Problem 66 - Calculus with Concepts in Calculus

Question

a. Show that there is exactly one value of \(x\) for which \(\cos x=2 x .\) (Hint: Let \(f(x)=\cos x-2 x\), and determine how many zeros \(f\) could have.) b. Use the Newton-Raphson method to approximate the value of \(x\) for which \(\cos x=2 x\). Continue until successive iterations obtained by the calculator are identical.

Step-by-Step Solution

Answer

There is exactly one value, and using the Newton-Raphson method, the approximation is iteratively refined.

Step 1: Define the Function

Let \( f(x) = \cos x - 2x \). We need to determine how many zeros this function has. A zero indicates where \( f(x) = 0 \), which corresponds to \( \cos x = 2x \).

Step 2: Analyze the Behavior of \(f(x)\)

We know \( \cos x \) oscillates between -1 and 1, while \( 2x \) is a linear function with slope 2. As \( x \to \infty \) or \( x \to -\infty \), \( \cos x \) remains bounded, but \( 2x \) diverges. Hence, \( f(x) = \cos x - 2x \to -\infty \) as \( x \to \infty \) or \( x \to -\infty \).

Step 3: Use Intermediate Value Theorem

To find the intersection, we look for \( x \) where \( -1 \leq 2x \leq 1 \), i.e., \( -\frac{1}{2} \leq x \leq \frac{1}{2} \). Given \( f(-\frac{1}{2}) = \cos(-\frac{1}{2}) + 1 \) and \( f(\frac{1}{2}) = \cos(\frac{1}{2}) - 1 \), use the Intermediate Value Theorem to confirm there is at least one root in \( (-\frac{1}{2}, \frac{1}{2}) \).

Step 4: Check \(f'(x)\) for Uniqueness

Compute \( f'(x) = -\sin x - 2 \). Since \( -\sin x \) ranges from -1 to 1, \( f'(x) \) is always negative (ranging between -3 to -1). \( f(x) \) is strictly decreasing, confirming there's exactly one zero (root).

Step 5: Use the Newton-Raphson Method

Starting with an initial guess (e.g., \( x_0 = 0 \)), apply the Newton-Raphson iteration formula: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] Compute using \( f(x) = \cos x - 2x \) and \( f'(x) = -\sin x - 2 \) to update \( x_n \) until \( x_{n+1} \approx x_n \).

Step 6: Iterative Calculation

Perform iterations: 1. \( x_0 = 0 \), \2. Calculates \( f(0) = \cos(0) - 0 = 1 \), \( f'(0) = -\sin(0) - 2 = -2 \). 3. Update: \( x_1 = 0 - \frac{1}{-2} = 0.5 \).4. Continuing: \( x_2 \) is found by substituting \( x_1 \) into the iteration formula. 5. Continue until \( x_n \approx x_{n+1} \).

Key Concepts

Cosine Function

The cosine function, denoted as \( \cos x \), is a fundamental trigonometric function. It is periodic with a period of \( 2\pi \), meaning its values repeat every \( 2\pi \) units. This function oscillates between -1 and 1, reaching maximum value (1) when \( x = 0, 2\pi, 4\pi, \ldots \), and minimum value (-1) when \( x = \pi, 3\pi, 5\pi, \ldots \). It is important to note that cosine function is even, which means \( \cos(-x) = \cos x \).

The graph of \( \cos x \) is wave-like and these properties help analyze equations like \( \cos x = 2x \), as part of the exercise. Since \( \cos x \) has a finite range and \( 2x \) can grow indefinitely, this helps us establish bounds when finding intersections between these functions.

Zeros of a Function

Zeros of a function occur where the function equals zero. For the function \( f(x) = \cos x - 2x \), zero points are critical because they represent values where \( \cos x = 2x \). To find zeros, we initially inspect the behavior of \( f(x) \) and its derivative.

Since \( \cos x \) is bounded and \( 2x \) diverges, a function formed by \( f(x) = \cos x - 2x \) will also diverge as \( x \to \pm \infty \). Observing its graph or using a theorem or method like Newton-Raphson can help us pinpoint the exact places where zeros occur.

Given the context, zeros are sought within specific bounds derived from when \( f(x) \) changes sign, in this case between \(-\frac{1}{2}\) and \(\frac{1}{2}\). Here, finding zeros is a central stepping stone to applying the iterative Newton-Raphson method more effectively.

Intermediate Value Theorem

The Intermediate Value Theorem is a cornerstone in calculus, especially when analyzing continuous functions. Simply put, if a continuous function, \( f \), takes on values \( f(a) \) and \( f(b) \) at points \( a \) and \( b \) such that \( f(a) eq f(b) \), the function takes on any intermediate value between \( f(a) \) and \( f(b) \) within \( (a, b) \).

This theorem affirms the presence of at least one root when checking \( f(x) = \cos x - 2x \), in our context. Specifically, when it’s observed that \( f(-\frac{1}{2}) = \cos(-\frac{1}{2}) + 1 \) and \( f(\frac{1}{2}) = \cos(\frac{1}{2}) - 1 \), if these are of opposite signs, the theorem assures us there is a point in \((-\frac{1}{2}, \frac{1}{2})\) where \( f(x) \) is zero. This helps substantiate the function's root before applying more precise numeric methods.

Derivative Analysis

Derivative analysis involves taking derivatives of a function to understand its behavior, such as identifying maxima, minima, or monotonicity. For \( f(x) = \cos x - 2x \), the derivative \( f'(x) = -\sin x - 2 \) tells us how the function behaves with changing \( x \).

The derivative indicates that \( f(x) \) is strictly decreasing, since \( -\sin x - 2 \) ranges from -3 to -1, which is consistently negative. A strictly decreasing function can only cross the x-axis once within a given interval, guaranteeing us a unique zero point within \((-\frac{1}{2}, \frac{1}{2})\).

This understanding is crucial when using iterative methods like Newton-Raphson, as it assures us that our starting guess will converge to a single root.