Problem 13 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The Taylor series at \( x=0 \) for \( x e^x \) is \( \sum_{n=1}^{\infty} \frac{x^n}{(n-1)!} \).

Step 1: Review known Taylor series

Recall the Taylor series expansion for the exponential function: The Taylor series for \( e^x \) at \( x=0 \) is given by: \[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \].

Step 2: Formulate the Taylor series for xe^x

To find the Taylor series for \( x e^x \), we multiply each term of the Taylor 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 3: Write the generalized term

From Step 2, we can see the pattern for the series. Simplifying further, we generalize the expression as: \[ x e^x = \sum_{n=1}^{\infty} \frac{x^n}{(n-1)!} \]. Notice that the start index is \( n=1 \) because \( x^0 = 1 \) affects only the constant term in the original series.

Key Concepts

Power Series

A power series is a series of the form \( \sum_{n=0}^{\infty} c_n (x-a)^n \), where \( c_n \) are coefficients and \( a \) is a constant. This infinite series can represent a wide variety of functions around a point \( a \). By using power series, we can express complex functions as polynomials, which makes them easier to manipulate and solve.

• The terms are powers of \( (x-a) \), expanding from a central point \( a \).
• Power series can converge to a function within a specific interval, known as its radius of convergence.

For example, the exponential function can be expressed as a power series centered at zero. This is particularly useful in calculus as it lets us differentiate and integrate the function term by term, maintaining the simplicity of working with polynomials.

Exponential Function

The exponential function, denoted as \( e^x \), is fundamental in mathematics, appearing frequently in calculus, differential equations, and beyond. Its unique property is that it is its own derivative, making it extremely useful in mathematical analysis.

• The function \( e^x \) can be expressed as an infinite series: \[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \]
• This series helps us compute approximate values of \( e^x \), especially when \( x \) is small, by using only the first few terms.

When dealing with functions like \( x e^x \), we can expand the exponential function as a series and then apply algebraic operations like multiplication to find the series expansion for the new function. This is demonstrated in step-by-step solutions, where multiplying the exponential series by \( x \) gives the Taylor series for \( x e^x \).

Series Expansion

Series expansion is a mathematical technique that represents functions as the sum of infinite terms derived from specific series. By breaking down a function into a sum of polynomial terms, it becomes easier to analyze and compute.

• In the context of Taylor or Maclaurin series, series expansion approximates a function near a point, making calculus operations more straightforward.
• The process involves calculating derivatives of the function and evaluating them at a specific point, typically zero for a Maclaurin series.

In this particular exercise, the concept of series expansion is applied to find the Taylor series for \( x e^x \). By multiplying each term of the series expansion of \( e^x \) by \( x \), we conveniently obtain the series expansion for \( x e^x \). Observing the pattern, we can generalize it to a formula that begins from \( n=1 \), reflecting the fact that the series starts without a constant term.