Problem 394 - Advanced Calculus

Question

Using Power series, show that $$ \log 2=1-(1 / 2)+(1 / 3)-(1 / 4)+\ldots $$

Step-by-Step Solution

Answer

Using power series, we can represent the natural logarithm as \(\ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n}\). To show that \(\log 2 = 1 -\frac{1}{2}+\frac{1}{3} -\frac{1}{4}+\ldots\), we substitute x with 1 in the above representation: \(\ln 2 = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{1}{n}\). Simplifying the series, we obtain \(\ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots\), which confirms that the given expression is correct.

Step 1: Find the power series representation of natural logarithm

To find the power series representation of the natural logarithm of x, we start with the definition of the natural logarithm as a power series: \[ \ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n} \] This is a known power series where x is the variable, and the series converges for -1 < x ≤ 1.

Step 2: Apply the power series to our problem

In our problem, we have \(\ln 2\), which can be rewritten as \(\ln(1+1)\). Now, we will replace x with 1 in the power series representation: \[ \ln 2 = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{1^n}{n} \]

Step 3: Simplify the power series

Since \(1^n = 1\) for any natural number n, we can simplify the series as: \[ \ln 2 = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{1}{n} \]

Step 4: Separate the terms in the series

To better visualize the series, it helps to separate the terms. For instance, for n = 1: \(-1^{1+1} * \frac{1}{1} = 1\) For n = 2: \(-1^{2+1} * \frac{1}{2} = -\frac{1}{2}\) For n = 3: \(-1^{3+1} * \frac{1}{3} = \frac{1}{3}\) And so on, the series goes as follows: \[ \ln 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots \] Now, we have shown that using power series, the natural logarithm of 2 is equivalent to the given expression.

Key Concepts

Natural Logarithm

The natural logarithm, often denoted as \( \ln(x) \), is a fundamental function in mathematics that arises frequently in calculus and complex analysis. It is specifically the logarithm to the base \( e \), where \( e \approx 2.71828 \), known as Euler's number, a key mathematical constant. The natural logarithm of a number \( x \) is defined as the area under the curve \( y = 1/t \) from 1 to \( x \). For any positive real number, it can be expressed using the power series representation for \( \ln(1 + x) \), given by:

\[\ln(1 + x) = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n}\]
This series converges for \( -1 < x \leq 1 \), meaning it successfully predicts values within this range. This representation is crucial, as it can be manipulated to approximate \( \ln(x) \) for different values of \( x \), helping calculate these logarithmic values without directly using numerical methods.

Convergence of Series

Series convergence is an important aspect when working with infinite series like power series. A series converges if its sequence of partial sums tends to a limit, meaning it approaches a specific value as more terms are added. For the power series representation of \( \ln(1+x) \), convergence is determined by the radius of convergence, which is the interval \( -1 < x \leq 1 \) for this particular series.

Here's why convergence matters:

• If a series converges, it can be used to approximate functions accurately within its radius.
• Outside its convergence interval, the series may diverge, leading to incorrect or meaningless results.

Understanding convergence ensures the series is applied in a valid and mathematically sound way. This concept is foundational when using series representations in calculus and ensures the correct utilization of the power of mathematical series in real-world problems.

Mathematical Series Representation

Mathematical series representation serves as a way to express functions as the sum of their terms, providing powerful tools for analysis and computation. A power series is one such representation which expresses functions like \( \ln(x) \) as the sum of infinitely many terms. By using the power series for \( \ln(1+x) \), we obtain:

\[\ln 2 = \sum_{n=1}^{\infty} (-1)^{n+1}\frac{1}{n}\]
This sum explains \( \ln(2) \) by showing how contributions from each term alternate in sign, creating an oscillating sequence. Each term contributes progressively smaller and oscillating values that sum to converge at the natural logarithm value for 2.

Mathematical series are not only useful for expressing functions but play a vital role in calculus, enabling easier computation of complex integrals, solving differential equations, and simulating functions. Understanding series representations, their convergence properties, and applicability allows mathematicians and engineers to derive solutions to applied and theoretical problems seamlessly.