Problem 27 - Essential Calculus Early Transcendentals

Question

Use a Maclaurin series in Table 1 to obtain the Maclaurin series for the given function. $$f(x)=\sin \pi x$$

Step-by-Step Solution

Answer

\( \sin(\pi x) = \sum_{n=0}^{\infty} (-1)^n \frac{(\pi x)^{2n+1}}{(2n+1)!} \).

Step 1: Recall the Maclaurin Series for Basic Functions

The Maclaurin series for a function \( f(x) \) is given by \( f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots \). Table 1 might provide some common Maclaurin series, such as \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \).

Step 2: Substitute \( \pi x \) for \( x \) in the Series for \( \sin x \)

Since \( f(x) = \sin(\pi x) \), substitute \( \pi x \) into the Maclaurin series for \( \sin x \): \( \sin(\pi x) = \pi x - \frac{(\pi x)^3}{3!} + \frac{(\pi x)^5}{5!} - \cdots \).

Step 3: Expand Each Term

Compute each term by expanding \( (\pi x)^n \) and simplifying: \( \pi x = \pi x \), \( (\pi x)^3 = \pi^3 x^3 \), \( (\pi x)^5 = \pi^5 x^5 \), etc.

Step 4: Simplify the Series

Combine all the terms: \( \sin(\pi x) = \pi x - \frac{\pi^3 x^3}{3!} + \frac{\pi^5 x^5}{5!} - \cdots \).

Step 5: Write the General Form

Write the general term for the series: \( \sin(\pi x) = \sum_{n=0}^\infty (-1)^n \frac{(\pi x)^{2n+1}}{(2n+1)!} \).

Key Concepts

Sin Function

The sine function, denoted as \( \sin x \), is a fundamental part of trigonometry. It relates the angle in a right triangle to the ratio of the length of the side opposite that angle to the hypotenuse. The sine function is periodic, which means it repeats its values in regular intervals or periods. In particular, the period of the sine function is \( 2\pi \).

In calculus and especially in series expansion, it's common to express the sine function as an infinite series. This is often represented by its Maclaurin series, which is a type of Taylor series centered at zero. The Maclaurin series for \( \sin x \) is:

• \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \)

This series gives us a powerful tool to approximate the sine function around \( x = 0 \), using an infinite sum of polynomial terms. This approximation is very useful for calculating sin values for small angles where producing exact trigonometric results might be complex or impossible.

Function Expansion

Function expansion is a mathematical technique where we express a function as a sum of simpler terms. In the context of the Maclaurin series, we expand a function into an infinite series of terms based on powers of \( x \). This method allows us to approximate complex functions using polynomials.

To expand a function using the Maclaurin series, we use derivatives evaluated at zero:

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

Each term in this expansion involves calculating the derivative of the function at \( x = 0 \), multiplying by a power of \( x \), and dividing by the factorial of the term number. Function expansion is particularly valuable because it transforms calculus problems into manageable algebraic expressions.

Understanding how to calculate these terms and why the expansion approximates the function well near the origin is crucial for mastering calculus concepts.

Series Substitution

Series substitution involves replacing each appearance of a variable in a series with another expression. This is essential when dealing with functions composed of other functions, which is quite common in calculus.

In this particular problem, the function \( \sin(\pi x) \) requires substituting \( \pi x \) into the known Maclaurin series for \( \sin x \). This process involves:

• Replacing \( x \) with \( \pi x \) in each term of the series.
• Generating new terms like \( \pi x - \frac{(\pi x)^3}{3!} + \frac{(\pi x)^5}{5!} - \cdots \).

Substituting another variable into a series allows us to address more complex functions using a series expansion that we already know.

This technique shows how flexible the concept of series can be, allowing us to expand and manipulate various functions beyond their standard form. Correctly implementing series substitution enhances problem-solving skills in calculus and higher-level mathematics.