chapter 3
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CHAPTER 3
NUMERICAL METHOD
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3.1 The Trapezium Rule
• OBJECTIVES : The objectives of this lesson
are to enable students to
• a) derive the trapezium rule by dividing the area
represented by dx into n ( n 7 )
trapezium each with width h.
• b) use the trapezium rule to approximate
b
a
)x(f
b
a
.dx)x(f
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Introduction
• What happens when a mathematical function cannot be
integrated ? In these cases a numerical method can be used
to find an approximate value for the integral
• As the definite integral is a number which represents the
area between y = f (x) , the x-axis and the lines
• x = a and x =b .
b
a
dxxf .)(
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• Therefore , even if cannot be
found, approximate value for can be
found by evaluating the appropriate area using
another method. The two common methods are the
trapezium rule and Simpson’s rule.
• However, here we will discuss the trapezium rule as
requested by our syllabus.
b
a
.dx)x(f
b
a
.dx)x(f
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The trapezium rule
• This method divides the area A under the curve
• y = f(x) is into n vertical strips. The width of each
strip is h.
• The area under the curve sum of areas of n
trapezia.
• Note : approximately equal to
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Let y0, y1 , y2, …., yn be the values of the function f (x) . These correspond to the (n+1) ordinates x0 , x1, x2, ……..,xn respectively.
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• Using
2
1 Area of trapezium = ( width x sum of parallel sides )
The area of the first trapezium is
A1 = 2
1 h ( y0 + y1 )
The areas of the second and the third trapezia are then
A2 = 2
1 h ( y1 + y 2 )
A3 = 2
1 h ( y 2 + y 3 )
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• Verify that if this process is continued then
A n-1 = h ( y n-2 + y n-1 )
A n = h ( y n-1 + y n )
• Now add all of these separate areas together. The approximate value of A is given by
• A A1 + A2 + A3 + … + A n-1 + A n
• h ( y 0 + y1 ) + h ( y1 + y 2 ) + h ( y2 + y
3 ) + … + h ( y n-2 + y n-1 ) + h ( y n-1 + y n )
• h ( y 0 + 2y1 +2y2 +2y3 +… + 2yn-1 + yn )
2
1
2
1
2
1
2
1
2
1
2
1
2
12
1
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• This method of approximating the area under a curve by n trapezia of equal width h is called the trapezium rule, and the result can be summarized as :
• dx h y 0 + y n + 2 ( y1
+ y2 + … + y n-1 )
• where h = and yr = f ( xr )
b
a
)x(f2
1
n
ab
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• Example 1
• Evaluate dx using 5 strips by the
trapezium rule.
• Notation: Working must have 4 or more decimal
places.
6
1ln x
Solution
The integration interval ( b- a ) = 6 –1 = 5 unitsSo h =
n
ab =
5
5 = 1
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The value of x at which y is calculated are :1, 2, 3, 4, 5, 6Tabulating the results as follows help the final calculation :
X y First and last ordinates Remaining ordinates
123456
y0
y1
y2
y3
y4
y5
0
1.792
0.6931.0991.3861.609
Totals 1.792 4.787
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6
1ln x
2
h dx ( y0 + y5) + 2 ( y1 +… +y4 )
=
2
1 1.792 + 2 ( 4.787 )
= 5.683
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Example 2
• Use the trapezium rule, with five
ordinates, to evaluate e x ² dx .
• Correct your answer to three decimal
places.
8.0
0
Solution:
For five ordinates, ( yo , y1 , y2 , y3 , y4 ) evenly spaced in the
range
0 ≤ x ≤ 0.8 , we need to divide this range into four equal parts
The integration interval ( b- a ) = 0.8 - 0 = 0.8 units
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• So, h = = = 0.2
• The value of x at which y is calculated are :
• 0 , 0.2, 0.4, 0.6 , 0.8
x Y First and last ordinates
Remaining ordinates
00.20.40.60.8
y0
y1
y2
y3
y4
1
1.8965
1.04081.17351.4333
Totals 2.8965 3.6476
n
ab 4
8.0
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• e x ² dx ≈ ( y0 + y4 ) + 2 ( y1 +… +y 3 ) = 2.8965 + 2 ( 3.6476 ) = 1.0192
= 1.019 ( three decimal
places )
8.0
0 2
h
2
2.0
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3.2 Solutions of non – linear equations
• OBJECTIVE : The objective of this lesson is to enable students to understand the method of finding roots using the iteration method by writing:
• f(x) = 0 in the form of x = g(x). The iteration scheme is xn+1 = g(xn) , n = 1,2,3….. The method fails when g(x) > 1 in the neighborhood of the roots of the equation.
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Introduction
• Many equations cannot be solved exactly, but various methods of finding approximate numerical solutions exist.
• The most commonly used methods have two main parts:
• (a) finding an initial approximate value
• (b) improving this value by an iterative process
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• Initial Values:• The initial value of the roots of f(x) = 0 can be
located approximately by either a graphical or an algebraic method.
• Graphical Method:• Either
(a) Plot ( or sketch ) the graph of y = f(x). The real roots are the points where the curve
cuts the x axis.• Or
(b) Rewrite f(x) = 0 in the form F(x) = G(x).Plot ( or sketch ) y =F(x) and y = G(x). The real roots are at the points where these graphs intersect.
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Example 1
• Find the approximate value of the equation ln x + x - 4 = 0 by using the graphical method.
y = 4-x
y
0 1 2 3 4 the intersection is at x 2.9
x
y
y = ln x
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• Algebraic Method
• Find two values a and b such that f(a) and
f(b) have different signs.
• At least one root must lie between a and b if
f(x) is continuous.
• If more than one root is suspected between
a and b, sketch a graph of y = f(x).
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• Iterative Method
• All iterative methods follow the same basic pattern. A sequence of approximations
• x1, x2 , x3 , x4 ….. is found , each one closer to
the root of f(x)=0.
• Each approximation is found from the one before it using a specified method.
• The process is continued until the required accuracy is reached.
• Two methods we have to discuss are Iteration Method and Newton-Raphson Method.
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• Iteration Method
• Rewrite the equation f(x) = 0 in the form x = g(x).• If the initial approximation is x1 , then calculate
• x2 = g(x1)• x3 = g(x2)• x4 = g(x3)
and so on …..
• This method fails if g (x) > 1 near the root. • So, the value of g(x1) should be < 1.
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Example 2
• Using the iteration method, find the solution of f(x) = x + ex near x = -0.6 to three decimal places.
Solution
Iteration method
Write the given equation as x = - ex x = g(x)g (x) = - ex
g(x) = - ex, g(-0.6) = -e-0.6 = 0.5488 ( < 1 )
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x1 = -0.6
x2 = -e-0.6 = -0.5488
x3 = -e-0.5488 = -0.5453
x4 = -e-0.5453 = -0.5797
x5 = -e-0.5797 = -0.5601
x6 = -e-0.5601 = -0.5712
x7 = -e-0.5712 = -0.5648
x8 = -e-0.5648 = -0.5684
x9 = -e-0.5684 = -0.5664
x10 = -e-0.5664= -0.5676
x11 = -e-0.5676 = -0.5668
• x12 = -e-0.5668 = -0.5673 the required solution is x = - 0.567
( three decimal places )#
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Example 3
• Show that the equation has a
root between 0.2 and 0.3.
• Taking 0.2 as first approximation find the
root of the equation ,giving your answer to
three significant figures by using the
iteration method.
0412 x
x
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• Solution
• f(x) = x2 - + 4
• f(0.2) =
• f(0.3) = = 0.756 ( positive )
• therefore f(x) has a root between x = 0.2 and x = 0.3.
x
1
42.0
1)2.0( 2 = - 0.96 ( negative )
43.0
1)3.0( 2
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0412 x
x 41 2 xx
4
12
x
x
g(x) = 12
2)4(
4
1
xx
g'(x) = )4()4( 222 xdx
dx
= )2()4( 22 xx =
22 )4(
2
x
x
g'(0.2) =
4)2.0(
)2.0(2
22
= 0.0245 ( <1 ) therefore, g(x) = iteration function
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• x = g(x)
• x1 = 0.2
• x2 = = 0.2475
• x3 = = 0.2462
• x4 = = 0.2463
the required solution is 0.246 (3 sig. figs)
4)2.0(
12
4)2475.0(
12
4)2462.0(
12
x3 and x
4 have same value when
round up to three sig.figs. Hence
we should stop the working.
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3.3 Solutions of non-linear equations
• OBJECTIVE : The objective of this lesson
is to enable students to
• find the root by the Newton-Raphson
method using the formula
• xn+1 = xn - , n = 1,2,3, …)('
)(
n
n
xf
xf
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Newton-Raphson Method
• If x1 is an approximation to root of f(x) =0 , then a better
approximation x2 is given by
.
• Repeat this process as required .
)(xf
)f(xxx
1
112
)(x f
)f(xxx
n
nn1 n
• The iteration is stop when m
nn xx 105.01
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Example 1
• Using the Newton-Raphson Method, find
the solution of f(x) = x + ex near x = - 0.5
to three decimal places.
• Solution
• Let f(x) = x + ex
• So f (x) = 1 + ex
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• x1 = -0.5
• = =
= -0.5663
• = = - 0.5671
• = = - 0.5671
• The required solution is x = -0.567
( three decimal places ) # .
)(x f
)f(xxx
1
112
(-0.5) f
f(-0.5)0.5-
)e (1
)e(-0.50.5-
0.5-
0.5-
3x(-0.5663) f
f(-0.5663)0.5663-
4x(-0.5671) f
f(-0.5671)0.5671-
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Example 1
Sketch the graph of y = ex and y = 2 – x
where x< 2, on the same axes.
Get the first approximation, x0 for the equation
ex = 2 – x where 0 < xo < 1.
Hence, by using Newton-Raphson method ,
solve the equation of e-x = x2
1
for x < 2 to three decimal places.
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2
1
Solution :
y
y = ex
1 x
y = 2 - x
00.4
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• Newton-Raphson method:
• e-x = = 2 – x
• ex = 2 – x ex – 2 + x = 0
• f(x) = ex – 2 + x f ’(x) = ex + 1
• x1 = first approximation = 0.4 ( from graph )
x2
1xe
1
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)('
)(1
nxf
nxfnxnx
)4.0('
)4.0(4.02 f
fx 0.4 - [
1
4.024.0
4.0
e
e] = = 0.4043
)4043.0('
)4043.0(40.03 f
fx = 0.4043 - [
1
4043.024043.0
4043.0
e
e] = 0.4433
)4433.0('
)4433.0(4433.04 f
fx = 0.4433 - [
1
4433.024433.0
4433.0
e
e] = 0.4428
)4428.0('
)4428.0(4428.05 f
fx = 0.4428 - [
1
4428.024428.0
4428.0
e
e] = 0.4428
The required solution is 0.443 ( three decimal places ).
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EXERCISES
• 1.)By taking 0.2 as first approximation find the root of the equation ,giving your answer to three
significant figures by using the Newton Raphson method.
•(Ans : 0.246)
0412 x
x
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• 2.) Show that the equations 2sin x – x = 0 has
a root between x = 1(radian) and x= 2 (radian).
Find the root of the equation by using
i) iteration method
ii) Newton Raphson method
Giving your answer to two decimal places.