Problem 25 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Find the Taylor series at \(x=0\) for the given function, either by using the definition or by manipulating a known series. $$ f(x)=\ln \left(\frac{x+1}{2 x+1}\right) $$

Step-by-Step Solution

Answer

The Taylor series at \( x=0 \) for \( \ln( \frac{x+1}{2x+1} ) \) is \( -x + \frac{3x^2}{2} - \frac{7x^3}{3} + \cdots \)

Step 1: Identify the Function and Expansion Point

We need to find the Taylor series for the function \( f(x)=\ln \left( \frac{x+1}{2x+1} \right) \) around \( x=0 \).

Step 2: Simplify the Function Using Logarithm Properties

Break down the function using the properties of logarithms. This helps in manipulating the function into a more suitable form: \[ f(x)=\ln \left( \frac{x+1}{2x+1} \right) = \ln(x+1) - \ln(2x+1) \]

Step 3: Find the Taylor Series for Individual Components

Find the Taylor series for each individual logarithmic function at \( x=0 \). For \( \ln(x+1) \), we use the known series: \[ \ln(x+1) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \] For \( \ln(2x+1) \), we can use the substitution \( u = 2x \) and expand around \( u = 0 \): \[ \ln(2x+1) = \ln(1+2x) = 2x - \frac{(2x)^2}{2} + \frac{(2x)^3}{3} - \cdots \] or \[ \ln(2x+1) = 2x - 2x^2 + \frac{8x^3}{3} - \cdots \]

Step 4: Subtract the Series

Subtract the series for \( \ln(2x+1) \) from the series for \( \ln(x+1) \): \[ f(x) = (x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots) - (2x - 2x^2 + \frac{8x^3}{3} - \cdots) \] \[ = x - 2x - \frac{x^2}{2} + 2x^2 + \frac{x^3}{3} - \frac{8x^3}{3} + \cdots \]

Step 5: Combine Like Terms

Combine like terms to simplify further: \[ f(x) = -x + \frac{3x^2}{2} - \frac{7x^3}{3} + \cdots \]

Key Concepts

Taylor Series

The Taylor series is a way to represent functions as infinite sums of terms. Each term is derived from the function’s derivatives at a single point. For this exercise, the Taylor series is centered at \(x=0\), also known as a Maclaurin series. If a function \(f(x)\) has a Taylor series expansion, it can be written as: f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \frac{f''''(0)x^4}{4!} + \text{...} Within this series, \(f'(0)\) is the first derivative of \(f(x)\) evaluated at \(x=0\), \(f''(0)\) is the second derivative evaluated at \(x=0\), and so forth. This process allows us to approximate complex functions using a series of simpler polynomial terms.

Logarithmic Properties

Logarithms have several key properties that can simplify complex expressions. One such property is the logarithm of a fraction, given by: \ln \frac{a}{b} = \ln(a) - \ln(b). This is incredibly useful, as it allows us to decompose the expression \(\ln \left(\frac{x+1}{2x+1}\right)\) into two separate logarithmic functions: \(\ln(x+1)\) and \(\ln(2x+1)\). This simplification is pivotal when seeking a Taylor series expansion since each term can be dealt with independently before combining them. These transformations help render the problem more manageable and structured.

Series Expansion

The series expansion technique is powerful in finding the Taylor series of functions. For the function \(\ln(x+1)\), the series expansion around \(x=0\) (or Maclaurin series) is well-known: \ln(x+1) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots. Each term in this series is derived from higher-order derivatives of \(\ln(x+1)\) evaluated at \(x=0\). For the function \(\ln(2x+1)\), a substitution \(u=2x\) transforms it, allowing us to use the series expansion: \ln(2x+1) = 2x - 2x^2 + \frac{8x^3}{3} - \cdots. This substitution simplifies the problem, making it easier to apply existing series formulas.

Calculus

Calculus plays a crucial role in understanding and deriving Taylor series expansions. By taking derivatives and evaluating them at specific points, we can construct infinite series that approximate functions. This process involves:

• Finding the derivative of the function repetitively.
• Evaluating these derivatives at \(x=0\).
• Formulating these coefficients into a polynomial series.

For logarithmic functions, this systematic approach allows us to break down complex expressions into their Taylor series components. These components are then combined, leading to the full series expansion that effectively captures the behavior of the original function near the point of expansion. Learning to navigate these steps fluidly through the lens of calculus deepens our understanding and equips us with tools to tackle advanced mathematical problems.