Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 19-24, find the Taylor series in \(x-\) a through the term \((x-a)^{3}\). $$ \cos x, a=\frac{\pi}{3} $$

Step-by-Step Solution

Answer

The Taylor series is \(\frac{1}{2} - \frac{\sqrt{3}}{2}(x - \frac{\pi}{3}) - \frac{1}{4}(x - \frac{\pi}{3})^2 + \frac{\sqrt{3}}{12}(x - \frac{\pi}{3})^3\).

Step 1: Understanding Taylor Series

The Taylor series for a function \( f(x) \) about a point \( a \) is given by:\[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots\]We need to find this series for \( \cos x \) when \( a=\frac{\pi}{3} \) through the \((x-a)^3\) term.

Step 2: Calculate Function and Derivative Values at \( a = \frac{\pi}{3} \)

First, we need to calculate the function and derivatives at \( a = \frac{\pi}{3} \):- \( f(x) = \cos x \) gives \( f\left(\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \).- \( f'(x) = -\sin x \), so \( f'\left(\frac{\pi}{3}\right) = -\sin\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} \).- \( f''(x) = -\cos x \), so \( f''\left(\frac{\pi}{3}\right) = -\cos\left(\frac{\pi}{3}\right) = -\frac{1}{2} \).- \( f'''(x) = \sin x \), so \( f'''\left(\frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \).

Step 3: Formulate Taylor Series Through \((x-a)^3\) Term

Using the values from the previous step, substitute them into the Taylor series formula:\[\cos x \approx \frac{1}{2} - \frac{\sqrt{3}}{2}(x - \frac{\pi}{3}) - \frac{1}{2} \frac{(x - \frac{\pi}{3})^2}{2} + \frac{\sqrt{3}}{2} \frac{(x - \frac{\pi}{3})^3}{6}\]Simplify each of the coefficients:\[\cos x \approx \frac{1}{2} - \frac{\sqrt{3}}{2}(x - \frac{\pi}{3}) - \frac{1}{4}(x - \frac{\pi}{3})^2 + \frac{\sqrt{3}}{12}(x - \frac{\pi}{3})^3\]

Step 4: Conclusion

The Taylor series for \( \cos x \) about \( a = \frac{\pi}{3} \) through the \((x-a)^3\) term is:\[\frac{1}{2} - \frac{\sqrt{3}}{2}(x - \frac{\pi}{3}) - \frac{1}{4}(x - \frac{\pi}{3})^2 + \frac{\sqrt{3}}{12}(x - \frac{\pi}{3})^3\]

Key Concepts

Cosine Function

The cosine function, often denoted as \( \cos x \), is a fundamental trigonometric function. It represents the x-coordinate of the point on the unit circle associated with a given angle. The cosine function is periodic, with a cycle repeating every \( 2\pi \) radians. This means that \( \cos x = \cos (x + 2\pi) \) for all \( x \). Understanding the behavior of the cosine function is essential in various fields, including physics, engineering, and computer graphics. In this exercise, we utilize derivatives of the cosine function to expand it into a Taylor series.

Calculus

Calculus is a field of mathematics that studies changes. It mainly involves two branches: differential calculus and integral calculus. Differential calculus concerns the concept of a derivative, which gives us the rate at which a particular quantity changes. In this problem, calculus is used to perform a Taylor series expansion. By differentiating the cosine function, we obtain various derivatives required to approximate the function around a specific point \( a = \frac{\pi}{3} \). These derivatives help form the polynomial expression of the cosine function.

Derivatives

The derivative of a function measures how the function value changes as its input changes. For the cosine function \( \cos x \), the first few derivatives are crucial for forming its Taylor series expansion. Here are the derivatives we use in this exercise:

• First Derivative: \( f'(x) = -\sin x \)
• Second Derivative: \( f''(x) = -\cos x \)
• Third Derivative: \( f'''(x) = \sin x \)

For \( x = \frac{\pi}{3} \), these derivatives are evaluated to yield specific numerical results. These values populate the Taylor series terms, helping us approximate \( \cos x \) around this point. By understanding derivatives, we can effectively translate the function's behavior into a series of polynomials.

Series Expansion

A series expansion, particularly a Taylor series, is a way to represent a function as an infinite sum of terms calculated from its derivatives at a single point. In this task, we approximate \( \cos x \) near \( \frac{\pi}{3} \) up to the third degree of \( (x-a)^3 \). This involves:

• Finding function and derivatives at \( a = \frac{\pi}{3} \).
• Substituting these values into the Taylor series formula.
• Simplifying coefficients to construct a polynomial.

The resulting polynomial \[\cos x \approx \frac{1}{2} - \frac{\sqrt{3}}{2}(x - \frac{\pi}{3}) - \frac{1}{4}(x - \frac{\pi}{3})^2 + \frac{\sqrt{3}}{12}(x - \frac{\pi}{3})^3\]provides a good approximation of \( \cos x \) around \( \frac{\pi}{3} \). Such series expansions are invaluable in simplifying complex mathematical expressions, especially when exact solutions are difficult to derive.