Problem 31 - Essential Calculus

Question

In Exercises 31-34, find the Maclaurin series for the function. (See Example \(7 .)\) $$ f(x)=x \sin x $$

Step-by-Step Solution

Answer

The Maclaurin series for the function \(f(x) = x \sin x\) is \(f(x) = \sum_{m = 1}^{\infty} (-1)^{m+1} \frac{(2m+1)! x^{2m+1}}{(2m+1)!}\).

Step 1: Identify the Function and Rule Needed

The function in the problem is \(f(x) = x \sin x\). The key here is to understand that a Maclaurin series involves the derivatives evaluated at zero. For this problem, we'll use the following rule: \[f^{(n)}(0) = \begin{cases}(-1)^{m+1}m! & \text{for } n = 2m + 1 \text{ (where } m = 0,1,2, \ldots) \0 & \text{for } n = \text{ even number}\end{cases}\]

Step 2: Apply the Maclaurin Series Formula

The standard formula for a Maclaurin series is: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots \] Given this, we need to substitute \(f^{(n)}(0)\) into the formula and implement the rule (from step 1) correctly. Moreover, we know that \(f(0) = 0\), \(f'(0) = 0\), \(f''(0) = 0\), \(f'''(0) = 1\), \(f''''(0) = 0\), \(f'''''(0) = 5\), and so on. So, the specific terms become zero for even numbers \(n\). Lastly, because \(\sin x\) is an odd function, it only has non-zero odd derivatives. We thus get the series: \[ f(x) = 0 + 0x + \frac{0x^2}{2!} + \frac{x^3}{3!} +\frac{0x^4}{4!} + \frac{5x^5}{5!} + \cdots \]

Step 3: Simplify the Series

Removing the terms that equal zero, our series simplifies to: \[ f(x) = x^3/3! + 5x^5/5! + \cdots \] This can be further simplified by noticing the pattern and constructing the generic series formula: \[ f(x) = \sum_{m = 1}^{\infty} (-1)^{m+1} \frac{(2m+1)! x^{2m+1}}{(2m+1)!} \]

Key Concepts

Taylor Series

The Taylor Series is a mathematical tool used to approximate functions by expressing them as an infinite sum of terms calculated from the derivatives of the function. This series is essentially the backbone for the more specific Maclaurin Series. The Taylor Series expands a function around a certain point, usually denoted by some constant such as \(a\). It has the general form:\[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \frac{f'''(a)(x-a)^3}{3!} + \cdots\]

• Each term's formula involves the \(n^{th}\) derivative of the function at point \(a\).
• It's designed to give an approximation that gets better with more terms.
• Special case: if \(a = 0\), it becomes the Maclaurin Series.

The Taylor Series is useful in calculus when you need to estimate a function near a certain point. It helps engineers and scientists model real-world phenomena by using polynomial approximations.

Series Expansion

Series Expansion involves expressing a function as a sum of components, which can reveal intricate details about the function's behavior. This concept is central in calculus and helps with understanding and simplifying complex functions. Two of the most common series expansions in calculus are the Taylor Series and the Maclaurin Series.When expanding functions into a series:

• Identify the function and appropriate expansions, like the Maclaurin or Taylor Series.
• Determine the derivatives required for the expansion, especially those that are zero which simplify the process.
• Ensure you recognize patterns, as they allow the creation of a simplified, generic form of the series.

In the provided Maclaurin series example for \(f(x) = x \sin x\), we discover a pattern: non-zero terms appear for odd powers of \(x\). By insightfully identifying these patterns, students can manage complex expansions efficiently with less computational effort.

Calculus

Calculus, a fundamental branch of mathematics, equips learners with tools to handle change and motion. The primary constructs are derivatives and integrals.

• A derivative signifies a function's rate of change, fundamental for Taylor and Maclaurin Series expansions where higher derivatives (at a specific point) convey how the function behaves locally.
• Calculus enables solving problems that entail rates (velocity, acceleration) or areas under curves.

The specific function \(f(x) = x \sin x\) in the Maclaurin Series example uses the derivative concept extensively. This function's series is constructed by evaluating multiple derivatives of \(f(x)\) at the point zero and combining them to form the series. Understanding derivatives in calculus is crucial, as they form the basis of more advanced mathematical constructs and applications in physics, engineering, and beyond.