Find Ly=B

L =y B

 l11 0 0    y1   b1 
 l 21 l 22 0   y2   b2 

l31 l32 l33   y3  b3 

STEP 4: Ux=y

U x= B

 1 u12 u13  x1    y1 
 0 1 u23  x2   y2 

 0 0 1  x3   y3 

BFMI / JUN 2020 / DBM30043

Apply LU Decomposition using Crout Method

STEP 3: Find Ly=B

L =y B

 3 0 0   y1   6 
 0 -4 0    y2   4 

3 y1  6  1 10 -13/6   y3  10    13  y3  10
y1 2  6 
 4 y2  4 y1  10 y2

y 2 1  13 
 6 
2  10(1)  y3  10

 108  y3
13

STEP 4: Ux=y

U x= B x1  2(0.25)  5 (8.31)  2 0 x2  3 x3  1 0  0  1x3   108
3 4 13
 1 5 
 -2  x1  0.5 13.85  2 3 x3  8.31
0 3  x1   2  x2  4  1
 1 3 x2   
0 0  4 x3    1  x1  15.35 x2  0.25
  108 
1   
13
 BFMI / JUN 2020 / DBM30043

Apply LU Decomposition using Doolittle Method

a11x1  a12 x2  a13x3  b1
a21x1  a22 x2  a23x3  b2
a31x1  a32 x2  a33x3  b3

STEP 1: CHANGE INTO MATRIX FORM

a11 a12 a13   x1  b1 
a21    b2 
a22 a 23   x 2   

a31 a32 a33   x3  b3 

BFMI / JUN 2020 / DBM30043

Apply LU Decomposition using Doolittle Method

STEP 2: Find LU

A= L U

 a11 a12 a13    1 0 0   u11 u12 u13 
 a 21 a 22 a 23   l 21 1 0   0 u22 u23 

 a31 a32 a33  l31 l32 1  0 0 u33 

 u11 u12 u13 
 l21u11 l21u12  U 22 l21u13  u23 
 l31u11 l31u12  l32u22 l31u13  l32u23  U33 

a11 = l11 a12 = u12 a13 = u13

a21  l u21 11 a22 = l21u12 + U22 a23 = l21u13 + u23
a31  l u31 11
a32 = l31u12 + l32 U22 a33 = l31u13 + l32u23 + 33

BFMI / JUN 2020 / DBM30043

Apply LU Decomposition using Doolittle Method

STEP 3: Find Ly=B

L =y B

 1 0 0    y1   b1 
 l 21 1 0   y2   b2 

l31 l32 1  y3  b3 

STEP 4: Ux=y

U x= B

 u11 u12 u13  x1    y1 
 0 u22 u23  x2   y2 

 0 0 u33  x3   y3 

BFMI / JUN 2020 / DBM30043

Apply LU Decomposition using Doolittle Method

Example 4: Solve this 3 simultaneous equation below using Doolittle method

3x  2 y  z  10
7x  y  6z  8
3x  2z  5

STEP 1: CHANGE INTO MATRIX FORM

 3 2 1  x  10
7 1 6  y  8 
 3 0 2   z   5 

BFMI / JUN 2020 / DBM30043

Apply LU Decomposition using Doolittle Method

STEP 2: Find LU

A= L U

 a11 a12 a13    1 0 0   u11 u12 u13 
 a 21 a 22 a 23   l 21 1 0   0 u22 u23 

 a31 a32 a33  l31 l32 1  0 0 u33 

 3 2  1  u11 u12 u13 

 7  1 6   l21u11 l21u12  U 22 l21u13  u23 

 3 0 2   l31u11 l31u12  l32u22 l31u13  l32u23  U33 

a11 = u11 a12 = 2 a13 = -1
3 = u11
a22 = l2731u122+ u22 6  l21u13  u23
a21 = l21u11 -1  u
7 = l21(3) 22 6   7 1  u23
 3
7   17 
21 = 3 u22 3
25
a31  l u31 11 u23  3
3  l313
l31  1 a31  l u31 11 0  l21u13  l32u23 2  l31u13  l32u23  u33

0  12  l32   17  2  11   6  25   u33
 3   17  3 
a 21 l u21 11
6  1
17 u33  17
l32  BFMI / JUN 2020 / DBM30043

Apply LU Decomposition using Doolittle Method

STEP 3: Find Ly=B

L =y B

 1 0 0    y1  180 
 7 1 0   y2  
   y3 
3 1 
 6   5 
 1 17 

1 = 10 7 6
3 1 + 2 = 8 1 + 17 2 + 3 = 5

7 6 46
2 = 8 − 3 10 3 = 5 − 10 − 17 − 3

46 7
2 = − 3 3 = 17

STEP 4: Ux=y

U x= B 17 25 46

 3 2 -1    17 − 3 + 3 = − 3 3 + 2 − = 10
 25    17 = 17 221 17 10 − 2 13 + (7)
3 x   10  = − 3 ÷ − 3
17 1 y    46  = 7 = 3
3 z    − 3
0 -  3
 7 = 13

 0 0 17   17 


CONSTRUCT SOLUTIONS OF POLYNOMIAL
EQUATIONS

BFMI / JUN 2020 / DBM30043

Fixed Point Iteration Method

Step in Fixed Point Iteration Method

STEP 1 Make x as a subject x = g(x)
STEP 2
Initial approximate value x0

 Use the given x0
 If x0 is not given, assume x0 = 1

Substitute x0 in x = g(x) to find x1
Substitute x1 in x = g(x) to find x2

STEP 3 Find Ixn+1- xnI

STEP 4 Repeat second and third step until
Ixn+1- xnI < (depends on the required accuracy)

BFMI / JUN 2020 / DBM30043

Fixed Point Iteration Method

Example 5: Solve 5xex = 1 using fixed point iteration method. Give your answer correct
to 3 decimal places. Consider xo= 1

Solution :

First step : Make x as a subject

5xex  1

x  1

5e x

Second step : Substitute x0 =1 in x = g(x) to find x1

x  1  0.074
1
5e(1)

Third step : Find Ixn+1- xnI <0.001 FALSE. More iterations
Ix1- x0I <0.001 needed because the

I 0.074-1 I < 0.001 value is still larger than
I -0.926 I <0.001 0.001

BFMI / JUN 2020 / DBM30043

Fixed Point Iteration Method

Forth step : Repeat second and third step till Ixn+1- xnI <0.001

n − +1 − < 0.001
+1 = 5
01 0.926
1 0.074 0.074 0.112
2 0.186 0.186 0.019
3 0.167 0.167 0.003
4 0.170 0.170 0.001
0.169

TRUE, The value
is smaller than

0.001

Therefore, the approximate root is x5 = 0.169

BFMI / JUN 2020 / DBM30043

Fixed Point Iteration Method

Example 6 : Given x3  6x  4 has an approximate rootx0  1. By using the fixed point iteration
method, find the root up to three decimal places

Solution :
First step : Make x as a subject
Second step : Substitute x0 =1 into 3 6x  4 to find x1

Third step : Find Ixn+1- xnI <0.001
Ix1- x0I <0.001

I 2.154 -1 I < 0.001
I 1.154 I < 0.001

BFMI / JUN 2020 / DBM30043

Fixed Point Iteration Method

Forth step : Repeat second and third step till Ixn+1- xnI <0.001

n xn1  3 6x  4 +1 − < 0.001

01 2.154 1.154
1 2.154 2.567 0.413
2 2.567 2.685 0.118
3 2.685 2.719 0.034
4 2.719 2.729 0.010
5 2.729 2.731 0.002
6 2.731 2.731 0.000

Therefore, the approximate root is x6 = 2.731

BFMI / JUN 2020 / DBM30043

Newton Raphson Method

Step in Newton Raphson Method (If x0 is not given)

STEP 1 Find x0 (change in sign)

x0  1 x1 y1
y2  x1 x2 y2

STEP 2 Differentiate f(x)

STEP 3 Substitute into table

STEP 4 Repeat second and third step until
Ixn+1- xnI < (depends on the required accuracy)

BFMI / JUN 2020 / DBM30043

Newton Raphson Method

Example 7 : Solve 3x3  2x  4  0 using the Newton Raphson method. Give your answer

correct to four decimal places.

Solution : Substitute the value of x,
f (x)  3x3  2x  4
First step : Find x0 (change in sign) f (0)  3(0)  2(0)  4  4
f (1)  3(1)3  2(1)  4  1
x 01

f (x)  4 1

Using the formula x0  1 x1 y1
y2  x1 x2 y2

x0  1 1 0 4
(4) 1 1

 0.8

Second step : Differentiate f (x)  3x3  2x  4  0

f ,(x)  9x2  2

BFMI / JUN 2020 / DBM30043

Newton Raphson Method

Third step :Substitute into table

f (x)  3x3  2x  4  0 f ,(x)  9x2  2 xn1  xn  f (xn )
f '(xn )

f (x0 )  3(0.8)3  2(0.8)  4 f , (x)  9(0.8)2  2 x1  0.8    0.864 
 0.864  7.76  7.76 

 0.9113

f (x1)  3(0.9113)3  2(0.9113)  4 f , (x1)  9(0.9113)2  2 x2  0.9113   0.0930 
 9.4742 
 0.0930  9.4742
 0.9015

BFMI / JUN 2020 / DBM30043

Newton Raphson Method

Fourth step: Repeat second and third step until Ixn+1- xnI < 0.0001

Ixn+1- xnI < 0.0001
Ix1- x0I <0.0001

I 0.9113 - 0.8 I < 0.0001

I 0.1113 I < 0.0001 FALSE

n xn1  xn  f (xn ) +1 − < 0.001
f '(xn )

0 0.8 0.9113 0.1113

1 0.9113 0.9015 0.0098

2 0.9015 0.9014 0.0001
3 0.9014 0.9014 0.0000

Therefore, the approximate root is x4 = 0.9014

BFMI / JUN 2020 / DBM30043

Newton Raphson Method

Example 8 : Given the function f (x)  x4  x  2 :
a) Show that there is a real root for f(x) =0 between x=1 and x= 1.5
b) Find the real roots of f(x)= 0 using Newton Raphson method , correct to
three decimal places

Solution :

First step : Find x0 (change in sign)

x 1 1.5 Substitute the value of x,
f (x)  2 1.5625 f (x)  x4  x  2
f (1)  14  (1)  4  2
f (1.5)  (1.5)4  (1)  2  1.5625

Since three is a change in the sign of f(x) between x= 1 and x =1.5, there is a real root for f(x)=0
between x=1 and x =1.5.

Using the formula x0  1 x1 y1
y2  x1 x2 y2

x0  1 0 2 BFMI / JUN 2020 / DBM30043
1.5625  (2) 1.5 1.5625

 1.281

Newton Raphson Method

Second step : Differentiate f (x)  x4  x  2
f '(x)  4x3 1

Third step : Substitute into table

f (x)  x4  x  2 f '(x)  4x3 1 xn1  xn  f (xn )
f '(xn )

f (x0 )  (1.281)4  (1.281)  2 f '(x0 )  4(1.281)3 1 x1  1.281    0.588 
 7.408 
 0.588  7.408

BFMI / JUN 2020 / DBM30043

Newton Raphson Method

Fourth step: Repeat second and third step until Ixn+1- xnI < 0.0001

Ixn+1- xnI < 0.0001
Ix1- x0I <0.0001

I 1.360-1.281I < 0.0001

I 0.079 I < 0.0001 FALSE

n xn1  xn  f (xn ) +1 − < 0.001
f '(xn )
0 1.281 0.0790
1 1.360 1.360 0.0070
2 1.353 0.0000
1.353

1.353

Therefore, the approximate root is x3 = 1.353

BFMI / JUN 2020 / DBM30043

Newton Raphson Method

Step in Newton Raphson Method (If x0 is given)

STEP 1 Differentiate f(x)
STEP 2
STEP 3 Substitute into table

Repeat second and third step until
Ixn+1- xnI < (depends on the required accuracy)

BFMI / JUN 2020 / DBM30043

Newton Raphson Method

Example 9 : Solve 3x3  2x  4  0using the Newton Raphson method. Give your answer

correct to three decimal places. Consider x0  1

Solution : First step : Differentiate f (x)  3x3  2x  4  0
f ,(x)  9x2  2

Second step : Substitute into

f (x)  3x3  2x  4 f '(x)  9x2  2 xn1  xn  f (xn )
f (x0 )  3(1)3  2(1)  4 f ' (x)  9(1)3  2 f '(xn )

xn 1  1   1 
1

 0.909

BFMI / JUN 2020 / DBM30043