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Chapter 9: Problem 12

Find the Maclaurin serics for the functions in Exercises \(11-24\) $$x e^{x}$$

Short Answer

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

Step by step solution

01

Understand the Maclaurin Series

The Maclaurin series is a special case of the Taylor series where the function is expanded around 0. It is given by the formula:\(f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots\)
02

Determine Function and Derivatives

We need to find the derivatives of \(f(x) = x e^{x}\). The first few derivatives are:- \(f(x) = x e^{x}\)- \(f'(x) = e^{x} + x e^{x}\)- \(f''(x) = 2e^{x} + x e^{x}\)- \(f'''(x) = 3e^{x} + x e^{x}\)This pattern continues, where the n-th derivative is \((n e^{x} + x e^{x})\).
03

Evaluate Derivatives at x=0

Evaluate the derivatives from step 2 at \(x = 0\):- \(f(0) = 0\)- \(f'(0) = 1\)- \(f''(0) = 2\)- \(f'''(0) = 3\)- In general, \(f^{(n)}(0) = n\) for \(n \geq 1\).
04

Apply Maclaurin Series Formula

Substitute the derivative values at \(x=0\) into the Maclaurin series formula:\[f(x) = 0 + 1 \cdot x + \frac{2}{2!} x^2 + \frac{3}{3!} x^3 + \ldots\]This simplifies to:\[f(x) = x + \frac{1}{1} x^2 + \frac{1}{2} x^3 + \ldots\]
05

Write General Term

Notice the pattern of the coefficients closely follows that of the exponential series. In general, the n-th term of \(f(x)\) can be written as:\[\frac{n}{n!} x^n\]Thus, the Maclaurin series expansion of \(x e^{x}\) is:\[\sum_{n=1}^{\infty} \frac{n}{n!} x^n\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Taylor Series
The Taylor series is a fundamental concept in calculus that allows us to represent functions as infinite sums of their derivatives evaluated at a single point. This concept is especially useful because it lets us approximate complex functions with simple polynomial terms. Think of it as breaking down a complicated function into a series of manageable parts.
  • The Taylor series for a function, \(f(x)\), expanded around a point \(a\), is given by: \[ f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \ldots \]
  • Here, \(f'(a), f''(a),\) and higher derivatives are the derivatives of \(f(x)\) evaluated at \(a\).
  • The Maclaurin series is a special type of Taylor series centered at \(a = 0\).
This series provides a powerful tool for analyzing functions that are otherwise difficult to handle, especially in engineering and physical sciences.
Derivatives
Derivatives are the backbone of calculus and play a crucial role in the Taylor and Maclaurin series. They describe how a function changes at any given point, essentially giving us the function's slope at that point.
  • Finding derivatives of a function involves applying rules like the power rule, product rule, and chain rule.
  • For a function \(f(x) = x e^{x}\), derivatives help us explore the function's behavior as it is expanded into a series.
  • Derivatives at zero, \(f^{(n)}(0)\), are specifically used in Maclaurin series.
In the case of \(f(x) = x e^{x}\), each derivative follows a pattern that becomes instrumental for writing the series expansion.
Exponential Functions
Exponential functions, like \(e^x\), are a particular type of mathematical function where a constant base is raised to a variable exponent. These functions grow very quickly and are immensely important in sciences, economics, and engineering.
  • The function \(e^x\) is unique because its derivative is itself, \(\frac{d}{dx}e^x = e^x\).
  • This characteristic simplifies series expansions, as often seen in exponential growth and decay models.
  • When combined with polynomials, as in \(x e^x\), the resulting function can represent more complex phenomena.
Understanding exponential functions is essential for exploring their representation in series, revealing their underlying simplicity and depth.
Series Expansion
Series expansion is a technique used to express a function as an infinite sum, making it easier to analyze and approximate. It helps transform complex functions into simpler polynomial forms.
  • The success of this method hinges on recognizing patterns in the terms and coefficients that appear in the expansion.
  • In Maclaurin series, each coefficient is determined from derivatives of the function evaluated at zero.
  • For the specific function \(x e^x\), the series expansion is given by: \[ \sum_{n=1}^{\infty} \frac{n}{n!} x^n \]
This expansion captures the essence of the original function in a form that can be easily manipulated, serving as a robust tool for continuous mathematical and computational exploration.

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Most popular questions from this chapter

The Taylor polynomial of order 2 generated by a twice-differentiable function \(f(x)\) at \(x=a\) is called the quadratic approximation of \(f\) at \(x=a\). In Exercises \(45-50,\) find the (a) linearization (Taylor polynomial of order 1 ) and (b) quadratic approximation of \(f\) at \(x=0.\) $$f(x)=\tan x$$

Prove that if \(\sum a_{n}\) converges absolutely, then \(\sum a_{n}^{2}\) converges.

Use any method to determine whether the series converges or diverges. Give reasons for your answer. $$\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\cdots$$

Show that neither the Ratio Test nor the Root Test provides information about the convergence of $$\sum_{n=2}^{\infty} \frac{1}{(\ln n)^{p}} \quad(p \text { constant })$$

The Taylor polynomial of order 2 generated by a twice-differentiable function \(f(x)\) at \(x=a\) is called the quadratic approximation of \(f\) at \(x=a\). In Exercises \(45-50,\) find the (a) linearization (Taylor polynomial of order 1 ) and (b) quadratic approximation of \(f\) at \(x=0.\) $$f(x)=\cosh x$$
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