# Solved Assignment by OecilKritingz

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Assignment # 1 spring 2012

Question 1
Find the real root of the equation correct to three decimal places by Bisection method.

Solution

We will take to points in which the root can lie
x0  1
x1  2

Root may lie between 0 and 1 so we will put it as
f (0)  2(0)3  0  2  0
f (0)  0  0  2  0 = -2
Now
f (1)  2(1)3  1  2  0
f (1)  2(1)3  1  2  0
2 1  2  1
So both have opposite signs 1 and -2 so root lies between these two points
Now
x0  2 And x1  1
As we know the formula of Bisection method
x  x1
x2  0
2
Putting the values in the equation we get
2  1 1
x2                   0.5
2       2
Now to perform next approximation we will calculate
f (0.5)  2(0.5)3  (0.5)  2  0
 2(0.125)  (0.5)  2
0.25  (0.5)  2
0.25  0.5  2  2.25
-
Now we will apply formula as

(2)  (0.5)
x3                 3.5
2
Now putting values in equation we get
(2)  (0.5)
x3                 3.5
2
Now the same as we have done before
f (3.5)  2(3.5)3  (3.5)  2  0
 85.75  (5.5)
85.75  5.5  91.25

Now for the last 3rd iteration
Follow the same above procedure do it yourself

Question 2
Use Regula-Fasli method to find the real root of the equation e x  3x  0
Correct to four decimal places after three successive approximations in 0  x  1

Solution

As it is an algebric equation so we can write this as in the form of
f ( x)  e x  3 x = 0
Now as we know the condition is 0  x  1
So first approximation will be 0 and 1 as root must lie between these two points 0 and 1
So
f ( x)  e x  3 x
f (0)  e0  3(0)  1
f (1)  e1  3(1)  2.71828  3  0.2817
Since f (0) f (1) 0,Soliesbetween and 1
1st iteration
x2  x1
x3  x2                        f ( x2 ).
f ( x2 )  f ( x1 )
letsay x1  0and  x2  1
x2  x1                            1 0                           1
x3  x2                        f ( x2 )                   (0.2817)  1           (0.2817)
f ( x2 )  f ( x1 )               ( 0.2817)  (1)                 (1.2817)
 0.7802
f ( x3 )  f (0.7802)  e0.7802  3(0.7802)  0.1587
sin ce f ( x3 )  0,soweneed tocalculatex4
Now f ( x3 ) f ( x1 )  0Thereforesolutionliesbetween 0.7802and 
2nd iteration
x3  x1                               (0.7802)  0
x4  x3                        f ( x3 )  (0.7802)                  (0.1587)
f ( x3 )  f ( x1 )                       (0.1587)  (1)
 0.6733
f ( x4 )  f (.6733)  e.6733  3(.6733)  0.0592
sin ce f ( x4 )  0,soweneed tocalculate x
Now f ( x4 ) f ( x2 )  0Thereforesolutionliesbetween.6733and 
3rd iteration
x4  x1
x5  x4                        f ( x4 )
f ( x4 )  f ( x1 )
.6733 0
x5 .6733                    (0.0592)
(0.0592)  (1)
x5  0.6357
f ( x5 )  f (0.6357)  e0.6357  3(0.6357)  0.0188
f ( x5 )  0.0188
sin ce f ( x5 )  0,soweneed tocalculatex
Now f ( x5 ) f ( x1 )  0Thereforesolutionliesbetween0.6357and 
4th iteration
x5  x1
x6  x5                        f ( x5 )
f ( x5 )  f ( x1 )
0.6357 0
x5  0.6357                     (0.0188)
(0.0188)  (1)
x6  0.6240
f ( x6 )  f (0.6240)  e0.6240  3(0.6240)  0.0056
f ( x6 )  0.0056
sin ce f ( x6 )  0,soweneed tocalculatex
Now f ( x7 ) f ( x1 )  0Thereforesolutionliesbetween0.6240and 
5th iteration
x6  x1
x7  x6                        f ( x6 )
f ( x6 )  f ( x1 )
0.6240 0
x7 0.6240                      ( 0.0056)
(0.0056)  (1)
x7  0.6205
f ( x7 )  f (0.6205)  e0.6205  3(0.6205)  0.0016
similarly 6th iteration
x8  0.6195

The required answer is up to 4th iteration so take only 4 iterations that’s enough

Other Question will upload soon stay blessed

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