Problem 51 - Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The binomial series for \(1/\sqrt{1 - x^2}\) gives the Maclaurin series for \(\sin^{-1} x = x + \frac{x^3}{6} + \frac{3x^5}{40} + \cdots \).

Step 1: Recognize the General Binomial Series Expansion

The general form of the binomial series expansion is given by \[ (1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \dots \]For the expression \(1/\sqrt{1-x^2}\), notice that it can be rewritten as \((1 - x^2)^{-1/2}\). Here, \(u = -x^2\) and \(n = -1/2\).

Step 2: Apply the Binomial Series to Expand

Substitute \(u = -x^2\) and \(n = -1/2\) into the binomial series expansion:\[(1 - x^2)^{-1/2} = 1 + \left(-\frac{1}{2}\right)(-x^2) + \frac{\left(-\frac{1}{2}\right)(-\frac{3}{2})}{2!}(-x^2)^2 + \dots\]Simplify this to:\[ 1 + \frac{1}{2}x^2 + \frac{3}{8}x^4 + \cdots \]

Step 3: Integrate to Find the Maclaurin Series for \(\sin^{-1} x\)

To find the Maclaurin series for \(\sin^{-1} x\), integrate the series for \(\frac{1}{\sqrt{1-x^2}}\):\[\sin^{-1} x = \int_0^x \frac{1}{\sqrt{1-t^2}} \, dt = \int_0^x \left(1 + \frac{1}{2}t^2 + \frac{3}{8}t^4 + \cdots\right) dt\]Integrate term by term:\[sin^{-1} x = x + \frac{1}{2}\left(\frac{x^3}{3}\right) + \frac{3}{8}\left(\frac{x^5}{5}\right) + \cdots = x + \frac{x^3}{6} + \frac{3x^5}{40} + \cdots\]

Step 4: Write the Final Maclaurin Series for \(\sin^{-1} x\)

The Maclaurin series for \(\sin^{-1} x\) is given by:\[\sin^{-1} x = x + \frac{x^3}{6} + \frac{3x^5}{40} + \frac{5x^7}{112} + \cdots \]

Key Concepts

Binomial Series

The binomial series is a powerful tool in calculus that allows us to expand expressions of the form \((1 + u)^n\) into an infinite series. This expansion is particularly useful for approximations and solving problems in physics and engineering. In the binomial series, the value \(u\) is a small number with \(n\) being any real number. The general expansion is given by:

• \( (1 + u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \dots \)

When we expand \(1 / \sqrt{1-x^2}\), it corresponds to \((1 - x^2)^{-1/2}\). This is achieved by setting \(u = -x^2\) and \(n = -1/2\) in the binomial series formula. Upon substituting, the expansion yields:

• \(1 + \frac{1}{2}x^2 + \frac{3}{8}x^4 + \cdots\)

This series helps approximate functions where direct computation may be difficult.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as \(\sin^{-1} x\), undo or reverse the process of the trigonometric functions like sine. These functions have a special role in calculus, particularly when dealing with integrals and series.
To derive the Maclaurin series for \(\sin^{-1} x\), we use the previously found expansion of \(1 / \sqrt{1-x^2}\). By integrating this series term by term, we find the expansion for \(\sin^{-1} x\) centered at 0:

• \(x + \frac{x^3}{6} + \frac{3x^5}{40} + \cdots\)

In this way, we transform the trigonometric inverse function into a series that makes solving complex integrals more manageable.

Integration Techniques

Integration is a core process in calculus, crucial for solving problems involving areas, volumes, and other quantities. In particular, the technique of integrating a series term by term facilitates the transformation of expanded series into new functions.
When faced with finding the Maclaurin series for \(\sin^{-1} x\), we integrate the series expansion \(1 / \sqrt{1-x^2}\). The definite integral from 0 to \(x\) of \(1 / \sqrt{1-t^2}\) leads to the series:

• \(x + \frac{x^3}{6} + \frac{3x^5}{40} + \cdots\)

This step-by-step integration advances our understanding of inverse functions and how they behave in different calculus contexts. Utilizing such techniques often simplifies the problem-solving process and enriches one's mathematical toolkit.