Problem 24 - Calculus: Early Transcendentals

Question

Find the Taylor series for \( f(x) \) centered at the given value of \( a. \) [Assume that \( f \) has a power series expansion. Do not show that \( R_n (x) \to 0.\)] Also find the associated radius of convergence. \( f(x) = \cos x, \) \( a = \pi/2 \)

Step-by-Step Solution

Answer

The Taylor series is \(-\sum_{n=0}^{\infty} \frac{(x-\pi/2)^{2n+1}}{(2n+1)!}(-1)^n\) and the radius of convergence is \( R = \infty \).

Step 1: Find the function derivatives

First, we need to calculate the derivatives of the function \( f(x) = \cos x \) at the point \( a = \pi/2 \). Compute the first few derivatives until a recognizable pattern can be established.\( f(x) = \cos x \) \( f'(x) = -\sin x \) \( f''(x) = -\cos x \) \( f'''(x) = \sin x \) \( f^{(4)}(x) = \cos x \)

Step 2: Evaluate the derivatives at the center

Substitute \( x = \pi/2 \) into each derivative to find the coefficients of the Taylor series.\( f(\pi/2) = \cos(\pi/2) = 0 \)\( f'(\pi/2) = -\sin(\pi/2) = -1 \)\( f''(\pi/2) = -\cos(\pi/2) = 0 \)\( f'''(\pi/2) = \sin(\pi/2) = 1 \)\( f^{(4)}(\pi/2) = \cos(\pi/2) = 0 \)Observe a pattern: Only odd derivatives contribute as even derivatives equal zero.

Step 3: Write the Taylor series

The Taylor series for \( f(x) \) about \( a = \pi/2 \) is given by: \[ \sum_{n=0}^{\infty} \frac{f^{(n)}(\pi/2)}{n!} (x-\pi/2)^n \]. Using the non-zero derivatives identified earlier, the series becomes:\[-\frac{1}{1!} (x-\pi/2)^{1} + \frac{1}{3!} (x-\pi/2)^3 - \frac{1}{5!} (x-\pi/2)^5 + \ldots\]The pattern is clear: it is alternating and only odd powers are present.

Step 4: Determine the radius of convergence

The function \( \cos(x) \) has a known radius of convergence of \( \infty \) because it is entire (analytic everywhere on the complex plane). The Taylor series for \( \cos(x) \) about any point converges for all \( x \). Thus, the radius of convergence \( R = \infty \).

Key Concepts

Radius of Convergence

When we talk about the radius of convergence for a Taylor series, we're referring to the distance within which the series converges to the function. For most power series, there's a specific range around the center of the series where it accurately represents the function. This is called its interval of convergence. The radius of convergence tells us how far from the center, along the real axis, our series is still valid.

However, some functions, like cosine, are entire. This means they're analytic across the entire complex plane. For the cosine function centered around any point, the radius of convergence is infinitely large. Therefore, the Taylor series for cosine will converge for all values of x.

• The radius of convergence is found using the formula involving the limit of the coefficients: \ \( R = \lim_{n \to \infty} \left| \frac{a_n}{a_{n+1}} \right| \). For cosine, this evaluation shows that the radius is infinite.
• An infinite radius means the series converges everywhere, making it a powerful representation.

Cosine Function

The cosine function, denoted as \( \cos(x) \), is a fundamental trigonometric function widely used in mathematics. It relates the angle to the horizontal coordinate on the unit circle.

In calculus, cosine frequently appears because of its periodic and symmetrical properties. One of its intriguing characteristics is that its derivatives create a repeating cycle:

• \( \cos(x) \)
• \( - \sin(x) \)
• \( - \cos(x) \)
• \( \sin(x) \)

This cycle repeats every four functions, making it predictable, which is particularly helpful when developing series expansions.

Derivatives

To build a Taylor series, we must calculate the derivatives of the function at the center point — in our case,\( a = \pi/2 \).

For any smooth function like cosine, these derivatives offer crucial insights into the function's behavior near that point. In our exercise, the derivatives of cosine around \( \pi/2 \) help to define the coefficients of its Taylor series. Let's see:

• \( \cos(x)'s \text{ derivatives are: }\)
• \( f(x) = \cos x \)
• \( f'(x) = -\sin x \)
• \( f''(x) = -\cos x \)
• \( f'''(x) = \sin x \)
• \( f^{(4)}(x) = \cos x \)

At \( x = \pi/2 \), the values of these derivatives produce a sequence crucial to the path of our series: odd derivatives yield non-zero contributions (either \(-1\) or \(1\)), while even derivatives result in zero, as cosine is zero at this point for even multiples. This alternation crafts the pattern for our series terms.

Alternating Series

An alternating series is one where the signs of the terms switch back and forth between positive and negative. This occurs naturally in the Taylor series for cosine function, specifically when centered around \( a = \pi/2 \).

The Taylor series for cosine in this setup generates terms that alternate in sign and only include odd powers. As a result, its expression is:

• \( -\frac{1}{1!}(x-\pi/2)^1 + \frac{1}{3!}(x-\pi/2)^3 - \frac{1}{5!}(x-\pi/2)^5 + \ldots \)

This alternation is both intriguing and significant:

• It's a hallmark of the trigonometric function's symmetry.
• This feature ensures precision in approximation, balancing terms out over its interval.
• The convergence behavior of an alternating series is often gentle and reliable if the terms decrease in magnitude.