Question

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. $$r=(-67)^{1 / 5}$$

Short answer

Question: Using Newton's method, approximate the value of \((-67)^{1/5}\) to five decimal places. Answer: \((-67)^{1/5} \approx -1.16172\)

Step-by-step solution

Find a Polynomial with Integer Coefficients

We need to find a polynomial with integer coefficients that have a root \(r = (-67)^{1/5}\). This can be done by raising both sides to the power of 5: $$\left((-67)^{1 / 5}\right)^{5} = (-67).$$ This gives us the polynomial: $$p(x) = x^5 + 67.$$ The root of this polynomial is \(r = (-67)^{1/5}\).

Apply Newton's Method

In order to apply Newton's method, first we differentiate the polynomial \(p(x)\). $$p'(x) = 5x^4.$$ Next, we will apply Newton's method using the formula: $$x_{n+1} = x_n - \frac{p(x_n)}{p'(x_n)}.$$ Here, we need to choose a suitable initial value \(x_0\). Since the problem calls for approximating \((-67)^{1/5}\), a decent initial value would be \(x_0 = -1\).

Iterate Until Convergence

Now we start the iterations with the initial value \(x_0 = -1\). We perform the iteration until the difference between the successive approximations is less than \(10^{-5}\). - First Iteration: \(x_1 = x_0 - \frac{p(x_0)}{p'(x_0)} = -1 - \frac{(-1)^5 + 67}{5(-1)^4} = -1.2\) - Second Iteration: \(x_2 = x_1 - \frac{p(x_1)}{p'(x_1)} = -1.2 - \frac{(-1.2)^5 + 67}{5(-1.2)^4} \approx -1.15975\) - Third Iteration: \(x_3 = x_2 - \frac{p(x_2)}{p'(x_2)} = -1.15975 - \frac{(-1.15975)^5 + 67}{5(-1.15975)^4} \approx -1.161748\) - Fourth Iteration: \(x_4 = x_3 - \frac{p(x_3)}{p'(x_3)} = -1.161748 - \frac{(-1.161748)^5 + 67}{5(-1.161748)^4} \approx -1.161724\) We see that the two successive approximations, \(-1.161748\) and \(-1.161724\), agree to five digits to the right of the decimal point after rounding. Thus, we stop here and can approximate: $$(-67)^{1 / 5} \approx -1.16172$$

Key concepts

Approximating Roots

In mathematics, the process of approximating roots is a method used to find a numerical approximation to the roots (solutions) of a real-valued function. One common application is to find roots of polynomials, such as square roots, cube roots, and roots of higher degree. Approximating roots is vital not only in theoretical mathematics but also in various fields like physics, engineering, and finance, where precise calculations are crucial.

Newton's method, also known as the Newton-Raphson method, is a powerful technique to find an approximation of a root. By starting with an initial guess and iterating the process, one can obtain increasingly accurate estimates of the root. The method uses tangent lines, or more specifically, the zeros of the tangent lines, to refine the approximations of the root. The formula for the Newton's method iteration is:
\[\begin{equation} x_{n+1} = x_n - \dfrac{p(x_n)}{p'(x_n)} \end{equation}\]
where \( p(x) \) is the polynomial equation and \( p'(x) \) is its derivative. For example, to approximate the cube root of 10 using Newton's method, one would set up the polynomial equation \( x^3 - 10 = 0 \) and iterate according to the formula until a satisfactory approximation is reached.

Polynomial with Integer Coefficients

When dealing with Newton's method for root approximation, the process often involves a polynomial with integer coefficients. A polynomial is an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents. A goal in these problems is to express the root one wishes to approximate as a root of such a polynomial.

For example, to approximate \((-67)^{1/5}\), one can use the polynomial \( p(x) = x^5 + 67 \), which clearly has \((-67)^{1/5}\) as its root. The integer coefficients make the differentiation step in Newton's method straightforward, providing a practical way to implement the successive iterations for finding the root's approximation. By manipulating the polynomial forms and the powers of the roots, one can always work with integer coefficients, which is essential to maintain accuracy during calculations.

Convergence in Iterations

Convergence refers to the behavior of the sequence of approximations produced by Newton's method; specifically, whether they settle down to a single number as the number of iterations increases. A sequence of approximations is said to converge if the difference between successive approximations becomes arbitrarily small as more iterations are performed.

In practice, we iterate the Newton's method formula until the change between consecutive approximations is less than a given tolerance level. As we refine our guess with each iteration, we reach a point where successive approximations are so close to one another that for practical purposes, they are considered to be the same, indicating convergence. In the provided solution, the use of the stopping criterion is when two successive approximations agree to five decimal places. This criterion provides a balance between computational efficiency and desired precision in the root approximation process.

Differentiation for Newton's Method

The use of differentiation in Newton's method is crucial. Differentiation is a fundamental concept in calculus that refers to finding the derivative of a function. The derivative of a function at any given point can be interpreted as the slope of the tangent line to the function's graph at that point. When applying Newton's method, the derivative shows up in the formula: \( x_{n+1} = x_n - \dfrac{p(x_n)}{p'(x_n)} \), where \( p'(x) \) is the derivative of the polynomial \( p(x) \).

The derivative provides the rate of change of the polynomial, which is used to predict where the function will intercept the x-axis as we adjust our guess. It is essential to accurately compute this derivative for Newton's method to work correctly. In our example, for the polynomial \( p(x) = x^5 + 67 \), the derivative is \( p'(x) = 5x^4 \). This step is straightforward due to the polynomial having integer coefficients, which simplifies the procedure of computing the derivative and thus applying the iterative Newton's method.