Chapter 4: Problem 12
Use a graphing utility to determine how many solutions the equation has, and then use Newton’s Method to approximate the solution that satisfies the stated condition. \(\sin x=x^{2} ; x >0\)
Short Answer
Expert verified
The equation has one solution for \( x > 0 \), and it's approximately \( x \approx 0.8768 \).
Step by step solution
01
Setting up the equation for graphing
First, we'll graph the functions \( y = \sin x \) and \( y = x^2 \) to find their points of intersection. This will help us determine how many solutions exist for the equation \( \sin x = x^2 \) when \( x > 0 \).
02
Graphing the functions
Using a graphing utility, plot the graphs of \( y = \sin x \) and \( y = x^2 \). Observe where these graphs intersect beyond the point where \( x > 0 \). You will notice these graphs intersect only once beyond \( x = 0 \). This indicates there's one solution for \( \sin x = x^2 \) in the given domain.
03
Choosing the initial guess for Newton's Method
Observe the graph to approximate where the intersection point occurs. Let's choose an initial guess for \( x \) close to the intersection; approximately \( x = 0.8 \).
04
Deriving the function for Newton's Method
For Newton's Method, we need a function \( f(x) \) such that \( f(x) = \sin x - x^2 \). The derivative \( f'(x) = \cos x - 2x \). Newton's formula is \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \).
05
Applying Newton's Method
Using \( x_0 = 0.8 \) as our initial guess, apply Newton's Method:1. Calculate \( f(x_0) = \sin(0.8) - (0.8)^2 \).2. Calculate \( f'(x_0) = \cos(0.8) - 2(0.8) \).3. Use the formula \( x_1 = 0.8 - \frac{f(0.8)}{f'(0.8)} \).4. Repeat this process until successive values of \( x \) converge.
06
Convergence and Final Solution
Continue with the iterative process, calculating further approximations: \( x_2, x_3, ... \) until \( |x_{n+1} - x_n| \) becomes very small or the values become stable. The approximation converges at approximately \( x \approx 0.8768 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Graphing Utility
To find out where the two functions intersect, we'll use a graphing utility, which is a tool or software program that helps you visualize functions. In this case, we're dealing with the functions \( y = \sin x \) and \( y = x^2 \). When you use a graphing utility:
- You'll plot both functions on the same coordinate plane.
- This visualization helps you see where the graphs cross each other, known as points of intersection.
- For the given function \( \sin x = x^2 \), we are interested in where they intersect for \( x > 0 \).
Function Intersection
When analyzing the equation \( \sin x = x^2 \), the goal is to find where these two functions intersect when graphed. An intersection point occurs at values of \( x \) where both functions return the same result. In this case:
- The intersection is where the value of \( \sin x \) exactly matches the value of \( x^2 \).
- Using the graphing utility, you observe that these functions only intersect once when \( x > 0 \).
Solution Approximation
Many mathematical equations might not have straightforward analytical solutions, which is where numerical methods like Newton's Method come in handy to approximate solutions:
- Newton's Method uses an iterative approach to zero in on the solution.
- You start with an initial guess (we chose \( x = 0.8 \) based on our graph), then refine that guess using the formula:
- \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \)