Chapter 7: Q. 75 (page 593)

In Exercises 75–78 use Newton’s method (see Example 8) to approximate a root for the given function with the specified value of $x_{0} .$ Terminate your sequence when $|x_{n + 1} - x_{n}| < 0.001 .$
$f (x) = x^{3} - 2, x_{0} = 1$

Short Answer

Expert verified

The approximate value of a root for the given function is $1.25992 .$

Step by step solution

Step 1. Given Information.

The given function is $f (x) = x^{3} - 2, x_{0} = 1 .$

Step 2. Find the root for the given function.

To find the root of the function, let the equation is $x_{k + 1} = x_{k} - \frac{f (x_{k})}{f' (x_{k})} .$

Now, $f (x_{k}) = {(x_{k})}^{3} - 2,$ so $f' (x_{k}) = 3 {(x_{k})}^{2} .$

Put the values in the equation,

$x_{k + 1} = x_{k} - \frac{{(x_{k})}^{3} - 2}{3 {(x_{k})}^{2}} .$

Step 3. Find the values of x1, x2.

To find the value of $x_{1}, put k = 0 :$

$x_{0 + 1} = x_{0} - \frac{{x_{0}}^{3} - 2}{3 {x_{0}}^{2}} x_{1} = 1 - \frac{1^{3} - 2}{3 (1)} x_{1} = 1 - \frac{- 1}{3} x_{1} = 1 + \frac{1}{3} x_{1} = \frac{4}{3} x_{1} = 1.33333$

Now, to find the value of $x_{2}, put k = 1 :$

$x_{1 + 1} = x_{1} - \frac{{x_{1}}^{3} - 2}{3 {x_{1}}^{2}} x_{2} = 1.33333 - \frac{{(1.33333)}^{3} - 2}{3 (1.33333)} x_{2} = 1.26388$

Step 4. Find the values of x3 and x4.

Now, let's find the values of $x_{3}, put k = 2 :$

$x_{2 + 1} = x_{2} - \frac{{x_{2}}^{3} - 2}{3 {x_{2}}^{2}} x_{3} = (1.26388) - \frac{{(1.26388)}^{3} - 2}{3 (1.26388)} x_{3} = 1.25993$

Let's find the values of $x_{4}, put k = 3 :$

$x_{3 + 1} = x_{2} - \frac{{x_{4}}^{3} - 2}{3 {x_{4}}^{2}} x_{4} = (1.25993) - \frac{{(1.25993)}^{3} - 2}{3 (1.25993)} x_{4} = 1.25992$

Step 5. Solve.

To approximate the root of the function, let's calculate $|x_{4} - x_{3}| = |1.25993 - 1.25992| = |0.00001|$

Since $|x_{4} - x_{3}| < 0.01 .$

Therefore, the approximate value of the root is $1.25992 .$

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In Exercises 75–78 use Newton’s method (see Example 8) to approximate a root for the given function with the specified value of $x_{0} .$ Terminate your sequence when $|x_{n + 1} - x_{n}| < 0.001 .$
$f (x) = x^{3} - 2, x_{0} = 1$

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Find the root for the given function.

Step 3. Find the values of x1, x2.

Step 4. Find the values of x3 and x4.

Step 5. Solve.

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In Exercises 75–78 use Newton’s method (see Example 8) to approximate a root for the given function with the specified value of x0.Terminate your sequence when xn+1-xn<0.001.f(x)=x3−2,x0=1

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Find the root for the given function.

Step 3. Find the values of x1, x2.

Step 4. Find the values of x3 and x4.

Step 5. Solve.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Pure Maths

Statistics

Discrete Mathematics

Applied Mathematics

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

In Exercises 75–78 use Newton’s method (see Example 8) to approximate a root for the given function with the specified value of $x_{0} .$ Terminate your sequence when $|x_{n + 1} - x_{n}| < 0.001 .$
$f (x) = x^{3} - 2, x_{0} = 1$