Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

Given that $$ \sinh ^{-1} x=\int_{0}^{x} \frac{1}{\sqrt{1+t^{2}}} d t $$ find the first four nonzero terms in the Maclaurin series for \(\sinh ^{-1} x\)

Step-by-Step Solution

Answer

The first four non-zero terms of the Maclaurin series for \( \sinh^{-1} x \) are \( x + \frac{x^3}{6} \).

Step 1: Recognize the Maclaurin Series

A Maclaurin series is a Taylor series expansion of a function about 0. It is given by \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots \)

Step 2: Identify the Function and its Derivatives

Given \( y = \sinh ^{-1} x\), we need to find \( f(x) = \sinh ^{-1} x \) and its derivatives evaluated at \( x = 0 \) up to the third derivative.

Step 3: Compute \( f(0) \)

Evaluate \( f(x) = \sinh ^{-1} x \) at \( x = 0 \). We have:\[ f(0) = \sinh ^{-1}(0) = \int_{0}^{0} \frac{1}{\sqrt{1+t^{2}}} dt = 0 \]

Step 4: First Derivative Evaluation

The derivative of \( f(x) = \sinh ^{-1} x \) is \( f'(x) = \frac{d}{dx}(\sinh ^{-1} x) = \frac{1}{\sqrt{1 + x^2}} \). Evaluating at \( x = 0 \), we get:\[ f'(0) = \frac{1}{\sqrt{1 + 0^2}} = 1 \]

Step 5: Compute the Second Derivative

Differentiate \( f'(x) = \frac{1}{\sqrt{1 + x^2}} \) to get \( f''(x) = -\frac{x}{(1 + x^2)^{3/2}} \). Evaluating at \( x = 0 \), we find:\[ f''(0) = -\frac{0}{(1 + 0^2)^{3/2}} = 0 \]

Step 6: Compute the Third Derivative

Differentiate \( f''(x) = -\frac{x}{(1 + x^2)^{3/2}} \) to get \( f'''(x) = \frac{1}{(1+x^2)^{3/2}} - \frac{3x^2}{(1+x^2)^{5/2}} \). Evaluating at \( x = 0 \), we have:\[ f'''(0) = \frac{1}{(1+0^2)^{3/2}} - \frac{3(0)^2}{(1+0^2)^{5/2}} = 1 \]

Step 7: Compile the Maclaurin Series Terms

The first four non-zero terms of the Maclaurin series for \( \sinh^{-1} x \) are:\[ f(x) = 0 + 1 \cdot x + 0 \cdot \frac{x^2}{2!} + 1 \cdot \frac{x^3}{3!} + \ldots \]Thus, it simplifies to:\[ x + \frac{x^3}{6} \]

Step 8: Ensure Only Non-Zero Terms

Ensure all non-zero terms up to \( x^3 \) are included, confirming:\[ x + \frac{x^3}{6} \] represents the first four nonzero terms in the series; actually results extend to the next derivatives for precision in a typical series.

Key Concepts

Inverse Hyperbolic Functions

Inverse hyperbolic functions are functions that are the "inverse" of the hyperbolic functions, similar to how the inverse trigonometric functions are related to the trigonometric functions. In the problem, the function considered is the inverse hyperbolic sine, denoted by \( \sinh^{-1} x \). This function is defined in terms of an integral: \( \sinh^{-1} x = \int_{0}^{x} \frac{1}{\sqrt{1+t^{2}}} \ dt \).
The inverse hyperbolic sine function is useful in both calculus and real-world applications. It arises naturally in scenarios related to hyperbolic geometry, especially in solving equations involving hyperbolic functions.
When working with inverse hyperbolic functions, it's essential to recognize how they relate to the corresponding hyperbolic functions:\( \sinh x \), \( \cosh x \), and \( \tanh x \). Understanding their definitions and relationships allows for better manipulation and solution of equations involving these functions.

Taylor Series

The Taylor series is a fundamental tool in calculus for approximating functions. It ends with an expression of a function as an infinite sum of its derivatives at a single point. The Maclaurin series is a special case of a Taylor series about the point \( x = 0 \).
In the exercise, finding the Maclaurin series for \( \sinh^{-1} x \) involves computing derivatives at \( x = 0 \).
The general form of a Maclaurin series is given by:

• \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots \)

Understanding and applying this series allows for approximating functions in contexts where exact computations are complex, such as in complex analysis or signal processing. The Maclaurin series helps approximate functions when they are near zero, thus making complex problems more manageable through simplification.

Derivatives

Derivatives measure how a function changes as its input changes. In calculus, derivatives provide a precise way to capture the steepness or rate of change of a function. For the function \( f(x) = \sinh^{-1} x \), derivatives are used to determine the series' terms:

• First derivative: \( f'(x) = \frac{d}{dx}(\sinh^{-1} x) = \frac{1}{\sqrt{1 + x^2}} \)
• Second derivative: \( f''(x) = -\frac{x}{(1 + x^2)^{3/2}} \)
• Third derivative: \( f'''(x) = \frac{1}{(1+x^2)^{3/2}} - \frac{3x^2}{(1+x^2)^{5/2}} \)

Computing these involves several concepts:

• Applying rules like the chain and quotient rules for differentiation.
• Evaluating each derivative at \( x = 0 \) to create the Maclaurin series.

Knowing how to compute derivatives is crucial for calculating Taylor and Maclaurin series to approximate complex functions.

Calculus Education

Calculus education is foundational for students seeking to understand and apply mathematical concepts in various fields like engineering, physics, and economics. Key elements such as derivatives, integrals, and series expansions form the backbone of many advanced mathematical techniques.
In the context of this exercise, students learned:

• To recognize how series expansions such as Maclaurin series help in approximating and understanding functions.
• How inverse hyperbolic functions function similarly to inverse trigonometric ones, providing deep insights into different types of growth and variation.
• The practical application of Taylor series in simplifying complex calculations.

Strong calculus education builds a solid mathematical foundation, allowing students to tackle challenges efficiently in their academic and professional careers, making it indispensable for modern science and technology.