Problem 41 - Essential Calculus Early Transcendentals

Question

\(\begin{array}{l}{\text { (a) Use the binomial series to expand } 1 / \sqrt{1-x^{2}}}. \\ {\text { (b) Use part (a) to find the Maclaurin series for } \sin ^{-1} x}.\end{array}\)

Step-by-Step Solution

Answer

1. Expand: \((1-x^2)^{-1/2} = 1 + \frac{1}{2}x^2 + \cdots\); 2. Integrate for \(\sin^{-1}x\): \(x + \frac{1}{6}x^3 + \cdots\).

Step 1: Understand the Binomial Series Formula

The binomial series formula is given by \( (1 + u)^n = \sum_{k=0}^{\infty} \binom{n}{k} u^k \), where \( \binom{n}{k} \) is the binomial coefficient. Here, \( u = -x^2 \) and \( n = -\frac{1}{2} \). We want to apply this formula to expand \( \frac{1}{\sqrt{1-x^2}} \).

Step 2: Write the Term for the Binomial Expansion

Substitute \( n = -\frac{1}{2} \) and \( u = -x^2 \) into the binomial series formula.\[ (1-x^2)^{-1/2} = \sum_{k=0}^{\infty} \binom{-1/2}{k} (-x^2)^k \]

Step 3: Calculate Binomial Coefficients

The binomial coefficient \( \binom{-1/2}{k} \) is calculated using the formula:\[ \binom{n}{k} = \frac{n(n-1)(n-2)...(n-k+1)}{k!} \]Substitute \( n = -\frac{1}{2} \):\[ \binom{-1/2}{k} = \frac{(-1/2)(-1/2-1)(-1/2-2)...(-1/2-k+1)}{k!} \]

Step 4: Expand the Series

Using the calculated coefficients, write out the first few terms of the series expansion:\( (1-x^2)^{-1/2} = 1 + \frac{1}{2}x^2 + \frac{3}{8}x^4 + \frac{5}{16}x^6 + \cdots\)

Step 5: Integrate to Find Maclaurin Series for \( \sin^{-1}x \)

The Maclaurin series of \( \sin^{-1}x \) can be found by integrating \( \frac{1}{\sqrt{1-x^2}} \). Integrate term by term:\( \int (1-x^2)^{-1/2} \, dx = x + \frac{1}{6}x^3 + \frac{3}{40}x^5 + \cdots + C. \)The constant \( C \) is zero because \( \sin^{-1}(0) = 0 \).

Step 6: Write the Maclaurin Series for \( \sin^{-1}x \)

Collect the series terms from the integration:\(\sin^{-1}x = x + \frac{1}{6}x^3 + \frac{3}{40}x^5 + \frac{5}{112}x^7 + \cdots\)This is the Maclaurin series for \( \sin^{-1}x \).

Key Concepts

Binomial Series

The binomial series is an essential tool in mathematics for expanding expressions that involve powers or squared terms. It allows us to express a function as an infinite sum of terms calculated from the coefficients of a binomial expansion. This method not only simplifies complex expressions but also provides a way to approximate functions with high precision.
A binomial series is represented using the formula:

• \( (1 + u)^n = \sum_{k=0}^{\infty} \binom{n}{k} u^k \)

In this context, \( n \) is a real number and \( u \) is a variable term. The notation \( \binom{n}{k} \) denotes the binomial coefficients, crucial for calculating each term in the series.
When dealing with problems related to calculus or series expansion, such as expanding \( \frac{1}{\sqrt{1-x^2}} \), we often utilize the binomial series. This helps to write a complex fraction into an understandable and analyzable series, aiding in further applications like integration or finding Maclaurin series.

Binomial Coefficients

Binomial coefficients are the building blocks of a binomial series. They show up in the expansion of expressions like \( (1 + u)^n \) and are represented as \( \binom{n}{k} \). These coefficients determine the weight or scaling factor of each term in the series.
To calculate a binomial coefficient, the formula used is:

• \( \binom{n}{k} = \frac{n(n-1)(n-2)...(n-k+1)}{k!} \)

In this formula, \( k! \) denotes the factorial of \( k \), which is the product of all positive integers up to \( k \). The numerator is a product that decreases from \( n \) by one in each step, for \( k \) factors.
These coefficients play a crucial role when expanding a series like \( (1-x^2)^{-1/2} \) since they help determine the contribution of each term involving \( x^k \). In the exercise example provided, when \( n \) is non-integer like \(-\frac{1}{2}\), the binomial coefficients become complex but follow the same procedural calculation to define the series expansion accurately.

Series Expansion

A series expansion is a way to express a function or expression as an infinite sum of terms. This can be particularly useful for approximating complex functions or expressions. The goal is to break down a complicated function into a series of simpler terms that are easier to analyze and manipulate.
Starting with an initial function, we identify specific terms (like \( u \) in a binomial series) and calculate each term using corresponding coefficients, such as binomial coefficients. These terms are then summed up to form the series.

• Example: Expanding \( \frac{1}{\sqrt{1-x^2}} \) results in a series: \( 1 + \frac{1}{2}x^2 + \frac{3}{8}x^4 + \cdots \)

Once we have the series, it can also be integrated term-by-term, allowing one to derive related series solutions, such as a Maclaurin series. For example, in the task of finding the Maclaurin series for \( \sin^{-1}x \), we began by expanding \( \frac{1}{\sqrt{1-x^2}} \) using a series expansion and then integrating each term to form the Maclaurin series.
This whole process makes complex functions accessible and useful for further mathematical exploration, whether through analysis or numerical approximation.