3. Centre Limit Theorem
• Given an equation of f(x) with an interval of
[a,b], you need to determine whether there
exist at least a real root in that interval
• CLT said that if f(a) and f(b) have opposite sign
(one is –ve and another is +ve) then there
exist at least a real root in that interval
a b
f(a) +ve f(b) -ve
Prepared by Dr. Suhaila Mohamad Yusuf
4. Bisection Method
f(x) = x3 – 3x2 + 8x - 5 c = (a + b) / 2 [0,1] ℇ=0.005
i a b f(a) f(b) c f(c)
0 0 1 -5 1 0.5 -1.625
1These f(c) > ℇ then,
0.5are from 1the -1.625
given Calculated from this stop!
1 Is this < ℇ? Yes,
0.75 -0.266
2 new interval!
0.75 interval
1 -0.266 No, next iteration!
1 equation
0.875 0.373
3 0.75 0.875 -0.266 0.373 0.8125 0.056
4 0.75 0.813How to choose 0.056new 0.7815
-0.266 the -0.103
5 0.782 0.813 interval?
-0.103 0.056 0.7975 -0.021
6 0.798 0.813 -0.021 0.056 0.8055 0.020
7 0.798 0.806 -0.021 0.02 0.802 0.002
Make sure these parts
0 Repeat all the steps
0.5 1 This is the
CAREFULLY until f(c) < ℇ
have opposite sign!
-ve f(x) -ve f(x) +ve f(x) root!!
We need to take this c value. CLT said that f(a) and f(b)
How about another one? should have opposite sign!
Prepared by Dr. Suhaila Mohamad Yusuf
5. False Position Method
f(x) = x3 – 3x2 + 8x - 5 c = [af(b) - bf(a)] / [f(b) – f(a)] [0,1] ℇ=0.005
i a b f(a) f(b) c f(c)
0 0 1 -5 1 0.833 -1.625
1 0 0.8333 -5
These are from the given 0.162 Is this < ℇ? Yes,
0.807
Calculated from this0.029 stop!
2 0 0.807
interval -5 0.029 No, next iteration!
0.802
equation 0.004
This is the
Remember how to choose the root!!
interval value? What does the CLT
said about interval?
Prepared by Dr. Suhaila Mohamad Yusuf
6. Secant Method
f(x) = sin (x) + 3x – e3 xi+2 = [xif(xi+1) – xi+1f(xi)] / [f(xi+1) – f(xi)]
x0 = 1 , x1 = 0 ℇ=0.0005
i xi xi+1 xi+2 f(xi+2)
0 1 0 0.4710 0.2652
1 0 0.4710
ℇ? Yes, stop!
These are from the given Is this <0.0295
0.3723
Calculated from this
-0.0012
2 0.4710 0.3723
No, next iteration!
0.3599
values of x0 and x1 equation
3 0.3723 interval. Take the latest 2
New 0.3599 0.3604 0.0000
values as next interval
Continue iteration
until f(xi+2) < ℇ This is the
root!!
Prepared by Dr. Suhaila Mohamad Yusuf
7. Newton’s Method
f(x) = x3 – sin x Xn+1 = xn – [f(xn) / f’(xn)] x0 = 1 ℇ=0.0005
n xn f(xn) f’(xn)
0 1 0.15853 2.45970
1 0.93555 0.01392 2.03239
This is from the given Calculated from the
2 values of x
0.92870 0.00015 1.98858
| < ℇ? Yes, stop!
0 derivative of f(x)
Is |xn+1 – xn0.92862Calculated from this
3 -0.00001 1.98807
No, next iteration!
4 0.92862 equation
Continue iteration
This < ℇ
until |xn+1 – xn|is the
root!!
Prepared by Dr. Suhaila Mohamad Yusuf
11. Power Method
• Aims to find dominant eigenvalue
(largest value of eigenvalue)
1 2 1
v(0) = (0,0,1)T
A 1 0 1
ε = 0.001
4 4 5
Prepared by Dr. Suhaila Mohamad Yusuf
12. Power Method
1 2 1 v(0) = (0,0,1)T A * v Abs max
A 1 0 1 between
ε = 0.001
4 4 5 these values
k (v(k))T (Av(k))T mk+1
0 0 0 1 -1 1 5 5
1 -0.2 0.2 1 -0.8 0.8 3.4 3.4
These values
ǁv(k+1)-v(k)ǁ < ε ?
2 -0.235 0.235 1 0.765 -0.7653.12
Note that ǁv(k+1)-0.755 is a 3.04
v(k)ǁ
3.12
3 -0.245 0.245 1 -0.755 3.04
No, next iteration
4 -0.248 0.248 1 -0.752
divided by
difference of two vectors.3.016
0.752 3.016
that value
5 -0.249 0.2492 1 2-0.751 0.751 2 3.008 3.008
v u (v1 u1 ) (v2 u2 ) ... (vn un )
6 -0.250 0.250 1 -0.750 to have 3.000
0.750 3.000
v (1) 7 v ( 0) -0.250( 0.0.250 ) 2 (1 .2 0) 2 (1 1) 2 values
2 0 0 these
Eigenvector Eigenvalue
Prepared by Dr. Suhaila Mohamad Yusuf
13. Shifted Power Method
• Aims to find smallest eigenvalue and
intermediate eigenvalue
1 2 1
v(0) = (0,1,0)T
A 1 0 1 ε = 0.001
4 4 5 λ1 = 3.0
Prepared by Dr. Suhaila Mohamad Yusuf
14. Shifted Power Method
B A I
A 3 .0 I
1 2 1 1 0 0
1 0 1 3.0 0 1 0
4 4 5 0 0 1
1 2 1 3. 0 0 0
1 0 1 0 3. 0 0
4 4 5 0 0 3 .0
2 2 1
1 3 1
4 4 2
Prepared by Dr. Suhaila Mohamad Yusuf
15. Shifted Power Method
2 2 1 v(0) = (0,1,0)T B * v Abs max
B 1 3 1 between
ε = 0.001
4 4 2 these values
k (v(k))T (Bv(k))T mk+1
0 0 1 0 2 -3 -4 -4
1 -0.5 0.75 1 1.5 -1.75 -3 -3
2 -0.5 0.583 1.166 1
These values -1.249 -2.332 -2.332
ǁv(k+1)-v(k)ǁ < ε ? 0.517
3 -0.5 Note that ǁv(k+1)-v(k)ǁ is a
1 1.072 -1.108 -2.144 -2.144
No, 4next iteration
-0.5 0.508 differencedivided by
1 of two vectors.
1.034 -1.051 -2.068 -2.068
5 -0.5 0.504 1 that value
1.016 -1.024 -2.032 -2.032
v u6 (v1 u1 ) 2 0.502 u2 )1 ... tovhave n-1.012
-0.5 (v2 2
1.008 u )
( n 2
-2.016 -2.016
7 -0.5 0.501 1 these values
1.004 -1.006 -2.008 -2.008
8 -0.5 0.5 1 1.002 -1.003 -2.004 -2.004
9 -0.5 0.5 1 1.000 -1.000 -2.000 -2.000
10 -0.5 10 1 Eigenvector
Prepared by Dr. Suhaila Mohamad Yusuf Shifted Eigenvalue
16. Shifted Power Method
• λshifted = -2.0
• λ3 = λshifted + λ1 = -2.0 + 3.0 = 1.0
1 2 1
• Intermediate λ2 A 1 0 1
λ1 + λ2 + λ3 = a11 + a22 + a33
3.0 + λ2 + 1.0 = 1 + 0 + 5 4 4 5
λ2 = 2.0 Caution!!! Use the
original matrix, A.
Not the shifted
Prepared by Dr. Suhaila Mohamad Yusuf
matrix, B.
18. Interpolation Approximation
Least Square
Newton Forward
Difference
Newton Backward
Difference
Newton Divided
Difference May need table
re-arrangement
Langrage
Prepared by Dr. Suhaila Mohamad Yusuf
19. Newton Forward Difference
k 0 1 2 3 4 5
xk 1.0 1.2 1.4 1.6 1.8 2.0
yk 0.5000 0.4545 0.4167 0.3846 0.3571 0.3333
Find y(1.1)
k xk yk ∆yk ∆2yk ∆3yk ∆4yk ∆5yk x=1.0 is
x=1.1 0 1.0 0.5000 -0.0455 0.0077 -0.0020 0.0009 -0.0007 chosen
located as ref.
1 1.2 0.4545 -0.0378 0.0057 value
This -0.0011 0.0002
here point
2 1.4 0.4167 -0.0321 0.0046 -0.0009 Repeat until because
3 1.6 0.3846 -0.0275 0.0037 this
minus last column of higher
value degree
4 1.8 0.3571 -0.0238
5 2.0 0.3333 To get this
value
Prepared by Dr. Suhaila Mohamad Yusuf
20. Newton Forward Difference
• h = 1.2 – 1.0 = 0.2 and
r = (x – x0) / h = (1.1 – 1.0) / 0.2 = 0.5
r( r 1) 2 r( r 1)( r 2 ) 3
p5 ( x ) y 0 r y0 y0 y0
2! 3!
r( r 1)( r 2 )( r 3) 4 r( r 1)( r 2 )( r 3)( r 4 ) 5
y0 y0
4! 5!
( 0.5 )( 0.5 1)
p5 (1.1) 0.5000 ( 0.5 )( 0.0455) ( 0.0077)
2
( 0.5 )( 0.5 1)( 0.5 2 ) ( 0.5 )( 0.5 1)( 0.5 2 )( 0.5 3)
( 0.0020) ( 0.0009)
6 24
( 0.5 )( 0.5 1)( 0.5 2 )( 0.5 3)( 0.5 4 )
( 0.0007)
120
0.5000 0.02275 0.0009625 0.000125 0.0000352 0.0000191
0.4761 Prepared by Dr. Suhaila Mohamad Yusuf
21. Newton Backward Difference
k 0 1 2 3 4 5
xk 1.0 1.2 1.4 1.6 1.8 2.0
yk 0.5000 0.4545 0.4167 0.3846 0.3571 0.3333
Find y(1.9)
k xk yk ∇yk ∇2yk ∇3yk ∇4yk ∇5yk x=2.0 is
0 1.0 0.5000 chosen
This value as ref.
1 1.2 0.4545 -0.0455 point
2 1.4 0.4167 -0.0378 0.0077 this
minus because
3 1.6 0.3846 value
-0.0321 0.0057 -0.0020
of higher
To get this degree
x=1.9 4 1.8 0.3571 -0.0275 0.0046 -0.0011 0.0009
located
5 2.0 0.3333 value
-0.0238 0.0037 -0.0009 0.0002 -0.0007
here
Repeat until
last column
Prepared by Dr. Suhaila Mohamad Yusuf
22. Newton Backward Difference
• h = 1.2 – 1.0 = 0.2 and
r = (x – x0) / h = (1.9 – 2.0) / 0.2 = -0.5
r (r 1) 2 r (r 1)(r 2) 3
p5 ( x) y5 r y5 y5 y5
2! 3!
r (r 1)(r 2)(r 3) 4 r (r 1)(r 2)(r 3)(r 4) 5
y5 y5
4! 5!
( 0.5)( 0.5 1)
p5 (1.9) 0.3333 ( 0.5)( 0.0238) (0.0037)
2
( 0.5)( 0.5 1)( 0.5 2) ( 0.5)( 0.5 1)( 0.5 2)( 0.5 3)
( 0.0009) (0.0002)
6 24
( 0.5)( 0.5 1)( 0.5 2)( 0.5 3)( 0.5 4)
( 0.0007)
120
0.3333 0.0119 0.0004625 0.00005625 0.000007869 0.0000191
0.3448 Prepared by Dr. Suhaila Mohamad Yusuf
23. Newton Divided Difference 1st step,
k 0 1 2 3 4 mark the
See this col like
xk 1.0 1.6 2.5 3.0 3.2
number? this start
yk 0.5000 0.3846 0.2857 0.2500 0.2381 from
Find y(1.3) f[xk].
1 2 3 4 5
k xk f[xk] f1[xk] f2[xk] f3[xk] f4[xk] Go to col xk and
5000 0
0..3846 (0.0.1923)
01099 1 1.0 0.5000 -0.1923 0.0549 -0.0137 0.0032 count down the
1.6 1.0
25 col according to
1 2 1.6
1 0.3846 -0.1099 0.0275 -0.0066
0.2857 0.3846 number on top of
2.5 1.6 2 2 2.5
3 0.2857 -0.0714 0.0170 the col. Then the
3 3.0 0.2500 last value of x
-0.0595
minus with the
4 3.2 0.2381 first value of x.
To fill in this col, we
Big problem is ‘divide
know that lower
value – upper value with what?’
Prepared by Dr. Suhaila Mohamad Yusuf
24. Newton Divided Difference
• Interpolation Polynomial expression
p4 ( x ) y 0 f [ x 0 , x1 ]( x x 0 ) f [ x 0 , x1 , x 2 ]( x x 0 )( x x1 )
f [ x 0 , x1 , x 2 , x 3 ]( x x 0 )( x x1 )( x x 2 )
f [ x 0 , x1 , x 2 , x 3 , x 4 ]( x x 0 )( x x1 )( x x 2 )( x x 3 )
• Assign the value into the polynomial
expression
p 4 (1.3 ) 0.5 ( 0.1923)(1.3 1.0 ) 0.0549(1.3 1.0 )(1.3 1.6 )
( 0.0137)(1.3 1.0 )(1.3 1.6 )(1.3 2.5 )
0.0032(1.3 1.0 )(1.3 1.6 )(1.3 2.5 )(1.3 3.0 )
0.5 0.05769 0.004941 0.0014796 0.00058752
0.4353
Prepared by Dr. Suhaila Mohamad Yusuf
25. Newton Divided Difference
k 0 1 2 3 4
xk 1.0 1.6 2.5 3.0 3.2
yk 0.5000 0.3846 0.2857 0.2500 0.2381
Find y(2.8)
• Re-arrange the table then do as previous
• How to re-arrange table? 2.8 is in here
k 0 1
Value of x before 2Decending value of 4
3
xk and after 2.8 1.6
1.0 2.5 3.0
remaining x 3.2
yk 0.5000 0.3846 0.2857 0.2500 0.2381
k 0 1 2 3 4
xk
yk Prepared by Dr. Suhaila Mohamad Yusuf
26. Langrage
k 0 1 2 3 4
xk 1.0 1.6 2.5 3.0 3.2
yk 0.5000 0.3846 0.2857 0.2500 0.2381
Find y(1.3)
n
pn ( x ) L 0 ( x )y 0 L1( x )y 1 .......L n ( x )y n L i ( x )y i
i 0
4
p4 ( x ) L i ( x )y i
i 0
p4 ( x ) L 0 ( x )y 0 L 1( x )y 1 L 2 ( x )y 2 L 3 ( x )y 3 L 4 ( x )y 4
Prepared by Dr. Suhaila Mohamad Yusuf
27. Langrage
Because it is L0, x0 is nowhere to
• Calculate L0 be found up here
(1.3 x1 )(1.3 x 2 )(1.3 x 3 )(1.3 x 4 )
L 0 ( 1.3 )
( x 0 x1 )( x 0 x 2 )( x 0 x 3 )( x 0 x 4 )
(1.3 1.6 )(1.3 2.5 )(1.3 3.0 )(1.3 3.2 )
0.2936
(1.0 1.6 )(1.0 2.5 )(1.0 3.0 )(1.0 3.2 )
Because it is L0, x0 is deducted
• With the same with other x
method, calculate L1,
L2 ,L3 ,L4
Prepared by Dr. Suhaila Mohamad Yusuf
28. Langrage
4
p 4 (1.3 ) L i (1.3 )y i
i 0
0.2936( 0.5000) 0.9613 0.3846) 0.6152( 0.2857)
(
0.7329( 0.2500) 0.3726( 0.2381)
0.4353
Prepared by Dr. Suhaila Mohamad Yusuf
29. Least Square
• Determine the appropriate linear polynomial
expression, p(x) = a0 + a1x based on the following data:
• Determine f(2.3)
x 1 2 3 4 5
f(x) 0.50 1.40 2.00 2.50 3.10
s 0 s1 a 0 v0
s1 s 2 a 1 v1
Prepared by Dr. Suhaila Mohamad Yusuf
30. Least Square
xk 0 xk 1 xk 2 fk xk 0 f k xk 1 f k
1 1 1 0.5 0.5 0.5
1 2 4 1.4 1.4 2.8
1 3 9 2.0 2.0 6.0
1 4 16 2.5 2.5 10.0
1 5 25 3.1 3.1 15.5
5 15 55 - 9.5 34.8
s 0 s1 a 0 v0 5 15 a 0 9.5
s1 s 2 a 1 v1 15 55 a 1 34.8
Prepared by Dr. Suhaila Mohamad Yusuf
31. Least Square
5 15 a 0 9.5
15 55 a 1 34.8
• Solution, a0 = 0.01 dan a1 = 0.63
• Therefore, the polynomial expression is p(x) =
0.01x + 0.63
• To determine f(2.3):
p( 2.3) 0.01( 2.3) 0.63 0.653
f ( 2.3) p( 2.3) 0.653
Prepared by Dr. Suhaila Mohamad Yusuf
32. CHAPTER 9
NUMERICAL
DIFFERENTIATION
Prepared by Dr. Suhaila Mohamad Yusuf
33. Forward Backward Forward Backward Central Forward Central
Difference Difference Difference Difference Difference Difference Difference
2-point 3-point 5-point
Formula Formula Formula
FIRST
DERIVATIVE
f‘(x)
SECOND
DERIVATIVE
f‘’(x)
NUMERICAL
INTEGRATION 3-point
Formula 5-point
Formula
Central
Difference Central
Difference
Prepared by Dr. Suhaila Mohamad Yusuf
34. SORRY!!! I DIDN’T PREPARE
ANYTHING. THIS CHAPTER IS TOO
EASY. MAKE SURE YOU KNOW
WHICH FORMULA TO USE..
Prepared by Dr. Suhaila Mohamad Yusuf
35. CHAPTER 10
NUMERICAL
INTEGRATION
Prepared by Dr. Suhaila Mohamad Yusuf
37. Trapezoidal Rule
• Approximate the following integral using the
Trapezoidal rule with h=0.5
4 x
dx
1
x 4
b a 4 1
N 6
h 05
.
Prepared by Dr. Suhaila Mohamad Yusuf
38. Trapezoidal Rule
4 x 4
xi dx f ( x) dx
i xi f ( xi ) 1
x 4 1
xi 4 h
f f6 2 f1 f2 f3 f4 f5
0 1.0 0.4472 2 0
1 1.5 0.6396 0.5
1.8614 2 4.8486
2 2.0 0.8165 2
3 2.5 0.9806 2.8896
4 3.0 1.1339
5 3.5 1.2780
6 4.0 1.4142
Total 1.8614 4.8486
prepared by Razana Alwee
1st and last In-between
values values
Prepared by Dr. Suhaila Mohamad Yusuf
39. Simpson’s 1/3 Rule
• Approximate the following integral using the
Trapezoidal rule with h=0.5
4 x
dx
1
x 4
b a 4 1
N 6
h 05
.
Prepared by Dr. Suhaila Mohamad Yusuf
40. Simpson’s 1/3 Rule 3 2
4 x h
dx f0 f6 4 f 2i 1 2 f 2i
1
x 4 3 i 1 i 1
0.5
xi 1.8614 4(2.8982) 2(1.9504)
i xi fi f ( xi ) 3
xi 4
0 1.0 0.4472 2.8925
1 1.5 0.6396
2 2.0 0.8165
3 2.5 0.9806
4 3.0 1.1339
5 3.5 1.2780
6 4.0 1.4142
Total 1.8614 2.8982 1.9504
1st and last Odd Even
Prepared by Dr. Suhaila Mohamad Yusuf
values column column
41. Simpson’s 3/8 Rule
• Approximate the following integral using the
Trapezoidal rule with h=0.25
4 x
dx
1
x 4
b a 4 1
N 12
h 0.25
Prepared by Dr. Suhaila Mohamad Yusuf
42. Simpson’s 3/8 Rule
4 3
4 x 3h
dx f0 f12 3 f 3i 2 f 3i 1 2 f 3i
1
x 4 8 i 1 i 1
xi
i xi fi f ( xi )
xi 4 3(0.25)
0 1.00 0.4472
1.8614 3(7.7191) 2(2.9174)
8
1 1.25 0.5455
2 1.50 0.6396 2.8925
3 1.75 0.7298
4 2.00 0.8165
5 2.25 0.9000
6 2.50 0.9806
7 2.75 1.0586
8 3.00 1.1339
9 3.25 1.2070
10 3.50 1.2780
11 3.75 1.3470
12 4.00 1.4142
Total 1.8614 7.7191 2.9174
‘Ganda
1st and Remaining
3’ (3,6,9)
last values values Prepared by Dr. Suhaila Mohamad Yusuf
43. Romberg Integration
Formula for Romberg Table
i hi Ri,1 Ri,2 Ri,3
1 b a h1
R1,1 f0 f1
2
2 1
h1
2
3 1
h2
2
1 2 i 2
4 j 1 Ri , j 1 Ri 1, j 1
Ri ,1 Ri hi f 2k Ri , j
2
1,1 1
k 1
1 4j 1
1
Prepared by Dr. Suhaila Mohamad Yusuf
44. Romberg Integration
• Use Romberg integration to approximate
4 x
dx
1
x 4
• Compute the Romberg table until
|Ri,j Ri,j-1|<0.0005
Prepared by Dr. Suhaila Mohamad Yusuf
45. Romberg Integration
How to calculate
4 x these?
dx b a 4 1
1 N 1
x 4 From the h here, we h 3
can know the N
h1 b a 4 1 3 (number of segments)
Integration starts from Therefore, we only have
1 to 4 and only has 1 f0 and f1, the first node
h1 and the last node
R1,1 f 0 f1 segment
2 1 4
h=3
3
0.4472 1.4142 f1
2 f0
4 1 4 4
2.7921. 1
1 4
0.4772 1
4 4
1.4142
Find the f0 = f(1) and f1
= f(4) using this f(x)by Dr. Suhaila Mohamad Yusuf
Prepared
46. Romberg Integration
i hi Ri,1 Ri,2 Ri,3
1 3 2.7921
2 1.5 h2 is half of h1
Now we need
3 to fill in the 2nd
row
This value needs to be
Fill in the Rombergcalculated using other
Table with answers that
you’ve got previously
formula
Prepared by Dr. Suhaila Mohamad Yusuf
47. Romberg Integration
How to calculate
4 x this?
dx Remember!! b a 4 1
1 N 2
x Every time you calculatewe R ,
4 From the h here, h 1.5
can know the N new xx
1 your f , f , ...(number ofbe changed!!
fx will segments)
h2 h1 1.5 0 1
2
Draw diagram to be safe!! , we have three
Integration starts from Therefore
1 1
1 to 4 and has 2 nodes f0 ,f1, and f2,
R2,1 R1,1 h1 f 2k
2 k 1
1
segments
1
R1,1 h1 ( f1 ) 1 h = 1.5 2.5 4
2
1 f0 f1 f2
R1,1 h1 ( f (2.5))
2 4 2 .5
1 0.9806
1
2.7921 3 0.9806 2 .5 4
2
2.8670 Find the f1 = f(2.5) using
Prepared by Dr. Suhaila Mohamad Yusuf this f(x)
48. Romberg Integration
• Calculate R2,2 using another formula
4 R2,1 R1,1 4 2.8670 2.7921
R2 , 2 2.8920
3 3
• Is |Ri,j Ri,j-1|<0.0005? Only compare with R in
the same row, not the same column
R2, 2 R2,1 2.8920 2.8670 0.025 0.0005
Prepared by Dr. Suhaila Mohamad Yusuf
49. Romberg Integration
i hi Ri,1 Ri,2 Ri,3
1 3 2.7921
2 1.5 2.8670 2.8920
3 Fill in0.75Romberg half ofwith answers that
the h3 is Table h2
you’ve got previously
Now we need
to fill in the 3rd
This value row to be
needs
calculated using other
formula
Prepared by Dr. Suhaila Mohamad Yusuf
50. Romberg Integration
How to calculate
4 x this?
dx b a 4 1
1 N 4
x 4 From the h here, we h 0.75
can know the N
1 (number of segments)
h3 h2 0.75
2 Integration starts from Therefore, we have five
1 2 1 to 4 and has 4 nodes f0 ,f1, f2 ,f3 and f4,
R3,1 R2,1 h2 f 2 k 1 segments
2 k 1
1 1 1.75 2.5 3.25 4
R2,1 h2 ( f1 f 3 )
2 h = 0.75
1 f0 f1 f2 f3 f4
R2,1 h2 ( f (1.75) f (3.25)) 4 1.75
2 0.7298
4 3.25
1
1.2070
1 1.75 4 1
3.25 4
2.8670 1.5 0.7298 1.2070
2 Find the f1 = f(1.75) and
2.8861 Prepared by Dr. Suhaila Mohamad Yusuf f3 = f(3.25) using this f(x)
51. Romberg Integration
• Calculate R3,2 using another formula
4 R3,1 R2,1 4 2.8861 2.8670
R3, 2 2.8925
3 3
• Is |Ri,j Ri,j-1|<0.0005? Only compare with R in
the same row, not the same column
R3, 2 R3,1 2.8925 2.8861 0.0064 0.0005
Prepared by Dr. Suhaila Mohamad Yusuf
52. Romberg Integration
• Calculate R3,3 using another formula
4 R3, 2 R2, 2 16 2.8925 2.8920
R3,3 2.8925
15 15
• Is |Ri,j Ri,j-1|<0.0005? Only compare with R in
the same row, not the same column
R3,3 R3, 2 2.8925 2.8925 0.0000 0.0005
Stop Iteration!!
Prepared by Dr. Suhaila Mohamad Yusuf
53. Romberg Integration
i hi Ri,1 Ri,2 Ri,3
1 3 2.7921
2 You 1.5
don’t need to generate
2.8670 2.8920
Trapezoidal Table!
3 0.75 2.8861 2.8925 2.8925
Just follow my way..
The solution of
Fill in the Romberg Table with answers that
integration is here
you’ve got previously
Prepared by Dr. Suhaila Mohamad Yusuf
