Problem 8 - Calculus

Question

Find the first four terms of the Taylor series for the function about the point \(a\). $$\sin x, \quad a=\pi / 4$$

Step-by-Step Solution

Answer

The first four terms are \( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}(x - \frac{\pi}{4}) - \frac{\sqrt{2}}{4}(x - \frac{\pi}{4})^2 - \frac{\sqrt{2}}{12}(x - \frac{\pi}{4})^3 \).

Step 1: Understand the Taylor Series Formula

The Taylor series for a function \( f(x) \) about a point \( a \) is given by \( f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \). For \( \sin x \), we need to find the derivatives and evaluate them at \( a = \frac{\pi}{4} \) to determine the first four terms.

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

Calculate the first few derivatives of \( \sin x \):- First derivative: \( f'(x) = \cos x \)- Second derivative: \( f''(x) = -\sin x \)- Third derivative: \( f'''(x) = -\cos x \)- Fourth derivative: \( f''''(x) = \sin x \)

Step 3: Evaluate Derivatives at \( a = \frac{\pi}{4} \)

Evaluate each derivative at \( x = \frac{\pi}{4} \):- \( f(\frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \)- \( f'(\frac{\pi}{4}) = \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \)- \( f''(\frac{\pi}{4}) = -\sin(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2} \)- \( f'''(\frac{\pi}{4}) = -\cos(\frac{\pi}{4}) = -\frac{\sqrt{2}}{2} \)

Step 4: Formulate the Taylor Series

Substitute the values from the previous step into the Taylor series formula:\[ \sin x \approx \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}(x - \frac{\pi}{4}) - \frac{\sqrt{2}}{2 \times 2!}(x - \frac{\pi}{4})^2 - \frac{\sqrt{2}}{2 \times 3!}(x - \frac{\pi}{4})^3 \]Simplify the factorial terms: \( 2! = 2 \) and \( 3! = 6 \).

Step 5: Simplify the Series

Rewrite the series with simplified terms:\[ \sin x \approx \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}(x - \frac{\pi}{4}) - \frac{\sqrt{2}}{4}(x - \frac{\pi}{4})^2 - \frac{\sqrt{2}}{12}(x - \frac{\pi}{4})^3 \]

Key Concepts

Sin Function

The sine function, denoted by \( \sin x \), is one of the fundamental trigonometric functions. It represents the ratio of the length of the opposite side to the hypotenuse in a right triangle. The function is periodic with a period of \( 2\pi \), meaning it repeats its pattern every \( 2\pi \) units. Often used in various fields, particularly in physics and engineering, the sine function helps describe wave patterns such as sound waves, light waves, and other oscillatory phenomena.

In mathematics, the sine function is used beyond simple triangles and finds its place in complex equations, particularly those involving calculus. This importance leads to its frequency in problems involving Taylor series, where an approximation is needed for complex equations.

Derivative Calculation

When working with Taylor series, calculating derivatives is essential. Derivatives are used to understand the rate of change of a function. For the sine function, the derivatives alternate between \( \sin x \) and \( \cos x \).

• The first derivative of \( \sin x \) is \( \cos x \).
• The second derivative is \( -\sin x \).
• The third derivative is \( -\cos x \).
• The fourth derivative returns to \( \sin x \).

This cyclic pattern of derivatives not only aids in calculating Taylor series but also highlights the inherent periodicity of trigonometric functions. Each step in derivative calculations provides an alternate function, reinforcing this result.

Series Approximation

A Taylor series is a powerful tool for approximating functions around a specific point \( a \). The series is an infinite sum of terms derived from the function's derivatives at \( a \). This transforms complex functions into the polynomial form.

For the sine function at point \( a = \frac{\pi}{4} \), the approximation begins with calculating the value of the function and its derivatives at this point. After evaluating the derivatives explained in the previous section, we substitute these values into the Taylor series formula:
\[ \sin x \approx \sin \left( \frac{\pi}{4} \right) + \cos \left( \frac{\pi}{4} \right)(x-\frac{\pi}{4}) + \frac{-\sin \left( \frac{\pi}{4} \right)}{2!}(x-\frac{\pi}{4})^2 + \frac{-\cos \left( \frac{\pi}{4} \right)}{3!}(x-\frac{\pi}{4})^3 \]
Simplifying the factorial expressions, this series provides a manageable way to estimate \( \sin x \) around \( a = \frac{\pi}{4} \).

Evaluation at a Point

Evaluating the Taylor series at a specific point \( a \) requires substituting both the function and its derivatives at that point into the series. For \( \sin x \) at \( \frac{\pi}{4} \), we calculated:

• \( \sin \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \)
• \( \cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \)
• \( -\sin \left( \frac{\pi}{4} \right) = -\frac{\sqrt{2}}{2} \)
• \( -\cos \left( \frac{\pi}{4} \right) = -\frac{\sqrt{2}}{2} \)

These results are then inserted into the Taylor series formula, demonstrating how each term contributes to the approximation. The method of evaluating at a point helps to tailor the series expansion to be as accurate as possible near that particular point \( a \), thereby allowing us to represent the function locally around \( \frac{\pi}{4} \) using polynomial terms.