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Using Newton's Method In Exercises \(3-6\) calculate two iterations of Newton's Method to approximate a zero of the function using the given initial guess. \(f(x)=\cos x, \quad x_{1}=1.6\)

Short Answer

Expert verified
The estimated zeros of the function \(f(x) = \cos x\) after two iterations of the Newton's Method starting with an initial guess of \(1.6\) are the values of \(x_{2}\) and \(x_{3}\) calculated in step 2 and step 3.

Step by step solution

01

Calculate the Derivative

The derivative of \(f(x) = \cos x\) is \(-\sin x\). This is a standard derivative that can be found in calculus textbooks or from knowledge of the derivative of trigonometric functions.
02

Apply Newton's Method (First Iteration)

Apply the formula for the first iteration: \(x_{2} = x_{1} - (f(x_{1}) / f'(x_{1}))\). Substitute \(x_{1} = 1.6\), \(f(x_{1}) = \cos(1.6)\), and \(f'(x_{1}) = -\sin(1.6)\) into the formula. Calculate the value of \(x_{2}\).
03

Apply Newton's Method (Second Iteration)

Repeat the process for the second iteration: \(x_{3} = x_{2} - (f(x_{2}) / f'(x_{2}))\). Substitute \(x_{2} = \text{the value from step 2}\), \(f(x_{2}) = \cos(\text{the value from step 2})\), and \(f'(x_{2}) = -\sin(\text{the value from step 2})\) into the formula. Calculate the value \(x_{3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Numerical Approximation
Numerical approximation is a cornerstone of computational mathematics, allowing us to find approximate solutions to equations that may be difficult or impossible to solve analytically. For example, Newton's Method is a powerful technique used to zero in on the roots of a real-valued function.

At its heart, the method relies on the idea that a curve can be approximated by its tangent line at a given point. If you select an initial guess close to the true root, you can iteratively improve this guess by following the tangent line to the x-axis, thereby obtaining a new and closer approximation to the actual root.

To succeed in numerical approximation, it's crucial to start with a reasonable initial guess, like the provided value of 1.6 in our exercise. If the guess is too far from the actual root, the method might converge slowly or not at all. A good understanding of the function's behavior greatly helps in this first step.
Trigonometric Functions
Trigonometric functions such as sine and cosine are fundamental in the study of mathematics, especially in areas dealing with waves, oscillations, and circular motion. The function cosine, denoted as \( \cos x \), represents the x-coordinate of a point on the unit circle, corresponding to an angle \( x \) measured in radians.

When trying to find the roots of the function \( f(x) = \cos x \)—which means solving for \( x \) when \( \cos x = 0 \)—knowledge of trigonometric identities and properties becomes crucial. For instance, since the cosine function is periodic with a period of \( 2\pi \) and symmetric about the y-axis, we can infer that it has infinitely many zeros at \( (2n+1) \frac{\pi}{2} \) for any integer \( n \).

In our exercise, we use this knowledge to approximate the root around our initial guess of 1.6, which we know is close to \( \frac{\pi}{2} \), a true zero of the \( \cos x \) function.
Derivatives
Derivatives are a measure of how a function changes as its input changes—essentially, they are the rate of change or slope of the function at any given point. Calculus students learn that \( \frac{d}{dx}(\cos x) = -\sin x \) through differentiation rules.

In Newton's Method, the derivative plays a crucial role in finding the slope of the tangent line at a given guess, which we need to find the next, improved guess. As demonstrated in the exercise, knowing the derivative of our function \( f(x) = \cos x \) as \( -\sin x \) allows us to apply the method effectively. This step is quite vital as any error in the calculation of the derivative can lead to incorrect subsequent approximations.

In straightforward functions like \( \cos x \) and \( \sin x \) where the derivatives are well-known, it simplifies the application of Newton's Method. For more complex functions, calculating the derivative and its value at specific points might require additional computation or techniques like symbolic differentiation.

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