Problem 36 - Thomas Calculus

Question

Find the first three nonzero terms of the Maclaurin series for each function and the values of \(x\) for which the series converges absolutely. \(f(x)=\left(1-x+x^{2}\right) e^{x}\)

Step-by-Step Solution

Answer

The first three nonzero terms are \(1 + 2x - \frac{1}{2}x^2\) and it converges absolutely for all real \(x\).

Step 1: Understand the Maclaurin Series

The Maclaurin series for a function is a Taylor series expansion of the function about 0. It is given by \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \] where \( f^{(n)}(0) \) is the nth derivative of \( f(x) \) evaluated at \( x = 0 \). We need to find the first three nonzero terms.

Step 2: Find the Derivatives

Let's find derivatives of \( f(x) = (1-x+x^2) e^x \). First, notice it's a product of \( u(x) = 1-x+x^2 \) and \( v(x) = e^x \).1. \( f'(x) = (u'(x)v(x) + u(x)v'(x)) = (e^x - xe^x + 2xe^x + (1-x+x^2)e^x)\) Simplify: \(f'(x) = (2-x+x^2)e^x\) 2. Find \(f''(x)\) similarly using the product rule.

Step 3: Evaluate Derivatives at x=0

Now, evaluate the derivatives at \( x = 0 \).1. \( f(0) = (1-0+0^2)e^0 = 1 \) 2. From previous derivative \( f'(x) = (2-x+x^2)e^x \), evaluate \( f'(0) = (2-0+0^2)e^0 = 2 \) 3. Find \( f''(0) \) using \( f''(x) = (-1 + 4x - x^2 + x^3)e^x \) evaluated at \( x = 0 \) which simplifies to \( -1 \).

Step 4: Write the First Three Terms

Using \(f(0)\), \(f'(0)\), and \(f''(0)\), write the first three nonzero terms of the Maclaurin series:1. First term: \( f(0) = 1 \)2. Second term: \( f'(0)x = 2x \) 3. Third term: \( \frac{f''(0)}{2!}x^2 = -\frac{1}{2}x^2 \)So, the series starts as \( 1 + 2x - \frac{1}{2}x^2 + \cdots \)

Step 5: Determine the Interval of Convergence

The original function \((1-x+x^2)e^x\) involves \(e^x\) which has a Maclaurin series that converges for all \(x\). The polynomial term \((1-x+x^2)\) is continuous everywhere, hence the series \(1 + 2x - \frac{1}{2}x^2 + \cdots \) converges absolutely for all real \(x\).

Key Concepts

Taylor series

The Taylor series is a method for representing functions as infinite sums of terms calculated from the values of their derivatives at a single point. When the point of expansion is zero, as in the case of a Maclaurin series, it gives us a simple and powerful tool to express functions. This series takes the form:

• Constant term: \( f(0) \)
• Linear term: \( f'(0)x \)
• Quadratic term: \( \frac{f''(0)}{2!}x^2 \)
• Cubic term: \( \frac{f'''(0)}{3!}x^3 \) ... and so on.

What makes the Taylor series so useful is its ability to approximate complicated functions with simple polynomials near the expansion point. The more terms we include, the better the approximation generally becomes.

derivatives

Derivatives play a crucial role in forming a Taylor series. A derivative of a function measures how the function changes as its input changes, and is denoted as \( f'(x) \) for the first derivative, \( f''(x) \) for the second derivative, and so on.
In the exercise, we have \( f(x) = (1-x+x^2) e^x \). By taking successive derivatives at \( x = 0 \), we find:

• \( f(0) = 1 \)
• \( f'(0) = 2 \)
• \( f''(0) = -1 \)

These help us obtain the Maclaurin series: \( 1 + 2x - \frac{1}{2}x^2 + \cdots \). As derivatives are central to understanding how a function behaves locally, they are equally essential in constructing its series representation.

interval of convergence

The interval of convergence is important for determining where a Taylor series is a valid representation of a function. It defines where the series will accurately approximate the function and is found by examining the behavior of the series as more terms are added.
In simpler terms, it tells us for which values of \( x \) the series converges.

• A function like \( e^x \) has a Maclaurin series that converges for all \( x \).
• For the provided series \( 1 + 2x - \frac{1}{2}x^2 + \cdots \), it can be shown that the series converges for all real \( x \).

This effective convergence is crucial for using series as approximations in a wide range of practical applications.

absolute convergence

Absolute convergence is a stronger form of convergence for series, where a series converges absolutely if the series formed by taking the absolute value of each term also converges. This type of convergence ensures that reordering the terms does not affect the sum of the series.
For our given series, the product of \( (1-x+x^2) \) and \( e^x \) characterizes convergence:

• Since \( e^x \) is known to converge for all real \( x \), and \( (1-x+x^2) \) is a polynomial, we can infer absolute convergence for all values of \( x \).

Such characteristics make the series not just a mathematical construct but a reliable tool for calculations across various domains of mathematics and physics.