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Lecture Note of Newton-Raphson Method

الكلية كلية الهندسة     القسم  الهندسة البيئية     المرحلة 4
أستاذ المادة وليد علي حسن       13/03/2017 06:57:34
University of Babylon
College of Engineering
Department of Environmental Engineering
Engineering Analysis I (ENAN 103)








Newton-Raphson Method
Undergraduate Leve, 3th Stage



Mr. Waleed Ali Tameemi
College of Engineering/ Babylon University
M.Sc. Civil Engineering/ the University of Kansas/ USA



2016-2017
3.0 – Open Methods
These methods are being used in estimating the roots of nonlinear equations. In any of the open methods, only one values of x (one roots) is assumed.

3.1 – Newton-Raphson Method
The following steps are required in estimating the root.
Evaluate f (x).
Assume an initial value for x_i.
Evaluate f(x_i) and f (x_i).
Calculate the estimated root (x_(i+1)) as following:
x_(i+1)=x_i-(f(x_i))/(f (x_i))
Calculate the error value, f(x_(i+1)), which represents the corresponding value of x_(i+1).
Compare the error value, |f(x_(i+1))|, with the desire accuracy (?).
If |f(x_(i+1))|? ? then x_(i+1) is the required root (accurate enough).
Otherwise go to step 3.

Ex1: Find a root for f(x)=x^3-x-1 with accuracy equal to ?=0.0001.
Solution:
Step 1:
f (x)=3x^2-1

Step 2:
Assume x_0=2

Step 3:
f(x_0 )=f(2)=2^3-2-1=5
f (x_0 )=f^ (2) =3?×2?^2-1=11

Step 4:
x_1=x_0-f(x_0 )/(f^ (x_0 ) )=2-5/11=1.5454

Step 5:
f(x_1)=f(1.5454)=?1.5454?^3-1.5454-1=1.1457

Step 6:
|f(x_1)| > ? ?(not enough accurat) go to step 3
Step # x_i x_(i+1) f(x_(i+1) ) |f(x_r)|- ?
0 2 1.545455 1.145755 -1.14476
1 1.545455 1.359615 0.153705 -0.1527
2 1.359615 1.325801 0.004625 -0.00362
3 1.325801 1.324719 4.66E-06 0.000995

The required root is equal to x=1.3247.

Ex2: Find a root for f(x)=3 cos?(x)-1+x with accuracy equal to ?=0.0001.
Solution:
Step 1:
f^ (x) =-3 sin?(x)+1

Step 2:
Assume x_0=2

Step 3:
f(x_0 )=f(2)=3 cos?(2)-1+2=-0.24844
f (x_0 )=f^ (2) ==-3 sin?(2)+2=-1.72789

Step 4:
x_1=x_0-f(x_0 )/(f^ (x_0 ) )=2-(-0.24844)/(-1.72789)=1.8562

Step 5:
f(x_1)=f(1.8562)=3 cos?(1.8562)-1+1.8562=0.01153

Step 6:
|f(x_1)| > ? ?(not enough accurat) go to step 3
Step # x_i x_(i+1) f(x_(i+1) ) |f(x_r)|- ?
1 2 1.856218 0.011532 -0.01053
2 1.856218 1.862356 1.6E-05 0.000984
3 1.862356 1.862365 3.16E-11 0.001
The required root is equal to x=1.8624.

Ex3: Find a root for f(x)=x^3+2x^2-2 with accuracy equal to ?=0.000001.
Solution:
Step 1:
f^ (x) =3x^2+4x

Step 2:
Assume x_0=1

Step 3:
f(x_0 )=f(1)=x^3+2x^2-2 =1
f (x_0 )=f (1)=3x^2+4x=7

Step 4:
x_1=x_0-f(x_0 )/(f^ (x_0 ) )=1-1/7=0.857143

Step 5:
f(x_1)=f(0.857143)=x^3+2x^2-2 =0.099125

Step 6:
|f(x_1)| > ? ?(not enough accurat) go to step 3
Step # x_i x_(i+1) f(x_(i+1) ) |f(x_r)|- ?
1 1 0.857143 0.099125 -0.09912
2 0.857143 0.839545 0.00141 -0.00141
3 0.839545 0.839287 3E-07 7E-07
The required root is equal to x=0.839287.

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Homework 3
Find a root for f(x)=3x+sin?x-e^x with accuracy equal to ?=0.00001.
Start with: x=0
Find a root for f(x)=x^3+x^2-3x-3 with accuracy equal to ?=0.0001.
Start with: x=1

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