Numerical Methods for Engineers
Numerical Methods for Engineers
7th Edition
ISBN: 9780073397924
Author: Steven C. Chapra Dr., Raymond P. Canale
Publisher: McGraw-Hill Education
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Chapter 7, Problem 3P

(a)

To determine

To calculate: The positive real root of f(x) by using Müller’s method if,

f(x)=x3+x24x4

(a)

Expert Solution
Check Mark

Answer to Problem 3P

Solution:

The positive real root of the given equation f(x) is 2.

Explanation of Solution

Given Information:

The given equation is,

f(x)=x3+x24x4

Use Müller’s method.

Formula used:

The expression for the new roots is,

x3=x2+2cb+b24ac

The expression for the error is,

error=| x3x2x3 |100

Calculation:

Recall the equation mentioned in the problem,

f(x)=x3+x24x4

Draw the plot of the equation.

Numerical Methods for Engineers, Chapter 7, Problem 3P , additional homework tip  1

From the above plot it is clear that one root is at about x=2.

Consider the initial guess be x0=1x1=1.5

, and x2=2.5 respectively to determine the positive real root of the equation.

f(x)=x3+x24x4

Thus, the values of f(x) at different initial value are,

f(1)=6 f(1.5)=4.375  f(2.5)=7.875

Now calculate the h0 and h1.

h0=x1x0 h1=x2x1

Substituting the value of x0=1x1=1.5

, and x2=2.5.

h0=0.5h1=1

Now calculate the δ0 and  δ1.

δ0=f(x1)f(x0)x1x0 δ1=f(x2)f(x1)x2x1

Substitute the above calculated values.

δ0=f(1.5)f(1)1.51 =4.375+60.5 =3.25

And,

δ1=f(2.5)f(1.5)2.51.5 =7.875+4.3751 =12.25

Calculate the value of constants a,b and c.

a=δ1δ0h1+h0 =12.253.251+0.5 =9.01.5 =6

For b,

b=ah1+δ1 =6×1+12.5=18.25

For c,

c=f(x2)=7.875

Thus, the new root is calculated as follows.

x3=x2+2cb+b24ac

Substitute the all values.

x3=2.5+2×(7.875)18.25+18.252(4)(6)(7.875) =2.50.520616=1.979384

Calculate the error estimate.

error=| 1.9793842.51.979384 |(100)=| 0.5206162 |(100)=26.03%

Because the error is large, so required new guesses for x0 is replaced by x1

, x1 is replaced by x2

, and x2 is replaced by x3.

Therefore, for the new iteration, x0=1.5

, x1=2.5

, x2=1.98.

Thus, the values of f(x) at different initial value are,

f(1.5)=4.375 f(2.5)=7.875 f(1.98)=0.23721

Now calculate the h0 and h1.

h0=x1x0 h1=x2x1

Substituting the value of x0=1.5x1=2.5

, and x2=1.98.

h0=1h1=0.52

Now calculate the δ0 and  δ1.

δ0=f(x1)f(x0)x1x0 δ1=f(x2)f(x1)x2x1

Substitute the above calculated values.

δ0=f(2.5)f(1.5)2.51.5 =7.875+4.3751 =12.25

And,

δ1=f(1.98)f(2.5)1.982.5 =0.237217.8750.52 =15.60

Calculate the value of constants a,b and c.

a=δ1δ0h1+h0 =15.6012.250.52+1 =3.350.48 =6.98

For b,

b=ah1+δ1 =6×(0.52)+15.60=12.48

For c,

c=f(x2)=0.23721

Thus, the new root is calculated as follows.

x3=x2+2cb+b24ac

Substitute the all values.

x3=1.98+2×(0.23721)12.78+12.482(4)(6.98)(0.23721) =1.98+0.037=2.017

Calculate the error estimate.

error=| 2.0171.9793842.017 |(100)=| 0.0372.017 |(100)=1.8%

Because the error is near about 1, so required new guesses for x0 is replaced by x1

, x1 is replaced by x2

, and x2 is replaced by x3

. And repeat the same process will get the root is 2.

i x3 error
0 1.979384 26.03
1 2.017 1.8
2 2 0.00241

Similarly repeat for all other roots.

To find the exact positive real root use MATLAB.

Write the following code in command window.

>> a=[1 1 -4 -4];

roots(a)

ans =

2.0000

-2.0000

-1.0000

Therefore, the only positive real root of the given equation f(x) is 2.

(b)

To determine

To calculate: The positive real root of f(x) by using Müller’s method if,

f(x)=x30.5x2+4x2

(b)

Expert Solution
Check Mark

Answer to Problem 3P

Solution:

The only positive real root of the given equation f(x) is 0.5.

Explanation of Solution

Given Information:

The given equation is,

f(x)=x30.5x2+4x2

Use Müller’s method.

Formula used:

The expression for the new roots is,

x3=x2+2cb+b24ac

The expression for the error is,

error=| x3x2x3 |100

Calculation:

Recall the equation mentioned in the problem,

f(x)=x3+x24x4

Draw the plot of the equation.

Numerical Methods for Engineers, Chapter 7, Problem 3P , additional homework tip  2

From the above plot it is clear that one root is at about x=0.5.

Consider the initial guess be x0=0.5x1=1

, and x2=1.5 respectively to determine the positive real root of the equation.

f(x)=x3+x24x4

Thus, the values of f(x) at different initial value are,

f(0.5)=0 f(1)=2.5 f(1.5)=6.25

Now calculate the h0 and h1.

h0=x1x0 h1=x2x1

Substituting the value of x0=0.5x1=1

, and x2=1.5.

h0=0.5h1=0.5

Now calculate the δ0 and  δ1.

δ0=f(x1)f(x0)x1x0 δ1=f(x2)f(x1)x2x1

Substitute the above calculated values.

δ0=f(0.5)f(1)0.51 =02.50.5 =5

And,

δ1=f(1.5)f(1)1.51 =6.25+2.50.5 =7.5

Calculate the value of constants a,b and c.

a=δ1δ0h1+h0 =7.550.5+0.5 =2.51.0 =2.5

For b,

b=ah1+δ1 =2.5(0.5)+7.5=8.75

For c,

c=f(x2)=6.25

Thus, the new root is calculated as follows.

x3=x2+2cb+b24ac

Substitute the all values.

x3=1.5+2×(6.25)8.75+8.752(4)(2.5)(6.25) =1.51.0=0.5

Calculate the error estimate.

error=| 0.51.50.5 |(100)=| 1.00.5 |(100)=200%

Because the error is large, so required new guesses for x0 is replaced by x1

, x1 is replaced by x2

, and x2 is replaced by x3.

Therefore, for the new iteration, x0=1

, x1=1.5

, x2=0.5.

Thus, the values of f(x) at different initial value are,

f(1)=2.5 f(1.5)=6.25f(0.5)=0

Now calculate the h0 and h1.

h0=x1x0 h1=x2x1

Substituting the value of x0=1

, x1=1.5

, x2=0.5.

h0=0.5h1=1

Now calculate the δ0 and  δ1.

δ0=f(x1)f(x0)x1x0 δ1=f(x2)f(x1)x2x1

Substitute the above calculated values.

δ0=f(1)f(1.5)11.5 =6.252.50.5 =7.5

And,

δ1=f(0.5)f(1.5)0.51.5 =06.251.0 =6.25

Calculate the value of constants a,b and c.

a=δ1δ0h1+h0 =6.257.51.0+0.5 =2.50.5 =2.5

For b,

b=ah1+δ1 =2.5(1)+6.25=3.75

For c,

c=f(x2)=0

Thus, the new root is calculated as follows.

x3=x2+2cb+b24ac

Substitute the all values.

x3=0.5+2×(0)3.75+3.752(4)(2.5)(0) =0.50=0.5

Calculate the error estimate.

error=| 0.50.50.5 |(100)=0%

Because the error is zero, the root is 0.5.

i x3 error
0 0.5 200%
1 0.5 0%

Similarly repeat for all other roots.

To find the exact positive real root use MATLAB.

Write the following code in command window.

>> a=[1 -0.5 4 -2];

roots(a)

ans =

-0.0000 + 2.0000i

-0.0000 - 2.0000i

0.5000 + 0.0000i

Therefore, the only positive real root of the given equation f(x) is 0.5.

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