Lecture Note of False-Point Position Method

University of Babylon
College of Engineering
Department of Environmental Engineering
Engineering Analysis I (ENAN 103)








False-Point Position 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
2.2 – False-Point Position Method
The following steps are required in estimating the equation root.
Assume two x-values(x_p,x_n) such that one of them gives positive y-value (f(x_p )=y_p) and the other gives negative y-value (f(x_n )=y_n).
f(x_p )×f(x_n )<0
Calculate the estimated root (x_r) as following:
x_r=x_n-(f(x_n ) ?×(x?_p-x_n))/(f(x_p )-f(x_n ) )
Calculate the error value, |f(x_r)|, which represents the corresponding value of x_r.
Compare the error value, |f(x_r)|, with the desire accuracy (?).
If |f(x_r)|? ? then x_r is the required root (accurate enough).
If |f(x_r)|> ? and f(x_r ) is positive then x_p=x_r and go to step 2.
If |f(x_r)|> ? and f(x_r ) is negative then x_n=x_r and go to step 2.

Ex1: Find a root for f(x)=e^(-x)-x with accuracy equal to ?=0.001.
Solution:
Step 1:
Assume
x_p=0 ? f(0)=e^(-0)-0=1
x_n=1 ? f(1)=e^(-1)-1=-0.632

Step 2:
x_r=x_n-(f(x_n ) ?×(x?_p-x_n))/(f(x_p )-f(x_n ) ) ? x_r=1-(-0.632×(0-1))/(1-(-0.632) )=0.6127

Step 3:
f(x_r )=f(0.6127)=e^(-0.6127)-0.6127=-0.0708

Step 4:
|f(x_r)|=|-0.0708|=0.0708> ? =0.001 ?(not enough accurat)

Step 5:
Since f(x_r )=-0.0708 (negative)
Then x_n=x_r=0.6127 ? f(0.6127)=e^(-0.6127)-0.6127=-0.0708
x_p=0 ? f(0)=e^(-0)-0=1 (from last step)
Start again from step 2

Step # x_p f(x_p ) x_n f(x_n ) x_r f(x_r ) |f(x_r)|- ?
1 0 1 1 -0.63212 0.6127 -0.07081 -0.06981
2 0.5 0.106531 0.6127 -0.07081 0.567699 -0.00087 0.00013

The required root is equal to x_r=0.5676.

Ex2: Find a root for f(x)=cos?x with accuracy equal to ?=0.00001.
Note: All measurements in this example are in radian.
Solution:
Step 1:
Assume
x_p=2 ? f(2)=sin?2=0.90929
x_n=4 ? f(4)=sin?4=-0.75680

Step 2:
x_r=x_n-(f(x_n ) ?×(x?_p-x_n))/(f(x_p )-f(x_n ) ) ? x_r=1-(-0.75680×(2-4))/(0.90929-(-0.75680) )=3.0915
Step 3:
f(x_r )=f(3.0915)=sin?3.0915=0.050044

Step 4:
|f(x_r)|=|0.050044|=0.050044> ? =0.00001 ?(not enough accurat)

Step 5:
Since f(x_r )=0.050044 (positive)
Then x_p=x_r=3.0915 ? f(3.0915)=0.050044
x_n=4 ? f(4)=sin?4=-0.75680 (from last step)
Start again from step 2
Step # x_p f(x_p ) x_n f(x_n ) x_r f(x_r ) |f(x_r)|- ?
1 2 0.909297 4 -0.7568 3.091528 0.050044 -0.05003
2 3.091528 0.050044 4 -0.7568 3.147875 -0.00628 -0.00627
3 3.091528 0.050044 3.147875 -0.00628 3.14159 2.3E-06 7.7E-06
The required root is equal to x_r=3.1415.

Ex3: Find a root for f(x)=x^3+x^2-3x-3 with accuracy equal to ?=0.0001.

Solution:
Step 1:
Assume
x_p=2 ? f(2)=2^3+2^2-3×2-3=3
x_n=1 ? f(1)=1^3+1^2-3×1-3=-4

Step 2:
x_r=x_n-(f(x_n ) ?×(x?_p-x_n))/(f(x_p )-f(x_n ) ) ? x_r=1-(-4×(2-1))/(3-(-4) )=1.57142
Step 3:
f(x_r )=f(1.57142)=?1.57142?^3+?1.57142?^2-3×1.57142-3=-1.36449

Step 4:
|f(x_r)|=|-1.36449|=1.36449> ? =0.0001 ?(not enough accurat)

Step 5:
Since f(x_r )=-1.36449 (negative)
Then x_n=x_r=1.57142 ? f(1.57142)=-1.36449
x_p=2 ? f(2)=2^3+2^2-3×2-3=3
Start again from step 2
Step # x_p f(x_p ) x_n f(x_n ) x_r f(x_r ) |f(x_r)|- ?
1 1 -4 2 3 1.571429 -1.36443 -1.36433
2 1 -4 1.571429 -1.36443 1.867257 1.395343 -1.39524
3 1.867257 1.395343 1.571429 -1.36443 1.717686 -0.13468 -0.13458
4 1.867257 1.395343 1.717686 -0.13468 1.730851 -0.01134 -0.01124
5 1.867257 1.395343 1.730851 -0.01134 1.731951 -0.00094 -0.00084
6 1.867257 1.395343 1.731951 -0.00094 1.732043 -7.8E-05 2.19E-05
The required root is equal to x_r=1.7320.

?
Homework 2
Find a root for f(x)=x^3-2x^2 with accuracy equal to ?=0.01.
Start with: x_p=3 and x_n=1.
Find a root for f(x)=cos?(x+1) with accuracy equal to ?=0.0001.
Start with: x_p=1 and x_n=0.5.

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