Problem 28 - Essential Calculus Early Transcendentals

Question

Use Newton's method to find the absolute maximum value of the function \(f(x)=x \cos x, 0 \leqslant x \leqslant \pi,\) correct to six decimal places.

Step-by-Step Solution

Answer

The absolute maximum value is found at \(x = \frac{\pi}{4}\) and is approximately 0.707107.

Step 1: Find the derivative of the function

To use Newton's method, we first need to find the derivative of the function. We have:\[ f(x) = x \cos x \]Using the product rule \((u'v + uv')\), where \(u = x\) and \(v = \cos x\), the derivative of \(f(x)\) is:\[ f'(x) = (x)' \cos x + x (-\sin x) = \cos x - x \sin x \]

Step 2: Start with an initial guess

Choose an initial guess \(x_0\) for Newton's method. A good choice might be the midpoint of the interval \([0, \pi]\), so take \(x_0 = \frac{\pi}{2}\).

Step 3: Apply Newton's method iteration formula

Newton's method uses the formula:\[ x_{n+1} = x_n - \frac{f'(x_n)}{f''(x_n)} \]We need to find the second derivative \(f''(x)\):\[ f''(x) = -\sin x - \sin x - x \cos x = -2\sin x - x \cos x \]Insert it into the formula and iterate until convergence:

Step 4: Iteration process

Iterate using the Newton's formula with the calculated derivatives:1. Compute \(f'(x_0) = \cos(\frac{\pi}{2}) - \frac{\pi}{2}\sin(\frac{\pi}{2}) = 0 - \frac{\pi}{2}\).2. Compute \(f''(x_0) = -2\sin(\frac{\pi}{2}) - \frac{\pi}{2}\cos(\frac{\pi}{2}) = -2\cdot1 - 0 = -2\).3. Substitute into the Newton's method formula: \[ x_1 = \frac{\pi}{2} - \frac{-\frac{\pi}{2}}{-2} = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4} \].Repeat this until the change is less than the required precision, say, until the change is less than \(10^{-6}\).

Step 5: Continue iterations until precision is met

Continue applying Newton's formula to the results of previous iterations until the result does not change within six decimal places:1. From \(x_1 = \frac{\pi}{4}\), calculate \(x_2, x_3, ... \) using the same formula and steps as before until the result stabilizes with precision smaller than \(10^{-6}\).

Step 6: Determine the absolute maximum value

Once we have convergence within six decimal places, evaluate the function at that \(x\)-value. This value will be the absolute maximum of \(f(x)\) over the given interval.

Key Concepts

Understanding Derivatives in Newton's Method

In Newton's method, derivatives play a crucial role in the process of finding extrema, like maximum or minimum points on a graph. Let's delve into how the derivative of a function aids in this iterative method.

A derivative, in simple terms, represents the rate at which a function's value changes concerning changes in the input. In mathematical notation for a function \(f(x)\), this is expressed as \(f'(x)\). If you're using Newton's method to solve for the extremum, you'll first need this derivative.

• The derivative gives us information about the slope of the tangent line to the function at any point \(x\).
• When the derivative is zero \((f'(x) = 0)\), it implies we've found a critical point, which might be an extremum.
• By utilizing derivatives, we formulate the iterative Newton's method to approach the desired extremum value systematically.

In our exercise, the derivative is derived using the product rule, yielding \(f'(x) = \cos x - x \sin x\). This expression is essential for determining the trajectory of our Newton's iterations.

Iteration Process in Newton’s Method

The iteration process in Newton's Method is a series of calculations that repeatedly refine an approximation of a solution until a desired level of precision is reached. This process is both systematic and iterative.

• Start with an initial guess, denoted as \(x_0\). For best results, this guess should be close to where you expect the maximum or minimum point.
• Use the iterative formula: \[ x_{n+1} = x_n - \frac{f'(x_n)}{f''(x_n)} \] to produce a sequence of better approximations \(x_1, x_2, ..., x_n\).
• Continue the iteration process until the change between successive approximations is smaller than a predefined threshold, ensuring the method has converged to a sufficient precision.

In practice, and in this specific exercise, starting with \(x_0 = \frac{\pi}{2}\), the iterations refine our value for \(x\) until the value stabilizes within six decimal places. Each step requires re-evaluating both first and second derivatives at the new approximation, fine-tuning our guess further.

Finding the Absolute Maximum

Determining the absolute maximum within a given interval can be significantly more efficient using Newton's method, especially when exacting precision is needed. The absolute maximum is essentially the greatest value a function can achieve within a specified range of \(x\).

• Once convergence is achieved via the iteration process, we evaluate the original function \(f(x)\) at the convergence point to find its real value.
• The absolute maximum value is determined by checking this computed value against other potential candidates within the interval, such as endpoints.
• In our case, when \(f(x) = x \cos x\) on \([0, \pi]\), once the iteration meets the precision criteria, the function value at this \(x\) marks the absolute maximum in that interval.

This meticulous approach aids in not only locating the maximum accurately but also confirms that the solution is valid within the defined parameters. Newton's method uses calculus, specifically derivatives, for rigorous and precise optimization within intervals efficiently.