Problem 10 - Calculus: Early Transcendentals

Question

Use the definition of a Taylor series to find the first four nonzero terms of the series for \( f(x) \) centered at the given value of \( a. \) $$ f(x) = \cos^2 x, \quad a = 0 $$

Step-by-Step Solution

Answer

The first four nonzero terms are: \( 1 - x^2 + \frac{x^4}{3} \).

Step 1: Identify Taylor Series Formula

The Taylor series for a function \( f(x) \) centered at \( a \) is given by: \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \] We will use this formula to find the first four nonzero terms by evaluating the needed derivatives at \( a = 0 \).

Step 2: Calculate Function and First Derivative

First, find \( f(0) \) and \( f'(x) \). Since \( f(x) = \cos^2 x \), using the identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \), we have:\[ f(x) = \frac{1 + \cos 2x}{2} \]. Therefore, \( f(0) = \frac{1 + 1}{2} = 1 \).Now differentiate: \( f'(x) = \frac{d}{dx}(\cos^2 x) = -2 \cos x \sin x = -\sin 2x \). Then \( f'(0) = -\sin 0 = 0 \).

Step 3: Calculate Second Derivative

Now find the second derivative: \( f''(x) = \frac{d}{dx}(-\sin 2x) = -2 \cos 2x \). Evaluating at \( x = 0 \), \( f''(0) = -2 \cdot 1 = -2 \).

Step 4: Calculate Third Derivative

The third derivative is: \( f'''(x) = \frac{d}{dx}(-2 \cos 2x) = 4 \sin 2x \). Substitute \( x = 0 \), \( f'''(0) = 4 \cdot 0 = 0 \).

Step 5: Calculate Fourth Derivative

Find the fourth derivative:\( f^{(4)}(x) = \frac{d}{dx}(4 \sin 2x) = 8 \cos 2x \). Then \( f^{(4)}(0) = 8 \).

Step 6: Construct Taylor Series

Substitute the derivative values found at \( a = 0 \) into the Taylor series formula.The Taylor series becomes:\[ f(x) \approx 1 + 0 \cdot x + \frac{-2}{2!} x^2 + 0 \cdot x^3 + \frac{8}{4!} x^4 + \cdots \]Simplifying gives:\[ f(x) \approx 1 - x^2 + \frac{x^4}{3} \]These are the first four nonzero terms.

Key Concepts

Calculating Derivatives

Derivatives are a fundamental concept in calculus, representing the rate at which a function changes at any given point. In this exercise, we compute the derivatives of the function \( f(x) = \cos^2 x \) one by one to construct its Taylor series. This is essential because each term in the series depends on a successive derivative evaluated at a specific point.When differentiating \( \cos^2 x \), we use the chain rule. This states that if a function is a composition of two functions, we take the derivative of the outside function and multiply it by the derivative of the inside function. Here, the outside function is the square, and the inside function is the cosine function.

• The first derivative \( f'(x) \) turns \( \cos^2 x \) into \( -\sin 2x \), as the derivative of \( \cos x \) is \( -\sin x \), and we use the double angle identity \( \sin 2x = 2 \sin x \cos x \). Evaluated at \( x = 0 \), this derivative is 0.
• The second derivative \( f''(x) \) becomes \( -2 \cos 2x \), with the constant factor from the double angle, evaluated as \(-2\) at \( x = 0 \).
• The third derivative \( f'''(x) \) results in \( 4 \sin 2x \), also evaluated at 0 as zero, fitting the pattern that every second derivative alters the function without contributing new terms at zero.
• The fourth derivative, \( f^{(4)}(x) \), evaluates to \( 8 \cos 2x \), resulting in 8 when centered around zero.

By calculating these derivatives, we lay the foundation for constructing a Taylor series centred at zero.

Cosine Function

The cosine function is integral to this exercise and a cornerstone of trigonometry. It periodically oscillates between 1 and -1, describing a wave-like pattern. In the Taylor series, the properties of the cosine function allow us to express even complex functions like \( \cos^2 x \) in polynomial forms around a given point.The identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \) simplifies dealing with \( \cos^2 x \). This identity is paramount for simplifying differentiation. Instead of differentiating the product, we break it down using the cosine's double angle identity, leading to more manageable derivatives.

• The first derivative involves simplifying using \( \sin 2x = 2 \sin x \cos x \) because directly differentiating \( \cos^2 x \) isn't straightforward.
• The evaluation of derivatives at zero uses the cosine's known values, leveraging that \( \cos 0 = 1 \) and \( \sin 0 = 0 \), simplifying most evaluations dramatically.

Understanding these transforms makes it feasible to derive expressions like \( \cos 2x \) or \( \sin 2x \), peeling back the complexity to simpler, repetitive patterns inherent in the cosine's properties.

Series Expansion

A series expansion such as a Taylor series is a way of expressing a function as a sum of its derivatives evaluated at a specific point. It creates an approximation with an infinite number of terms, each involving a higher-order derivative divided by the factorial of its order. This method is particularly useful because it allows us to approximate complex functions with simple polynomials.To construct such a series:

• First, identify the function and its derivatives as done previously with \( \cos^2 x \).
• Determine the order of the polynomial desired; in this case, first four non-zero terms.
• The series is centered at a point, \( a \), which we've chosen as zero, making it a Maclaurin series, a special case of the Taylor series where \( a = 0 \).

Each term is computed as \(\frac{f^n(a)}{n!}(x-a)^n\). For \( x = \cos^2 x \), plugging in the calculated derivatives results in:

• The zero order term: \(f(0)\), which is 1.
• The second order term: \(\frac{-2}{2!} x^2 = -x^2\).
• The fourth order term: \(\frac{8}{4!} x^4 = \frac{x^4}{3}\).

These computations yield \( f(x) \approx 1 - x^2 + \frac{x^4}{3} \), illustrating series expansion's power in simplifying and approximating functions.