Exam 4 Study Guide: Power Series Applications and Integrals

Exam 4 Study Guide Overview

The following study guide covers material for Exam 4, scheduled for July 27th at 1:00 PM (7/27 1M). The primary focus of this unit is based on Topics 9.7 through 9.10, specifically emphasizing the use of power series to represent and approximate integrals. Numerical precision and efficiency in calculation are highlighted as critical skills for this exam.

Power Series Representations of Indefinite Integrals

The ability to find power series representations for indefinite integrals involves converting the integrand into a known power series (often a Maclaurin series) and then performing term-by-term integration. The general procedure is as follows:

Identify a known power series for the function inside the integral.
Apply any necessary substitutions (e.g., replacing $x$ with $x^2$ ).
Multiply by any external factors (e.g., dividing by $x$ or multiplying by $x^2$ ).
Integrate each term using the Power Rule for integration, ensuring to include the constant of integration, $C$ .

Exercise 45 through 50: Indefinite Integral Examples

Exercise 45: Consider the integral $\int \frac{1}{1+x^3} \,dx$ . Using the geometric series representation where $\frac{1}{1-r} = \sum_{n=0}^{\infty} r^n$ , we substitute $r = -x^3$ to find the power series for the integrand. The resulting series is $\sum_{n=0}^{\infty} (-1)^n x^{3n}$ . Integrating term-by-term yields $C + \sum_{n=0}^{\infty} \frac{(-1)^n x^{3n+1}}{3n+1}$ .

Exercise 46: Consider the integral $\int \frac{x^2}{1+x^4} \,dx$ . Similar to Exercise 45, we find the series for the denominator using $r = -x^4$ , which is $\sum_{n=0}^{\infty} (-1)^n x^{4n}$ . Multiplying by the numerator's $x^2$ gives $\sum_{n=0}^{\infty} (-1)^n x^{4n+2}$ . Term-by-term integration results in $C + \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+3}}{4n+3}$ .

Exercise 47: Consider $\int \sin(x^2) \,dx$ . Using the Maclaurin series for $\sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$ , substitute $x^2$ for $x$ to obtain $\sum_{n=0}^{\infty} \frac{(-1)^n (x^2)^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+2}}{(2n+1)!}$ . Integration results in $C + \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+3}}{(4n+3)(2n+1)!}$ .

Exercise 48: Consider $\int x \tan^{-1}(x) \,dx$ . The series for $\tan^{-1}(x)$ is $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}$ . Multiplying by $x$ yields $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+2}}{2n+1}$ . Integration gives $C + \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+3}}{(2n+3)(2n+1)}$ .

Exercise 49: Consider $\int \frac{\ln(1+x)}{x} \,dx$ . Given $\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n}$ , dividing by $x$ results in $\sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^{n-1}}{n}$ . Integrating term-by-term leads to $C + \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n^2}$ .

Exercise 50: Consider $\int \frac{\sin(x)}{x} \,dx$ . Dividing the standard sine series by $x$ results in $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n+1)!}$ . Integration resulting in the Sine Integral series is $C + \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)(2n+1)!}$ .

Numerical Approximation and Accuracy

For Exercises 51 through 56, the goal is to use power series to obtain an approximation of a definite integral with four decimal places of accuracy. This requires identifying the terms necessary to meet the error tolerance and using the least amount of terms possible.

The Alternating Series Estimation Theorem (AST Error Thm)

When working with alternating power series, the error introduced by truncating the series at the $n^{th}$ term is guaranteed to be less than or equal to the absolute value of the first neglected term. If an alternating series is defined as $S = \sum_{n=0}^{\infty} (-1)^n a_n$ , then the remainder $|R_n|$ satisfies:

$|R_n| = |S - S_n| \leq |a_{n+1}|$

To achieve four decimal places of accuracy (an error strictly less than $0.00005$ or $0.5 \times 10^{-4}$ ), the calculation must proceed until the next term in the series (the $(n+1)^{th}$ term) is smaller than that threshold.

Exercise 51 through 56: Definite Integral Approximations

Exercise 51: $\int_0^{0.5} e^{-x^2} \,dx$ . The expansion for $e^{-x^2}$ is $\sum \frac{(-1)^n x^{2n}}{n!}$ . The integrated series is evaluated from $0$ to $0.5$ . We must find the smallest $n$ such that the magnitude of the next integrated term is less than $0.0001$ .

Exercise 52: $\int_0^{0.1} x^2 e^x \,dx$ . This is done by multiplying the series for $e^x$ by $x^2$ , integrating to obtain terms of the form $\frac{x^{n+3}}{(n+3)n!}$ , and evaluating at the upper bound of $0.1$ .

Exercise 53: $\int_0^{0.5} \cos(x^2) \,dx$ . Using the cosine series $\sum \frac{(-1)^n x^{4n}}{(2n)!}$ , the integrated form is $\sum \frac{(-1)^n x^{4n+1}}{(4n+1)(2n)!}$ . We evaluate this series until terms fall below the required error threshold at $x = 0.5$ .

Exercise 54: $\int_0^{0.1} \sin(x^2) \,dx$ . Similar to 53, using the sine expansion $\sum \frac{(-1)^n x^{4n+2}}{(2n+1)!}$ . Because the upper bound is so small ( $0.1$ ), very few terms (often just one or two) are needed to reach four decimal places of accuracy.

Exercise 55: $\int_0^{0.5} x \cos(x^3) \,dx$ . Using $\cos(u)$ where $u = x^3$ , the terms become $x \cdot \frac{(-1)^n (x^3)^{2n}}{(2n)!} = \frac{(-1)^n x^{6n+1}}{(2n)!}$ . Integration results in $\frac{(-1)^n x^{6n+2}}{(6n+2)(2n)!}$ . Evaluation occurs at $x = 0.5$ .

Exercise 56: $\int_0^{0.5} \tan^{-1}(x^3) \,dx$ . The series for $\tan^{-1}(u)$ results in $\sum \frac{(-1)^n x^{6n+3}}{2n+1}$ . Integrating yields $\sum \frac{(-1)^n x^{6n+4}}{(6n+4)(2n+1)}$ . Apply the AST Error Theorem to determine the number of terms required to maintain accuracy.

Essential Maclaurin Series Reference

Success on the exam requires familiarity with the following fundamental series, as noted in the provided formula sheet instructions:

Exponential Function: $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$
Sine Function: $\sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$
Cosine Function: $\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$
Inverse Tangent Function: $\tan^{-1}(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \dots$
Natural Logarithm: $\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$
Geometric Series: \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \dots \text{ for } |x| < 1

To find power series representations for indefinite integrals, the integrand is converted into a known power series, typically a Maclaurin series, followed by performing term-by-term integration. The general steps involved include: 1. Identify a known power series for the function within the integral. 2. Make any necessary substitutions, such as replacing $x$ with $x^2$ . 3. Multiply by any external factors, for example, dividing by $x$ or multiplying by $x^2$ . 4. Integrate each term using the Power Rule for integration, ensuring to include the constant of integration, $C$ . For instance, in the integral $\int \frac{1}{1+x^3} \,dx$ , the geometric series representation $\frac{1}{1-r} = \sum<em>{n=0}^{\infty} r^n$ is applied by substituting $r = -x^3$ , resulting in the series $\sum</em>{n=0}^{\infty} (-1)^n x^{3n}$ , which can then be integrated term-by-term. This method allows for the systematic construction of power series representations for various types of functions being integrated.