Problem 14 - Thomas Calculus

Question

Use power series operations to find the Taylor series at \(x=0\) for the functions in Exercises \(13-30\) . $$x^{2} \sin x$$

Step-by-Step Solution

Answer

The Taylor series for \(x^2 \sin x\) is \(x^3 - \frac{x^5}{3!} + \frac{x^7}{5!} - \frac{x^9}{7!} + \cdots\).

Step 1: Identify the Taylor Series for Base Functions

The Taylor series for \( \sin x \) around \( x=0 \) is known to be \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \). Our goal is to find the Taylor Series of the function \(x^2 \sin x\).

Step 2: Multiply by the Power Series of \(x^2\)

To find the Taylor series for \(x^2 \sin x\), multiply each term of the Taylor series of \(\sin x\) by \(x^2\). This yields:\[x^2 \sin x = x^2 \left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \right)\]

Step 3: Expand Each Term

Multiply \(x^2\) by each term of the series:\[x^3 - \frac{x^5}{3!} + \frac{x^7}{5!} - \frac{x^9}{7!} + \cdots\]This represents the Taylor series for \(x^2 \sin x\).

Step 4: Write Down the General Form

From the Pattern Observed, the general term is:\[(-1)^n \frac{x^{2n+3}}{(2n+1)!}\]for \( n = 0, 1, 2, \ldots \) which is derived directly from the series expansion terms.

Key Concepts

Power Series

A power series is a type of infinite series that represents functions as sums of powers of a variable. In simple terms, it is an infinite polynomial. The series takes the form: \[ \sum_{n=0}^{\infty} a_n(x-c)^n \]where \(a_n\) are coefficients and \(c\) is the center of the series. For the function \(x^2 \sin x\), our center is \(x=0\), making it a Maclaurin series, which is just a specific type of Taylor series centered at zero.Power series are quite versatile and beneficial because they:

• Help in approximating complex functions.
• Provide a natural extension of polynomial concepts.
• Can be differentiated and integrated term by term.

These properties make power series a powerful tool in calculus for function analysis and approximation.

Sine Function

The sine function, usually denoted as \( \sin x \), is a fundamental trigonometric function. It provides the y-coordinate of the point where a line from the origin at an angle \(x\) meets the unit circle. One of the fascinating aspects of the sine function is its periodicity—it repeats itself every \(2\pi\) units.For calculus and series, the sine function is known for simplifying into a power series representation:\[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \]This series allows us to express \( \sin x \) as an infinite polynomial, where each successive term involves higher powers of \(x\). This form is particularly useful:

• In approximating \( \sin x \) for values of \(x\) close to zero.
• In applications like physics where sinusoidal functions appear frequently.

Understanding the series expansion of the sine function helps in a broad array of mathematical applications.

Series Expansion

Series expansion is the process of representing a function as a sum of almost infinite simpler terms, usually centered around a specific point like \(x=0\). The Taylor series, specifically, expands a function into an infinite series of terms calculated from the function's derivatives at a single point.When we deal with functions like \(x^2 \sin x\), series expansion often involves a few steps:

• Identifying the known series for base functions, such as \( \sin x \) or \( e^x \).
• Modifying or multiplying these series to fit the desired function, here involving \(x^2\).
• Combining these steps effectively to generate a valid series for complex expressions.

By regularly practicing series expansions, solving calculations, and understanding patterns, we can simplify complex function approximations significantly.

General Term

In the context of power and Taylor series, the general term describes the pattern governing each term's structure in the series. For the function \(x^2 \sin x\), determining its general term involves observing the multiplication and pattern of the sine series terms by \(x^2\).From the exercise, the general term is formulated as:\[ (-1)^n \frac{x^{2n+3}}{(2n+1)!} \]This pattern arises from:

• Recognizing the alternating sign \((-1)^n\) which matches the sine function's alternating property.
• The exponent of \(x\), which grows by two plus the initial exponent \(3\) as \(n\) increases.
• The factorial component \((2n+1)!\) derived from the series terms of \(\sin x\).

Understanding the general term helps in not only writing the entire series clearly but also in computations regarding convergence and approximation characteristics of the series.