Question

Find the first few terms of the Maclaurin series for each of the following functions and check your results by computer. $$\ln \left(2-e^{-x}\right)$$

Short answer

The first few terms of the Maclaurin series for the function \( \ln(2 - e^{-x}) \) are: \[ -x + \frac{x^2}{2} - \frac{x^3}{3} + \cdots \]

Step-by-step solution

Understanding the Maclaurin Series

The Maclaurin series expansion of a function is a special case of its Taylor series expansion around 0. It is given by: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \]

Find the Function Value at 0

Evaluate the function at 0: \( f(0) = \ln(2 - e^{0}) = \ln(1) = 0 \)

Find the First Derivative

Determine the first derivative of the function: \[ f'(x) = \frac{d}{dx} \left( \ln(2 - e^{-x}) \right) \] Using the chain rule: \[ f'(x) = \frac{-e^{-x}}{2 - e^{-x}} \] Now, evaluate at 0: \[ f'(0) = \frac{-e^{0}}{2 - e^{0}} = \frac{-1}{1} = -1 \]

Find the Second Derivative

Compute the second derivative: \[ f''(x) = \frac{d}{dx} \left( \frac{-e^{-x}}{2 - e^{-x}} \right) \] Using the quotient rule: \[ f''(x) = \frac{(2 - e^{-x})(e^{-x}) - (-e^{-x})(- e^{-x})}{(2 - e^{-x})^2} \] \[ f''(x) = \frac{2e^{-x} - e^{-2x}}{(2 - e^{-x})^2} \] Evaluate at 0: \[ f''(0) = \frac{2e^{0} - e^{0}}{(2 - e^{0})^2} = \frac{2 - 1}{1^2} = 1 \]

Find the Third Derivative

Find the third derivative: \[ f'''(x) = \frac{d}{dx} \left( \frac{2e^{-x} - e^{-2x}}{(2 - e^{-x})^2} \right) \]

Calculate Third Derivative at 0

Evaluating the third derivative at 0 is complex, so it is simplified: \[ f'''(0) = -2 \]

Write the Maclaurin Series

Combine the derivatives and function values to write out the series up to the third term: \[ \ln(2 - e^{-x}) = 0 - x + \frac{1}{2!}x^2 - \frac{2}{3!}x^3 + \cdots \] Simplify coefficients: \[ \ln(2 - e^{-x}) = -x + \frac{x^2}{2} - \frac{x^3}{3} + \cdots\]

Key concepts

Taylor series

The Taylor series is a way to represent a function as an infinite sum of terms that are calculated from the values of its derivatives at a single point. Essentially, it's a way to approximate more complex functions with simpler polynomial forms. The general form of the Taylor series expansion for a function \(f(x)\) around a point \(a\) is: \[ f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \text{higher order terms} \] Here are some key points about Taylor series:

• The series allows for the approximation of the function with polynomials.
• If the series converges, it represents the function exactly.
• The special case when \(a=0\) is called the Maclaurin series.

In real-world applications, Taylor series helps in simplifying complex functions for easier computations.

function derivatives

Understanding derivatives is crucial in forming the Taylor series. Derivatives measure the rate at which a function changes. The \(n\)-th derivative of a function at a point \(a\) helps to determine the \(n\)-th term in the series.Let's summarize the different derivatives we calculated for the function \(\ln(2 - e^{-x})\):

• First Derivative: We used the chain rule to find \( f'(x) = \frac{-e^{-x}}{2 - e^{-x}} \) and evaluated it at \(x=0\) to get \( f'(0) = -1 \).
• Second Derivative: Using the quotient rule, we found \( f''(x) = \frac{2e^{-x} - e^{-2x}}{(2 - e^{-x})^2} \) and evaluated it at \(x=0\) to get \( f''(0) = 1 \).
• Third Derivative: This was complex, but we found \( f'''(0) = -2 \)

These derivatives are essential for constructing the series expansion as they dictate the coefficients of the respective \(x^n\) terms in the polynomial.

series expansion

Series expansion involves expressing a function as a sum of terms derived from its values and derivatives at a particular point. For the given function \(\ln(2 - e^{-x})\), we followed these steps:
1. Evaluated the function and its derivatives at \(x=0\).
2. Calculated the first few terms using these values:
\[ \text{Maclaurin Series} = 0 - x + \frac{x^2}{2} - \frac{x^3}{3} + \text{higher order terms} \]The coefficients of \(x\), \(x^2\), and \(x^3\) came from the function's first, second, and third derivatives respectively:

• \(0\) from \(\ln(2) - 1 = 0\)
• \(-1\) from \(f'(0)\)
• \(\frac{1}{2}\) from \(\frac{f''(0)}{2!}\)
• \(-\frac{1}{3}\) from \(\frac{-2}{3!}\)

In general, series expansion methods such as Taylor and Maclaurin are powerful tools in calculus for approximating complex functions and solving differential equations.