Use simple fixed-point iteration to locate the root of f ( x ) = sin ( x ) − x Use an initial guess of x 0 = 0.5 and iterate until ε a ≤ 0.01 % . Verify that the process is linearly convergent as described in Box 6.1.

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

100%

Chapter 6, Problem 1P

Use simple fixed-point iteration to locate the root of

$f ( x ) = sin ( x ) − x$

Use an initial guess of $x 0 = 0.5$ and iterate until $ε a ≤ 0.01 %$ . Verify that the process is linearly convergent as described in Box 6.1.

Expert Solution & Answer

To determine

To calculate: The root of the function $f(x)=sin(x)−x$ by the use of simple fixed-point iteration with $x0=0.5$ as the initial condition and iterate until $εa≤0.01%$ . Also, verify that the process is linearly convergent.

Answer to Problem 1P

Solution:

The root of the function $f(x)=sin(x)−x$ is 0.7686.

Explanation of Solution

Given:

The function, $f(x)=sin(x)−x$ .

The initial condition, $x0=0.5$ and iterate until $εa≤0.01%$ .

Formula used:

The simple fixed-point iteration formula for the function $x=g(x)$ ,

$xi+1=g(xi)$

And, formula for approximate error is,

$εa=|xi+1−xixi+1|100%$

Calculation:

Consider the function,

$f(x)=sin(x)−x$

The function can be formulated as fixed-point iteration as,

$0=sin(x)−xx=sin(x)xi+1=sin(xi)$

Use initial guess of $x0=0.5$ , the first iteration is,

$x0+1=sin(x0)x1=sin(0.5)=sin(0.7071)=0.6496$

Therefore, the approximate error is,

$εa=|0.6496−0.50.6496|×100%=|0.14960.6496|×100%=|0.2303|×100%=23.03%$

Use $x1=0.6496$ , the second iteration is,

$x1+1=sin(x1)x2=sin(0.6496)=sin(0.80598)=0.7215$

Therefore, the approximate error is,

$εa=|0.7215−0.64960.7215|×100%=|0.07190.7215|×100%=|0.09965|×100%=9.965%$

Use $x2=0.7215$ , the second iteration is,

$x2+1=sin(x2)x3=sin(0.7215)=sin(0.8494)=0.7509$

Therefore, the approximate error is,

$εa=|0.7509−0.72150.7509|×100%=|0.02940.7509|×100%=|0.03915|×100%=3.915%$

Similarly, all the iteration can be summarized as below,

$i$	$xi$	$εa=\|xi+1−xixi+1\|100%$
0	0.5
1	0.6496	23.03%
2	0.7215	9.965%
3	0.7509	3.915%
4	0.7621	1.47%
5	0.7662	0.535%
6	0.7678	0.208%
7	0.7683	0.0651%
8	0.76852	0.029%
9	0.7686	0.01%

Since, the approximate error in the ninth iteration is 0.01%. So, stop the iteration.

Hence, the root of the function is 0.7686.

Now, to verify that the process is linearly convergent, the condition to be satisfied is $|g′(x)|<1$ for $x=0.7686$ .

The fixed-point iteration is,

$xi+1=sin(xi)$

Therefore,

$g(x)=sin(x)$

Differentiate the above function with respect to x,

$g′(x)=ddx[sin(x)]=cos(x)ddx(x)=cos(x)(−12x)=−cos(x)2x$

Therefore, $|g′(x)|$ at $x=0.7686$ is,

$|g′(0.7686)|=|−cos(0.7686)20.7686|=|−0.63972×0.8767|=|−0.3648|=0.3648$

Since, $|g′(0.7686)|<1$ . Hence, it is verified that the process is linearly convergent.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Please help solve the following whilst showing all working out. Is part of exam revision questions but no solution is given

please help me with this question with working out thanks

Page < 1 of 2 - ZOOM + 1) a) Find a matrix P such that PT AP orthogonally diagonalizes the following matrix A. = [{² 1] A = b) Verify that PT AP gives the correct diagonal form. 2 01 -2 3 2) Given the following matrices A = -1 0 1] an and B = 0 1 -3 2 find the following matrices: a) (AB) b) (BA)T 3) Find the inverse of the following matrix A using Gauss-Jordan elimination or adjoint of the matrix and check the correctness of your answer (Hint: AA¯¹ = I). [1 1 1 A = 3 5 4 L3 6 5 4) Solve the following system of linear equations using any one of Cramer's Rule, Gaussian Elimination, Gauss-Jordan Elimination or Inverse Matrix methods and check the correctness of your answer. 4x-y-z=1 2x + 2y + 3z = 10 5x-2y-2z = -1 5) a) Describe the zero vector and the additive inverse of a vector in the vector space, M3,3. b) Determine if the following set S is a subspace of M3,3 with the standard operations. Show all appropriate supporting work.

Answer 2

Textbook Question

100%

Chapter 6, Problem 1P

Use simple fixed-point iteration to locate the root of

$f ( x ) = sin ( x ) − x$

Use an initial guess of $x 0 = 0.5$ and iterate until $ε a ≤ 0.01 %$ . Verify that the process is linearly convergent as described in Box 6.1.

Expert Solution & Answer

To determine

To calculate: The root of the function $f(x)=sin(x)−x$ by the use of simple fixed-point iteration with $x0=0.5$ as the initial condition and iterate until $εa≤0.01%$ . Also, verify that the process is linearly convergent.

Answer to Problem 1P

Solution:

The root of the function $f(x)=sin(x)−x$ is 0.7686.

Explanation of Solution

Given:

The function, $f(x)=sin(x)−x$ .

The initial condition, $x0=0.5$ and iterate until $εa≤0.01%$ .

Formula used:

The simple fixed-point iteration formula for the function $x=g(x)$ ,

$xi+1=g(xi)$

And, formula for approximate error is,

$εa=|xi+1−xixi+1|100%$

Calculation:

Consider the function,

$f(x)=sin(x)−x$

The function can be formulated as fixed-point iteration as,

$0=sin(x)−xx=sin(x)xi+1=sin(xi)$

Use initial guess of $x0=0.5$ , the first iteration is,

$x0+1=sin(x0)x1=sin(0.5)=sin(0.7071)=0.6496$

Therefore, the approximate error is,

$εa=|0.6496−0.50.6496|×100%=|0.14960.6496|×100%=|0.2303|×100%=23.03%$

Use $x1=0.6496$ , the second iteration is,

$x1+1=sin(x1)x2=sin(0.6496)=sin(0.80598)=0.7215$

Therefore, the approximate error is,

$εa=|0.7215−0.64960.7215|×100%=|0.07190.7215|×100%=|0.09965|×100%=9.965%$

Use $x2=0.7215$ , the second iteration is,

$x2+1=sin(x2)x3=sin(0.7215)=sin(0.8494)=0.7509$

Therefore, the approximate error is,

$εa=|0.7509−0.72150.7509|×100%=|0.02940.7509|×100%=|0.03915|×100%=3.915%$

Similarly, all the iteration can be summarized as below,

$i$	$xi$	$εa=\|xi+1−xixi+1\|100%$
0	0.5
1	0.6496	23.03%
2	0.7215	9.965%
3	0.7509	3.915%
4	0.7621	1.47%
5	0.7662	0.535%
6	0.7678	0.208%
7	0.7683	0.0651%
8	0.76852	0.029%
9	0.7686	0.01%

Since, the approximate error in the ninth iteration is 0.01%. So, stop the iteration.

Hence, the root of the function is 0.7686.

Now, to verify that the process is linearly convergent, the condition to be satisfied is $|g′(x)|<1$ for $x=0.7686$ .

The fixed-point iteration is,

$xi+1=sin(xi)$

Therefore,

$g(x)=sin(x)$

Differentiate the above function with respect to x,

$g′(x)=ddx[sin(x)]=cos(x)ddx(x)=cos(x)(−12x)=−cos(x)2x$

Therefore, $|g′(x)|$ at $x=0.7686$ is,

$|g′(0.7686)|=|−cos(0.7686)20.7686|=|−0.63972×0.8767|=|−0.3648|=0.3648$

Since, $|g′(0.7686)|<1$ . Hence, it is verified that the process is linearly convergent.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Textbook Question

100%

Chapter 6, Problem 1P

Use simple fixed-point iteration to locate the root of

$f ( x ) = sin ( x ) − x$

Use an initial guess of $x 0 = 0.5$ and iterate until $ε a ≤ 0.01 %$ . Verify that the process is linearly convergent as described in Box 6.1.

Expert Solution & Answer

To determine

To calculate: The root of the function $f(x)=sin(x)−x$ by the use of simple fixed-point iteration with $x0=0.5$ as the initial condition and iterate until $εa≤0.01%$ . Also, verify that the process is linearly convergent.

Answer to Problem 1P

Solution:

The root of the function $f(x)=sin(x)−x$ is 0.7686.

Explanation of Solution

Given:

The function, $f(x)=sin(x)−x$ .

The initial condition, $x0=0.5$ and iterate until $εa≤0.01%$ .

Formula used:

The simple fixed-point iteration formula for the function $x=g(x)$ ,

$xi+1=g(xi)$

And, formula for approximate error is,

$εa=|xi+1−xixi+1|100%$

Calculation:

Consider the function,

$f(x)=sin(x)−x$

The function can be formulated as fixed-point iteration as,

$0=sin(x)−xx=sin(x)xi+1=sin(xi)$

Use initial guess of $x0=0.5$ , the first iteration is,

$x0+1=sin(x0)x1=sin(0.5)=sin(0.7071)=0.6496$

Therefore, the approximate error is,

$εa=|0.6496−0.50.6496|×100%=|0.14960.6496|×100%=|0.2303|×100%=23.03%$

Use $x1=0.6496$ , the second iteration is,

$x1+1=sin(x1)x2=sin(0.6496)=sin(0.80598)=0.7215$

Therefore, the approximate error is,

$εa=|0.7215−0.64960.7215|×100%=|0.07190.7215|×100%=|0.09965|×100%=9.965%$

Use $x2=0.7215$ , the second iteration is,

$x2+1=sin(x2)x3=sin(0.7215)=sin(0.8494)=0.7509$

Therefore, the approximate error is,

$εa=|0.7509−0.72150.7509|×100%=|0.02940.7509|×100%=|0.03915|×100%=3.915%$

Similarly, all the iteration can be summarized as below,

$i$	$xi$	$εa=\|xi+1−xixi+1\|100%$
0	0.5
1	0.6496	23.03%
2	0.7215	9.965%
3	0.7509	3.915%
4	0.7621	1.47%
5	0.7662	0.535%
6	0.7678	0.208%
7	0.7683	0.0651%
8	0.76852	0.029%
9	0.7686	0.01%

Since, the approximate error in the ninth iteration is 0.01%. So, stop the iteration.

Hence, the root of the function is 0.7686.

Now, to verify that the process is linearly convergent, the condition to be satisfied is $|g′(x)|<1$ for $x=0.7686$ .

The fixed-point iteration is,

$xi+1=sin(xi)$

Therefore,

$g(x)=sin(x)$

Differentiate the above function with respect to x,

$g′(x)=ddx[sin(x)]=cos(x)ddx(x)=cos(x)(−12x)=−cos(x)2x$

Therefore, $|g′(x)|$ at $x=0.7686$ is,

$|g′(0.7686)|=|−cos(0.7686)20.7686|=|−0.63972×0.8767|=|−0.3648|=0.3648$

Since, $|g′(0.7686)|<1$ . Hence, it is verified that the process is linearly convergent.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

100%

Chapter 6, Problem 1P

Use simple fixed-point iteration to locate the root of

$f ( x ) = sin ( x ) − x$

Use an initial guess of $x 0 = 0.5$ and iterate until $ε a ≤ 0.01 %$ . Verify that the process is linearly convergent as described in Box 6.1.

Expert Solution & Answer

To determine

To calculate: The root of the function $f(x)=sin(x)−x$ by the use of simple fixed-point iteration with $x0=0.5$ as the initial condition and iterate until $εa≤0.01%$ . Also, verify that the process is linearly convergent.

Answer to Problem 1P

Solution:

The root of the function $f(x)=sin(x)−x$ is 0.7686.

Explanation of Solution

Given:

The function, $f(x)=sin(x)−x$ .

The initial condition, $x0=0.5$ and iterate until $εa≤0.01%$ .

Formula used:

The simple fixed-point iteration formula for the function $x=g(x)$ ,

$xi+1=g(xi)$

And, formula for approximate error is,

$εa=|xi+1−xixi+1|100%$

Calculation:

Consider the function,

$f(x)=sin(x)−x$

The function can be formulated as fixed-point iteration as,

$0=sin(x)−xx=sin(x)xi+1=sin(xi)$

Use initial guess of $x0=0.5$ , the first iteration is,

$x0+1=sin(x0)x1=sin(0.5)=sin(0.7071)=0.6496$

Therefore, the approximate error is,

$εa=|0.6496−0.50.6496|×100%=|0.14960.6496|×100%=|0.2303|×100%=23.03%$

Use $x1=0.6496$ , the second iteration is,

$x1+1=sin(x1)x2=sin(0.6496)=sin(0.80598)=0.7215$

Therefore, the approximate error is,

$εa=|0.7215−0.64960.7215|×100%=|0.07190.7215|×100%=|0.09965|×100%=9.965%$

Use $x2=0.7215$ , the second iteration is,

$x2+1=sin(x2)x3=sin(0.7215)=sin(0.8494)=0.7509$

Therefore, the approximate error is,

$εa=|0.7509−0.72150.7509|×100%=|0.02940.7509|×100%=|0.03915|×100%=3.915%$

Similarly, all the iteration can be summarized as below,

$i$	$xi$	$εa=\|xi+1−xixi+1\|100%$
0	0.5
1	0.6496	23.03%
2	0.7215	9.965%
3	0.7509	3.915%
4	0.7621	1.47%
5	0.7662	0.535%
6	0.7678	0.208%
7	0.7683	0.0651%
8	0.76852	0.029%
9	0.7686	0.01%

Since, the approximate error in the ninth iteration is 0.01%. So, stop the iteration.

Hence, the root of the function is 0.7686.

Now, to verify that the process is linearly convergent, the condition to be satisfied is $|g′(x)|<1$ for $x=0.7686$ .

The fixed-point iteration is,

$xi+1=sin(xi)$

Therefore,

$g(x)=sin(x)$

Differentiate the above function with respect to x,

$g′(x)=ddx[sin(x)]=cos(x)ddx(x)=cos(x)(−12x)=−cos(x)2x$

Therefore, $|g′(x)|$ at $x=0.7686$ is,

$|g′(0.7686)|=|−cos(0.7686)20.7686|=|−0.63972×0.8767|=|−0.3648|=0.3648$

Since, $|g′(0.7686)|<1$ . Hence, it is verified that the process is linearly convergent.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

100%

Answer 6

Textbook Question

100%

Answer 7

Expert Solution & Answer

To determine

To calculate: The root of the function $f(x)=sin(x)−x$ by the use of simple fixed-point iteration with $x0=0.5$ as the initial condition and iterate until $εa≤0.01%$ . Also, verify that the process is linearly convergent.

Answer to Problem 1P

Solution:

The root of the function $f(x)=sin(x)−x$ is 0.7686.

Explanation of Solution

Given:

The function, $f(x)=sin(x)−x$ .

The initial condition, $x0=0.5$ and iterate until $εa≤0.01%$ .

Formula used:

The simple fixed-point iteration formula for the function $x=g(x)$ ,

$xi+1=g(xi)$

And, formula for approximate error is,

$εa=|xi+1−xixi+1|100%$

Calculation:

Consider the function,

$f(x)=sin(x)−x$

The function can be formulated as fixed-point iteration as,

$0=sin(x)−xx=sin(x)xi+1=sin(xi)$

Use initial guess of $x0=0.5$ , the first iteration is,

$x0+1=sin(x0)x1=sin(0.5)=sin(0.7071)=0.6496$

Therefore, the approximate error is,

$εa=|0.6496−0.50.6496|×100%=|0.14960.6496|×100%=|0.2303|×100%=23.03%$

Use $x1=0.6496$ , the second iteration is,

$x1+1=sin(x1)x2=sin(0.6496)=sin(0.80598)=0.7215$

Therefore, the approximate error is,

$εa=|0.7215−0.64960.7215|×100%=|0.07190.7215|×100%=|0.09965|×100%=9.965%$

Use $x2=0.7215$ , the second iteration is,

$x2+1=sin(x2)x3=sin(0.7215)=sin(0.8494)=0.7509$

Therefore, the approximate error is,

$εa=|0.7509−0.72150.7509|×100%=|0.02940.7509|×100%=|0.03915|×100%=3.915%$

Similarly, all the iteration can be summarized as below,

$i$	$xi$	$εa=\|xi+1−xixi+1\|100%$
0	0.5
1	0.6496	23.03%
2	0.7215	9.965%
3	0.7509	3.915%
4	0.7621	1.47%
5	0.7662	0.535%
6	0.7678	0.208%
7	0.7683	0.0651%
8	0.76852	0.029%
9	0.7686	0.01%

Since, the approximate error in the ninth iteration is 0.01%. So, stop the iteration.

Hence, the root of the function is 0.7686.

Now, to verify that the process is linearly convergent, the condition to be satisfied is $|g′(x)|<1$ for $x=0.7686$ .

The fixed-point iteration is,

$xi+1=sin(xi)$

Therefore,

$g(x)=sin(x)$

Differentiate the above function with respect to x,

$g′(x)=ddx[sin(x)]=cos(x)ddx(x)=cos(x)(−12x)=−cos(x)2x$

Therefore, $|g′(x)|$ at $x=0.7686$ is,

$|g′(0.7686)|=|−cos(0.7686)20.7686|=|−0.63972×0.8767|=|−0.3648|=0.3648$

Since, $|g′(0.7686)|<1$ . Hence, it is verified that the process is linearly convergent.

Answer 8

Expert Solution & Answer

To determine

To calculate: The root of the function $f(x)=sin(x)−x$ by the use of simple fixed-point iteration with $x0=0.5$ as the initial condition and iterate until $εa≤0.01%$ . Also, verify that the process is linearly convergent.

Answer to Problem 1P

Solution:

The root of the function $f(x)=sin(x)−x$ is 0.7686.

Explanation of Solution

Given:

The function, $f(x)=sin(x)−x$ .

The initial condition, $x0=0.5$ and iterate until $εa≤0.01%$ .

Formula used:

The simple fixed-point iteration formula for the function $x=g(x)$ ,

$xi+1=g(xi)$

And, formula for approximate error is,

$εa=|xi+1−xixi+1|100%$

Calculation:

Consider the function,

$f(x)=sin(x)−x$

The function can be formulated as fixed-point iteration as,

$0=sin(x)−xx=sin(x)xi+1=sin(xi)$

Use initial guess of $x0=0.5$ , the first iteration is,

$x0+1=sin(x0)x1=sin(0.5)=sin(0.7071)=0.6496$

Therefore, the approximate error is,

$εa=|0.6496−0.50.6496|×100%=|0.14960.6496|×100%=|0.2303|×100%=23.03%$

Use $x1=0.6496$ , the second iteration is,

$x1+1=sin(x1)x2=sin(0.6496)=sin(0.80598)=0.7215$

Therefore, the approximate error is,

$εa=|0.7215−0.64960.7215|×100%=|0.07190.7215|×100%=|0.09965|×100%=9.965%$

Use $x2=0.7215$ , the second iteration is,

$x2+1=sin(x2)x3=sin(0.7215)=sin(0.8494)=0.7509$

Therefore, the approximate error is,

$εa=|0.7509−0.72150.7509|×100%=|0.02940.7509|×100%=|0.03915|×100%=3.915%$

Similarly, all the iteration can be summarized as below,

$i$	$xi$	$εa=\|xi+1−xixi+1\|100%$
0	0.5
1	0.6496	23.03%
2	0.7215	9.965%
3	0.7509	3.915%
4	0.7621	1.47%
5	0.7662	0.535%
6	0.7678	0.208%
7	0.7683	0.0651%
8	0.76852	0.029%
9	0.7686	0.01%

Since, the approximate error in the ninth iteration is 0.01%. So, stop the iteration.

Hence, the root of the function is 0.7686.

Now, to verify that the process is linearly convergent, the condition to be satisfied is $|g′(x)|<1$ for $x=0.7686$ .

The fixed-point iteration is,

$xi+1=sin(xi)$

Therefore,

$g(x)=sin(x)$

Differentiate the above function with respect to x,

$g′(x)=ddx[sin(x)]=cos(x)ddx(x)=cos(x)(−12x)=−cos(x)2x$

Therefore, $|g′(x)|$ at $x=0.7686$ is,

$|g′(0.7686)|=|−cos(0.7686)20.7686|=|−0.63972×0.8767|=|−0.3648|=0.3648$

Since, $|g′(0.7686)|<1$ . Hence, it is verified that the process is linearly convergent.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

To calculate: The root of the function $f(x)=sin(x)−x$ by the use of simple fixed-point iteration with $x0=0.5$ as the initial condition and iterate until $εa≤0.01%$ . Also, verify that the process is linearly convergent.

Answer 12

Answer to Problem 1P

Solution:

The root of the function $f(x)=sin(x)−x$ is 0.7686.

Answer 13

Explanation of Solution

Given:

The function, $f(x)=sin(x)−x$ .

The initial condition, $x0=0.5$ and iterate until $εa≤0.01%$ .

Formula used:

The simple fixed-point iteration formula for the function $x=g(x)$ ,

$xi+1=g(xi)$

And, formula for approximate error is,

$εa=|xi+1−xixi+1|100%$

Calculation:

Consider the function,

$f(x)=sin(x)−x$

The function can be formulated as fixed-point iteration as,

$0=sin(x)−xx=sin(x)xi+1=sin(xi)$

Use initial guess of $x0=0.5$ , the first iteration is,

$x0+1=sin(x0)x1=sin(0.5)=sin(0.7071)=0.6496$

Therefore, the approximate error is,

$εa=|0.6496−0.50.6496|×100%=|0.14960.6496|×100%=|0.2303|×100%=23.03%$

Use $x1=0.6496$ , the second iteration is,

$x1+1=sin(x1)x2=sin(0.6496)=sin(0.80598)=0.7215$

Therefore, the approximate error is,

$εa=|0.7215−0.64960.7215|×100%=|0.07190.7215|×100%=|0.09965|×100%=9.965%$

Use $x2=0.7215$ , the second iteration is,

$x2+1=sin(x2)x3=sin(0.7215)=sin(0.8494)=0.7509$

Therefore, the approximate error is,

$εa=|0.7509−0.72150.7509|×100%=|0.02940.7509|×100%=|0.03915|×100%=3.915%$

Similarly, all the iteration can be summarized as below,

$i$	$xi$	$εa=\|xi+1−xixi+1\|100%$
0	0.5
1	0.6496	23.03%
2	0.7215	9.965%
3	0.7509	3.915%
4	0.7621	1.47%
5	0.7662	0.535%
6	0.7678	0.208%
7	0.7683	0.0651%
8	0.76852	0.029%
9	0.7686	0.01%

Since, the approximate error in the ninth iteration is 0.01%. So, stop the iteration.

Hence, the root of the function is 0.7686.

Now, to verify that the process is linearly convergent, the condition to be satisfied is $|g′(x)|<1$ for $x=0.7686$ .

The fixed-point iteration is,

$xi+1=sin(xi)$

Therefore,

$g(x)=sin(x)$

Differentiate the above function with respect to x,

$g′(x)=ddx[sin(x)]=cos(x)ddx(x)=cos(x)(−12x)=−cos(x)2x$

Therefore, $|g′(x)|$ at $x=0.7686$ is,

$|g′(0.7686)|=|−cos(0.7686)20.7686|=|−0.63972×0.8767|=|−0.3648|=0.3648$

Since, $|g′(0.7686)|<1$ . Hence, it is verified that the process is linearly convergent.

Answer 14

Ch. 6 - 6.1 Use simple fixed-point iteration to locate the...Ch. 6 - 6.2 Determine the highest real root of...Ch. 6 - Use (a) fixed-point iteration and (b) the...Ch. 6 - Determine the real roots of f(x)=1+5.5x4x2+0.5x3:...Ch. 6 - 6.5 Employ the Newton-Raphson method to determine...Ch. 6 - Determine the lowest real root of...Ch. 6 - 6.7 Locate the first positive root of Where x...Ch. 6 - 6.8 Determine the real root of, with the modified...Ch. 6 - 6.9 Determine the highest real root of:...Ch. 6 - 6.10 Determine the lowest positive root...

Use simple fixed-point iteration to locate the root of f ( x ) = sin ( x ) − x Use an initial guess of x 0 = 0.5 and iterate until ε a ≤ 0.01 % . Verify that the process is linearly convergent as described in Box 6.1.

Videos

Answer to Problem 1P

Explanation of Solution

Want to see more full solutions like this?

Chapter 6 Solutions