Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the terms through \(x^{5}\) in the Maclaurin series for \(f(x) .\) Hint: It may be easiest to use known Maclaurin series and then perform multiplications, divisions, and so on. For example, \(\tan x=(\sin x) /(\cos x) .\) \(f(x)=\frac{1}{1-\sin x}\)

Step-by-Step Solution

Answer

The Maclaurin series for \( f(x) = \frac{1}{1-\sin x} \) is \( 1 + x + \frac{7x^3}{6} + \frac{3x^5}{40} \).

Step 1: Recall the Maclaurin Series for Sine Function

The Maclaurin series for \( \sin x \) is given by \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots \). This series will be used in the next steps to evaluate the expansion of \( 1 - \sin x \).

Step 2: Expand the Base Function

Write the expression \( 1 - \sin x \) using the series: \( 1 - (x - \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots) = 1 - x + \frac{x^3}{6} - \frac{x^5}{120} + \cdots \).

Step 3: Use the Geometric Series for Inversion

The function \( f(x) = \frac{1}{1 - \sin x} \) can be expanded using the geometric series \( \frac{1}{1-u} = 1 + u + u^2 + u^3 + \cdots \), with \( u = \sin x \). Substitute \( u = x - \frac{x^3}{6} + \frac{x^5}{120} \) into the series expansion.

Step 4: Calculate Terms Using Series Substitution

Substitute and simplify the series: - The first few terms substituting only the necessary power of \( x \) yields 1. \( u = x \), so \( u^2 = x^2 \), \( u^3 = x^3 \). 2. \( 1 + (x - \frac{x^3}{6} + \frac{x^5}{120}) + (x - \frac{x^3}{6})^2 + \cdots \).

Step 5: Simplify Each Term

Collect and simplify terms up to \( x^5 \): - Term up to \( x \) is 1.- Term \( x \) is \( x \).- Term \( x^2 \) is 0 (since there is no \( x^2 \) from substitution).- Term \( x^3 \) is \( \frac{x^3}{6} + x^3 \).- Term \( x^4 \) is 0.- Term \( x^5 \) is \( \frac{3x^5}{40} \).

Step 6: Collect the Series Terms

The resulting series up to \( x^5 \) is \( f(x) = 1 + x + \frac{x^3}{6} + x^3 + \frac{3x^5}{40} + \cdots \). Combine like terms for the \( x^3 \): \( f(x) = 1 + x + \frac{7x^3}{6} + \frac{3x^5}{40} \).

Key Concepts

sine function

The sine function, a fundamental function in trigonometry, is essential for many mathematical applications, including the study of waves and oscillations. It's expressed as \( \sin x \) and is periodic, meaning it repeats its values in regular intervals. In the context of series, it plays a crucial role by allowing expansion into a Maclaurin series, which is a type of Taylor series centered at zero. The Maclaurin series for \( \sin x \) is given as:

• \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots \)

Understanding this series is key for approximating \( \sin x \) when \( x \) is near zero. In our exercise, this series helped simplify the expression \( \frac{1}{1 - \sin x} \). By breaking down the complex function into manageable terms, it becomes easier to perform calculations.

geometric series

The geometric series is a powerful tool in mathematics, especially useful when dealing with functions that can be expressed as fractions. Defined as:

• \( \frac{1}{1-u} = 1 + u + u^2 + u^3 + \cdots \)

this series converges when \(|u| < 1\). In the given problem, we utilized the geometric series expansion to evaluate the expression \( \frac{1}{1 - \sin x} \). By letting \( u = \sin x \), we were able to substitute the series expansion of the sine function into the geometric series, allowing us to further simplify and compute the series expansion of the original function up to the desired power of \( x^5 \). This geometric approach is particularly useful when trying to simplify complex expressions iteratively.

series expansion

Series expansion, specifically the Maclaurin series, is a method of expressing a function as an infinite sum of terms. Each term is derived from the function's derivatives at a single point. For a function \( f(x) \) with an expansion at \( x=0 \), the Maclaurin series is shown as:

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

This method enables us to approximate functions with polynomials, simplifying them for analysis or computation. Our exercise task was to find the series expansion of \( \frac{1}{1 - \sin x} \) using the known series of the sine function. By applying the series expansion method, we derived a polynomial that effectively approximates the function for small values of \( x \). This approach is crucial in calculus, helping bridge the gap between complex functions and more straightforward algebraic forms.

calculus problem solving

Calculus problem solving often involves manipulating series and functions to find solutions to complex problems. Approaches can include differentiation, integration, and using series expansions. In this exercise, we solved a calculus problem by:

• Recognizing known series, like the Maclaurin series for \( \sin x \), and using it to simplify the target function.
• Employing the geometric series for manipulating expressions in the form of \( \frac{1}{1 - u} \).
• Applying algebraic techniques to collect and simplify terms up to a specific degree — in this case, up to \( x^5 \).

This type of problem solving requires a strong understanding of calculus concepts and the ability to apply them creatively and efficiently to reach a solution. By breaking down complex tasks into manageable steps, calculus provides the tools needed to tackle a wide range of mathematical challenges.