001-Numerical Solution of Non Linear Equations

8/13/2019 001-Numerical Solution of Non Linear Equations

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Numerical solution of non linear equations.

Curve fitting

Numerical solution of ordinary differential equations.

Numerical solution of partial differential equations.

Numerical solution of linear equations.

Numerical integrations.


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The Main Menu PreviousNext


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An equation in which one or more terms have a variable of

degree 2 or higher is called a nonlinear equation. The non linear

equation f (x) = 0 and we find the roots of this equation.

Non Linear equation

System of non linear equations

A nonlinear system of equations contains at least one

nonlinear equation.

A system has one

solution. A system has twosolutionsA system has manysolutions.

A system has no

solution.


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It is one of the first numerical methods to find the root of anonlinear equation f(x) = 0 also called binary-searchmethod. The method is

based on finding the root between two points.At least one root exists between the two points if the function is real,

continuous, and changes sign. If the function does not change sign between

the two points then there may not be any root for the equation between the

two points.The steps to apply the bisection method to find the root of the equation

are:

1- Choose aand bfor the root such that f(a) f(b) < 0, or in other words, f (x)changes sign between aand b.2- Estimate the root xnsuch that xn= (an+ bn) / 2.If f(xn) = 0, then xnis the root of the equation. If f(xn) 0, f(xn) > 0 ,find the

new estimate such that the mid point between a , xn. If f(xn) 0, f(xn) < 0,

find the new estimate such that the mid point between xn, b.


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It is the numerical method to find the root of a nonlinear equation

f(x) = 0 also called method.

The steps to apply the simple iteration method to find the root of the

equation are:

1- Choose x0.

2- The function f(x) = 0 is transformed into equation x = g(x).

3- Test the convergence

4- Compute xn+1= g(xn)

5- If ; end

1)(xg 0

xx n1n


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It is the numerical method to find the root of a nonlinear equation

f(x) = 0.

The secant method is defined by the recurrence relation:)f(x

)f(x)f(x

xxxx n

1nn

1nnn1n

It requires two initial values, x0and x1, which should ideally be

chosen to lie close to the root.The steps to apply the Secant method to find the root of the equation

are:1. Select x0and x1

2. Evaluate f(x0) and f(x

1)

3. Evaluate xn+1

4. If ; endxx n1n


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It is the best numerical method and fast way to find the root of a

nonlinear equation f(x) = 0.

This method is defined by the recurrence relation:

)(xf

)f(xxxn

nn1n

)(xf

)f(xxx

n

nn1n

xx n1n

1- From f (x) get f\ (x)

2. Choose x0

3. Evaluate

4. If ; end

The steps to apply the Newton-Raphson method to find the root of

the equation are:

and test the convergence 1.)](xf[

)(xf)f(x2

n

00


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Solve, using the Bisection method to find the positive root of the

equation:f(x) = x3+ 4 x210 = 0.

Let x =a = 1 f(1) = ,

x = b =2 f(2) = +, then the root of this equation lies between [1, 2].

f(xn)bnann

2.3751.5210

-1.796871.251.5110.162111.3751.51.252

-0.848391.31251.3751.253

-0.0033961.3649902351.3652343751.3647609412

2

bax nnn


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Solve, using the simple iteration method to find the positive root :

f(x) = x

sin x = 0.25 correct to three decimal places.

f(x) = xsin x - 0.25 = 0.

1.5210x

0.2530.841- 0.091- 0.25f(x)

1362.02.1cos)( 0

xg

Let x0=1.2, x = sin x + 0.25.

Then x = g(x) = sin x + 0.25.

Test the convergence

cosx(x)g

43210n

1.1721.1721.1731.1751.182xn+1

Then xn+1= g(xn) =sin xn+0.25.


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Solve, using the Secant method to find the positive root :

f(x) = 0.18x3

0.5 x2-2x +1 correct to four decimal places,where x0= 4, x1= 6.

)f(x

)f(x)f(x

xxxx n

1nn

1nnn1n

)125.018.0()125.018.0()125.018.0(

xxxx 231

2

11323

1nnn1n

nnn

nnn

nnn

xxxxxxxxx

xn+1f(xn)xnf(xn-1)xn-1n

4.52109.88006-3.480041

4.7303-1.62874.52109.8800624.8513-0.59674.7303-1.62874.52103

4.8368-0.08154.8513-0.59674.73034

4.8373-0.00324.8368-0.08154.85135


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1.x1.08(x)f

2,xx0.54(x)f 2

54321x

1-3.48-4.64-3.56-1.32f(x)

133.0)435.4(

)86.3)(7225.1(

)](xf[

)(xf)f(x22

n

00

Solve, using the Newton - Raphson method to find the positive root :

f(x) = 0.18x3

0.5 x2-2x +1 correct to four decimal places,

Let x0= 4.5,

Test the convergence

.2xx0.54

1x2x0.5x0.18xxn

2

n

n

2

n

3

nn1n

)(xf)f(xxx

n

nn1n

210n

4.83734.83824.8884xn+1


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Let there be given a system of two equations in two unknowns,

F (x, y) = 0,

G (x, y) = 0.

Let x = x0, y = y0be the approximate values of the root of this system.

By using Newton-Raphsons method

The general formula to get the required root is:

,

GG

FF

GG

FF

xx

)y,(xyx

yx

)y,(xy

y

n1n

nn

nn

.

GG

FFGG

FF

yy

)y,(xyx

yx

)y,(xx

x

n1n

nn

nn


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Find the root of the system,

F (x, y) = 4

x2

y2= 0,

G (x, y) = 1exy = 0.

Given that (x0, y0) = (1, -1.7). Two steps are required. Correct to 4D.

We get:

x,2x

FFx

,2

FF y

yy

,exGG xx

1.y

GGy

For (x0, y0) = (1, -1.7), we get:

,2Fx ,4.3F y ,7183.2Gx 1,Gy 0.11,F 0.0813.G


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Then,

0043.1

17183.2

4.32

10183.0

4.311.0

1

GG

FF

GG

FF

xx

)y,(xyx

yx

)y,(xy

y

01

nn

00

.7299.1

17183.2

4.32

0183.07183.2

11.02

7.1

GG

FF

GG

FF

yy

)y,(xyx

yx

)y,(xx

x

01

00

00

,0086.2Fx

For (x1, y1) = (1.0043, -1.7299), we get:

,4598.3F y,7300.2Gx 1,Gy ,00120.F .0010.0G

,0042.1

17300.2

4598.30086.2

10001.04598.30012.0

0043.1x2

7296.1

17300.2

4598.30086.2

0001.07300.2

0012.00086.2

7299.1y2

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001-Numerical Solution of Non Linear Equations

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