Roots of Equation - Philadelphia UniversityCh5+Ch6.pdf · Roots of equation Bracketing Methods: ......

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Philadelphia University

Engineering Faculty Mechanical Engineering Department

Eng. Laith Batarseh

Roots of Equation

Roots of equation

Methods to find roots of equation

Bracketing

Methods

Graphical methods

Bisection method

False position method

Open

Methods

Simple fixed point iteration

Newton – Raphson

Secant method

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Roots of equation

Bracketing Methods:

Uses the fact that a function typically changes sign in

the vicinity of a root

Two initial guesses for the root are required

Open Methods:

Single initial guesses for the root are required

Bracketing Methods

Example : f(x) = sin(10x)+cos(3x)

Graphical Methods

0 1 2 3 4 5-2

-1

0

1

2f(x)

x

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Bracketing Methods

Advantages

easy to conduct

Disadvantages

Not precise enough

In multi – roots case such as triangular functions , it

may mislead you when the chosen range of data used

to draw the function passes some roots.

Graphical Methods

Bracketing Methods

Philosophy

function

changes its sign

when it passes

from one side

of the root to

the other side.

Graphical

representation

Mathematical

translation

f(xl) f(xu)<0

Bisection method

0 0.1 0.2 0.3 0.4 0.5-1

-0.5

0

0.5

1

1.5

2

xl

xu

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Bisection method

The following procedures are used to find root of equation by

Bisection method :

Step 1: Chose lower xl and upper xu guesses for the root such that the function

changes sign over the interval. You can check this out by insuring

that f(xl) f(xu)<0

Step 2: An estimate of root xr is determined by:

Step3: make the following evaluations to determine the next iteration:

If f(xl) f(xr)<0 Set xu = xr and return to step 2

If f(xl) f(xr)>0 Set xl = xr and return to step 2

If f(xl) f(xr)=0 stop the iteration. You find the true root

Procedures

2

ulr

xxx

Bisection method

To terminate the iteration, we have to define a termination criteria.

Because bisection method is a numerical method and the true value is

not found in typical ways, the use of approximation error is

convenient to this case:

Termination criteria

%100 new

r

old

r

new

ra

x

xx

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Bisection method

Find the root of the following function :

Solution:

Graphical representation

Example #1

201100

)( 15.0 xex

xf

-15 -10 -5 0 5 10 15-20

-10

0

10

20

30

40

As you can see, the

exact root of the given

equation is hard to find

so numerical method

must be used

Bisection method

Solution: 1st iteration

Step 1: assume xl = -5 and xu =1 ( f(-5) f(1) < 0)

Step2:

Step3: f(-5) f(-2) < 0 → xu = xr → xu = -2

Example #1 cont

51.2)2(,34.2)5(22

51

2

ff

xxx ul

r

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Bisection method

Solution: 2nd iteration

Step 1: assume xl = -5 and xu =-2

Step2:

Step3: f(-5) f(-3.5) < 0 → xu = xr → xu = -3.5

Example #1 cont

273.0)5.3(,34.2)5(5.32

52

2

ff

xxx ul

r

Bisection method

Solution: 3rd iteration

Step 1: assume xl = -5 and xu =-3.5

Step2:

Step3: f(-5) f(-4.25) >0 → xl = xr → xl = -4.25

Example #1 cont

982.0)25.4(,34.2)5(25.42

55.3

2

ff

xxx ul

r

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Bisection method

Solution: 4th iteration

Step 1: assume xl = -4.25 and xu = -3.5

Step2:

Step3: f(-5) f(-4.25) >0 → xl = xr → xl = -3.875

Example #1 cont

353.0)875.3(,982.0)25.4(875.32

25.45.3

2

ff

xxx ul

r

Bisection method

Solution: 5th iteration

Step 1: assume xl = -3.875 and xu = -3.5

Step2:

Step3: f(-5) f(-4.25) <0 → xl = xr → xu = -4.25

Example #1 cont

-8.4786)6875.3(,3425.0)875.3(6875.32

875.35.3

2

ff

xxx ul

r

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Bisection method

Solution: iteration summary

Example #1 cont

Iteration xl xu xr εa

1 -5 1 -2 ------

2 -5 -2 -3.5 42.86

3 -5 -3.5 -4.25 17.41

4 -4.25 -3.5 -3.875 9.68

5 -3.875 -3.5 -3.6875 5.08

Bisection method

Find the root of the following function:

Solution:

Graphical representation

Example #2

1)(2 xxf

-1.5 -1 -0.5 0 0.5 1 1.5-1

-0.5

0

0.5

As you can see, the

roots of the given

equation are -1 and +1.

we will aim to find the

second root and you

can try to find the other

root.

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Bisection method


Example #2

Iteration xl xu xr εa εt

1 0 1.2 0.6 ---- 40

2 0.6 1.2 0.9 33.3 10

3 0.9 1.2 1.05 14.3 5

4 0.9 1.05 0.975 7.7 2.5

5 0.975 1.05 1.0125 3.7 1.25

Bracketing Methods

This method is based on a graphical insight

Graphical presentation


x

f(x)

xl

xu

f(xl )

f(xu )

xr

xr,true

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Bracketing Methods

This method is based on a graphical insight

Mathematical formula

Implementation :

Step 1: Chose lower xl and upper xu guesses for the root such that the functionchanges sign over the interval. You can check this out by insuringthat f(xl) f(xu)<0

Step 2: An estimate of root xr is determined by:

Step3: make the following evaluations to determine the next iteration:

If f(xl) f(xr)<0 Set xu = xr and return to step 2

If f(xl) f(xr)>0 Set xl = xr and return to step 2

If f(xl) f(xr)=0 stop the iteration. You find the true root


ul

uluur

xfxf

xxxfxx


Find the root of the following function :

Solution: 1st iteration

Step 1: assume xl = -5 and xu =1 ( f(-5) f(1) < 0)

Step2:

Step3: f(-5) f(-3.4797) < 0 → xu = xr → xu = -3.4797

Example #3

201100

)( 15.0 xex

xf

-3.479715

5111

ff

fxr

.

.

.

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Example #3 cont

Iteration xl xu xr εa

1 -5 1 -3.3307 ------

2 -5 -3.3307 -3.6449 8.5

3 -5 -3.6449 -3.6664 0.58

4 -5 -3.6664 -3.6679 0.106

5 -5 -3.6679 -3.668 0.00272

Open Methods

It is also called the one-point iteration or successive iteration .

It depends on change the function to be in the form : x = g(x).

This transformation could be done by:

Algebraic manipulation

Simply add x to both sides of equation

Examples

you can use any method but its preferred to try the first one and then if

failed you can go to the second method

Simple fixed – point iteration

2

3032

22

x

xxx xxxx sin0sin

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Open Methods

Methodology and error

methodology :

Error


ii xgx 1

%1001

1

i

iia

x

xx

Open Methods

Example#4 :

Find the roots of the following function:

Solution:

step1: then

step 2: define the iterative formula :

step 3: start iteration : assume xi =0 then xi+1 = 1


xexfx

xxexxe 0 x

exg

ix

i ex

1

.

.

.

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Example #4 cont

i xi εa (%)

0 0 ----

1 1.000000 100

2 0.367879 171.8

3 0.692201 46.9

4 0.500473 38.3

5 0.606244 17.4

Error increased


Separate the iteration formula into two functions : y1 = x, y2 = g(x)

Draw both functions and the intersect point between them will be the root:

Convergence

x

y

xr

y1y2

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Convergence examples

Open Methods


methodology :

Error

Newton – Raphson Method

%1001

1

i

iia

x

xx

i

iii

xf

xfxx

'1

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Open Methods

Example#5 :


Solution:

step1:

step 2: define the iterative formula :

step 3: start iteration : assume xi =0 then xi+1 = 1


xexfx

1exf' x

1e

xexx

i

i

x

i

x

i1i

.

.

.



Example #5 cont

i xi εa (%)

0 0 ----

1 0.500000000 100

2 0.566311003 11.8

3 0.567143165 0.0000220

4 0.567143290 <10-8

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In some cases, the Newton – Raphson method shows slow rate of

convergence. The following example Illustrates that:

Given data

Solution

Pitfalls

i xi

0 0.5

1 51.65

2 46.486

3 41.8362

4 37.65285

.

.

.

∞ 1.0000000

1xxf 10 5.0x0

9

i

10

ii1i

x10

1xxx

Open Methods


Methodology :

Notes:

Although two initial guesses are required but it is still open method because it

dose not need change in sign in the function like the bracketing methods

Its derived by substitution of the first derivative evaluated from the backward

divided difference into the Newton – Raphson formula

Error

Secant method

%1001

1

i

iia

x

xx

i1i

i1iii1i

xfxf

xxxfxx

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Open Methods

Example#6 :


Solution:

step1: define the initial iterations xi-1 and xi.

x-1 = 0 → f(x-1) = 1.000000

x0 = 1 → f(x0) = -0.63212

step2:

Secant method

xexfx

61270.063212.01

1063212.011

x

Open Methods

Example#6 cont :

2nd iteration :

step3: define the 2nd iterations xi-1 and xi.

x0 = 1→ f(x0) = -0.63212

x1 = 0.61270→ f(x1) =-0.07081

step4:

Secant method

56384.00708.063212.0

61270.010708.061270.02

x

.

.

.

Roots of Equation - Philadelphia UniversityCh5+Ch6.pdf · Roots of equation Bracketing Methods: ......

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