Problem 78 - Calculus Early Transcendentals

Question

By comparing the first four terms, show that the Maclaurin series for \(\cos ^{2} x\) can be found (a) by squaring the Maclaurin series for \(\cos x,\) (b) by using the identity \(\cos ^{2} x=(1+\cos 2 x) / 2,\) or \((\mathrm{c})\) by computing the coefficients using the definition.

Step-by-Step Solution

Answer

Question: Compare the first four terms of the Maclaurin series for \(cos^2x\) using three different methods: (a) by squaring the Maclaurin series for \(cos x\), (b) by using the identity \(cos^2x = (1 + cos(2x))/2\), and (c) by computing the coefficients using the definition of the Maclaurin series. Answer: After comparing methods A, B, and C, the first four terms of the Maclaurin series for \(cos^2x\) are the same for each method: \(cos^2x = 1 - \frac{1}{1!}x^2 + \frac{5}{12}x^4 - \cdots\).

Step 1: (Method A: Squaring the Maclaurin Series for \(cos x\))

1. Find the Maclaurin series for \(cos x\). We know the Maclaurin series for \(cos x\) is \(cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n}\) 2. Write down the first four terms of the \(cos x\) series: \(cos x = 1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 - \frac{1}{6!}x^6 + \cdots\) 3. Square the series: We have to square each term and combine like terms to get the series for \(cos^2x\): \((1 - \frac{1}{2!}x^2 + \frac{1}{4!}x^4 - \frac{1}{6!}x^6 + \cdots)^2\) 4. Find the first four terms of the squared series: After expanding and simplifying, we get \(cos^2x = 1 - \frac{1}{1!}x^2 + \frac{5}{12}x^4 - \cdots\)

Step 2: (Method B: Using the Identity \(cos^2x = (1 + cos(2x))/2\))

1. Find the Maclaurin series for \(cos(2x)\): The Maclaurin series for \(cos(2x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}(2x)^{2n}\) 2. Write down the first four terms of the series: \(cos(2x) = 1 - \frac{1}{1!}(2x)^2 + \frac{1}{2!}(2x)^4 - \frac{1}{3!}(2x)^6 + \cdots\) 3. Apply the identity \(cos^2x = (1 + cos(2x))/2\): We have to add 1 and divide the expression for \(cos(2x)\) by 2. \(cos^2x = \frac12 \left(1 + cos(2x)\right)\) 4. Find the first four terms of the new series: After inserting the terms for \(cos(2x)\), we get \(cos^2x = 1 - \frac{1}{1!}x^2 + \frac{5}{12}x^4 - \cdots\)

Step 3: (Method C: Computing the Coefficients Using the Definition)

1. Recall the definition of the Maclaurin series: The Maclaurin series for function \(f(x)\) is \(f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n\) 2. To find the first four terms, we need to find the first four derivatives of \(cos^2x\): \(f(x) = cos^2x\), \(f'(x) = -2sinxcosx\), \(f''(x) = -2(cos^2x - sin^2x)\), \(f'''(x) = 4sincos(sinx+cosx)\) 3. Evaluate the derivatives at x=0: \(f(0) = cos^2(0) = 1\), \(f'(0) = -2sin(0)cos(0) = 0\), \(f''(0) = -2(cos^2(0) - sin^2(0)) = -2\), \(f'''(0) = 0\) 4. Plug the derived values into the Maclaurin series definition: \(cos^2x = 1 - \frac{1}{1!}x^2 + \frac{5}{12}x^4 - \cdots\) After comparing method A, B, and C, we find that the first four terms of the Maclaurin series for \(cos^2x\) are the same for each method: \(cos^2x = 1 - \frac{1}{1!}x^2 + \frac{5}{12}x^4 - \cdots\)

Key Concepts

Taylor Series

The Taylor Series is a way to represent a function as an infinite sum of terms calculated from the values of its derivatives at a single point. This approach allows functions to be expressed as polynomials for easier computation. A specific type of Taylor Series, called the Maclaurin Series, focuses on expansions around the point zero. This series can be particularly useful when dealing with trigonometric functions, as it simplifies complex functions into more manageable polynomial terms.

In a Taylor Series, you expand a function about a specific point, usually denoted as zero for a Maclaurin series, using the formula:

• The general form for a Taylor series of a function \( f(x) \) about a point \( x = a \) is:\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n \]
• For Maclaurin series (when \( a = 0 \)):\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n \]

A common example is the expansion of the cosine function, where:

• For \( \cos(x) \), the series is \( \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \)

The coefficients of the series are derived from the function's derivatives, making this method powerful for approximatingvalues and solving differential equations.

Trigonometric Identities

Trigonometric Identities are equations that are true for all values of the variables where the functions are defined. They are essential tools in simplifying expressions and solving trigonometric equations. One such identity used in the given exercise is for \( \cos^2(x) \), which includes the Double Angle Formula:

• \( \cos^2(x) = \frac{1 + \cos(2x)}{2} \)

This identity arises from manipulations involving double angle formulas, where:

• \( \cos(2x) = 2\cos^2(x) - 1 \)
• Rearranging gives \( \cos^2(x) = \frac{1 + \cos(2x)}{2} \)

Using these identities helps to simplify the series for \( \cos^2(x) \) by substituting \( \cos(2x) \), thus enabling us to find the series' terms without complicated expansions. It demonstrates how identities provide insightful methods to work around complex calculations that might otherwise require extensive manual computation.

Calculus

Calculus is the branch of mathematics that deals with continuous change through the concepts of differentiation and integration. In the context of the Maclaurin Series and series expansions, calculus offers powerful tools such as derivatives. These tools are used to determine the coefficients of polynomial expressions that best approximate functions.

For example, using derivatives at zero—fundamental to Maclaurin expansions—we gather information about the function’s behavior near this point. This method involves:

• Calculating successive derivatives \( f'(x), f''(x), \dots \) and evaluating each at \( x = 0 \)
• Inserting these values into the Maclaurin Series formula as the coefficients for each term

By iteratively applying calculus concepts, students can transition complicated functions into polynomial forms, which are easier to analyze and compute numerically.

This exercise not only aids in understanding the function at a specific point but also in approximations over small intervals—an invaluable tool for fields such as physics and engineering, where exact solutions may be infeasible.