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Use Newton's method to find the point of intersection of the graphs to four decimal places of accuracy by solving the equation \(f(x)-g(x)=0 .\) Use the initial estimate \(x_{0}\) for the \(x\) -coordinate. f(x)=\sin x, g(x)=\frac{1}{5} x, \quad x_{0}=2

Short Answer

Expert verified
The point of intersection of the graphs f(x) = sin(x) and g(x) = (1/5)x to four decimal places of accuracy is approximately (1.8955, 0.3791) using Newton's method with an initial estimate \(x_0 = 2\).

Step by step solution

01

Find expression for difference of functions f(x) and g(x)

We need to find f(x) - g(x). Given f(x) = sin(x) and g(x) = (1/5)x, the expression for f(x) - g(x) is: \(h(x) = f(x) - g(x) = \sin x - \frac{1}{5}x\)
02

Find the derivative of h(x)

Next, we need to find the derivative of h(x) with respect to x. To do this, we differentiate both sin(x) and (1/5)x with respect to x: \(h'(x) = \frac{d(\sin x - \frac{1}{5}x)}{dx} = \cos x - \frac{1}{5}\)
03

Apply Newton's Method

Now we will use the Newton-Raphson formula to find the point of intersection using the initial estimate x_0 = 2: \(x_{n+1} = x_n - \frac{h(x_n)}{h'(x_n)}\) We will iterate this process until the point of intersection is found to four decimal places of accuracy.
04

Iteration 1

Start with the initial estimate x_0 = 2 and plug it into the Newton-Raphson formula: \(x_1 = x_0 - \frac{h(x_0)}{h'(x_0)} = 2 - \frac{\sin(2)-\frac{1}{5}(2)}{\cos(2)-\frac{1}{5}} = 1.8579\)
05

Iteration 2

Continue with the second iteration using x_1: \(x_2 = x_1 - \frac{h(x_1)}{h'(x_1)} = 1.8579 - \frac{\sin(1.8579)-\frac{1}{5}(1.8579)}{\cos(1.8579)-\frac{1}{5}} = 1.8953\)
06

Iteration 3

Continue with the third iteration using x_2: \(x_3 = x_2 - \frac{h(x_2)}{h'(x_2)} = 1.8953 - \frac{\sin(1.8953)-\frac{1}{5}(1.8953)}{\cos(1.8953)-\frac{1}{5}} = 1.8955\) Since x_3 and x_2 are the same to four significant decimal places, we can stop the iteration process here. Now we have found x-coordinate of the point of intersection.
07

Finding the point of intersection

To find the y-coordinate of the point of intersection, we can plug the x-coordinate into either function, f(x) or g(x). Here, we will use g(x): \(y = g(1.8955) = \frac{1}{5}(1.8955) = 0.3791\) So the point of intersection is approximately (1.8955, 0.3791).

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Key Concepts

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

Point of Intersection
A point of intersection is where two or more graphs meet on a coordinate plane. In simpler terms, it’s the spot where the graphs share the same coordinates.

For any two given functions, say \( f(x) \) and \( g(x) \), the point of intersection is found where their values are equal: \( f(x) = g(x) \). This can also be written as \( h(x) = f(x) - g(x) = 0 \). When using Newton's Method to find this point, our main goal is to solve this equation.

In our specific example with functions \( f(x) = \sin x \) and \( g(x) = \frac{1}{5}x \), the intersection is where these two equations yield the same y-value. Thus, solving \( \sin x - \frac{1}{5}x = 0 \) helps us identify the exact points on the graph where these functions intersect.
Graph Functions
Graph functions visually represent mathematical equations on an axis, allowing us to see how the functions behave.

The graph of a function \( f(x) = \sin x \) displays a wavy pattern because sine functions oscillate between -1 and 1. In contrast, the graph of \( g(x) = \frac{1}{5} x \) forms a straight line with a gentle slope intersecting the y-axis at 0.

By drawing both graphs, we can often estimate their intersection points visually. However, for precise intersections (such as exact to four decimal places), mathematical methods like Newton's Method become essential. The graphing doesn’t just help in spotting intersections, but also in understanding the overall behavior of functions, which is invaluable for predicting and validating results.
Derivative of Functions
The derivative of a function gives us the rate at which a function is changing at any point. It’s like observing the slope of the function’s graph.

In the context of Newton's Method, we need derivatives to calculate the tangent line at specific points. For our example, we find the derivative of \( h(x) = \sin x - \frac{1}{5}x \), which results in \( h'(x) = \cos x - \frac{1}{5} \).

This derivative helps us understand how \( h(x) \) changes, and it’s crucial for iterating toward the intersection point. By knowing how quickly \( h(x) \) is increasing or decreasing, the Newton-Raphson method can adjust towards zero more efficiently, thus zeroing in on the intersection point.
Iterative Process
An iterative process involves repeating a series of steps over and over to gradually reach a desired outcome. Newton's Method is such a process, where we repeatedly apply a formula to get closer to the solution.

Starting with an initial guess, \( x_0 = 2 \) in this case, we use the formula \( x_{n+1} = x_n - \frac{h(x_n)}{h'(x_n)} \) to find better approximations. Each cycle uses the result from the last to improve accuracy until changes become negligible—here, until reaching four decimal places.

This repetitive nature is both the strength and beauty of the method: it continually corrects and hones in, like narrowing in on a target, ensuring the solution is as accurate as necessary.

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Most popular questions from this chapter

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