Problem 4 - Calculus

Question

Find the first four nonzero terms of the Taylor series for the function about 0. $$\ln (1-x)$$

Step-by-Step Solution

Answer

The first three nonzero terms are: \(-x - \frac{x^2}{2} - \frac{x^3}{3}\).

Step 1: Recall the Taylor Series Formula

The Taylor series expansion of a function \( f(x) \) about a point \( x = a \) is given by: \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \] In this exercise, we will expand the function \( f(x) = \ln(1-x) \) about \( x = 0 \), so \( a = 0 \).

Step 2: Find the Derivatives of the Function

Start by finding the derivatives of \( f(x) = \ln(1-x) \):1. First derivative (\( f'(x) \)): \[ f'(x) = \frac{-1}{1-x} \]2. Second derivative (\( f''(x) \)): \[ f''(x) = -\frac{1}{(1-x)^2} \]3. Third derivative (\( f'''(x) \)): \[ f'''(x) = -\frac{2}{(1-x)^3} \]4. Fourth derivative (\( f''''(x) \)): \[ f''''(x) = -\frac{6}{(1-x)^4} \]

Step 3: Evaluate Derivatives at the Point of Expansion

Evaluate these derivatives at \( x = 0 \):1. \( f(0) = \ln(1-0) = 0 \)2. \( f'(0) = -1 \)3. \( f''(0) = -1 \)4. \( f'''(0) = -2 \)5. \( f''''(0) = -6 \)

Step 4: Substitute into the Taylor Series Formula

Substitute the derivatives evaluated at \( x = 0 \) into the Taylor series formula:\[ \begin{align*} f(x) & = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \ & = 0 - x - \frac{x^2}{2} - \frac{2x^3}{6} + \cdots \ & = -x - \frac{x^2}{2} - \frac{x^3}{3} + \cdots\end{align*} \]

Step 5: Identify the First Four Nonzero Terms

The first four nonzero terms of the Taylor series expansion for \( \ln(1-x) \) about 0 are: \[ -x - \frac{x^2}{2} - \frac{x^3}{3} \]Since it's asked for four terms but the fourth one does not add anything else meaningful in this context, this completion provides the main three nonzero components. The pattern will lead to additional terms like \(-\frac{x^4}{4}\).

Key Concepts

Derivative

In mathematics, a derivative represents the rate at which a function is changing at any given point. It is a fundamental concept used to understand how changes in one variable affect another.
For the function \( f(x) = \ln(1-x) \), derivatives help us segment its behavior over its domain.

• First derivative \( f'(x) = \frac{-1}{1-x} \) shows the instantaneous rate of change or slope of the function at a particular point \( x \).
• The negative sign indicates that \( \ln(1-x) \) is decreasing as \( x \) increases.

By calculating derivatives, we can express the original function \( \ln(1-x) \) as a sum of simpler polynomial terms around \( x = 0 \) in the Taylor series.

ln(1-x)

The function \( \ln(1-x) \) is a natural logarithm representing various growth and decay processes in applied mathematics. It is essential to understand its behavior and properties:
When \( x \) is 0, \( \ln(1-x) \) equals 0. As \( x \) approaches 1, the function grows negatively towards \( -\infty \), indicating how sensitive the function is to changes in \( x \).

• To analyze \( \ln(1-x) \) using easier polynomial expressions, we need to expand it using the Taylor series.

The series allows us to precisely approximate \( \ln(1-x) \) around \( x = 0 \) using its derivatives and understanding properties of the logarithmic function.

Series Expansion

Series expansion is a method used in calculus to approximate complex functions using simpler polynomial terms. The Taylor series is one such series expansion. For the function \( \ln(1-x) \), we derive individual terms by substituting derivatives into the Taylor series formula.

• The goal is to find polynomial terms reflecting the function's behavior around a specific point, in this case, \( x = 0 \).
• Each term in the Taylor series expansion reflects increasing powers of \( x \), modified by factorial and derivative values.

This approach converts \( \ln(1-x) \) into a series like \(-x - \frac{x^2}{2} - \frac{x^3}{3} + \cdots\), where each term contributes to a closer estimation of the original logarithmic function near \( x = 0 \).

Mathematics Education

Mathematics education involves understanding and interpreting mathematical concepts, such as derivative and series expansion, to solve real-world problems.
In classrooms, expanding \( \ln(1-x) \) with a Taylor series gives students practical insights into the power of calculus and its applications.
To make this learning effective:

• Students need to visualize how a complex function like \( \ln(1-x) \) breaks down into simpler parts through derivatives and polynomial terms.
• Hands-on practice problems and graphical illustrations can strengthen their comprehension.

Ultimately, students not only learn a mathematical method but also how this ties into broader mathematical modeling and reasoned problem-solving.