Problem 5 - Calculus: Early Transcendentals

Question

Use the definition of a Taylor series to find the first four nonzero terms of the series for \( f(x) \) centered at the given value of \( a. \) \( f(x) = xe^x, \) \( a = 0 % \)

Step-by-Step Solution

Answer

The first four nonzero terms are: \( x + x^2 + \frac{x^3}{2} + \frac{x^4}{6} \).

Step 1: Understand the Problem

We need to find the first four nonzero terms of the Taylor series for the function \( f(x) = xe^x \), centered at \( a = 0 \). The Taylor series for a function \( f(x) \) centered at \( a \) is given by the formula: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^n(a)}{n!}(x-a)^n \]. We will compute the derivatives of \( f(x) \), evaluate them at \( x = 0 \), and find the first four nonzero terms of the series.

Step 2: Compute the Derivatives

First, compute the derivatives of \( f(x) = xe^x \).- \( f(x) = xe^x \)- First derivative, \( f'(x) = e^x + xe^x \)- Second derivative, \( f''(x) = e^x + e^x + xe^x = 2e^x + xe^x \)- Third derivative, \( f'''(x) = 2e^x + e^x + xe^x = 3e^x + xe^x \)- Fourth derivative, \( f^{(4)}(x) = 3e^x + e^x + xe^x = 4e^x + xe^x \)

Step 3: Evaluate Derivatives at \( x = 0 \)

Evaluate each of the derivatives at \( x = 0 \):- \( f(0) = 0 \cdot e^0 = 0 \)- \( f'(0) = e^0 + 0 \cdot e^0 = 1 \)- \( f''(0) = 2e^0 + 0 \cdot e^0 = 2 \)- \( f'''(0) = 3e^0 + 0 \cdot e^0 = 3 \)- \( f^{(4)}(0) = 4e^0 + 0 \cdot e^0 = 4 \)

Step 4: Apply Taylor Series Formula

Plug in the derivatives into the Taylor series formula to find the first four nonzero terms:- The first term when \( n = 0 \) is \( \frac{f(0)}{0!}(x-0)^0 = 0 \).- The second term when \( n = 1 \) is \( \frac{f'(0)}{1!}(x-0)^1 = 1 \times x = x \).- The third term when \( n = 2 \) is \( \frac{f''(0)}{2!}(x-0)^2 = \frac{2}{2}x^2 = x^2 \).- The fourth term when \( n = 3 \) is \( \frac{f'''(0)}{3!}(x-0)^3 = \frac{3}{6}x^3 = \frac{x^3}{2} \).- The fifth term when \( n = 4 \) is \( \frac{f^{(4)}(0)}{4!}(x-0)^4 = \frac{4}{24}x^4 = \frac{x^4}{6} \).

Step 5: Write Out the First Four Nonzero Terms

The first four nonzero terms of the Taylor series for \( f(x) = xe^x \) are:\[ x + x^2 + \frac{x^3}{2} + \frac{x^4}{6} \]

Key Concepts

Maclaurin Series

A Maclaurin series is a special case of the Taylor series. It represents a function as an infinite sum of terms based on the function's derivatives evaluated at zero. This is useful when approximating functions near the origin. In the context of the original exercise, we used a Maclaurin series because we centered our series at zero (i.e., setting \( a = 0 \) in the Taylor series formula).

The formula for the Maclaurin series is:

• \( f(x) = \sum_{n=0}^{\infty} \frac{f^n(0)}{n!}x^n \)

This means we compute the derivatives of \( f(x) \), evaluate them at zero, and plug them into this formula to find the terms of the series.

For \( f(x) = xe^x \), this series helps to express the function as a sum of polynomial terms, making complex calculations easier, especially when working close to \( x = 0 \).

Derivatives

Derivatives represent the rate of change of a function. They are essential in forming Taylor and Maclaurin series, as these series heavily depend on the values of derivatives.

To compute a derivative, you find how a function changes as you make infinitesimally small changes to its input. In the exercise, we calculated several derivatives of \( f(x) = xe^x \) to obtain the series terms.

• The first derivative of \( xe^x \) was \( e^x + xe^x \).
• The second derivative was \( 2e^x + xe^x \).
• The third derivative was \( 3e^x + xe^x \).
• And the fourth derivative was \( 4e^x + xe^x \).

Calculating these derivatives accurately is crucial for constructing the series correctly, as they determine the coefficients of the terms in the series.

Power Series

A power series is an infinite series of the form \( \sum_{n=0}^{\infty} c_n x^n \), where \( c_n \) are constants known as coefficients. These series represent functions as sums of powers of \( x \) and are foundational in calculus for approximating functions.

In our problem, the Taylor series expands \( f(x) = xe^x \) into a power series centered at zero. For each derivative evaluated, a term in the power series is constructed:

• The zeroth term, \( 0 \), is ignored since it contributes nothing to the series.
• The first term, \( x \), comes from \( f'(0) \).
• The second term, \( x^2 \), arises from \( f''(0) \).
• The third term, \( \frac{x^3}{2} \), and the fourth term, \( \frac{x^4}{6} \), follow similarly from \( f'''(0) \) and \( f^{(4)}(0) \), respectively.

These simplify computations for functions by using polynomials, especially near the expansion center. Thus, power series are incredibly valuable in analytical calculations and approximations.

Exponential Functions

An exponential function is written in the form \( e^x \), where \( e \) is approximately 2.718. This type of function is vital across numerous scientific fields due to its unique properties like constant growth rates and natural logarithmic relationships.

In the exercise, \( f(x) = xe^x \) involves an exponential function multiplied by \( x \), demonstrating how exponential functions can be combined with other algebraic elements to yield complex results.

• The derivative of exponential functions is particularly useful: the derivative of \( e^x \) is itself \( e^x \).
• In our exercise, this property simplifies the differentiation process, as seen in calculating derivatives of \( xe^x \).

Understanding exponential functions not only aids in solving calculus problems but also sheds light on natural phenomena like population growth and radioactive decay. These functions, thus, serve as key tools in mathematical modeling and problem-solving.