Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 19-24, find the Taylor series in \(x-\) a through the term \((x-a)^{3}\). $$ \sin x, a=\frac{\pi}{6} $$

Step-by-Step Solution

Answer

The Taylor series for \( \sin x \) at \( a = \frac{\pi}{6} \) up to \((x-a)^3\) 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: Recall the Formula for Taylor Series

The Taylor series for a function \( f(x) \) about a point \( a \) is given by \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \). Our task is to find the series up to the term \( (x-a)^3 \).

Step 2: Find Derivatives of \(\sin x\)

The first few derivatives of \( \sin x \) are: - \( f(x) = \sin x \)- \( f'(x) = \cos x \)- \( f''(x) = -\sin x \)- \( f'''(x) = -\cos x \)

Step 3: Evaluate Derivatives at \( a = \frac{\pi}{6} \)

Evaluate each derivative at \( x = \frac{\pi}{6} \):- \( f(a) = \sin(\frac{\pi}{6}) = \frac{1}{2} \)- \( f'(a) = \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \)- \( f''(a) = -\sin(\frac{\pi}{6}) = -\frac{1}{2} \)- \( f'''(a) = -\cos(\frac{\pi}{6}) = -\frac{\sqrt{3}}{2} \)

Step 4: Substitute in Taylor Series Formula

Substitute these values into the Taylor series formula:\[\sin x \approx \frac{1}{2} + \frac{\sqrt{3}}{2}(x - \frac{\pi}{6}) - \frac{1}{2}\frac{(x - \frac{\pi}{6})^2}{2} - \frac{\sqrt{3}}{2}\frac{(x - \frac{\pi}{6})^3}{6}\]

Step 5: Simplify the Expression

Simplify the expression to obtain:\[\sin x \approx \frac{1}{2} + \frac{\sqrt{3}}{2}(x - \frac{\pi}{6}) - \frac{(x - \frac{\pi}{6})^2}{4} - \frac{\sqrt{3}}{12}(x - \frac{\pi}{6})^3\]

Key Concepts

Understanding Derivatives

Derivatives are a fundamental concept in calculus that help us understand how a function changes. Essentially, a derivative represents the rate at which a function is changing at any given point. For example, when working with the sine function, derivatives allow us to examine changes in the wave form at specific points.

It's interesting to note how the derivatives of the sine function are cyclical. The derivative of \( \sin x \) is \( \cos x \). Continuing this process, the derivative of \( \cos x \) is \( -\sin x \), and the derivative of \( -\sin x \) is \( -\cos x \). This cycle repeats, making the derivatives predictable.

When computing a Taylor series, you need these derivatives. Each derivative provides a piece of the puzzle to construct the series around a specific point. This helps make accurate approximations of the function behavior around that point.

Exploring the Sin Function

The sine function, denoted as \( \sin x \), is one of the fundamental trigonometric functions. It maps an angle to the y-coordinate of the corresponding point on the unit circle. This function has a range that spans from -1 to 1, creating its characteristic wave-like pattern.

The sine function is periodic, meaning it repeats values at regular intervals. The period for \( \sin x \) is \( 2\pi \), which is the length of the interval over which the function completes one full cycle. This property is important when analyzing the function or predicting its behavior.

When approached using Taylor series, the sine function is expanded using derivatives calculated at a specific point, called the point of expansion. This allows a polynomial approximation of the sine function that can be essential in simplifying complex calculations.

Point of Expansion in Taylor Series

The point of expansion, or expansion point, is a crucial part of constructing a Taylor series. This is the point \( a \) around which the function is approximated. The Taylor series expansion uses this point to determine the function values and derivative values needed for the approximation.

In the context of the problem, the point of expansion is \( a = \frac{\pi}{6} \). At this point, the sine function and its derivatives have been calculated, forming the basis of constructing the series up to the \( (x-a)^3 \) term.

Choosing the right expansion point helps provide more accurate approximations near that point. If a different point of expansion was chosen, the values of derivatives and the resulting Taylor series would differ, altering the closeness of the approximation to the original function near that new point.