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Lecture 2 Numerical Analysis
2Eng. Malek Abuwarda
Roots of Equations
Bracketing methodsBisection Method
False position method
Open methods
Simple fixed point iteration
Newton Raphson
Secant Method
Modified Newton Raphson
System of nonlinear equations
Roots of polynomials
Using computers
Mullers Method
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Lecture 2 Numerical Analysis
3Eng. Malek Abuwarda
Roots of Equations
What is the root of equation?It is the value of the equations variable which makes the equation equal to
zero.
Direct method
2( ) 0f x ax bx c= + + =
What is the root for the equation below?
2 4
2
b b acx
a
=
We can use the quadratic formula
( ) xf x e x=
Another example
( ) tan( )f x x x= or
Can you find the roots??
Exact solution but it is not
always available.
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Lecture 2 Numerical Analysis
4Eng. Malek Abuwarda
Roots of Equations
Graphical Solution
2( )f x ax b=
What is the root for the equation below?
f(x)
x
roots
f(x)=0 f(x)=0
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Lecture 2 Numerical Analysis
5Eng. Malek Abuwarda
Graphical Solution
ExampleThe parachutist velocity is
What is the drag coefficient c needed to reach
a velocity of 40 m/s if m=68.1 kg, t =10 s, g=9.8 m/s2
)(tm
c
e1c
mgv
=
40)1(38.667)(
)1()(
146843.0 =
=
c
tm
c
ec
cf
vec
mgcf
-8.40120
-2.26916
6.06712
17.6538
34.1154
f(c)c
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Lecture 2 Numerical Analysis
6Eng. Malek Abuwarda
Graphical Solution Graphical techniques are of limited
practical value because they are not
precise. However, graphical methods
can be utilized to obtain rough
estimates of roots.
Graphical methods are important
tools for understanding the properties
of the function.
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Lecture 2 Numerical Analysis
7Eng. Malek Abuwarda
Graphical Solution
Use of computer graphics tolocate roots
( ) sin 20 cos3x x x= +
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Lecture 2 Numerical Analysis
8Eng. Malek Abuwarda
Bracketing Methods
Two initial guesses (xl andxu) are required for the root which
bracketthe root (s).
If one root of a real and continuous function, f(x)=0, is bounded by
values xl, xu then f(xl).f(xu)
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Lecture 2 Numerical Analysis
9Eng. Malek Abuwarda
Bracketing Methods
Bisection Method
Generally, iff(x) is real and continuous in the intervalxl toxuand
f(xl).f(xu)
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Lecture 2 Numerical Analysis
10Eng. Malek Abuwarda
Bisection Method
Step 1: Choose lowerxland upperxuguesses for the root such that:
f(xl).f(xu)
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Lecture 2 Numerical Analysis
11Eng. Malek Abuwarda
Bisection Method
Example
0.146843667.38( ) (1 ) 40cf c ec
=
What is the root of the equation below?
From the graphical solution we can identify that
the root lies in the interval (12 , 16)
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Lecture 2 Numerical Analysis
12Eng. Malek Abuwarda
Bisection Method
1612 13 14 15-0.425
1.569
-2.268
6.067
1514.87514.75
-0.425
0.059
-0.062 -0.184
(14+16)/2=15 -0.425 (14, 15)
(12+16)/2=14 1.569 (14, 16)
16 -2.268 (12, 16)
12 6.067 -
(14.75+14.875)/2=14.812 -0.062 (14.75, 14.812)
(14.75+15)/2=14.875 -0.184 (14.75, 14.875)
(14.5+15)/2=14.75 0.059 (14.75, 15)
(14+15)/2=14.5 0.552 (14.5, 15)
c (c) Interval
(14.75+14.812)/2=14.781 -0.001
1514 14.25 14.5 14.75-0.425
1.569
0.059
0.552