Problem 12 - Thomas Calculus

Question

Find the Maclaurin series for the functions \(x e^{x}\)

Step-by-Step Solution

Answer

The Maclaurin series for \( x e^x \) is \( \sum_{n=1}^{\infty} \frac{x^n}{(n-1)!} \).

Step 1: Know the Maclaurin Series Formula

The Maclaurin series for a function \( f(x) \) is given by \( f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots \). This is a specific case of the Taylor series centered at \( x = 0 \). We'll use this formula to find the series for \( x e^x \).

Step 2: Understand the Exponential Function Series

Recall the Maclaurin series for \( e^x \), which is \( e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \). We'll use this series to help determine the series for \( x e^x \).

Step 3: Multiply the Known Series by \( x \)

To find the series for \( x e^x \), multiply each term of the Maclaurin series for \( e^x \) by \( x \): \[ x e^x = x \left(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \right) = x + x^2 + \frac{x^3}{2!} + \frac{x^4}{3!} + \cdots \].

Step 4: Write the Resulting Maclaurin Series

The resulting Maclaurin series for \( x e^x \) is: \[ x e^x = x + x^2 + \frac{x^3}{2} + \frac{x^4}{6} + \cdots \]. This corresponds to the general series \( \sum_{n=1}^{\infty} \frac{x^{n}}{(n-1)!} \).

Key Concepts

Taylor series

The Taylor series is an incredibly useful mathematical tool. It allows us to express functions as infinite sums of terms calculated from their derivatives at a single point. The general form of a Taylor series for a function \( f(x) \) centered around a point \( a \) is given by:\[f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)(x - a)^2}{2!} + \frac{f'''(a)(x - a)^3}{3!} + \cdots\]When the series is centered at \( a = 0 \), the Taylor series becomes a Maclaurin series:\[f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots\]In simpler terms, the Taylor series is like a mathematical microscope. It reveals the behavior of a function near a specific point by examining its derivatives.

• Each term in the series is derived from the function's derivatives at the center point.
• As more terms are added, the series provides a more accurate representation of the function.

This makes Taylor series particularly powerful for approximating complex functions with polynomials.

Exponential function

The exponential function \( e^x \) stands out because of its unique properties and frequent occurrence in mathematics, science, and engineering. It is defined as:\[e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\]This formula expresses \( e^x \) as an infinite sum, showcasing its continuous and smooth nature. The exponential function has a special characteristic: it is its own derivative. This property means that the slope of \( e^x \) is the same as its value at any point, making it exponential.

• The exponential function grows rapidly and is crucial for modeling growth processes, like population growth or radioactive decay.
• In mathematics, it's a cornerstone for continuous compounding in finance and solutions to differential equations.

Additionally, within the context of series expansion, \( e^x \) typically serves as a starting point for generating related series via multiplication or other transformations, as seen with \( x e^x \).

Series expansion

Series expansion is a method used to express complex functions as sum of simpler terms. Each term often increases in degree, making it easier to analyze the function's behavior incrementally. Series expansions are especially useful because:

• They convert a function into a form that is often simpler to compute, manipulate, and analyze.
• For functions like polynomials, they provide approximations that can be as accurate as needed by considering more terms.

For instance, in the problem involving \( x e^x \), we start with the known series for \( e^x \) and expand it to get:\[e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\]By multiplying each term in this series by \( x \), we find the series expansion for \( x e^x \):\[x e^x = x + x^2 + \frac{x^3}{2!} + \frac{x^4}{3!} + \cdots\]This method effectively uses the principle of series expansion to transform the original exponential function into a form suitable for various applications, such as solving differential equations or evaluating integrals in a more straightforward manner.