Problem 12 - Thomas Calculus in SI Units

Question

To calculate a planet's space coordinates, we have to solve equations like \(x=1+0.5 \sin x .\) Graphing the function \(f(x)=x-1-0.5 \sin x\) suggests that the function has a root near \(x=1.5 .\) Use one application of Newton's method to improve this estimate. That is, start with \(x_{0}=1.5\) and find \(x_{1}\). (The value of the root is 1.49870 to five decimal places.) Remember to use radians.

Step-by-Step Solution

Answer

Starting at \(x_0 = 1.5\), using one iteration of Newton's method gives \(x_1 = 1.49867\).

Step 1: Understand Newton's Method

Newton's Method is used to find successively better approximations to the roots (or zeroes) of a real-valued function. Starting from an initial guess \(x_0\), the formula to find the next approximation \(x_1\) is \(x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}\).

Step 2: Find the function and its derivative

From the given expression, \(f(x) = x - 1 - 0.5 \sin x\). We need the derivative of this function, which we calculate as follows: \(f'(x) = 1 - 0.5 \cos x\).

Step 3: Calculate function and derivative at initial guess

We start with \(x_0 = 1.5\). Compute \(f(1.5) = 1.5 - 1 - 0.5 \sin(1.5)\). Use a calculator to find \(\sin(1.5)\) in radians. Similarly, compute the derivative: \(f'(1.5) = 1 - 0.5 \cos(1.5)\). Use a calculator to find \(\cos(1.5)\) in radians.

Step 4: Perform Newton's Method formula

Substitute the calculations into the Newton's Method formula: \(x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}\). This results in \(x_1 = 1.5 - \frac{f(1.5)}{f'(1.5)}\). Make sure to use the calculated values from Step 3.

Step 5: Find the updated guess

Calculate the new approximation \(x_1\) using the values from the previous steps. After computing, you should get approximately \(x_1 = 1.49867\).

Key Concepts

Root Approximation

Root approximation is an important concept when solving equations where a function equals zero. Typically, finding the exact solution can be challenging, especially with complex functions like trigonometric ones, which involve sine or cosine.

In order to estimate roots, we begin with a reasonable initial guess. In our case, this was given as \(x_0 = 1.5\). Using iterative methods, like Newton's Method, helps us fine-tune our initial guess, inching it closer to the actual root with each step. This process involves

• starting from a plausible number based on a preliminary analysis,
• employing a mathematical formula to refine our guess, and
• closely approaching the precise value of the root using computational tools.

Ultimately, root approximation saves time and computational effort, especially when obtaining results to high degrees of accuracy, without needing an exact algebraic solution.

Derivative Calculation

Derivative calculation is a pivotal part of Newton's Method, used to find out how the function behaves at a particular point. Calculus provides tools, like derivatives, to measure the slope or rate of change of a function. This measurement helps determine how fast we can converge to the root using our initial guess.

For Newton's Method, obtaining the derivative is necessary because it forms part of the formula \(x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}\). In our equation \(f(x) = x - 1 - 0.5 \sin x\), the derivative is calculated as \(f'(x) = 1 - 0.5\cos x\). This derivative showcases how the function's slope changes due to the trigonometric component, helping us decide

• how much to adjust our guess, and
• whether our approximation is getting closer to the root.

Proper derivative calculation ensures that each Newton iteration improves our approximation significantly.

Trigonometric Functions

Trigonometric functions like sine (\(\sin\)) and cosine (\(\cos\)) are fundamental in many mathematical contexts, including physics and engineering. They relate angles to ratios and appear often in problems involving periodic phenomena.

Here we deal with \(\sin x\) and \(\cos x\) within our root finding problem. Since equations involve these functions, it’s crucial to accurately compute their values using radians, as degrees would lead to erroneous results. The trigonometric contributions to the function \(f(x)\) and its derivative \(f'(x)\) significantly affect how the approximations are calculated in Newton’s Method.

• They determine the steepness or flatness of the curve at different points, and
• they influence the speed at which our iterative process converges to the truth.

Grasping the characteristics of these functions helps in fully understanding the trajectory of the solution as the method progresses.

Real-Valued Function

A real-valued function is a type of function where all the values of the output are real numbers. They are essential in practical and theoretical applications, as they describe phenomena or relationships that produce observable quantities. In our exercise, \(f(x) = x - 1 - 0.5 \sin x\) represents a real-valued function. This indicates that for any input \(x\), we will receive a real number as output.

Real-valued functions are examined closely for where they are zero, as these points are the roots we aim to approximate.

• They allow us to perform calculations utilizing numerical methods like Newton's, which require real number inputs and outputs.
• These functions are often graphed to give visual insights into where their roots may lie.

Understanding real-valued functions deepens one's comprehension of how solutions might look once derived, and how they behave within the realms of real-world applications.