Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the Taylor series in \(x-a\) through the term \((x-a)^{3}\). \(\sin x, a=\frac{\pi}{6}\)

Step-by-Step Solution

Answer

The Taylor series for \(\sin x\) about \(\frac{\pi}{6}\) is \(\frac{1}{2} + \frac{\sqrt{3}}{2}(x-\frac{\pi}{6}) - \frac{1}{4}(x-\frac{\pi}{6})^2 - \frac{\sqrt{3}}{12}(x-\frac{\pi}{6})^3\).

Step 1: Understanding the Taylor series formula

The Taylor series of a function \(f(x)\) about \(x = a\) is given by: \[f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \, ...\] We are tasked with finding this series for the function \(\sin x\) about \(a = \frac{\pi}{6}\).

Step 2: Calculate \(f(a)\)

For \(f(x) = \sin x\) and \(a = \frac{\pi}{6}\), we find \(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\). So, \(f(a) = \frac{1}{2}\).

Step 3: Calculate the first derivative \(f'(x)\) and evaluate at \(x = a\)

The first derivative of \(\sin x\) is \(\cos x\). Evaluating this at \(x = \frac{\pi}{6}\) gives \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\). So, \(f'(a) = \frac{\sqrt{3}}{2}\).

Step 4: Calculate the second derivative \(f''(x)\) and evaluate at \(x = a\)

The second derivative of \(\sin x\) is \(-\sin x\). Evaluating this at \(x = \frac{\pi}{6}\) gives \(-\sin\left(\frac{\pi}{6}\right) = -\frac{1}{2}\). So, \(f''(a) = -\frac{1}{2}\).

Step 5: Calculate the third derivative \(f'''(x)\) and evaluate at \(x = a\)

The third derivative of \(\sin x\) is \(-\cos x\). Evaluating this at \(x = \frac{\pi}{6}\) gives \(-\cos\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{2}\). So, \(f'''(a) = -\frac{\sqrt{3}}{2}\).

Step 6: Substitute into the Taylor series formula

Using the Taylor series formula, substitute the derivatives: \[\sin(x) \approx \frac{1}{2} + \frac{\sqrt{3}}{2}(x - \frac{\pi}{6}) - \frac{1}{2}\frac{(x - \frac{\pi}{6})^2}{2} - \frac{\sqrt{3}}{2}\frac{(x - \frac{\pi}{6})^3}{6}\] Simplifying yields: \[\sin(x) \approx \frac{1}{2} + \frac{\sqrt{3}}{2}(x-\frac{\pi}{6}) - \frac{1}{4}(x-\frac{\pi}{6})^2 - \frac{\sqrt{3}}{12}(x-\frac{\pi}{6})^3\]

Key Concepts

Sin Function

The sine function is a fundamental concept in trigonometry often denoted as \( \sin x \). It describes the y-coordinate of a point on the unit circle that corresponds to an angle \( x \).
Understanding the behaviour of the sine function is crucial for solving problems involving periodic phenomena, such as sound waves or circular motion.

• Range and Domain: The sine function has a range of \([-1, 1]\) and is defined for all real numbers \(x\).
• Periodicity: It is periodic with a period of \(2\pi\), meaning \( \sin(x) = \sin(x + 2\pi n) \) for any integer \(n\).
• Symmetry: The sine function is odd, which means that \( \sin(-x) = -\sin(x) \).

Exploring the Taylor series of the sine function helps us approximate its value near a certain point. This is useful because direct computation might be difficult or impossible in some circumstances.

Derivatives

Derivatives are a basic concept in calculus, measuring how a function changes as its input changes. For the sine function, the derivatives show how it bends and oscillates. Each derivative of \( \sin x \) follows a cyclic pattern:

• First Derivative: The first derivative of \( \sin x \) is \( \cos x \), indicating the rate of change of \( \sin x \). It tells us how steep the sine curve is at a particular point.
• Second Derivative: The second derivative of \( \sin x \) is \( -\sin x \). This reflects the curvature of \( \sin x \). If the second derivative is negative, the function is concave down.
• Third Derivative: The third derivative is \( -\cos x \), which continues the pattern and further describes the function’s periodic nature.

Using derivatives to construct the Taylor series allows us to approximate \( \sin x \) around any point like \( x = \frac{\pi}{6} \). This involves evaluating the derivative at that point to predict small changes in the function's behavior.

Series Expansion

Series expansions, like the Taylor series, are powerful tools for approximating complex functions with polynomials. The Taylor series specifically approximates a function around a point \( a \), using derivatives.
This series is crucial because it turns complicated functions into simpler polynomial forms.

• General Formula: The Taylor series expansion of a function \( f(x) \) around a point \( a \) is given by:\[ f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots\]
• Terms Explanation: Each term in the Taylor series represents the value of the function, a slope, or a curvature contribution, utilizing higher order derivatives.
• Convergence: More terms in the series yield a better approximation. However, beyond a certain point, additional terms provide minimal improvement for functions like \( \sin x \).

For \( \sin x \), using derivatives through the third order provides a near-perfect approximation around the point \( \frac{\pi}{6} \), as detailed in the solution of the exercise.