Problem 65 - Thomas Calculus

Question

Series for sin \(^{-1} x\) 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 the binomial series for \((1-x^2)^{-1/2}\) term by term.

Step 1: Recall the Binomial Series

The binomial series expansion for \( (1 - x^2)^{-1/2} \) is given by:\[(1-x^2)^{-1/2} = 1 + \frac{1}{2}x^2 + \frac{1 \cdot 3}{2 \cdot 4} x^4 + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} x^6 + \cdots\]This series converges for \(|x| < 1\).

Step 2: Integrate the Series Term by Term

To find \( \sin^{-1} x \), we integrate the series for \((1-x^2)^{-1/2}\) term by term:\[\sin^{-1} x = \int (1-x^2)^{-1/2} \, dx = \int \left(1 + \frac{1}{2}x^2 + \frac{1 \cdot 3}{2 \cdot 4} x^4 + \cdots\right) \, dx\]

Step 3: Integrate Each Term Separately

Perform the integration for each term:- The integral \( \int 1 \, dx = x \).- The term \( \frac{1}{2} x^2 \) integrates to \( \frac{1}{6} x^3 \).- The term \( \frac{1 \cdot 3}{2 \cdot 4} x^4 \) integrates to \( \frac{1 \cdot 3}{2 \cdot 4 \cdot 5} x^5 \).- In general, \( \int \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} x^{2n} \, dx = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n) \cdot (2n+1)} x^{2n+1} \).

Step 4: Sum the Resultant Series

The series from integrating the binomial series combines to give the infinite series for \( \sin^{-1} x \):\[\sin^{-1} x = x + \sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \cdot \frac{x^{2n+1}}{2n+1}\]This is the required series for \( \sin^{-1} x \) when \(|x| < 1\).

Key Concepts

Binomial Series

The binomial series is an essential tool in calculus and algebra that allows us to express functions in a power series form, particularly when dealing with expressions of the form \((1 - x^2)^{-1/2}\).
This series expands such functions using a formula that involves coefficients derived from the binomial theorem.
The expansion for \((1 - x^2)^{-1/2}\) is given by:

• \((1-x^2)^{-1/2} = 1 + \frac{1}{2}x^2 + \frac{1 \cdot 3}{2 \cdot 4} x^4 + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} x^6 + \cdots\)

This series is infinite and converges for values of \(|x| < 1\), meaning it provides accurate approximations within this interval.
Binomial series are particularly useful when dealing with inverse trigonometric functions, as it allows us to work in terms of powers rather than complicated expressions.

Integration Techniques

Integration is the process of finding the antiderivative of a function.
In the given exercise, the integration of the binomial series is crucial to retrieve the series expansion for \( \sin^{-1} x \).To integrate the series \((1-x^2)^{-1/2}\), we apply term-by-term integration.
This means each term of the series is integrated individually, using basic integration rules.

• For a constant term, the integral is straightforward, such as \( \int 1 \, dx = x \).
• Power terms are integrated by applying the rule \( \int x^n \, dx = \frac{x^{n+1}}{n+1} \).

The integration is done step-by-step for each term in the series:

• The integral of \( \frac{1}{2}x^2 \) becomes \( \frac{1}{6}x^3 \) by increasing the power and dividing by the new exponent.
• Similarly, \( \frac{1 \cdot 3}{2 \cdot 4} x^4 \) integrates to \( \frac{1 \cdot 3}{2 \cdot 4 \cdot 5} x^5 \).

Through this method, we transform the binomial series into an infinite series that represents \( \sin^{-1} x \). Integration of series is an elegant approach to find antiderivatives, especially when dealing with series expansions.

Infinite Series

An infinite series is a sum of infinitely many terms.
In mathematics, these series are foundational because they can represent complex functions succinctly and powerfully within specified intervals.
In the exercise, the infinite series derived from the binomial expansion and integration provides us with the representation of \( \sin^{-1} x \): \[\sin^{-1} x = x + \sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \frac{x^{2n+1}}{2n+1}\]This infinite series is specific to the domain \(|x| < 1\), which ensures convergence and accuracy.
Infinite series are beneficial in calculating functions, estimating integral values, and solving different mathematical problems without closing forms.

Convergence of Series

In mathematics, convergence of a series refers to the condition where the sum of its terms approaches a certain value as more terms are added.
Convergence ensures the series represents a meaningful mathematical entity and results in an accurate approximation of functions.
For the binomial series used in the exercise, it converges for \(|x| < 1\).This implies:

• The terms of the series get progressively smaller and approach zero as \(n\) increases.
• The overall sum approaches the actual function, in this case, \( \sin^{-1} x \), for values within a specified range (interval), ensuring precision.

Understanding convergence is vital because it dictates whether the infinite series can be utilized for practical calculation or analysis.
Without convergence, an infinite series could yield divergent or non-realistic outcomes, making the understanding of this concept essential in calculus and advanced mathematical studies.