Problem 8 - Calculus: Early Transcendentals

Question

Find the Taylor polynomials \( T_3(x) \) for the function \( f \) centered at the number \( a \) Graph \( f \) and \( T_3 \) on the same screen. \( f(x) = x \cos x, \) \( a = 0 \)

Step-by-Step Solution

Answer

The Taylor polynomial is \( T_3(x) = x - \frac{x^3}{2} \).

Step 1: Find Derivatives

To find the Taylor polynomial, we first need the derivatives of the function. Let's find the first few derivatives of \( f(x) = x \cos x \) evaluated at \( a = 0 \). We will need \( f'(x), f''(x), \) and \( f'''(x) \).1. \( f(x) = x \cos x \)2. \( f'(x) = \cos x - x \sin x \)3. \( f''(x) = -2 \sin x - x \cos x \)4. \( f'''(x) = -3 \cos x + x \sin x \)Evaluate these derivatives at \( x = 0 \):- \( f(0) = 0 \cdot \cos 0 = 0 \)- \( f'(0) = \cos 0 - 0 \cdot \sin 0 = 1 \)- \( f''(0) = -2 \sin 0 - 0 \cdot \cos 0 = 0 \)- \( f'''(0) = -3 \cos 0 + 0 \cdot \sin 0 = -3 \)

Step 2: Form the Taylor Polynomial

The Taylor polynomial of degree 3 centered at \( x = 0 \) is given by: \[ T_3(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} \]Substitute the values found in Step 1:- \( T_3(x) = 0 + 1 \cdot x + \frac{0 \cdot x^2}{2} + \frac{-3 \cdot x^3}{6} \)- Simplified, this gives \( T_3(x) = x - \frac{x^3}{2} \)

Key Concepts

Function Derivatives

Understanding derivatives is essential when working with Taylor polynomials and is foundational in calculus. A derivative represents the rate of change of a function with respect to a variable. For the function given, \( f(x) = x \cos x \), we calculated its derivatives to be used in forming the Taylor polynomial. Here’s how we did it:

• The first derivative, \( f'(x) = \cos x - x \sin x \), indicates how \( f(x) \) changes as \( x \) changes.
• The second derivative, \( f''(x) = -2 \sin x - x \cos x \), tells us how the rate of change itself changes.
• The third derivative is \( f'''(x) = -3 \cos x + x \sin x \), which continues this pattern of evaluating change.

By finding these derivatives and evaluating them at \( a = 0 \), we gather crucial information for approximating \( f(x) \) through a polynomial. Derivative evaluations like \( f'(0) = 1 \) and \( f'''(0) = -3 \) help in understanding the behavior of the function around the point \( a = 0 \). The calculations show how the function starts at a point, how it turns, and eventually, how it trends upwards or downwards.

Polynomial Approximation

Polynomial approximations, particularly Taylor polynomials, are extremely useful in calculus and real-world applications, providing an approach to approximate complex functions with polynomials that are simpler to compute.

• In this exercise, we derived a third-degree Taylor polynomial to approximate \( f(x) = x \cos x \) near \( a = 0 \).
• We combined the value of the function and its derivatives at this point to obtain a polynomial \( T_3(x) = x - \frac{x^3}{2} \).
• This polynomial uses the value and behavior of the function at specific points to provide a local approximation.

Polynomial approximations become increasingly accurate as the degree of the polynomial increases. For small values near \( a \), the third-degree Taylor polynomial \( T_3(x) \) effectively represents the original complex function \( f(x) = x \cos x \) by using a simpler expression, facilitating easier computations and analysis for students and scientists alike.

Third-degree Taylor Polynomial

Creating a Taylor polynomial of the third degree involves using derivatives of a function up to the third order, evaluated at a particular point \( a \). In this case, the function \( f(x) = x \cos x \) and the center point \( a = 0 \). We employ the formula:\[ T_3(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} \]

• Substituting the derivatives we evaluated at \( x=0 \), it simplifies to \( T_3(x) = x - \frac{x^3}{2} \).
• This expression represents the third-degree Taylor polynomial, offering an approximation of \( f \) using only a cubic polynomial.

The proximity of the polynomial’s degree directly affects its approximation accuracy around the center point \( a \). For points close to zero, \( x - \frac{x^3}{2} \) approximates the behavior of \( x \cos x \) effectively—it retains the initial slope and captures some curvature. Understanding how to derive these polynomials gives insight into how various functions behave and allows for efficient estimations in computational scenarios.