Problem 6 - 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. \( f(x) = e^{-x} \sin x, \) \( a = 0 \)

Step-by-Step Solution

Answer

The Taylor polynomial is \( T_3(x) = x + \frac{x^3}{3} \). It approximates the function well near 0.

Step 1: Find derivatives of the function

To find the Taylor polynomial, we first need the derivatives of the function. The derivatives of \( f(x) = e^{-x} \sin x \) are found using the product rule and chain rule. Calculate: 1. \( f'(x) = -e^{-x}\sin x + e^{-x}\cos x \) 2. \( f''(x) = -e^{-x}\cos x - e^{-x}\sin x - e^{-x}\sin x + e^{-x}\cos x = -2e^{-x}\sin x \) 3. \( f'''(x) = -2(-e^{-x}\cos x - e^{-x}\sin x) = 2e^{-x}(\cos x + \sin x) \)

Step 2: Evaluate derivatives at center

The Taylor series is centered at \( a = 0 \). Evaluate the derivatives at \( x = 0 \). 1. \( f(0) = e^{0} \times 0 = 0 \) 2. \( f'(0) = -e^{0} \times 0 + e^{0} \times 1 = 1 \) 3. \( f''(0) = -2e^{0} \times 0 = 0 \) 4. \( f'''(0) = 2e^{0} \times 1 = 2 \)

Step 3: Construct the Taylor Polynomial

The third-order Taylor polynomial \( T_3(x) \) at \( x = 0 \) is given by: \[ T_3(x) = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 \]Substitute the values: \[ T_3(x) = 0 + \frac{1}{1!} x + \frac{0}{2!} x^2 + \frac{2}{3!} x^3 = x + \frac{x^3}{3} \]

Step 4: Graph both functions

Graph the original function \( f(x) = e^{-x} \sin x \) and the Taylor polynomial \( T_3(x) = x + \frac{x^3}{3} \) on the same set of axes. This will show how well the Taylor series approximates \( f(x) \) near \( x = 0 \).

Key Concepts

Taylor Series

The Taylor series is a fundamental concept in calculus, used to approximate functions. It does this by expressing a function as an infinite sum of terms that are calculated from the values of its derivatives at a specific point. For a function \( f(x) \) centered at \( a \), the nth order Taylor polynomial is given by:\[ T_n(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^n(a)}{n!}(x-a)^n \]This polynomial gives us a linearly dependent approximation of the function near the point \( a \). As more terms are included, the approximation becomes more accurate. However, in practical applications, a finite number of terms is often sufficient to get a good estimate near the point of expansion. In this exercise, we're focusing on finding the third-order Taylor polynomial \( T_3(x) \), which provides a cubic approximation of the function \( f(x) = e^{-x} \sin{x} \) around \( x = 0 \).
The approximation includes derivatives of the function evaluated at the center.

Product Rule

The product rule is a differentiation technique used in calculus when finding the derivative of the product of two functions. If you have two functions \( u(x) \) and \( v(x) \), their product \( y = u(x)v(x) \) can be differentiated using the product rule.The formula for the product rule is: \[ y' = u'(x)v(x) + u(x)v'(x) \]In this problem, the function \( f(x) = e^{-x} \sin{x} \) is the product of \( e^{-x} \) and \( \sin{x} \). Applying the product rule here means:- Differentiate \( e^{-x} \) to get \( -e^{-x} \).- Differentiate \( \sin{x} \) to get \( \cos{x} \).- Substitute these into the formula:\[ f'(x) = -e^{-x} \sin{x} + e^{-x} \cos{x} \]This rule simplifies finding derivatives of complex products of functions and is essential for calculating Taylor polynomials in such scenarios.

Chain Rule

The chain rule is a crucial concept in calculus, used to differentiate compositions of functions. When one function is nested inside another, it helps to "chain" them together. For a composite function \( y = f(g(x)) \), the chain rule states that:\[ y' = f'(g(x)) \cdot g'(x) \]In this context, the function \( f(x) = e^{-x} \sin{x} \) requires using the chain rule during differentiation due to the exponential \( e^{-x} \) term. Here, \( g(x) = -x \) when considering \( e^{-x} \). - Differentiate \( e^{-x} \) as follows using the chain rule: - First, treat \( e^u \) (where \( u = -x \)) and find its derivative as \( e^u \). - Then multiply by the derivative of the internal function \( -x \), which is \( -1 \). - Resulting in \( e^{-x} \cdot (-1) = -e^{-x} \).This practice streamlines finding derivatives when themes of nested functions appear, enabling more intricate Taylor series expansions.

Derivative Evaluation

Derivative evaluation is a crucial step when forming Taylor polynomials. It involves calculating the derivatives of the function at a specific center point \( a \). For our function \( f(x) = e^{-x} \sin{x} \) centered at \( a = 0 \), you begin with finding derivatives:- Zero-order: \( f(0) = e^{0} \cdot \sin{0} = 0 \)- First-order: \( f'(0) = -e^{0} \cdot \sin{0} + e^{0} \cdot \cos{0} = 1 \)- Second-order: \( f''(0) = -2e^{0} \cdot \sin{0} = 0 \)- Third-order: \( f'''(0) = 2e^{0} \cdot (\cos{0} + \sin{0}) = 2 \)By evaluating these, we establish the coefficients for each corresponding term in the Taylor polynomial. These values are plugged into the Taylor polynomial equation to generate \( T_3(x) = x + \frac{x^3}{3} \), which approximates \( f(x) \) near the chosen center. Derivative evaluation thus bridges function behavior and polynomial representation.