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Locating a planet To calculate a planet's space coordinates, we have to solve equations like \(x=1+0.5 \sin x .\) Graphing the function \(f(x)=x-1-0.5 \sin x\) suggests that the function has a root near \(x=1.5 .\) Use one application of Newton's method to improve this estimate. That is, start with \(x_{0}=1.5\) and find \(x_{1}\) . (The value of the root is 1.49870 to five decimal places.) Remember to use radians.

Short Answer

Expert verified
The improved estimate using one application of Newton's method is approximately 1.222.

Step by step solution

01

Understand Newton's Method Formula

Newton's method is an iterative technique to find successively better approximations to the roots of a real-valued function. The update formula is given by \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \), where \( f'(x) \) is the derivative of \( f(x) \).
02

Define the Function and its Derivative

The given function is \( f(x) = x - 1 - 0.5 \sin x \). To apply Newton's method, we first need the derivative of \( f(x) \). So, we find \( f'(x) = 1 - 0.5 \cos x \).
03

Evaluate Function and Derivative at Initial Guess

Use the initial guess \( x_0 = 1.5 \) to evaluate \( f(x_0) \) and \( f'(x_0) \). Calculate \( f(1.5) = 1.5 - 1 - 0.5 \sin(1.5) \approx 0.259 \) and \( f'(1.5) = 1 - 0.5 \cos(1.5) \approx 0.929 \).
04

Apply Newton's Method Formula

With \( f(x_0) \approx 0.259 \) and \( f'(x_0) \approx 0.929 \), apply Newton's formula: \( x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 1.5 - \frac{0.259}{0.929} \approx 1.222 \).

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Key Concepts

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

Root Finding
Root finding is a fundamental concept in mathematics where we solve equations to find the values of the variable that make the equation true, particularly where the function equals zero. Consider an equation like \( f(x) = 0 \). The "root" of the function is the value \( x \) that makes this equation true. In practical applications such as locating a planet's position in space, finding these roots becomes essential to solving the equations related to its trajectory.When we graph a function, roots are the points where the graph intersects the x-axis. In the given problem, the function \( f(x) = x - 1 - 0.5\sin x \) has a root where this intersection occurs. Initially, we estimate the root using graphical insights, which suggests a root near \( x = 1.5 \). However, for accurate calculations, numerical methods like Newton's method help hone in on a more precise estimate.
Iterative Techniques
Iterative techniques are methods that approximate solutions through repetition, refining estimates step by step. Newton’s Method is a classic example, where through iteration, we achieve increasingly accurate estimates of a function’s root. In each iteration, we use the previous estimate to calculate a better approximation.In the problem provided, Newton’s Method involves:
  • Starting with an initial guess, \( x_0 \).
  • Applying the update formula \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \).
Iterative techniques like these are powerful because they often converge quickly to a solution, provided the initial guess is sufficiently close to the actual root. It’s important to ensure convergence by choosing an appropriate starting point and checking conditions such as the function being differentiable.
Derivatives
Derivatives play a crucial role in Newton’s Method as they offer information about the rate at which the function's value is changing. For root finding, understanding how \( f(x) \) changes is essential as it influences the estimate of where the root is.For the function \( f(x) = x - 1 - 0.5\sin x \), its derivative is \( f'(x) = 1 - 0.5\cos x \). This derivative measures how the function’s output changes with a small change in \( x \). In Newton’s Method, the derivative helps refine the estimate by indicating the slope of the tangent line at the point \( x_n \).Calculating the derivative correctly ensures that the iteration steps accurately adjust the estimates towards the root, such as improving the initial guess from \( x_0 = 1.5 \) to a more accurate \( x_1 \). Without precise derivatives, this method could fail to converge or lead to incorrect results.

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