/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 40 Estimating roots The values of v... [FREE SOLUTION] | 91Ó°ÊÓ

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Estimating roots The values of various roots can be approximated using Newton's method. For example, to approximate the value of \(\sqrt[3]{10},\) we let \(x=\sqrt[3]{10}\) and cube both sides of the equation to obtain \(x^{3}=10,\) or \(x^{3}-10=0 .\) Therefore, \(\sqrt[3]{10}\) is a root of \(p(x)=x^{3}-10,\) which we can approximate by applying Newton's method. Approximate each value of \(r\) by first finding a polynomial with integer coefficients that has a root \(r\). Use an appropriate value of \(x_{0}\) and stop calculating approximations when two successive approximations agree to five digits to the right of the decimal point after rounding. $$r=2^{1 / 3}$$

Short Answer

Expert verified
Question: Use Newton's method to approximate the value of \(r=2^{1/3}\) to five decimal places. Answer: Using Newton's method and iterating until two successive approximations agree to five decimal places, we find that the approximation for \(r=2^{1/3}\) is approximately \(1.25992\).

Step by step solution

01

Identify the function and its derivative

Let's consider the polynomial \(p(x) = x^3 - 2\), whose root is \(2^{1/3}\). We want to find its root using the Newton's method. For applying Newton's method, we also need the derivative of the function, which is given by: \(p'(x) = 3x^2\).
02

Choose an initial approximation, \(x_0\)

We need to choose an appropriate initial approximation \(x_0\) for Newton's method. Since \(2^{1/3}\) is greater than 1 but less than 2, let's choose \(x_0 = 1.5\) as our initial approximation.
03

Apply Newton's method

To apply Newton's method, we need to use the following iterative formula: \(x_{n+1} = x_n - \frac{p(x_n)}{p'(x_n)}\) We start with \(x_0=1.5\), and continue calculating successive approximations until two successive approximations agree to five digits to the right of the decimal point after rounding. \(x_1 = x_0 - \frac{p(x_0)}{p'(x_0)} = 1.5 - \frac{(1.5)^3-2}{3(1.5)^2}\) After calculating this expression, we obtain \(x_1 \approx 1.25926\).
04

Repeat the process until convergence

Now, let's perform the same calculation with \(x_1\) as our new guess: \(x_2 = x_1 - \frac{p(x_1)}{p'(x_1)} = 1.25926 - \frac{(1.25926)^3-2}{3(1.25926)^2}\) After calculating this expression, we obtain \(x_2 \approx 1.25993\). Since the two successive approximations \(x_1 \approx 1.25926\) and \(x_2 \approx 1.25993\) agree up to only four decimal places (not five), we should continue this process. After more iterations (you can perform these calculations using a calculator or software like Excel), we find that two successive approximations agree to five digits to the right of the decimal point: \(x_6 \approx 1.25992\) and \(x_7 \approx 1.25992\) Therefore, we can approximate the value of \(r=2^{1/3}\) as \(\approx 1.25992\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Estimating roots
Newton's Method is a powerful tool for estimating roots of polynomial equations like finding cube roots. The process involves converting the expression for the root into an equation with a convenient form. For example, to find the cube root of 10, we write it as \(x = \sqrt[3]{10}\). Cubing both sides, we get \(x^3 = 10\), which transforms into the function \(p(x) = x^3 - 10 = 0\). This is a great form that allows us to use iterative methods for finding where the function equals zero, essentially finding the root. When estimating roots using such polynomial equations:
  • Convert the root to a polynomial equation \(p(x) = 0\).
  • Choose an initial approximation close to the estimated root.
  • Use iterative methods to refine this approximation.
  • Continue this process until the desired precision is reached.
As seen in the exercise, estimation focuses on gradually improving guesses to zero in on the precise root value.
Iterative method
An iterative method like Newton's Method involves refining an initial guess through repeated calculations. Here, the key formula in Newton's method is: \[ x_{n+1} = x_n - \frac{p(x_n)}{p'(x_n)} \]This process starts with an initial approximation \(x_0\). Each iteration aims to move closer to the actual root.Steps in the iterative process:
  • Calculate the function's value and its derivative at the current estimate.
  • Use the Newton's formula to find the next approximation.
  • Check if the successive approximations match to a desired precision.
In the exercise, starting at \(x_0 = 1.5\), iterations bring us closer to precise estimate \(r = 2^{1/3}\). This step-by-step correction continues until the values stabilize at the desired degree of accuracy. Iterative methods work well by making small corrections, leading to the most accurate solution with each iteration.
Derivative of a function
The derivative of a function is at the heart of Newton's Method because it helps determine how to adjust our guess. In essence, when using Newton's Method to find roots, the derivative \(p'(x)\) represents the slope of the tangent to the function \(p(x)\) at any point \(x\). It informs us about the rate of change or how steep the function is around that point.Here's how the derivative fits in:
  • The derivative is used to calculate the next guess in the iterative formula.
  • A steeper slope (higher derivative value) can lead to larger changes in the estimate, thus focusing on correction size.
  • If the derivative is zero, it suggests a horizontal tangent, meaning adjustments are minimized as it's a flat region in the function.
For the function \(p(x) = x^3 - 2\), the derivative \(p'(x) = 3x^2\) was derived in the exercise. This function provides the necessary information to perform needed adjustments at each iteration. Understanding the derivative's role helps in comprehending how Newton's Method effectively hones in on root approximations efficiently.

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Most popular questions from this chapter

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