Problem 25 - Calculus Single Variable

Question

In each of Exercises 23-34, derive the Maclaurin series of the given function \(f(x)\) by using a known Maclaurin series. $$ f(x)=x^{2} \sin (x / 2) $$

Step-by-Step Solution

Answer

The Maclaurin series for \(f(x) = x^2 \sin(x/2)\) is \(\frac{x^3}{2} - \frac{x^5}{48} + \frac{x^7}{3840} - \cdots\).

Step 1: Recall the Maclaurin series for the known function

The Maclaurin series for \( \sin(x) \) is given by \( \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \).

Step 2: Substitute \(x/2\) into the series

Since we have \(\sin(x/2)\), substitute \(x/2\) into the Maclaurin series for \(\sin(x)\): \[ \sin(x/2) = \frac{x}{2} - \frac{(x/2)^3}{3!} + \frac{(x/2)^5}{5!} - \frac{(x/2)^7}{7!} + \cdots \].Simplifying each term of the series gives:\[ \sin(x/2) = \frac{x}{2} - \frac{x^3}{48} + \frac{x^5}{3840} - \frac{x^7}{645120} + \cdots \].

Step 3: Multiply by \(x^2\) to get \(f(x)\)

Multiply each term of the series for \(\sin(x/2)\) by \(x^2\) to find the Maclaurin series for \(f(x) = x^2 \sin(x/2)\):\[ f(x) = x^2 \left( \frac{x}{2} - \frac{x^3}{48} + \frac{x^5}{3840} - \cdots \right) = \frac{x^3}{2} - \frac{x^5}{48} + \frac{x^7}{3840} - \cdots \].

Key Concepts

Function Series Expansion

Function series expansion is a powerful calculus tool, helping us represent complex functions as infinite sums of simpler terms.
For functions that have derivatives of all orders, such as polynomials, exponentials, or trigonometric functions, we can express them as power series. The Maclaurin series, a specific type of Taylor series, allows us to express functions as expanded series centered at zero.
In general, the Maclaurin series of a function \( f(x) \) can be written as:

• \( f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots \).

This approach is particularly useful because it transforms difficult functions into series where we can more easily compute values, derivatives, and integrals.
In the given exercise, the series expansion is employed to express the function \( f(x) = x^2 \sin(x/2) \) utilizing the known Maclaurin series expansion of \( \sin(x) \).
By manipulating and expanding \( \sin(x/2) \), we simplify the series to reveal how the complex function can be written as the sum of its simpler components.

Trigonometric Function Series

Trigonometric functions often appear in calculus and can be tricky to handle in their original forms. Functions like sine and cosine can be expanded using series, enabling transformations and simplifications.
The series for \( \sin(x) \) is a well-known trigonometric series represented as:

• \( \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \).

This series is derived through the concept of derivatives, considering the odd powers of \( x \) because sine is an odd function.In the exercise, substituting \( x/2 \) into this series, we transformed the original trigonometric function into:

• \( \sin(x/2) = \frac{x}{2} - \frac{(x/2)^3}{3!} + \frac{(x/2)^5}{5!} - \cdots \).

Our task was to reframe \( \sin(x/2) \) to leverage it in the function \( f(x) \).
This approach highlights how the manipulation of a known trigonometric series can simplify expressing other related trigonometric functions and aid in complex calculus problems.

Calculus Problem Solving

In calculus, solving problems through function series expansion is a remarkable skill. It offers insight into the stability and behavior of a function across different values of \( x \). Using a series expansion involves strategic substitution and manipulation of known series. This approach is not only practical but also elegant.The process begins by identifying a known series that closely resembles the function in question.
In this exercise, we utilized sin(x)'s Maclaurin series as our foundation.

• Step 1: We substituted to modify for relevant variables (\( x/2 \) in place of \( x \)) and adjusted each term accordingly.
• Step 2: Rewriting the series by introducing variables, we tailored it to the specific function \( f(x) \).
• Step 3: We expanded and simplified by multiplying the overall series by another term (\( x^2 \) in this case), adapting to fit the function's entire structure.

By systematically organizing and executing these steps, we unraveled a complex function into a manageable series.
This process not only provides the solution but also deepens understanding of function interactions and the beauty within mathematical problem-solving.