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

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.

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 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.