Problem 32 - Thomas Calculus

Question

Find the first four nonzero terms in the Maclaurin series for the functions in Exercises \(29-34 .\) $$\cos ^{2} x \cdot \sin x$$

Step-by-Step Solution

Answer

The first three nonzero terms are: \( x - 2x^3 + \frac{x^5}{4} \).

Step 1: Identify the Maclaurin Series

Recall that the Maclaurin series for a function is the Taylor series centered at 0. We need to find the series expansion of each component of \( \cos^2 x \cdot \sin x \). The Maclaurin series for \( \sin x \) is \( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \) and for \( \cos x \) is \( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \).

Step 2: Express \( \cos^2 x \)

Since \( \cos^2 x = (\cos x)^2 \), square the series for \( \cos x \):\[(1 - \frac{x^2}{2} + \frac{x^4}{24})^2 = 1 - x^2 + \frac{x^4}{12}\] (For this exercise, higher order terms are ignored as they contribute to higher powers than needed.)

Step 3: Multiply by \( \sin x \)

Multiply the series for \( \cos^2 x \) by the series for \( \sin x \):\[(1 - x^2 + \frac{x^4}{12})(x - \frac{x^3}{6}) = x - x^3 + \frac{x^5}{12} - x^3 + \frac{x^5}{6}\] Simplify the terms.

Step 4: Simplify and Collect Terms

Combine like terms from the product:\[x - 2x^3 + \frac{x^5}{4}\]

Step 5: Extract the First Four Nonzero Terms

List the first four nonzero terms from the simplified series. Since we have fewer than four, stop after extracting the last nonzero term. The series is:\[x - 2x^3 + \frac{x^5}{4}\]

Key Concepts

Taylor Series

The Taylor series is a powerful tool for representing functions as infinite sums of terms. Derived from the derivatives of a function at a single point, it offers mathematicians and engineers a way to approximate complex functions. Imagine expanding a function into an infinite polynomial. Each added term increases the precision of the approximation.

• The general Taylor series is expressed as: \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \frac{f^{(3)}(a)(x-a)^3}{3!} + \cdots \]
• Here, \(f(a)\) is the function value at \(x = a\), and the subsequent terms involve higher order derivatives evaluated at \(x = a\).

When \(a = 0\), the Taylor series becomes the Maclaurin series. This makes it particularly useful for series expansion around zero.

Trigonometric Series

Trigonometric series involve functions like sine and cosine, often expressed in power series. Such series expansions are crucial in approximating values of trigonometric functions for calculations. They appear frequently in solving differential equations and in Fourier analysis.

For example:

• The Maclaurin series for \(\sin x\) begins as: \[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \]
• Similarly, \(\cos x\) starts as: \[ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \]

To find the series for \(\cos^2 x \cdot \sin x\), each function is expanded separately. Thereafter, it's necessary to multiply their series together and simplify, using properties of such series. This process allows us to handle complex trigonometric expressions by breaking them down into manageable parts.

Series Expansion

Series expansion is critical in calculus and mathematical analysis, transforming complex expressions into polynomial-like forms. This simplification enables easier computation and understanding of functions in various mathematical fields.

• During expansion, terms are identified by powers of \(x\), ranging from constants to higher powers.
• Each term in the series provides a more precise approximation of the original function within a specific range.

In our example, series expansion required combining series for \(\cos^2 x\) and \(\sin x\). This began by squaring the Maclaurin series for \(\cos x\), then multiplying it by the series for \(\sin x\). Like terms were simplified to construct the expanded series: \[ x - 2x^3 + \frac{x^5}{4} \]. The resulting expression provides an insightful glimpse of the original function's behavior near zero.