Problem 16 - Thomas Calculus

Question

Find the Maclaurin series for the functions \(\sin \frac{x}{2}\)

Step-by-Step Solution

Answer

The Maclaurin series for \( \sin \frac{x}{2} \) is \( \frac{1}{2}x - \frac{1}{48}x^3 + \cdots \).

Step 1: Recall the Maclaurin Series Formula

The Maclaurin series is a special case of the Taylor series centered at zero. The formula for a Maclaurin series is \[ f(x) = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \cdots \] where \(f^n(0)\) is the \(n\)-th derivative of \(f(x)\) evaluated at \(x=0\).

Step 2: Define the Function and Its Derivatives

Given the function \( f(x) = \sin \frac{x}{2} \), calculate its derivatives: - First derivative: \( f'(x) = \frac{1}{2} \cos \frac{x}{2} \)- Second derivative: \( f''(x) = -\frac{1}{4} \sin \frac{x}{2} \)- Third derivative: \( f'''(x) = -\frac{1}{8} \cos \frac{x}{2} \)Continue finding higher derivatives as needed.

Step 3: Evaluate Derivatives at x = 0

Evaluate each derivative at \(x = 0\):- \( f(0) = \sin 0 = 0 \)- \( f'(0) = \frac{1}{2} \cos 0 = \frac{1}{2} \)- \( f''(0) = -\frac{1}{4} \sin 0 = 0 \)- \( f'''(0) = -\frac{1}{8} \cos 0 = -\frac{1}{8} \)

Step 4: Build the Maclaurin Series Using Evaluated Derivatives

Now substitute the evaluated derivatives into the Maclaurin series formula:\[ \sin \frac{x}{2} = 0 + \frac{\frac{1}{2}}{1!}x - \frac{\frac{1}{8}}{3!}x^3 + \cdots \]This simplifies to:\[ \sin \frac{x}{2} = \frac{1}{2}x - \frac{1}{48}x^3 + \cdots \]

Step 5: Write the General Term for the Series

Observe the pattern of the terms to derive the general term:The series alternates between positive and negative terms with odd powers of \(x\). The general term can be represented as:\[ \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{((2n+1)!) 2^{2n}} \]

Key Concepts

Taylor series

The Taylor series is a mathematical concept that involves expanding a function into an infinite sum of terms calculated from the values of its derivatives at a single point. The Maclaurin series is a special case of the Taylor series, where the center of expansion is zero.
The general formula for a Taylor series centered at a point \( a \) is:

• \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots \)

In the case of the Maclaurin series, we set \( a = 0 \), leading to a simpler formula:

• \( f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \ldots \)

This series provides a powerful way to approximate functions using polynomials, making them easier to compute and analyze for small values of \( x \).
In practice, few terms are often sufficient for a good approximation, particularly for well-behaved functions like polynomials and certain trigonometric and exponential functions.

trigonometric functions

Trigonometric functions, like sine and cosine, are fundamental in both pure and applied mathematics. They relate the angles of a triangle to the lengths of its sides in a right-angled triangle. In calculus and series expansion, these functions are often expanded into an infinite series to facilitate calculations.
The basic trigonometric functions include:

• Sine: \( \sin x \)
• Cosine: \( \cos x \)
• Tangent: \( \tan x \)

When working with the Maclaurin series, functions such as \( \sin \frac{x}{2} \) can be expanded using their derivatives at zero.
Sine and cosine are periodic functions, with series expansions that involve alternating terms. For instance, the Maclaurin expansion for sine series:

• \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \)

Understanding these expansions is crucial for solving many practical problems in physics and engineering, where trigonometric functions arise naturally.

derivatives

Derivatives are a key concept in calculus, representing the rate of change of a function concerning one of its variables. For a function \( f(x) \), the derivative \( f'(x) \) is defined as the limit:

• \( \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)

In the context of a Taylor or Maclaurin series, derivatives allow us to capture the slope, curvature, and higher-order behaviors of functions. Each derivative evaluated at a point contributes a term to the polynomial approximation.
For \( \sin \frac{x}{2} \), the derivatives are:

• First derivative: \( f'(x) = \frac{1}{2} \cos \frac{x}{2} \)
• Second derivative: \( f''(x) = -\frac{1}{4} \sin \frac{x}{2} \)
• Third derivative: \( f'''(x) = -\frac{1}{8} \cos \frac{x}{2} \)

Each of these derives contributes significantly to our understanding of the function's behavior near the point \( x = 0 \). Calculating and evaluating higher-order derivatives are crucial steps in developing series expansions for more complex functions.

series expansion

Series expansion is the process of expressing a complex function as an infinite sum of simpler terms. These terms often involve powers of a variable \( x \). By converting functions into power series, it becomes possible to simplify complex mathematical operations like integration and differentiation.
In many cases, series expansions offer approximations that are sufficient for practical applications, especially when only a few terms of the series are considered. For the function \( \sin \frac{x}{2} \), the series expansion gives us a polynomial:

• \( \sin \frac{x}{2} \approx \frac{1}{2}x - \frac{1}{48}x^3 + \cdots \)

Notably, the general term for the series is:

• \( \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{((2n+1)!) 2^{2n}} \)

This technique is widely used across physics, engineering, and computer science to address real-world problems where exact solutions are impractical or impossible to find. Through series expansion, we can gain deeper insights into the nature of complex functions.