Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

Recall that $$ \sin ^{-1} x=\int_{0}^{x} \frac{1}{\sqrt{1-t^{2}}} d t $$ Find the first four nonzero terms in the Maclaurin series for \(\sin ^{-1} x\).

Step-by-Step Solution

Answer

The first four nonzero terms are \( x + \frac{x^3}{3} \).

Step 1: Understand the Maclaurin Series Definition

The Maclaurin series is a special case of the Taylor series, centered at 0. It expresses a function as an infinite sum of terms calculated from the values of its derivatives at a single point. For function \( f(x) \), it is given by:\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \]

Step 2: Calculate the First Derivative

We start by differentiating the given integral representation of \( \sin^{-1}x \). By the Fundamental Theorem of Calculus, the derivative of \( \sin^{-1}x \) is:\[ \frac{d}{dx}(\sin^{-1}x) = \frac{1}{\sqrt{1-x^2}} \]

Step 3: Compute First Few Derivatives at x = 0

Calculate the first few derivatives of \( f(x) = \sin^{-1} x \) and evaluate them at \( x = 0 \):- \( f(0) = \sin^{-1}(0) = 0 \)- \( f'(x) = \frac{1}{\sqrt{1-x^2}} \), so \( f'(0) = 1 \)- \( f''(x) = \frac{x}{(1-x^2)^{3/2}} \), so \( f''(0) = 0 \)- \( f'''(x) = \frac{1 + 2x^2}{(1-x^2)^{5/2}} \), so \( f'''(0) = 2 \)- \( f^{(4)}(x) = \frac{6x + 6x^3}{(1-x^2)^{7/2}} \), so \( f^{(4)}(0) = 0 \)

Step 4: Formulate the Series Terms

Using the derivatives from the previous step, plug them into the Maclaurin series formula:\[\sin^{-1}x \approx 0 + 1 \cdot x + \frac{0}{2!} \cdot x^2 + \frac{2}{3!} \cdot x^3 + \frac{0}{4!} \cdot x^4 + \cdots\]

Step 5: Simplify the Maclaurin Series

The given terms simplify to:\[\sin^{-1}x \approx x + \frac{x^3}{3} + \cdots\]Thus, the first four nonzero terms of the Maclaurin series are \( x + \frac{x^3}{3} + \cdots \).

Key Concepts

Inverse Sine Function

The inverse sine function, often represented as \( \sin^{-1}x \) or \( \arcsin x \), is one of the inverse trigonometric functions. Its primary purpose is to retrieve the angle whose sine is a given number. For instance, if \( \sin(\theta) = x \), then \( \theta = \sin^{-1}(x) \). This function is useful for determining angles when the value of the sine is known. Its range is typically restricted to \( [-\frac{\pi}{2}, \frac{\pi}{2}] \) to make it a true function, preventing it from providing multiple values for a single input. One interesting aspect of \( \sin^{-1}x \) is its representation with an integral. It can be expressed as:\[\sin^{-1} x = \int_{0}^{x} \frac{1}{\sqrt{1-t^{2}}} \, dt\]This integral form is essential in calculus for evaluating the function’s derivatives, forming the backbone of its expansion in series and other advanced calculations.

Calculus

Calculus is a branch of mathematics focusing on the study of rates of change and accumulation. Its two fundamental concepts are differentiation and integration.

• Differentiation: This is the process used to compute the derivative of a function, which represents its rate of change. Derivatives are central to understanding how functions behave, especially for analyzing graphs, finding tangents, and solving problems involving motion or change.
• Integration: Conversely, integration merges rates of change to determine accumulate quantities, such as areas under curves or total change over time. It is pivotal in finding exact solutions for areas, volumes, and solving differential equations.

The Fundamental Theorem of Calculus connects these two branches, showing that differentiation and integration are inverse operations. The theorem provides a powerful tool for evaluating definite integrals, which is crucial in evaluating expressions like the inverse sine function. Understanding these concepts allows students to move from solving basic problems to tackling more complex challenges in mathematics and applied fields.

Taylor Series

The Taylor series is a crucial mathematical tool, representing functions as infinite sums of terms based on their derivatives at a particular point. This approximation can be centered around any point, but when the point is zero, it’s called the Maclaurin series. The general form of a Taylor series for a function \( f(x) \) about a point \( a \) is:\[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots\]The Maclaurin series, being a special case of the Taylor series, simplifies this by using \( a = 0 \):\[f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots\]For \( \sin^{-1}x \), the Maclaurin series expansion provides an efficient approximation for this inverse trigonometric function. By calculating the derivatives at zero and substituting them into the series formula, we derive the initial terms \( x + \frac{x^3}{3} + \cdots \). This series expansion is valuable for computations that require a high degree of precision yet must be expressed in simple algebraic forms.