Problem 11 - Calculus Early Transcendentals: Pearson New International Edition

Question

Obtain the power series in \(x\) for \(\ln [(1+x) /(1-x)]\) and specify its radius of convergence. Hint: $$ \ln [(1+x) /(1-x)]=\ln (1+x)-\ln (1-x) $$

Step-by-Step Solution

Answer

The power series is \( 2 \sum_{n=1}^{\infty} \frac{x^{2n-1}}{2n-1} \) with a radius of convergence of 1.

Step 1: Expand \( \ln(1+x) \) using Power Series

Recall the power series expansion for \( \ln(1+x) \) is given by \( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n} \). We will use this series formula to expand the function.

Step 2: Expand \( \ln(1-x) \) using Power Series

Similarly, the power series expansion for \( \ln(1-x) \) is \( -\sum_{n=1}^{\infty} \frac{x^n}{n} \). This series accounts for the negative sign in front of the log function.

Step 3: Combine the Series

Now, combine the two series from Steps 1 and 2: \( \ln(1+x) - \ln(1-x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n} + \sum_{n=1}^{\infty} \frac{x^n}{n} \).

Step 4: Simplify the Combined Series

The series simplifies by combining like terms: each term \((-1)^{n+1} \frac{x^n}{n} + \frac{x^n}{n}\) simplifies to \(\frac{2x^{2n-1}}{2n-1}\) for odd \(n\). Hence, the series becomes \( 2 \sum_{n=1}^{\infty} \frac{x^{2n-1}}{2n-1} \).

Step 5: Determine the Radius of Convergence

For both \( \ln(1+x) \) and \( \ln(1-x) \), the series converge when \(|x|<1\). Thus, the radius of convergence of the series \( \ln \frac{1+x}{1-x} \) is 1.

Key Concepts

Radius of Convergence

The radius of convergence is a crucial concept when working with power series. It tells us the range of values for which a power series converges to a finite value. In simpler terms, it's the "safe zone" for our series, where it behaves nicely and gives us meaningful results. For power series like the expansions of logarithmic functions, knowing where and how far the series converges is essential to avoid incorrect calculations or interpretations.
To determine the radius of convergence, one often checks the interval within which the terms of the series approach zero as more terms are added. This involves examining the limit of the ratio or root of terms as the series progresses. For the functions \( \ln(1+x) \) and \(-\ln(1-x) \), we find that they both converge when \(|x|<1\). This means that the values of \(x\) that make each term of the series go to zero lie strictly between -1 and 1.

• Converges within \(-1 < x < 1\)
• Radius of Convergence \(R = 1\)

Understanding the radius helps in utilizing series effectively in calculus and other applications, ensuring accuracy and reliability in computations.

Logarithmic Functions

Logarithmic functions are fundamental in calculus and algebra. The function \( \ln(x) \) denotes the natural logarithm, a critical function that reveals properties of growth and multiplication. It simplifies complex multiplication into addition through its unique properties.
For a given value of \(x\), \( \ln(x) \) is defined as the exponent to which the base \(e\) (approximately 2.718) needs to be raised to produce \(x\). This makes it extremely useful in scientific calculations where exponential growths or decays occur, such as in population dynamics or radioactive decay.
In our exercise, involving \( \ln\left(\frac{1+x}{1-x}\right) \), we use the logarithmic identity:
\[ \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \]
This helps transform a complex expression into a simpler form by decomposing it into two separate logarithmic functions. We then tackle each function using power series expansion, a useful method for dealing with complicated functions to make them easier to analyze and understand.

Series Expansion

Series expansion is vital in mathematical analysis because it allows functions to be represented as an infinite sum of simpler terms. This is especially useful when dealing with complex functions such as logarithmic operations. Power series expansion provides a way to simplify complex functions into an easily workable form.
To expand a function into a series, we express it as a sum of terms derived from polynomials. For instance, the function \( \ln(1+x) \) can be expressed as:

• \[ \ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n} \]
• Likewise, \( \ln(1-x) = -\sum_{n=1}^{\infty} \frac{x^n}{n} \)

This transformation allows for better manipulation and understanding of the original function, making it possible to solve otherwise complex calculus problems.
In our task of combining these expansions for \( \ln\left(\frac{1+x}{1-x}\right) \), the series terms are adjusted, combined, and simplified to represent \( 2 \sum_{n=1}^{\infty} \frac{x^{2n-1}}{2n-1} \), which makes working with the function more feasible for deriving solutions or further analysis.