Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 19-24, find the Taylor series in \(x-\) a through the term \((x-a)^{3}\). $$ 2-x+3 x^{2}-x^{3}, a=-1 $$

Step-by-Step Solution

Answer

The Taylor series is \(7 - 10(x+1) + 6(x+1)^2 - (x+1)^3\).

Step 1: Determine the Function and Point of Expansion

The given function is \(f(x) = 2 - x + 3x^2 - x^3\) and the point of expansion is \(a = -1\). We want to find the Taylor series expansion about \(x = -1\) through the term \((x - (-1))^3\), which is \((x + 1)^3\).

Step 2: Calculate Derivatives

Calculate the first few derivatives of \(f(x)\). The first derivative \(f'(x)\) is \(-1 + 6x - 3x^2\). The second derivative \(f''(x)\) is \( 6 - 6x\). The third derivative \(f'''(x)\) is \(-6\).

Step 3: Evaluate the Derivatives at \(a = -1\)

Evaluate each derivative at \(x = a = -1\): \(f(-1) = 2 + 1 + 3(-1)^2 + (-1)^3 = 7\), \(f'(-1) = -1 + 6(-1) - 3(-1)^2 = -10\), \(f''(-1) = 6 + 6 = 12\), \(f'''(-1) = -6\).

Step 4: Substitute into Taylor Series Formula

The Taylor series of a function \(f(x)\) about \(x = a\) is given by: \[f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3\]. Substitute the values found: \[7 - 10(x+1) + \frac{12}{2}(x+1)^2 - \frac{6}{6}(x+1)^3\].

Step 5: Simplify the Expression

Simplify the expression: \[7 - 10(x+1) + 6(x+1)^2 - 1(x+1)^3\], which simplifies further to \[7 - 10(x+1) + 6(x+1)^2 - (x+1)^3\].

Key Concepts

Calculus

Calculus is a branch of mathematics that focuses on change. It provides tools to analyze the variation and cumulative effects in functions. While it sounds complex, calculus contains foundational concepts that are widely used in science and engineering.

• Differentiation: This concept helps determine the rate of change of a function. It is found using derivatives, which measure how a function changes at any given point.
• Integration: This calculates the area under curves and accumulation of quantities. It is essentially the reverse of differentiation.

Understanding calculus is crucial to solve problems involving non-constant rates of change. In our exercise, calculating derivatives and using them in a Taylor series expansion is a practical application of calculus.

Derivatives

Derivatives are all about measuring how a function changes. When we compute a derivative, we are finding the function's rate of change or its "slope" at any particular point. This provides a powerful method to forecast how a model behaves over time.

• For the function given in the exercise, the first derivative tells us how the function is changing at each point.
• Further derivatives, such as the second and third, give deeper insights into the function’s behavior—such as acceleration and curvature.

By evaluating derivatives at specific points, like in the Taylor series expansion step, we can model more complex functions and predict their outcomes without requiring extensive observation.

Polynomial Expansion

Polynomial expansion is at the heart of many calculus problems. It involves expressing complex functions as sums of simpler polynomial terms, which are easier to handle and analyze.

• In Taylor series, we expand a function into an infinite sum of terms calculated from the values and derivatives of the function at a single point.
• The exercise simplifies this expansion process by limiting the expansion to up to the cubic term, making it more manageable yet insightful.

This technique helps in approximating functions that might otherwise be hard to compute, offering a closer look at the behavior of these functions within a specified range.

Mathematical Analysis

Mathematical analysis is widely recognized for its role in understanding limits, derivatives, and integrals, and the behavior of series and functions.

• The core aim is to provide a firm understanding of complex mathematical concepts and their underlying logic.
• In our specific exercise, mathematical analysis allows us to derive the Taylor series from the polynomial. We determine each term’s contribution through detailed evaluations of derivatives.

By breaking down problems through analysis, it not only gives answers but also clarifies the "why" and "how" behind mathematical phenomena. This forms the backbone of modern calculus and helps forge a deeper comprehension of advanced mathematical ideas.