Problem 25 - Thomas Calculus

Question

Use power series operations to find the Taylor series at \(x=0\) for the functions in Exercises \(11-28 .\) $$e^{x}+\frac{1}{1+x}$$

Step-by-Step Solution

Answer

The Taylor series at \(x=0\) for \(e^{x} + \frac{1}{1+x}\) is \(2 + \frac{3}{2}x^2 - \frac{5}{6}x^3 + \cdots\).

Step 1: Recall Taylor Series of Standard Functions

First, we need to recall the Taylor series expansions of the standard functions involved, evaluated at \( x = 0 \). The Taylor series for \( e^x \) at \( x = 0 \) is:\[e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \ = \sum_{n=0}^{\infty} \frac{x^n}{n!}.\]Next, the Taylor series for \( \frac{1}{1+x} \) at \( x = 0 \) is:\[\frac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots = \sum_{n=0}^{\infty} (-1)^n x^n.\]

Step 2: Add the Series of Each Function

Now, we need to find the series for the given function by adding the series of each function:\[ e^x + \frac{1}{1+x} = \left( \sum_{n=0}^{\infty} \frac{x^n}{n!} \right) + \left( \sum_{n=0}^{\infty} (-1)^n x^n \right). \]This can be expressed as a single sum:\[ \sum_{n=0}^{\infty} \left( \frac{1}{n!} + (-1)^n \right) x^n. \]

Step 3: Write Out the First Few Terms

To better understand the resulting series, write out the first few terms:- For \( n=0 \): \( \left( \frac{1}{0!} + (-1)^0 \right) x^0 = (1 + 1) = 2 \).- For \( n=1 \): \( \left( \frac{1}{1!} + (-1)^1 \right) x = (1 - 1) x = 0 \).- For \( n=2 \): \( \left( \frac{1}{2!} + (-1)^2 \right) x^2 = \left( \frac{1}{2} + 1 \right) x^2 = \frac{3}{2} x^2 \).- For \( n=3 \): \( \left( \frac{1}{3!} + (-1)^3 \right) x^3 = \left( \frac{1}{6} - 1 \right) x^3 = -\frac{5}{6} x^3 \).Thus, the first few terms of the series are \( 2 + \frac{3}{2} x^2 - \frac{5}{6} x^3 + \cdots \).

Key Concepts

Power Series Operations

Power series operations involve the methods used to manipulate and combine infinite series of powers of a variable, typically represented by the variable \( x \). When dealing with power series, it's essential to understand the basic operations that can be applied:

• Addition and Subtraction: You can add or subtract power series term by term. This is straightforward since each power of \( x \) in one series aligns with the same power of \( x \) in the other series.
• Multiplication: To multiply power series, you use the so-called Cauchy product, which involves taking each term from one series and multiplying it with each term from the other series. This requires careful computation to ensure all terms of the resulting series are accounted for.
• Substitution and Composition: More complex operations involve substituting one series into another or composing series, which can lead to more challenging calculations.

When combining power series, as in the case of \( e^x + \frac{1}{1+x} \), you focus on aligning like terms and ensuring the operations respect the rules of power series addition.

e^x Expansion

The expansion of \( e^x \) into a Taylor series is one of the most fundamental concepts in calculus. At \( x = 0 \), this expansion is derived by evaluating the function and its derivatives at zero and is given as:\[e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots = \sum_{n=0}^{\infty} \frac{x^n}{n!}\]

• Basis: The series is based on the fact that the derivative of \( e^x \) is itself, meaning each derivative of \( e^x \) evaluated at zero equals one.
• Convergence: This series converges for all real numbers \( x \), making it a powerful tool for approximating the exponential function.

The series allows you to understand functions built from \( e^x \) as an infinite series of terms, each scaled by the appropriate factorial, which simplifies many complex calculus problems.

1/(1+x) Expansion

The Taylor expansion for \( \frac{1}{1+x} \) when centered at \( x = 0 \) springs from the geometric series formula for \( |x| < 1 \):\[\frac{1}{1+x} = 1 - x + x^2 - x^3 + \cdots = \sum_{n=0}^{\infty} (-1)^n x^n\]

• Alternating Signs: This series alternates in sign with each successive term, reflecting the characteristics of an alternating series.
• Convergence Radius: It converges for \(|x| < 1\), indicating where the power series accurately depicts the function.

Understanding this series is crucial when looking to expand expressions involving rational functions into power series, which are common in various mathematical and physical applications.

Sum of Series

The sum of two series involves adding each pair of corresponding terms from the two series together. In the given problem, you are tasked with finding the Taylor series for the sum \( e^x + \frac{1}{1+x} \). To sum these series:

• Align Terms: For each power of \( x \), sum the coefficients from both series. This aligns terms like \( x^0, x^1, \) etc.
• Example Terms: As shown in the solution:
  • For \( n=0 \), the sum is \(2\) because \(1 + 1 = 2\).
  • For \( n=1 \), both series cancel out, leaving \(0\).
  • Continuing, you find terms like \(\frac{3}{2} x^2\) and \(-\frac{5}{6} x^3\), calculated by similarly adding respective terms.

This process is essential for understanding how complex functions can be broken down into simpler components, which are manageable and analyzable through their series expressions.