Question

Suppose \(f\) is twice differentiable on \([a, b], f(a)<0, f(b)>0, f^{\prime}(x) \geq 8>0\), and \(0 \leq f^{\prime \prime}(x) \leq M\) for all \(x \in[a, b]\). Let \(\xi\) be the unique point in \((a, b)\) at which \(f(\xi)=0 .\) Complete the details in the following outline of Newton's method for computing \(\xi\) (a) Choose \(x_{1} \in(\xi, b)\), and define \(\left\\{x_{n}\right\\}\) by $$ x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} $$ Interpret this geometrically, in terms of a tangent to the graph of \(f\). (b) Prove that \(x_{n+1}

Short answer

The geometric interpretation of Newton's method at every step is finding the x-intercept of the tangent line to the graph of the function f at the point (x_n, f(x_n)). This x-intercept, denoted as x_{n+1}, serves as the new approximation for the root ξ.

Step-by-step solution

(a) Geometric Interpretation

At every step \(n\), the iterative formula for Newton's method is given by: $$ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} $$ This formula can be interpreted as finding the \(x\)-intercept of the tangent line to the graph of \(f\) at the point \((x_n, f(x_n))\). At each step, we take the tangent line to \(f\) at the current approximation \(x_n\) and use its \(x\)-intercept \(x_{n+1}\) as the new approximation for \(\xi\).

(b) Proof of \(x_{n+1} < x_n\) and Convergence

First, we need to show that the new approximation \(x_{n+1}\) is always less than \(x_n\): $$ x_{n+1} - x_n = -\frac{f(x_n)}{f'(x_n)} < 0 $$ Since \(f'(x_n) \geq 8 > 0\) and \(f(x) < 0\) for \(x \in (\xi, b]\) because \(f(a) < 0\) and \(f(b) > 0\), we have: $$ x_{n+1} < x_n $$ Now, in order to show that the sequence \(\{x_n\}\) converges to \(\xi\), observe that \(f'(x) \geq 8 > 0\) for all \(x \in [a,b]\). Since \(f'(x)\) is continuous and positive, \(f(x)\) is strictly increasing on \([a,b]\). Also because \(f(a) < 0\) and \(f(b) > 0\), by the intermediate value theorem, there exists a unique \(\xi \in (a, b)\) such that \(f(\xi) = 0\) and \(f(x) < 0\) for \(x \in [a, \xi)\) and \(f(x) > 0\) for \(x \in (\xi, b]\). Since \(x_{n+1}<x_n\) for all \(n \in \mathbb{N}\) and \(x_1 \in (\xi, b)\), we have \(\{x_n\}\) is a decreasing sequence bounded below by \(\xi\). Therefore, \(\{x_n\}\) converges, and since \(f(x_n)\to 0\) as \(n\to \infty\), we have \(\lim_{n\to\infty} x_n = \xi\).

(c) Taylor's Theorem

Using Taylor's theorem, we can represent \(f\) at the point \(x_{n+1}\) using \(f(x_n)\) and its derivatives as follows: $$ f(x_{n+1}) = f(x_n) + f'(x_n)(x_{n+1}-x_n)+\frac{f''(t_n)}{2}(x_{n+1}-x_n)^2 $$ for some \(t_n \in (\xi, x_n)\). Since \(x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)}\) and \(f(x_{n+1})=0\) because \(x_{n+1}\) approximates the root, we get: $$ 0 = f(x_n) + f'(x_n)(x_{n+1}-x_n) + \frac{f''(t_n)}{2}(x_{n+1}-x_n)^2 $$ Solving for \(x_{n+1} - \xi\) gives: $$ x_{n+1} - \xi = \frac{f''(t_n)}{2f'(x_n)}(x_n - \xi)^2 $$

(d) Upper Bound

Let \(A = \frac{M}{28}\), where \(M\) is the upper bound of \(f''(x)\) on the interval \([a, b]\). Then we have: $$ 0 \leq \frac{f''(t_n)}{2f'(x_n)} \leq \frac{M}{28} \cdot \frac{1}{8} = A $$ Using the inequality derived in part (c), we get: $$ 0 \leq x_{n+1} - \xi \leq A(x_n - \xi)^2 \implies 0 \leq x_{n+1} - \xi \leq \frac{1}{A}\left[A(x_1 - \xi)\right]^{2n} $$

(e) Fixed Point of the Function \(g\)

Consider the function \(g(x) = x - \frac{f(x)}{f'(x)}\). Observe that Newton's method is equivalent to finding a fixed point of this function, as for \(\xi\), we have: $$ \lim_{n \rightarrow \infty} g(x_n) = \xi $$ Now we want to analyze how \(g'(x)\) behaves for \(x\) near \(\xi\). Differentiating \(g(x)\) yields: $$ g'(x) = 1 - \frac{f'(x) f'(x) - f(x) f''(x)}{f'(x)^2} = \frac{f(x)f''(x)}{f'(x)^2} $$ Because \(f(\xi) = 0\), we see that \(g'(\xi) = 0\).

(f) Newton's Method for \(f(x) = x^{1 / 3}\)

Consider the function \(f(x) = x^{1/3}\). Then we have \(f'(x) = \frac{1}{3}x^{-2/3}\) and \(f''(x) = -\frac{2}{9}x^{-5/3}\). Applying Newton's method, we get the iteration formula: $$ x_{n+1} = x_n - \frac{x_n^{1/3}}{\frac{1}{3}x_n^{-2/3}} $$ Notice that Newton's method does not converge for this function; the iteration formula leads to an infinite cycle of values for \(x_n\). This is because the function \(f(x) = x^{1/3}\) does not satisfy the conditions required for Newton's method (e.g., \(f''(x)\) is unbounded).

Key concepts

Twice Differentiable Functions

Understanding twice differentiable functions is essential when discussing Newton's Method. A function is twice differentiable when it possesses both a first and second derivative that exist and are continuous. This property is crucial because it allows for accurate approximations of the function's behavior using derivatives.
For the problem provided, it's given that the function \(f\) is twice differentiable within the interval \([a, b]\). That means we can compute both \(f'(x)\) and \(f''(x)\). This information aids us in applying Taylor's Theorem, as the second derivative provides insights into the curvature of \(f\).
The second derivative \(f''(x)\) plays a vital role when determining the accuracy of the approximation in Newton's Method. Having an upper bound ensures control over the behavior of \(f\) within the specified interval. This allows for predictable, safe convergence towards the solution or root of the function, known as \(\xi\).

Taylor's Theorem

Taylor's Theorem is a powerful tool in numerical analysis and calculus, allowing functions to be expressed as an infinite sum of terms calculated from the values of its derivatives at a single point. Building on the twice differentiable function \(f\), Taylor's Theorem helps approximate \(f(x)\) near any point \(x_0\).
When applied to Newton's Method, Taylor's Theorem assists in demonstrating convergence. Specifically, it is used to determine the relationship between consecutive iterations by approximating \(f\) around each point \(x_n\). This approximation facilitates calculating \(x_{n+1}\) based on \(x_n\), as evidenced by:
\[ x_{n+1} - \xi = \frac{f''(t_n)}{2f'(x_n)}(x_n - \xi)^2\]Here, \(t_n\) is some point between \(\xi\) and \(x_n\), illustrating how each iteration gives a better approximation of the root.
Thus, Taylor's Theorem provides a critical link in proving the quadratic convergence of Newton's method, by relating the approximate error squared at each iteration.

Fixed Points

Fixed points are fundamental in understanding iterative processes like Newton's Method. A fixed point of a function \(g\) is an input \(x\) such that \(g(x) = x\). This concept arises naturally in Newton's Method, where the aim is to find a root of \(f\) by iterating the process:
\[ x_{n+1} = g(x_n) = x_n - \frac{f(x_n)}{f'(x_n)}\]The goal is for the iteration \(x_{n+1}\) to become constant, which corresponds to \(x_n\) being a fixed point of \(g\). In this context, achieving \(g(\xi) = \xi\) means we have located the root \(\xi\) of \(f\).
As each iteration brings \(x_n\) closer to \(\xi\), the sequence converges. The behavior of the derivative \(g'(x)\) near \(\xi\) is critical for convergence analysis. If \(g'(\xi) = 0\), the rate of convergence is faster, indicating Newton's method's efficient approach to finding the fixed point.

Convergence Analysis

Convergence analysis in Newton's Method examines the conditions under which the sequence \( \{x_n\} \) converges to the root \( \xi \). Understanding convergence involves determining how quickly \( x_n \) approaches the actual root. For the problem discussed, several conditions ensure convergence:

• The function \(f\) is twice differentiable.
• The first derivative \(f'(x)\) is non-zero, ensuring that the tangent lines used in the iterations are not horizontal.
• The second derivative \(f''(x)\) is bounded within the interval.

These conditions provide a foundation for quadratic convergence in Newton's Method, where each step approximately squares the accuracy of our approximation.
A practical way to assess convergence is through the inequality derived from Taylor's Theorem:
\[ 0 \leq x_{n+1} - \xi \leq \frac{1}{A}\left[A(x_1 - \xi)\right]^{2n}\]Here, \(A\) relates to the upper bound of \(f''(x)\), and this inequality demonstrates how successive iterations yield a rapidly shrinking distance to the root. Convergence analysis is crucial for understanding both the efficiency and reliability of Newton's Method in practical applications.