Question

Using the Newton-Raphson procedure find, correct to three decimal places, the root nearest to 7 of the equation \(4 x^{3}+2 x^{2}-200 x-50=0\).

Short answer

1.044

Step-by-step solution

Understand the Newton-Raphson Formula

The Newton-Raphson method uses the formula: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]where \( x_n \) is the current approximation, \( f(x_n) \) is the function value at \( x_n \), and \( f'(x_n) \) is the derivative of the function at \( x_n \).

Define the Function and Its Derivative

The given function is \[ f(x) = 4x^3 + 2x^2 - 200x - 50 \]Its derivative is \[ f'(x) = 12x^2 + 4x - 200 \]

Choose an Initial Guess

Since the root is near 7, let's choose \( x_0 = 7 \) as the initial guess.

First Iteration

Calculate \( x_1 \) using \[ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} \]Thus, we find:\[ f(7) = 4(7)^3 + 2(7)^2 - 200(7) - 50 = 1372 \]\[ f'(7) = 12(7)^2 + 4(7) - 200 = 418 \]Now, \[ x_1 = 7 - \frac{1372}{418} \approx 3.717 \]

Second Iteration

Using \( x_1 \approx 3.717 \), calculate \( x_2 \):\[ f(3.717) = 4(3.717)^3 + 2(3.717)^2 - 200(3.717) - 50 \approx 156.458 \]\[ f'(3.717) = 12(3.717)^2 + 4(3.717) - 200 \approx 65.057 \]Thus, \[ x_2 = 3.717 - \frac{156.458}{65.057} \approx 1.310 \]

Third Iteration

Using \( x_2 \approx 1.310 \), calculate \( x_3 \):\[ f(1.310) = 4(1.310)^3 + 2(1.310)^2 - 200(1.310) - 50 \approx -50.087 \]\[ f'(1.310) = 12(1.310)^2 + 4(1.310) - 200 \approx -192.442 \]Thus, \[ x_3 = 1.310 - \frac{-50.087}{-192.442} \approx 1.050 \]

Fourth Iteration

Using \( x_3 \approx 1.050 \), calculate \( x_4 \):\[ f(1.050) = 4(1.050)^3 + 2(1.050)^2 - 200(1.050) - 50 \approx -1.142 \]\[ f'(1.050) = 12(1.050)^2 + 4(1.050) - 200 \approx -185.386 \]Thus, \[ x_4 = 1.050 - \frac{-1.142}{-185.386} \approx 1.044 \]

Check Convergence

Since the change between \( x_3 \) and \( x_4 \) is very small, and the value is correct to three decimal places, \( x_4 \approx 1.044 \) is a suitable approximation.

Key concepts

Root-Finding Algorithm

The Newton-Raphson method is a widely used root-finding algorithm in mathematical analysis. It helps us locate the roots of a real-valued function. A root of a function is a value of the variable that makes the function equal to zero. This algorithm is iterative, meaning it refines an initial guess to get progressively closer to the actual root.
The formula used can be given by: \[\begin{equation} x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \frac{}{} \text{ where} x_n\text{ is the current guess, }f(x_n)\text{ is the function value, and }f'(x_n)\text{ is the derivative value of the function.}\end{equation}\]
- Start with an initial guess, usually near where you suspect the root to be. - Use the Newton-Raphson formula to compute a better approximation. - Repeat the process until the change between successive approximations is very small.
Applications of the Newton-Raphson method include finding solutions to polynomials and other complex functions, especially when a closed-form solution is difficult or impossible to obtain.

Numerical Methods

Numerical methods are techniques used to solve mathematical problems that are too complex for analytical solutions. These methods involve iterative processes, calculations, and approximations. The Newton-Raphson method is a prime example of a numerical method used in calculus for finding roots of functions.
Unlike exact methods, numerical methods provide approximate solutions but with a high degree of accuracy.

Some key points:

• They often involve repeated iterations to converge to an accurate solution.
• Each iteration refines the previous approximation.
• They are particularly useful for solving differential equations, integral equations, and systems of equations.

In essence, numerical methods are indispensable tools in engineering, physics, economics, and many other fields where complicated equations frequently arise.

Derivatives in Calculus

In the Newton-Raphson method, derivatives play a crucial role. A derivative represents the rate of change of a function with respect to its variable. For a function given by \[ f(x) \], its first derivative, noted as \[ f'(x) \], gives us the slope of the function at any point \[ x \]. In the context of the root-finding algorithm, the derivative \[ f'(x) \] helps determine how steep the function is at a particular point, which is essential for refining root approximations.
- For the function: \[ f(x) = 4x^3 + 2x^2 - 200x - 50 \]The derivative is: \[ f'(x) = 12x^2 + 4x - 200 \]
During each iteration, we compute both \[ f(x_n) \] and \[ f'(x_n) \] to update our approximation using the formula: \[\begin{equation} x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \frac{}{} \text{ where } x_n\text{ is the current approximation}\text{and } x_{n+1}\text{ is the updated approximation using the root-finding formula}\end{equation}\]This knowledge of derivatives is vital for refining our guess on where the function crosses the x-axis.

Iteration Process

The iteration process in the Newton-Raphson method is a sequence of approximations that progressively get closer to the actual root. Starting with an initial guess, the method uses the formula mentioned earlier to compute the next approximation.Here’s how the iteration process typically unfolds:

• 1. **Initial Guess**: Start with an initial guess \(x_0\), close to where you think the root might be.
• 2. **First Iteration**: Calculate \(x_1\) using: \( x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} \).
• 3. **Second Iteration**: Use \(x_1\) to compute \(x_2\), and so on.
• 4. **Convergence Check**: Continue iterating until the change between successive approximations is very small, or until the desired degree of accuracy is achieved.

For example, if we begin with \(x_0 = 7\), our successive approximations \(x_1, x_2, x_3\), etc., will get closer and closer to the actual root.The iteration process is highly effective, but it requires:

• A good initial guess: A poor initial guess may lead to divergence.
• Convergence Criteria: Ensure the difference between successive values tends towards zero.

Understanding the iteration process can greatly aid in solving complex functions accurately and efficiently.