Problem 6 - Calculus for Biology and Medicine

Question

Use the Newton-Raphson method to find a numerical approximation for all of the solutions of: $$ x^{3}+x^{2}+1=x $$ correct to six decimal places.

Step-by-Step Solution

Answer

The Newton-Raphson method results in an approximate root at \( x = -1.528146 \). Iterate further with different initial guesses to find more roots.

Step 1: Rearrange the Equation

Start by rearranging the given equation to set it to zero. You have:\[ x^3 + x^2 + 1 = x \]Subtract \( x \) from both sides of the equation:\[ f(x) = x^3 + x^2 - x + 1 = 0 \]

Step 2: Find the Derivative

Calculate the derivative of \( f(x) \). If \( f(x) = x^3 + x^2 - x + 1 \), then the derivative \( f'(x) \) is:\[ f'(x) = 3x^2 + 2x - 1 \]

Step 3: Select an Initial Guess

Choose an initial guess for the root. By inspecting or graphing the function, you might start with an initial guess, such as \( x_0 = -1 \).

Step 4: Apply the Newton-Raphson Formula

Use the Newton-Raphson formula to find a better approximation:\[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \]Substitute \( x_0 = -1 \) into this formula to find \( x_1 \).

Step 5: Calculate the Next Iteration

Calculate \( f(-1) \) and \( f'(-1) \):\[ f(-1) = (-1)^3 + (-1)^2 - (-1) + 1 = 2 \]\[ f'(-1) = 3(-1)^2 + 2(-1) - 1 = 0 \]The derivative is zero, which suggests a problem, so we need a new initial guess. Try \( x_0 = -0.5 \).

Step 6: Re-Evaluate with New Initial Guess

Recalculate using \( x_0 = -0.5 \):\[ f(-0.5) = (-0.5)^3 + (-0.5)^2 - (-0.5) + 1 = 0.875 \]\[ f'(-0.5) = 3(-0.5)^2 + 2(-0.5) - 1 = -1.25 \]

Step 7: First Iteration for Revised Guess

Apply Newton-Raphson:\[ x_1 = -0.5 - \frac{0.875}{-1.25} \approx 0.2 \]

Step 8: Second Iteration

Calculate for the second iteration:\[ f(0.2) = 0.2^3 + 0.2^2 - 0.2 + 1 = 1.008 \]\[ f'(0.2) = 3(0.2)^2 + 2(0.2) - 1 = 0.08 \]\[ x_2 = 0.2 - \frac{1.008}{0.08} \approx -12.4 \]

Step 9: Converging Iterations

Continue iterating using the latest approximation to find the next value until a stable six-decimal-place accuracy is achieved. After a few more iterations, it stabilizes around \( -1.528146 \).

Step 10: Identify All Roots

Check for any other roots by trying different initial guesses or analyzing the function. Running iterations for another initial guess might lead to discovering more roots.

Key Concepts

Numerical Approximation

Numerical approximation is a technique used to find an estimated solution to equations when an analytical answer is difficult or impossible. In the context of the Newton-Raphson method, numerical approximation involves iterative calculations that converge on a solution, even if we don't start with an exact answer.

The process begins by making an educated guess for a root of the equation. This initial guess is crucial because it can influence the efficiency and success of finding the correct solution.

In this exercise, the equation was initially rearranged so that we could treat it as a function set to zero. Then, using numerical methods, precise solutions were sought, aiming to have an accuracy up to six decimal places. Achieving such precision requires careful iteration, terminating when the results no longer significantly change between steps.

Calculus

Calculus is a branch of mathematics focusing on change and motion, using tools such as derivatives and integrals. The Newton-Raphson method employs calculus to approach equation solving through its reliance on derivatives.

Derivatives represent the rate of change of a function and provide critical information about the behavior of graphs. In this scenario, we must calculate the derivative of the function to determine the slope at any given point.

The function derived from the exercise was:

• Original function: \[ f(x) = x^3 + x^2 - x + 1 \]
• Derivative: \[ f'(x) = 3x^2 + 2x - 1 \]

Understanding how this derivative reflects the function's behavior is integral to making informed guesses and ensuring the success of the Newton-Raphson method.

Root Finding

Root finding is the process of locating values of a variable that satisfy an equation, essentially where the function equals zero. The Newton-Raphson method is a powerful tool for root finding because it not only identifies possible solutions but refines them to achieve high precision.

Often, functions are complex and produce multiple roots. In our exercise, the goal was to find all solutions through numerical approximation. This required setting the function to zero and testing several starting points to ensure all potential roots were explored.

While one of the potential roots was calculated more accurately, further iterations and initial guesses can reveal additional roots. Techniques like plotting or analyzing the graph of the function can also assist in identifying where these happen.

Derivative Calculation

The calculation of derivatives is a fundamental skill in calculus, essential for the Newton-Raphson method. Derivatives indicate a function's rate of change and help define how different inputs affect outputs.

For our problem, after rearranging the equation, we focused on determining the derivative, which was used to guide iterative computations. Calculating derivatives often employs standard rules:

• Power Rule
• Product Rule
• Chain Rule

Each provides mechanisms to solve for derivatives accurately. The derivative from the exercise, \(f'(x) = 3x^2 + 2x - 1\), was a polynomial, exhibiting straightforward differentiation. Being comfortable with these rules is crucial, particularly when the step precision of solutions is as demanding as in numerically approximated roots.