Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The Taylor series up to the \( (x-a)^3 \) term is \( f(x) = 3 + 5(x-1) + 4(x-1)^2 + (x-1)^3 \).

Step 1: Identify the Function and Point for Expansion

The given function is \( f(x) = 1 + x^2 + x^3 \) and we need to find the Taylor series of this function around the point \( a = 1 \).

Step 2: Calculate the Function at the Point

Evaluate \( f(a) = f(1) \). This gives \( f(1) = 1 + 1^2 + 1^3 = 3 \).

Step 3: Determine the First Derivative and Evaluate at the Point

The first derivative of the function is \( f'(x) = 2x + 3x^2 \). Evaluate this at \( a = 1 \), giving \( f'(1) = 2(1) + 3(1)^2 = 5 \).

Step 4: Calculate the Second Derivative and Evaluate at the Point

The second derivative is \( f''(x) = 2 + 6x \). We evaluate this at \( a = 1 \), resulting in \( f''(1) = 2 + 6(1) = 8 \).

Step 5: Determine the Third Derivative and Evaluate at the Point

The third derivative is \( f'''(x) = 6 \). Since this derivative is constant, \( f'''(1) = 6 \).

Step 6: Construct the Taylor Series up to the Cube Term

The Taylor series of \( f(x) \) around \( a \) is given by \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 \]. Substituting the values, we have: \[ f(x) = 3 + 5(x-1) + \frac{8}{2}(x-1)^2 + \frac{6}{6}(x-1)^3 \]. Simplifying, the series is \[ f(x) = 3 + 5(x-1) + 4(x-1)^2 + (x-1)^3 \].

Key Concepts

Calculus

Calculus is the mathematical study dealing with continuous change. It plays a central role in the understanding of motion, change, and other complex phenomena in mathematics and science. At its core, calculus is about analyzing and understanding functions.
Through calculus, we use the concept of a limit to define and work with derivatives and integrals. Limits help us understand the behavior of functions as they approach certain points, particularly where they might not be explicitly defined. This concept is crucial when developing series expansions like Taylor series, which are used to approximate functions through polynomials.
In this exercise, calculus provides the tools to approximate the function around a specific point. This is done by deriving the Taylor series, where we calculate derivatives and evaluate these derivatives at a particular point, leading towards creating a polynomial that closely replicates the function in a localized manner.

Derivative

The derivative of a function measures its rate of change. It illustrates how a function's output changes as its input changes. This concept is fundamental in calculus, helping you understand the behavior of functions.

• The first derivative tells us about the slope or the steepness at any given point of the function. For example, in this problem, the first derivative of the function is calculated as \( f'(x) = 2x + 3x^2 \), and evaluated at \( a = 1 \), resulting in \( f'(1) = 5 \).
• The second derivative provides information about the function's curvature or how the rate of change is changing, giving insight into concavity. In this exercise, the second derivative is \( f''(x) = 2 + 6x \), and at \( a=1 \), we find \( f''(1) = 8 \).
• Higher derivatives, like the third derivative here \( f'''(x) = 6 \), continue this pattern, providing further insights into the function's behavior.

Each of these derivatives plays an integral role in constructing the Taylor series, where they help form the coefficients of the polynomial terms that approximate the function.

Polynomial Functions

Polynomial functions are mathematical expressions consisting of variables raised to whole number exponents, combined using addition, subtraction, and multiplication. These functions are crucial in calculus because they are smooth and continuous, making them ideal for approximating other complex functions.
A basic polynomial function is expressed as \( a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \), depending on the degree of the polynomial. In the exercise at hand, the function \( 1 + x^2 + x^3 \) is already represented in polynomial form.
When implementing a Taylor series, polynomial functions are constructed to approximate a given function in a specific neighborhood of a point, \( x = a \). Here, through derivatives, we build a new polynomial that mimics the original function closely within a certain interval, allowing for simpler calculations and analysis of the function's behavior near that point.

Series Expansion

Series expansion is a method used to represent functions as an infinite sum of terms calculated from the values of their derivatives at a certain point. The Taylor series, a specific kind of series expansion, approximates a function using a polynomial.
To form a Taylor series, you take the derivatives of a function at a specific point, then use these derivatives to generate polynomial terms. Each term in the series depends on the function's derivative and the factorial of the term's order. The Taylor series formula is:
\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \]
In the problem given, the Taylor series of the function \( f(x) = 1 + x^2 + x^3 \) about \( a=1 \) is developed using derivatives up to the third order. This provides an approximation, \( f(x) \), that includes cubic terms, thereby presenting a strong local approximation of the original function near \( x = 1 \).
Series expansions like the Taylor series are crucial across various fields of mathematics and engineering, enabling complex functions to be studied and utilized in practical applications.