Problem 65 - Thomas Calculus

Question

Series for \(\sin ^{-1} x \quad\) Integrate the binomial series for \(\left(1-x^{2}\right)^{-1 / 2}\) to show that for \(|x|<1,\) \begin{equation}\sin ^{-1} x=x+\sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot(2 n-1)}{2 \cdot 4 \cdot 6 \cdot \cdots \cdot(2 n)} \frac{x^{2 n+1}}{2 n+1}.\end{equation}

Step-by-Step Solution

Answer

The series for \(\sin^{-1}(x)\) is derived by integrating \((1-x^2)^{-1/2}\) and simplifying each term.

Step 1: Expand the Binomial Series

The binomial series for \((1-x^2)^{-1/2}\) is the expansion of the expression using the formula for a binomial series: \[(1-x^2)^{-1/2} = \sum_{n=0}^{\infty} \binom{-1/2}{n} (-x^2)^n.\] The general term of the series is given by \((-1/2 \cdot -3/2 \cdot \ldots \cdot \,(-(2n-1)/2)x^{2n}) / n!\). This simplifies to \( \binom{2n}{n} \frac{x^{2n}}{4^n} \).

Step 2: Integrate Term by Term

Now, integrate the series term by term to find the antiderivative. Start with the expression \(\int (1-x^2)^{-1/2} \, dx\). The integration of the series \( \sum_{n=0}^{\infty} \binom{2n}{n} \frac{x^{2n}}{4^n}\) term by term gives: \[\sum_{n=0}^{\infty} \int \binom{2n}{n} \frac{x^{2n}}{4^n} \, dx.\] This results in \[\sum_{n=0}^{\infty} \binom{2n}{n} \frac{x^{2n+1}}{(2n+1)4^n}.\]

Step 3: Simplify the General Term

Simplify the general term of the series. The binomial coefficient can be expressed as \(\frac{(2n)!}{(n!)^2}\), and notice that \(4^n = 2^{2n}\). Thus, the term becomes \[\frac{(2n)! \, x^{2n+1}}{(n!)^2 (2n+1) 2^{2n}}.\] Recognizing further simplification by rewriting it as: \[\frac{1 \cdot 3 \cdot 5 \cdot \, \ldots \, \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \, \ldots \, \cdot (2n)} \frac{x^{2n+1}}{2n+1}.\]

Step 4: Combine and Conclude

Combine the results from previous steps to write the series representation of \(\sin^{-1}(x)\). The initial term from the integration is \(x\), and adding the series gives: \[\sin^{-1}(x) = x + \sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdot \, \ldots \, \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \, \ldots \, \cdot (2n)} \frac{x^{2n+1}}{2n+1},\] where the convergence is ensured for \(|x|<1\).

Key Concepts

Binomial Series

The binomial series is a way to expand expressions of the form \( (1-x)^k \), where \( k \) can be any real number, using an infinite series. This is an extension of the binomial theorem, which deals with cases where \( k \) is a positive integer. The binomial series is quite useful because it allows us to represent functions involving square roots or other fractional exponents as a series, which can be easier to work with in calculus.

• The general formula for the binomial series is: \( (1-x)^k = \sum_{n=0}^{\infty} \binom{k}{n} (-x)^n \).
• The binomial coefficient \( \binom{k}{n} \) is calculated as: \( \frac{k(k-1)(k-2)\ldots(k-n+1)}{n!} \).
• For our problem, the expression \( (1-x^2)^{-1/2} \) is expanded using this series, providing a starting point for further calculations.

Understanding this concept helps in analyzing and manipulating expressions to find their antiderivatives or power expansions effectively.

Integration

Integration is the process of finding the antiderivative of a function, which essentially means the opposite of differentiation. In the context of this exercise, we use integration to find a function whose derivative matches the given series expansion derived from the binomial series. When integrating the series term by term, remember:

• Each term of a power series can be integrated independently within limits where the series converges.
• The constant of integration can be determined based on boundary conditions or initial values if provided.

Applying integration term by term to the binomial series results in each term being integrated independently, leading to an expanded series that represents the antiderivative function. Learning to integrate such series terms systematically simplifies the task of finding solutions to complex expressions in calculus.

Power Series Expansion

The power series expansion allows us to write a function as an infinite sum of terms, which are powers of a variable. This method is particularly useful when dealing with inverse trigonometric functions, such as \( \sin^{-1}(x) \), since they can be expressed as infinite series.Key aspects include:

• A power series takes the general form: \( \sum_{n=0}^{\infty} a_n x^n \), where \( a_n \) represents the coefficients and \( x^n \) the powers of \( x \).
• Each term in the series is derived from expanding, simplifying, and sometimes integrating another series, as shown in the exercise.
• In the case of the inverse sine function, the series expansion helps reveal its behavior for \( |x|<1 \).

Understanding power series expansion provides a valuable tool for approximating and working with functions that would otherwise be difficult to manipulate, helping to tackle problems involving convergence and function representation.