Problem 26 - Calculus Single Variable

Question

The given function has one real root. Approximate it by making an initial estimate \(x_{1}\) and applying the Newton-Raphson Method until an integer \(n\) is found such that \(\left|x_{n}-x_{n-1}\right|<5 \times 10^{-7}\). State \(x_{1}, x_{2}, x_{3}, \ldots, x_{n}\). $$ f(x)=2 x+\cos (x) $$

Step-by-Step Solution

Answer

The Newton-Raphson method with initial guess \(x_1 = 0\) converges to \(-0.332157409\) after 6 iterations.

Step 1: Understand the Newton-Raphson Method

The Newton-Raphson method is used to approximate the roots of a real-valued function. It uses the formula \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\) to find successive approximations. The iteration stops when the change between estimates is smaller than a given threshold.

Step 2: Compute the Derivative

To apply the Newton-Raphson method, we first need the derivative of the function \(f(x) = 2x + \cos(x)\). The derivative is \(f'(x) = 2 - \sin(x)\).

Step 3: Make an Initial Estimate

Choose an initial approximation \(x_1\). A reasonable initial guess is often helpful. For this problem, let's start with \(x_1 = 0\).

Step 4: Apply Newton-Raphson Formula

Compute the next approximation using \(x_2 = x_1 - \frac{f(x_1)}{f'(x_1)}\). Substitute \(x_1 = 0\) into the functions:\[f(0) = 2(0) + \cos(0) = 1\]\[f'(0) = 2 - \sin(0) = 2\]So, \(x_2 = 0 - \frac{1}{2} = -0.5\).

Step 5: Iterate Until Convergence

Continue applying the formula to get \(x_3, x_4, \ldots\) until the change between \(x_n\) and \(x_{n-1}\) is less than \(5 \times 10^{-7}\). - For \(x_3\): \[f(-0.5) = 2(-0.5) + \cos(-0.5) \approx -0.37758256\] \[f'(-0.5) = 2 - \sin(-0.5) \approx 2.47942554\] \[x_3 = -0.5 - \frac{-0.37758256}{2.47942554} \approx -0.348208857\] - For \(x_4\): \[f(-0.348208857) \approx -0.037254546\] \[f'(-0.348208857) \approx 2.34174763\] \[x_4 = -0.348208857 - \frac{-0.037254546}{2.34174763} \approx -0.332351708\] - For \(x_5\): \[f(-0.332351708) \approx -0.000821355\] \[f'(-0.332351708) \approx 2.326412092\] \[x_5 = -0.332351708 - \frac{-0.000821355}{2.326412092} \approx -0.332157511\] - For \(x_6\): \[f(-0.332157511) \approx -1.2298 \times 10^{-7}\] \[f'(-0.332157511) \approx 2.326232458\] \[x_6 = -0.332157511 - \frac{-1.2298 \times 10^{-7}}{2.326232458} \approx -0.332157409\]

Step 6: Check for Convergence

Check the difference \(|x_6 - x_5|\). It is \(|-0.332157409 - (-0.332157511)| \approx 1.02 \times 10^{-7}\), which is less than \(5 \times 10^{-7}\).

Key Concepts

Function derivative

The derivative of a function gives us critical information about its rate of change. In this exercise, the function involved was \( f(x) = 2x + \cos(x) \). To apply the Newton-Raphson method, computing the derivative \( f'(x) \) is essential. The derivative tells us how steep the function is at any point, which directly influences our root approximations. Differentiating \( f(x) \) gives \( f'(x) = 2 - \sin(x) \). This allows us to substitute into the Newton-Raphson formula, yielding more accurate results with each iteration. Understanding derivatives is a cornerstone in calculus, because they help us predict how curves behave.

Root approximation

Root approximation involves finding values where a given function equals zero. This is achieved through successive guesses that get incrementally closer to the true root. In our problem, we started with an initial guess, \( x_1 = 0 \). The Newton-Raphson method refines this guess, adjusting based on the function's slope (derivative) at that point. Subsequent approximations like \( x_2, x_3, \text{and} \ x_4 \) were computed using \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \). Although calculations may seem complex, this strategy systematically moves us toward the actual root by narrowing down the possibilities.

Iterative methods

Iterative methods are techniques for progressively approaching a solution through repeated cycles. In this method, we commence with an initial guess and successively refine it through iterations. Each iteration applies the Newton-Raphson formula: \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \).

• First Iteration: Initial guess \( x_1 = 0 \) leads to \( x_2 = -0.5 \).
• Subsequent Iterations: We continue calculating \( x_3, x_4, \text{and beyond} \) until a predefined accuracy is achieved.

Iterative methods like Newton-Raphson are powerful for solving complex equations since they don’t require you to explicitly know the root beforehand. Instead, they home in on the solution iteratively, a step at a time, yielding successive root approximations.

Convergence criteria

Convergence criteria define when an iterative process should terminate, signifying that the approximation is sufficiently close to the actual solution. In our scenario, this criterion was set as \(|x_n - x_{n-1}| < 5 \times 10^{-7}\). This threshold ensures that the difference between consecutive approximations is minimal, indicating that subsequent values won’t significantly change.

• Convergence Check: For the sequence, \( x_6 \) and \( x_5 \) were compared and found to differ by only \( 1.02 \times 10^{-7} \).

When the change between iterates is below the predefined criterion, it is safe to conclude that the method has converged, and the current approximation is highly reliable. Making convergence criteria explicit helps avoid unnecessary calculations, ensuring efficiency and effectiveness in mathematical problem-solving.