Problem 4 - Calculus: One and Several Variables

Question

Expand \(f(x)\) in powers of \(x,\) basing your calculations on the geometric series $$\frac{1}{1-x}=1+x+x^{2}+\cdots+x^{n}+\cdots$$ $$f(x)=\ln (1-x)$$

Step-by-Step Solution

Answer

The power series expansion of \(f(x)=\ln(1-x)\) is: \[ f(x) = \ln(1-x) = -(x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots + \frac{x^{n+1}}{n+1} + \cdots) \]

Step 1: Expressing \(\ln(1 - x)\) as a series

We know that the derivative of \(\ln(1 - x)\) is \(-\frac{1}{1 - x}\). To express it as a series, we can use the provided formula for the geometric series: \[ -\frac{1}{1 - x} = - (1 + x + x^2 + \cdots + x^n + \cdots) \] Now, to obtain the expansion for \(\ln(1 - x)\), we can integrate the series term-by-term.

Step 2: Integrating the series

To find the series representation of \(f(x)\), we'll integrate the expression above term-by-term: \[ \int -\frac{1}{1 - x} dx = \int - (1 + x + x^2 + \cdots + x^n + \cdots) dx \] This gives: \[ \ln(1 - x) + C = - (x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots + \frac{x^{n+1}}{n+1} + \cdots) \] We know that \(f(0) = \ln(1 - 0) = 0\), which means \(C = 0\). So, the power series expression for \(f(x)\) is: \[ f(x) = \ln(1 - x) = -(x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots + \frac{x^{n+1}}{n+1} + \cdots) \] Thus, we have expanded the function \(f(x) = \ln(1 - x)\) in powers of \(x\) using the geometric series.

Key Concepts

Geometric Series

The geometric series is a simple and widely used concept in mathematics. It is a sum of terms that all are powers of a particular variable, usually denoted as \( x \). In the geometric series formula \( \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots + x^n + \cdots \), each term represents successive powers of \( x \). This series converges, or adds up to, a finite value for values of \( x \) that are between \(-1\) and \(1\).

• The series starts with the coefficient \(1\).
• Each following term is the previous term multiplied by \(x\).
• As long as the absolute value of \(x\) is less than one, it will converge to \(\frac{1}{1-x}\).

Using the geometric series, we can represent complex functions, like \(\ln(1-x)\). Learning to recognize and manipulate geometric series is fundamental for expanding functions into power series.

Integration of Series

Integration of series is a fascinating method where we integrate term-by-term. The beauty of this method comes from its flexibility in handling functions represented as an infinite series like our geometric series.

• Each term is integrated separately.
• Integration changes the power \(n\) to \(n+1\) and divides the term by the new power.

For example, if we take the infinite series from the geometric series and integrate it term-by-term, we end up with:\[- (x + \frac{x^2}{2} + \frac{x^3}{3} + \cdots)\]This results in the series representation of \( \ln(1-x)\). This process integrates the geometric series to achieve the power series expansion of logarithmic functions.

Logarithmic Functions

Logarithmic functions are an essential type of mathematical function with broad applications in solving exponential equations and modeling many natural phenomena. In our case, we focused on representing the natural logarithm function, \( \ln(1 - x) \), as a power series.

• Natural logarithms work with the base \(e\), an important constant in mathematics approximately equal to 2.718.
• \( \ln(1-x) \) can be expanded into a power series using integration techniques to express it as a sum of powers of \(x\).

Through integration of the derived series \( -\frac{1}{1-x} \), we obtained \[\ln(1-x) = -(x + \frac{x^2}{2} + \frac{x^3}{3} + \ldots)\]This expansion is important as it provides a way to approximate \( \ln(1-x) \) for values \( x \) between \(-1\) and \(1\), allowing easier computation and understanding of logarithmic behavior.