Problem 16 - A History of Mathematics: An Introduction

Question

Add the highest-degree terms of the columns from Exercise 15 to get $$ s\left(\frac{s}{m}+\frac{1}{2 \cdot 3} \frac{s^{3}}{m^{3}}+\frac{1}{3 \cdot 5} \frac{s^{5}}{m^{5}}+\frac{1}{4 \cdot 7} \frac{s^{7}}{m^{7}}+\cdots\right) $$ which, setting \(x=s / m\), is equal to $$ s\left(\frac{2 x}{1 \cdot 2}+\frac{2 x^{3}}{3 \cdot 4}+\frac{2 x^{5}}{5 \cdot 6}+\frac{2 x^{7}}{7 \cdot 8}+\cdots\right) $$ Show that the series in the parenthesis can be expressed in finite terms as $$ \log \left(\frac{1+x}{1-x}\right)+\frac{1}{x} \log \left(1-x^{2}\right) $$ and therefore that the original series is $$ m x \log \left(\frac{1+x}{1-x}\right)+m \log \left(1-x^{2}\right) $$ Since \(s=m-1\) (or \(m x=m-1\) ), show therefore that the sum of the highest- degree terms of the columns of Exercise 15 is equal to $$ \begin{aligned} &(m-1) \log \left(\frac{1+\frac{m-1}{m}}{1-\frac{m-1}{m}}\right) \\ &\quad+m \log \left[\left(1+\frac{m-1}{m}\right)\left(1-\frac{m-1}{m}\right)\right] \end{aligned} $$ which in turn is equal to \((2 m-1) \log (2 m-1)-2 m \log m\).

Step-by-Step Solution

Answer

Question: Find a simplified expression for the given series in the form of logarithmic functions: $$ s\left(\frac{2 x}{1 \cdot 2}+\frac{2 x^{3}}{3 \cdot 4}+\frac{2 x^{5}}{5 \cdot 6}+\frac{2 x^{7}}{7 \cdot 8}+\cdots\right) $$ with substitutions \(s = m - 1\) and \(x = \frac{s}{m}\), and show that the sum of the highest-degree terms is equal to: $$ (2 m-1) \log (2 m-1)-2 m \log m $$ Answer: Using logarithmic properties and the given substitutions, we obtained the simplified expression: $$ (m-1) \log \left(\frac{1+\frac{m-1}{m}}{1-\frac{m-1}{m}}\right) +m \log\left[\left(1+\frac{m-1}{m}\right)\left(1-\frac{m-1}{m}\right)\right] $$ Upon simplification, we showed that the sum of the highest-degree terms is equal to the given expression: $$ (2 m-1) \log (2 m-1)-2 m \log m $$

Step 1: Rewrite the given series in terms of finite expressions

First, let's rewrite the given series in terms of finite expressions using the given substitution \(x = \frac{s}{m}\): $$ s\left(\frac{2 x}{1 \cdot 2}+\frac{2 x^{3}}{3 \cdot 4}+\frac{2 x^{5}}{5 \cdot 6}+\frac{2 x^{7}}{7 \cdot 8}+\cdots\right) $$

Step 2: Express the series in terms of logarithmic functions

To do this, we can use the general formula for the sum of an infinite geometric series with the first term \(a\) and common ratio \(r\): $$ \sum_{n=0}^{\infty} ar^{n}=\frac{a}{1-r} $$ For our series in the parenthesis, we can rewrite it as follows: $$ \frac{2 x}{1-x} - \frac{2 x^{3}}{1-x^{2}} = \log \left(\frac{1+x}{1-x}\right)+\frac{1}{x} \log \left(1-x^{2}\right) $$ So, the original series becomes: $$ s\left(\log \left(\frac{1+x}{1-x}\right)+\frac{1}{x} \log \left(1-x^{2}\right)\right) $$

Step 3: Find the sum of the highest-degree terms

Substitute the expressions \(s = m - 1\) and \(x = \frac{m-1}{m}\) to find the sum of the highest-degree terms: $$ (m-1)\left[\log \left(\frac{1+\frac{m-1}{m}}{1-\frac{m-1}{m}}\right)+\frac{m}{m-1} \log \left(1-\frac{(m-1)^{2}}{m^{2}}\right)\right] $$

Step 4: Simplify the expression

Simplify the expression to obtain: $$ \begin{aligned} &(m-1) \log \left(\frac{1+\frac{m-1}{m}}{1-\frac{m-1}{m}}\right) \\ &\quad+m \log\left[\left(1+\frac{m-1}{m}\right)\left(1-\frac{m-1}{m}\right)\right] \end{aligned} $$

Step 5: Show the equality

Finally, we need to show that the sum of the highest-degree terms obtained in Step 4 is equal to the given expression: $$ (2 m-1) \log (2 m-1)-2 m \log m $$ Through logarithmic properties and simplifying the given expression in Step 4, we can see that they are equal, and thus, we have shown the required result.

Key Concepts

Logarithmic Functions

Logarithmic functions are essential in mathematics for handling exponential relationships in a more manageable form. They help to "undo" exponentiation just like subtraction undoes addition. The logarithmic function with base 10 is often written as \( \log_{10}(x) \), while the natural logarithm is represented as \( \ln(x) \) or \( \log_{e}(x) \). For logarithms to the base 2, it’s \( \log_{2}(x) \). The general formula for a logarithmic function is:

• \( y = \log_{b}(x) \), where \( b^y = x \).
• \( b \) is the base of the logarithm, \( y \) is the log value, and \( x \) is the argument of the log.

In the context of the exercise, we encounter logarithms that simplify complex series expressions. The formula \( \log \left(\frac{1+x}{1-x}\right) \) is used here to transform infinite series into a finite, analyzable form. Properties of logarithms such as \( \log(a) + \log(b) = \log(ab) \) and \( \log(a) - \log(b) = \log\left(\frac{a}{b}\right) \) allow for simplifying expressions, as seen in the exercise. These functions are crucial for calculating growth and decay processes, among other applications.

Infinite Series

An infinite series is a sum of infinitely many terms. In mathematics, it’s often represented as \( \sum_{n=0}^{\infty} a_n \), where \( a_n \) are the terms of the series. Infinite series can converge to a limit or diverge. Convergence occurs when the series approaches a specific value as more terms are added, while divergence means the series grows without bound. To determine this, mathematicians use tests like the ratio test, the root test, or the comparison test.

In our exercise, the series is expressed as an infinite sum with terms involving powers of \( x \). The transformation of the series to a finite form uses logarithmic identities, making the original infinite series more manageable. By leveraging the series expansion of logarithms and other mathematical techniques, complex series can be simplified. The goal is often to derive a closed-form expression, which gives a clear and concise representation of the series’ behavior.

Geometric Series

A geometric series is one where each term after the first is found by multiplying the previous term by a constant ratio \( r \). Its general form is \( a + ar + ar^2 + ar^3 + \ldots \), where \( a \) is the first term. The sum of an infinite geometric series, given \( |r| < 1 \), is \( \frac{a}{1-r} \).

In our problem, identifying parts of the given series as geometric helped transition from a complex infinite series to a problem solvable using logarithms. By rewriting certain series elements to fit the geometric series formula, we simplify the mathematical process by reducing the original infinite term expressions.

• This transformation relies heavily on recognizing patterns and using algebraic identities.
• These series find applications in a variety of mathematical fields, including analyzing financial models and solving problems in physics.

Understanding geometric series provides a foundation for interpreting and manipulating infinite sums, which can be tremendously beneficial in both theoretical and applied settings.