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Chapter 3: Problem 21

What can you say about the sequence of approximations obtained using Newton's method if your initial estimate, through a stroke of luck, happens to be the root you are seeking?

Short Answer

Expert verified
If the initial estimate in Newton's method is the exact root, then the sequence of approximations will always be equal to the exact root, and no further iterations are needed. This is because all subsequent approximations, using the iterative formula \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\), will follow the same pattern and be equal to the root \(r\).

Step by step solution

01

Understanding Newton's Method

Newton's method is an iterative numerical technique that helps find the roots (or solutions) of a real-valued function. Given a function \(f(x)\) and an initial guess \(x_0\), the method approximates the root by using the tangent line at the point \((x_0, f(x_0))\). The tangent line intersects the \(x\)-axis at the point \((x_1, 0)\), and we use this new \(x\)-value as our next approximation. The iterative formula can be expressed as: \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\) where \(x_n\) is the nth approximation, \(f(x_n)\) is the function value at \(x_n\), and \(f'(x_n)\) is the derivative value at \(x_n\).
02

Analyzing Newton's Method with the Exact Root as Initial Estimate

Now consider the initial estimate \(x_0\) to be the exact root of the function \(f(x)\), i.e., \(x_0 = r\), where \(r\) is a root, such that \(f(r) = 0\). We will now analyze how the iterative process behaves under this condition and determine the sequence of approximations. Using the Newton's method formula with the initial estimate as the given root, we have: \(x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}\) Since \(x_0 = r\) and \(f(r) = 0\), we get: \(x_1 = r - \frac{0}{f'(r)} = r\) This means that the first approximation is also equal to the exact root, and since \(x_1 = r\), we can further analyze the subsequent approximations in the sequence.
03

Concluding the Sequence of Approximations

As we saw earlier, \(x_1 = r\), which implies that the subsequent approximations will follow the same pattern. Using the iterative formula: \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\) Since \(x_n = r\) for all \(n\), it follows that: \(x_{n+1} = r - \frac{f(r)}{f'(r)} = r - \frac{0}{f'(r)} = r\) Hence, all approximations in the sequence will be equal to the exact root \(r\). When the initial estimate happens to be the root we are seeking in Newton's method, the sequence of approximations obtained will always be equal to the exact root, and thus no further iterations are needed.

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Most popular questions from this chapter

Estimate the value of the radical accurate to four decimal places by using three iterations of Newton's method to solve the equation \(f(x)=0\) with initial estimate \(x_{0}\). \(\sqrt[3]{7} ; \quad f(x)=x^{3}-7 ; x_{0}=2\)

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