Problem 35 - Essential Calculus Early Transcendentals

Question

Use a Maclaurin series in Table 1 to obtain the Maclaurin series for the given function. $$f(x)=\sin ^{2} x\left[\text { Hint: Use } \sin ^{2} x=\frac{1}{2}(1-\cos 2 x).\right]$$

Step-by-Step Solution

Answer

The Maclaurin series is \( f(x) = x^2 - \frac{x^4}{3} + \frac{x^6}{90} + \cdots \).

Step 1: Use the Trigonometric Identity

The given function is \( f(x) = \sin^2 x \). We use the provided trigonometric identity: \( \sin^2 x = \frac{1}{2} (1 - \cos 2x) \). Thus, we rewrite \( f(x) \) as \( f(x) = \frac{1}{2} (1 - \cos 2x) \).

Step 2: Break Down the Expression

The expression is \( \frac{1}{2} (1 - \cos 2x) \). We need to find the Maclaurin series for \( 1 \) and \( \cos 2x \) separately and then combine them.

Step 3: Identify the Maclaurin Series for Each Component

For the constant \( 1 \), the Maclaurin series is simply \( 1 \). For \( \cos 2x \), recall the Maclaurin series for \( \cos x \): \[ \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \]Substitute \( x \) with \( 2x \):\[ \cos 2x = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \frac{(2x)^6}{6!} + \cdots \]

Step 4: Calculate the Maclaurin Series for \( \cos 2x \)

Now substitute and simplify the terms:\[ \cos 2x = 1 - \frac{4x^2}{2} + \frac{16x^4}{24} - \frac{64x^6}{720} + \cdots \] Simplifying these:\[ \cos 2x = 1 - 2x^2 + \frac{2x^4}{3} - \frac{x^6}{45} + \cdots \]

Step 5: Substitute into the Expression for \( f(x) \)

We now substitute back into the expression for \( f(x) \):\[ f(x) = \frac{1}{2} (1 - \cos 2x) = \frac{1}{2} (1 - (1 - 2x^2 + \frac{2x^4}{3} - \frac{x^6}{45} + \cdots)) \]Combining terms, the simplified series is:\[ f(x) = \frac{1}{2} (2x^2 - \frac{2x^4}{3} + \frac{x^6}{45} + \cdots) \] Finally, it becomes:\[ f(x) = x^2 - \frac{x^4}{3} + \frac{x^6}{90} + \cdots \]

Step 6: Conclusion

The Maclaurin series for \( f(x) = \sin^2 x \) is:\[ f(x) = x^2 - \frac{x^4}{3} + \frac{x^6}{90} + \cdots \]

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental tools in mathematics that simplify expressions involving trigonometric functions like sine, cosine, and tangent. In this exercise, we used the identity \( \sin^2 x = \frac{1}{2} (1 - \cos 2x) \). This specific identity is derived from the double-angle formulas, allowing us to express \( \sin^2 x \) in terms of \( \cos \). This transformation is crucial as it simplifies the process of finding the Maclaurin series by breaking down complex trigonometric functions into simpler components we can work with more easily.
When tackling problems involving trigonometric identities:

• Identify the relevant identity that simplifies the expression.
• Apply the identity to rewrite the function in a more manageable form.
• Check the manipulation to ensure no mathematical errors have been introduced.

This step is essential before proceeding to series expansion, setting up a foundation that allows us to plug functions directly into known series.

Series Expansion

Series expansion is a way to express a function as an infinite sum of terms. The Maclaurin series is a type of Taylor series centered at zero, and it's particularly useful for approximating functions near this point. In this example, our goal was to find the Maclaurin series for \( f(x) = \sin^2 x \).
To accomplish this:

• We first rewrote the function using a trigonometric identity: \( f(x) = \frac{1}{2} (1 - \cos 2x) \).
• We then expanded each component, starting with the simpler constant 1, and then \( \cos 2x \) using its known Maclaurin series: \( \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \).
• Substitute back values, simplify, and expand to approximate the expression.

Series expansion is particularly powerful as it turns complex, non-linear functions into polynomials, making it easier to calculate and understand their behavior around a particular point.

Calculus

Calculus plays a pivotal role in deriving series expansions. It helps us find derivatives which are used to express functions as infinite series. In the context of the Maclaurin series, we particularly rely on the function's value and the essence of derivatives at zero to define each term.
Steps involving calculus include:

• Understanding the basis for the expansion: a Maclaurin series formula \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots \).
• Applying derivative rules to find derivatives of functions involved, such as the cosine function's derivatives at zero.
• Combining calculus-based logic with our series expansion knowledge to derive functions into usable polynomial-like series.

Calculus enables us to comprehensively navigate and translate functions into series, ensuring precise expansions for approximations and further mathematical analysis. Understanding these basics can deepen one's ability in tackling similar problems across mathematical fields.