Problem 53 - Thomas Calculus

Question

Replace \(x\) by \(-x\) in the Taylor series for \(\ln (1+x)\) to obtain a series for \(\ln (1-x)\) . Then subtract this from the Taylor series for \(\ln (1+x)\) to show that for \(|x|<1\), \begin{equation} \ln \frac{1+x}{1-x}=2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\cdots\right). \end{equation}

Step-by-Step Solution

Answer

Subtract the series for \(\ln(1-x)\) from \(\ln(1+x)\) to obtain \(2(x + \frac{x^3}{3} + \frac{x^5}{5} + \cdots)\), confirming the result.

Step 1: Recall the Taylor Series for \(\ln(1+x)\)

The Taylor series for \(\ln(1+x)\) is given by \(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\) for \(|x| < 1\). This series is an alternating series derived from the power series expansion of the natural logarithm function.

Step 2: Substitute \(x\) with \(-x\) in the Series

We substitute \(x\) with \(-x\) in the Taylor series for \(\ln(1+x)\). This gives us the series for \(\ln(1-x)\), which is \(-x - \frac{x^2}{2} - \frac{x^3}{3} - \frac{x^4}{4} - \cdots\). Notice that the signs of each term have changed, as substituting \(-x\) affects the even powers and their coefficients.

Step 3: Subtract \(\ln(1-x)\) Series from \(\ln(1+x)\) Series

Subtract the series for \(\ln(1-x)\) from \(\ln(1+x)\). This means calculating \((x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots) - (-x - \frac{x^2}{2} - \frac{x^3}{3} - \cdots)\). The resulting series becomes \(2x + \frac{2x^3}{3} + \frac{2x^5}{5} + \cdots\) because the \(x^2\) terms cancel out and the odd terms get doubled.

Step 4: Simplify the Subtracted Series

The series \(2x + \frac{2x^3}{3} + \frac{2x^5}{5} + \cdots\) can be written more compactly as \(2(x + \frac{x^3}{3} + \frac{x^5}{5} + \cdots)\). This simplification highlights the pattern in the series derived from subtracting \(\ln(1-x)\) from \(\ln(1+x)\).

Step 5: State the Final Result

Thus, the operation confirms that for \(|x|<1\), \(\ln \frac{1+x}{1-x}=2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\cdots\right)\). This shows the derivation and fits the condition stated in the problem.

Key Concepts

Natural Logarithm Series

One of the intriguing properties of mathematical functions is how certain patterns and expansions define their behavior. In the case of natural logarithms, the function \(\ln(1+x)\) can be expressed as a Taylor series. This series helps us understand the logarithmic behavior close to zero, fitting neatly as \(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\) for \|x| < 1\.
What makes this series incredibly useful is that it provides a polynomial approximation of the natural logarithm, which allows for practical and theoretical applications in calculus and beyond. But why do these terms alternate in sign? This is because the natural logarithm's derivatives alternate in sign when evaluated at zero, leading to an **alternating series** when expanded into a power series.
Understanding this series allows us to explore more complex expressions. Here, replacing \(x\) with \(-x\) in this series forms the basis for understanding how logarithmic expressions can be expanded into series that fit between different values of \(x\) and \(-x\).

Power Series Expansion

A power series is a series of the form \sum_{n=0}^{∞} a_n (x-c)^n\, where each power term involves some variable \(x\) subtracted or offset by some center point \(c\). In many cases, such as the Taylor series for \(\ln(1+x)\), \(c\) is zero, simplifying the form to \sum_{n=0}^{∞} a_n x^n\.
This expansion framework is pivotal because it enables mathematicians to use polynomials to approximate more complex functions, offering insights into how functions behave near specific points like \(x=0\).
In our exercise scenario, we utilized the concept of power series expansion by replacing \(x\) with \(-x\) to derive another series and then subtract these series to find the simplified series \(2(x + \frac{x^3}{3} + \frac{x^5}{5} + \cdots)\).
This isn't just about numbers; it is a way of understanding the intricate nature of function behavior using relatively simple building blocks—polynomials.

Alternating Series

An alternating series is a series where the terms alternate in sign; that is, they follow a pattern of becoming positive and negative alternately. The natural logarithm series \(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\) is a classic example.
These series are significant because they converge under certain conditions, providing accurate approximations of functions. For example, one of the key features of alternating series is how quickly they converge, especially compared to non-alternating series. This property is immensely useful for computations and approximation to functions.
In the context of our problem, the alternating nature allowed us to see the resulting simplified series after subtracting one expanded \ln(1-x)\ series from another (the \ln(1+x)\ series).
Understanding alternating series is crucial for anyone delving into calculus and series because it outlines how the convergence and behavior of series can depend fundamentally on the alternating sign. This is yet another blend of creativity and analytical rigor in mathematics.