Problem 9 - University Calculus: Early Transcendentals

Question

Show that if \(h>0,\) applying Newton's method to $$f(x)=\left\\{\begin{array}{ll}\sqrt{x}, & x \geq 0 \\\\\sqrt{-x}, & x<0\end{array}\right.$$ leads to \(x_{1}=-h\) if \(x_{0}=h\) and to \(x_{1}=h\) if \(x_{0}=-h .\) Draw a picture that shows what is going on.

Step-by-Step Solution

Answer

Newton's method iterates to -h if starting at h, and to h if starting at -h.

Step 1: Understand Newton's Method

Newton's method is used to find successively better approximations to the roots (or zeroes) of a real-valued function. For a given function \( f(x) \), and its derivative \( f'(x) \), the iterative formula is given by:\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \].This means we need to calculate \( f'(x) \) for the given piecewise function \( f(x) \).

Step 2: Decompose the Function by Interval and Determine Derivative

The function \( f(x) \) is defined differently for \( x \geq 0 \) and \( x < 0 \). Calculate \( f'(x) \):- For \( x \geq 0, \ f(x) = \sqrt{x} \). Using the chain rule, \( f'(x) = \frac{1}{2\sqrt{x}} \).- For \( x < 0, \ f(x) = \sqrt{-x} \). Similarly, \( f'(x) = -\frac{1}{2\sqrt{-x}} \).

Step 3: Apply Newton's Method for \( x_0 = h \)

Substitute \( x_0 = h \) into the Newton's formula using \( f(x) = \sqrt{x} \) and \( f'(x) = \frac{1}{2\sqrt{x}} \):\[x_1 = h - \frac{\sqrt{h}}{\frac{1}{2\sqrt{h}}} = h - 2h = -h.\]Thus, when \( x_0 = h \), Newton's method updates the next approximation to \( -h \).

Step 4: Apply Newton's Method for \( x_0 = -h \)

Substitute \( x_0 = -h \) into the Newton's formula using \( f(x) = \sqrt{-x} \) and \( f'(x) = -\frac{1}{2\sqrt{-x}} \):\[x_1 = -h - \frac{\sqrt{h}}{-\frac{1}{2\sqrt{h}}} = -h + 2h = h.\]Thus, when \( x_0 = -h \), Newton's method updates the next approximation to \( h \).

Step 5: Visual Representation

The function \( f(x) \) is not differentiable at \( x = 0 \) but exhibits symmetrical behavior around the y-axis. When plotted, the function shows a mirrored square root curve on both sides of the y-axis. Newton's method moves initial positive guesses to their negative mirror and vice versa quickly due to the derivative's form, as confirmed by Steps 3 and 4.

Key Concepts

Understanding Piecewise Functions

A piecewise function is a type of function that is defined by multiple sub-functions, where each sub-function applies to a specific interval of the main function's domain. This allows piecewise functions to model situations where a rule or behavior changes over different ranges of the input variable. In this exercise, the function given is:

• \( f(x) = \sqrt{x} \) for \( x \geq 0 \)
• \( f(x) = \sqrt{-x} \) for \( x < 0 \)

These sub-functions allow the main function to switch behaviors at different intervals, such as transitioning from real to imaginary values or simply changing the form of an expression. When a piecewise function is involved in mathematical methods like Newton's method, it's crucial to handle each piece according to its domain carefully in calculations, especially when finding derivatives.

Deriving with Care: Derivative Calculation

Derivative calculation is at the core of understanding how a function behaves and changes. For piecewise-defined functions, calculating the derivative involves evaluating each segment separately. For the given function:

• For \( x \geq 0 \), where \( f(x) = \sqrt{x} \), the derivative is calculated utilizing the power rule, leading to \( f'(x) = \frac{1}{2\sqrt{x}} \).
• For \( x < 0 \), where \( f(x) = \sqrt{-x} \), the derivative involves an additional negative sign from the inside function, resulting in \( f'(x) = -\frac{1}{2\sqrt{-x}} \).

The distinct forms of the derivative in each interval reflect the piecewise nature of the function, which impacts how methods like Newton's update the approximations. Calculating derivatives accurately is essential for employing numerical methods effectively, as they determine the direction and magnitude of the iterations.

Root Finding through Numerical Approximation

Numerical approximation is a powerful tool especially when analytic solutions are difficult or impossible to find. Newton's Method is one such technique used for approximating roots of functions. This iterative method makes use of both the function and its derivative to update guesses for the roots through the formula:\[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\]For the piecewise function considered in the example, starting with different initial guesses (\(x_0 = h\) or \(x_0 = -h\)) led to surprising results:

• When \(x_0 = h\), the next estimate is \(x_1 = -h\), flipping the sign.
• Conversely, when \(x_0 = -h\), \(x_1 = h\), again flipping the sign.

These changes illustrate the sensitivity of the approximation based on the derivative's value and demonstrate how Newton's method harnesses iterative refinement to approach solutions, even in complex functions like piecewise definitions.