BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 4.7, Problem 27E
To determine

To find: The Newton’s method fails for the given equation with any approximation x10 with sketch the graph.

Expert Solution

Explanation of Solution

Given:

The equation is x13=0.

The initial approximation is x1=0.5.

Result Used:

The Newton’s method is xn+1=xnf(xn)f(xn).

Graph:

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 4.7, Problem 27E

From the Figure 1, it is observed that the value gets doubled.

Calculation:

Let f(x)=x13, then f(x)=0 by the given condition.

Calculate the derivative of f(x).

Let f(x)=df(x)dx.

f(x)=d(x13)dx=13x23

Substitute the value of f(x) and f(x) in the above mentioned result to obtain xn+1.

xn+1=xnxn1313xn23=xn3xn=2xn

Set n=1, x1=0.5 and obtain x2 from the above expression.

x2=2×0.5=1

Set n=2, x2=1 and obtain x3 from the above expression.

x3=2×(1)=2

Similarly, proceed in the above manner.

Observation:

From the above calculation, each successive approximation gives the value twice than previous value.

Thus, if we consider the sequence of the Newton’s approximation fails to converge to the root 0. Therefore, the Newton’s method fails.

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