Problem 69 - Calculus: Early Transcendentals

Question

Use multiplication of division of power series to find the first three nonzero terms in the Maclaurin series for each function. \( y = \frac {x}{\sin x} \)

Step-by-Step Solution

Answer

First three non-zero terms are: 1 + \frac{1}{6}x^2.

Step 1: Identify Known Series

The Maclaurin series for \( \sin x \) is given by \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \). We need this to find the Maclaurin series for \( y = \frac{x}{\sin x} \).

Step 2: Multiply by the Inverse Series

To find \( \frac{x}{\sin x} \), consider it as \( x \cdot \frac{1}{\sin x} \). We first need the series expansion for \( \frac{1}{\sin x} \). Suppose the expansion \( \frac{1}{\sin x} = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots \), we multiply it with the series for \( \sin x \) and match like powers of \(x\) to find the coefficients. Setting \( 1 = (x) (a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots), \) we compare terms for a solution.

Step 3: Solve for Coefficients

We need the coefficients by multiplying and matching terms. Set \[ x (a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \cdots) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots. \] Comparing same powers: 1. At \(x^0\): No constant terms exist, so match alternatively. 2. At \(x^1\): Compare \( a_0 \cdot x = x \quad \Rightarrow a_0 = 1. \) 3. At \(x^3\): Compare \( -a_1 \cdot x^3 = -\frac{1}{6} \quad \Rightarrow a_1 = \frac{1}{6} . \)

Step 4: Write Down Simplified Series

Now use the coefficients found to write the series to the desired number of terms: 1. At \(x^1\): \( x, \) \( a_0 = 1 \), so first term is plain \(x\). 2. At \(x^3\): we have \( \frac{1}{6} x^3 \). 3. For \(x^5\): Further terms match higher order still. Safe to truncate here.

Step 5: Form the Complete Series

Keep combining terms, results in \( \frac{x}{\sin x} = 1 + \frac{1}{6}x^2 + \cdots . \) Continue only until relevant order. Remembering the goal is only the first three terms total. Given this arrangement, can see terms where comparable constantly achieved.

Key Concepts

Power Series

A power series is a way of representing a function as an infinite sum of terms, calculated from the powers of a variable. It typically takes the form:

• \[\sum_{n=0}^{\infty} a_n x^n\]

Here, each term in the series involves a coefficient \(a_n\) and a power of \(x^n\). Power series are crucial in mathematics, as they help approximate functions that are otherwise difficult to compute directly.
The Maclaurin series is a type of power series where each term is derived from the derivatives of a function evaluated at zero.
For instance, the Maclaurin series representation of \(\sin x\) is:

• \[\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\]

This is important because it provides a foundation for analyzing and approximating functions like \(\frac{x}{\sin x}\). Using such a series allows us to explore and solve problems about functions that are otherwise complex or not easily represented by simpler algebraic expressions.

Coefficients

Coefficients are the numerical factors in the terms of the power series. For example, in the term \(a_n x^n\), \(a_n\) is the coefficient. They play a critical role by dictating the series' behavior and shape, impacting its graph and convergence.
To find the coefficients for a series such as \(\frac{1}{\sin x}\), you align it with another series (in this case, \(\sin x\)) and match the coefficients of like powers of \(x\).
During this process, you often start with known terms:

• At \(x^0\), make corrections as needed;
• At \(x^1\), ensure coefficients match intentionally, leading to meaningful interpretations like \(a_0 = 1\);
• At \(x^3\), move to further similar steps, finding specific values, such as \(a_1 = \frac{1}{6}\).

Finding these coefficients involves both intuition and systematic comparison with the original function's derivatives, helping maintain alignment in the series.

Series Expansion

Series expansion is the process of expressing a function as an infinite series. In mathematics, this allows one to see the influence of each term and helps approximate complex functions with polynomials. Each function has unique expansions based on its Maclaurin series.
For example, the function \(\frac{x}{\sin x}\) requires expansion in terms of \(x\) for better analysis.
To achieve this:

• Multiply the series for \(\sin x\) by its inverse.
• Align and compute each of the resulting terms accurately.
• Verify that computations accurately reflect the initial conditions and desired approximation.

The expansion can reveal significant insight into a function, allowing us to view the function's structure more clearly and predict its behavior effectively. In both practical and theoretical applications, series expansions serve as a bridge from abstract function descriptions to usable numerical forms.

Sin x

The function \(\sin x\) is one of the fundamental trigonometric functions. Its nature as a periodic function makes it highly useful in various applications from physics to engineering.
Using the Maclaurin series:

• The core approximate value of \(\sin x\) is refined by further terms like \(-\frac{x^3}{3!}+\frac{x^5}{5!}\).
• This particular series does not end and can infinitely reveal more about \(\sin x\) through accurate approximations by increasing terms.
• In the context of the exercise \(\frac{x}{\sin x}\), understanding the series of \(\sin x\) is crucial as it lays the groundwork for further processes.

Whether studying waveforms, simple harmonic motion, or oscillations, \(\sin x\) and its properties are densely informative and useful in deriving and simplifying functions like \(\frac{x}{\sin x}\). The Maclaurin series representation here greatly simplifies studying its effects and representations in different analytical conditions.