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The equation \(x+\tan x=0\) has an infinite number of positive roots \(r_{1}, r_{2}, r_{3}, \ldots, r_{n}\) slightly larger than \(\left(n-\frac{1}{2}\right) \pi\) Use the Newton-Raphson method to find \(r_{1}\) and \(r_{2}\) lo three decimal place accuracy.

Short Answer

Expert verified
The first two positive roots of the equation \(x + \tan x = 0\) are approximately \(r_1 \approx 4.493\) and \(r_2 \approx 7.725\), found using the Newton-Raphson method with initial approximations \(0.5\pi\) and \(1.5\pi\), respectively.

Step by step solution

01

Write the given equation and the formula for the Newton-Raphson method

First, let's write down the given equation: \[x + \tan x = 0\] Now, let's write down the formula for the Newton-Raphson method: \[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\] where \(f(x) = x + \tan x\). We need to find the first two positive roots of this equation, \(r_1\) and \(r_2\), with three decimal place accuracy.
02

Find the derivative of the given function

In order to find the derivative of the given function, we can apply the sum rule and the chain rule: \(f'(x) = 1 + \sec^2 x\) Now we have the derivative of the function.
03

Set up the initial approximations for the roots and calculate using Newton-Raphson method

We are given that: \(r_n \approx \left(n-\frac{1}{2}\right) \pi\) For \(r_1\), let \(n=1\), so the initial approximation is: \(x_0 \approx \left( 1 - \frac{1}{2} \right) \pi \approx 0.5\pi\) Similarly, for \(r_2\), let \(n=2\), so the initial approximation is: \(x_0 \approx \left( 2 - \frac{1}{2} \right) \pi \approx 1.5\pi\) Now, plug the initial approximations into the Newton-Raphson formula and iterate until the approximations converge to three decimal places. For finding \(r_1\): - \(x_0 \approx 0.5\pi\) - Continue iterating using Newton-Raphson formula: \(x_{n+1} = x_n - \frac{x_n + \tan x_n}{1 + \sec^2 x_n}\) For finding \(r_2\): - \(x_0 \approx 1.5\pi\) - Continue iterating using Newton-Raphson formula: \(x_{n+1} = x_n - \frac{x_n + \tan x_n}{1 + \sec^2 x_n}\)
04

Iterating the Newton-Raphson method and finding the root approximations

After iterating the Newton-Raphson method for several steps, we find the following approximations for \(r_1\) and \(r_2\): \(r_1 \approx 4.493\) \(r_2 \approx 7.725\) These are the approximate values of the first two positive roots of the equation \(x + \tan x = 0\) with three decimal place accuracy.

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Key Concepts

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

Root Approximation
Root approximation is a critical concept in numerical analysis, particularly when solving equations where the exact solution is difficult to find analytically. One of the most common methods for root approximation is the Newton-Raphson method, an iterative technique that successively approximates the roots of a real-valued function.

In practice, the method starts with an initial guess and refines this approximation by applying a specific formula. The Newton-Raphson formula is
\[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\]

In the given exercise, we use this technique to find the positive roots of the trigonometric equation \(x + \tan x = 0\). The root approximation begins with a rough estimate, which is refined through iterations of the Newton-Raphson method until it reaches the desired level of accuracy.
Derivative of a Function
The derivative of a function is a fundamental concept in calculus that describes the rate at which the function's value changes with respect to a change in its input value. For the Newton-Raphson method to work, we require the derivative of the function for which we're seeking the root.

The given function in the exercise is \(f(x) = x + \tan x\), and its derivative, based on calculus rules for differentiation, is

\[f'(x) = 1 + \sec^2 x\]

This derivative is used in the denominator of the Newton-Raphson formula and plays a crucial role in the convergence to the root. Errors in calculating the derivative can significantly affect the accuracy of the root approximation.
Convergence of Iterative Methods
The convergence of iterative methods is related to how quickly and accurately a method can reach the desired solution. In context of the Newton-Raphson method, convergence depends on several factors including the function itself, the accuracy of derivatives, and the initial guess.

The iterative method converges to a root if successive approximations approach the actual root as the number of iterations increase. However, if the function isn't well-behaved, or if the initial guess is far from the actual root, the method may diverge. For the exercise, by carefully choosing initial approximations and applying the Newton-Raphson formula, we can ensure the convergence towards the roots \(r_1\) and \(r_2\) with the required accuracy.
Trigonometric Equations
Trigonometric equations involve trigonometric functions like sine, cosine, tangent, and their reciprocals. Solving these equations analytically can be complex, especially when seeking precise numeric solutions.

In our exercise, the equation \(x + \tan x = 0\) represents a trigonometric equation featuring the tangent function. Traditional methods for solving trigonometric equations may not yield roots easily, especially when the roots do not correspond to standard angles. The Newton-Raphson method provides a way to approximate these roots numerically with a desired precision, demonstrating its utility in handling equations that stump analytical methods.

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Most popular questions from this chapter

A tour boat heads out on a 100 -kilometre sight-seeing trip. Given that the fund costs of operating the boat total 2500 dollar per hour, that the cost of fuel varies directly with the square of the speed of the boat, and at 10 kilometres per hour the cost of the fuel is 400 dollar per hour, find the speed test minimizes the boat owner's expenses. Is the speed that minimizes the owner's expenses dependent on the length of the trip?

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At what rate is the volume of a sphere changing at the instant when the surface area is increasing at the rate of 4 square centimeters per minute and the radius is increasing at the rate of 0.1 centimeters per minute?

Use a CAS to find the oblique asymptotes. Then use a graphing utility to draw the 2 graph of \(f\) and is asymptotes, and thereby confirm your findings. $$f(x)=\frac{5 x^{3}-3 x^{2}+4 x-4}{x^{2}+1}$$

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