Problem 70 - Essential Calculus

Question

Find the Maclaurin series for $$ f(x)=\ln \frac{1+x}{1-x} $$ and determine its radius of convergence. Use the first four terms of the series to approximate \(\ln 3\).

Step-by-Step Solution

Answer

The Maclaurin series for the function is \(4x + \frac{2x^3}{3} + \frac{2x^5}{5} +...\). The radius of convergence is 1. Using the first four terms of this series, we can approximate \(\ln (3)\) as approximately 2.111.

Step 1 - Find the Maclaurin series for the function

Firstly, notice that \(f(x)=2(\ln(1+x)-\ln(1-x))\). Now, recall the Maclaurin series expansion of \(\ln(1+x)\) which is given by: \(\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...\)We will use this formula twice, first with \(x\) and then with \(-x\), so we get: \(f(x)=2((x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...)-(-x - \frac{(-x)^2}{2} + \frac{(-x)^3}{3} - \frac{(-x)^4}{4} +...) = 2 (2x + \frac{x^3}{3} + \frac{x^5}{5} +...)= 4x + \frac{2x^3}{3} + \frac{2x^5}{5} +...\)

Step 2 - Determine the radius of convergence

The original power series for \(\ln(1+x)\) converges if \(-1 < x < 1\). Because we utilize this series in our \(f(x)\) function twice, once with \(x\) and once with \(-x\), the radius of convergence remains the same, which is \(1\).

Step 3 - Approximate \(\ln 3\)

Next, note that this series is an approximation for \(\ln\frac{2}{2-2x}\) and therefore \(\ln(3)\) = \(\ln\frac{2}{2-2*(1/2)}\) can be approximated using the first four terms of our series with \(x = 1/2\) plugged in: \(\ln(3) \approx 4*(1/2) + \frac{2(1/2)^3}{3} + \frac{2(1/2)^5}{5} \approx 2 + \frac{1}{12} + \frac{1}{80} \approx 2.111\)

Key Concepts

Calculus

In the world of mathematics, calculus stands tall as a fundamental branch that deals extensively with change and motion, encompassing two major sub-fields: differential calculus and integral calculus. Differential calculus focuses on the concept of a derivative, which measures how a function changes as its input changes. Integral calculus is concerned with the accumulation of quantities, such as areas under a curve, and it is linked to the concept of an integral.

Understanding calculus is crucial when dealing with series expansions such as the Maclaurin series. This series, named after the Scottish mathematician Colin Maclaurin, is a special case of the Taylor series and is used to approximate functions around the point zero (also known as an expansion around the origin). The Maclaurin series takes the form \( f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + ... \), showcasing the core idea of calculus: breaking down complicated functions into simpler polynomial terms that can be more readily analyzed and used for approximation.

Radius of Convergence

Diving into the depths of the series, the radius of convergence is a measure that tells us how far out from the center point a power series will still hold true (converge). More specifically, it's the distance from the center point, expressed as a particular value of \(x\), within which the series will converge. Outside of this radius, the series may diverge or simply not be valid.

For the Maclaurin series mentioned in the example, the radius of convergence is crucial because it defines the interval for which our series approximation accurately represents the function. Determining this radius involves the Ratio Test or other convergence tests, which in this case, shows that the series for the natural logarithm \(\( \ln(1+x) \)\) converges when \(-1 < x < 1\). Thus, the radius of convergence for our function \(f(x)\) is found to be 1, signifying that we can reliably use this series to approximate \(f(x)\) when \(|x| < 1\).

Natural Logarithm

At the heart of exponential functions lies the natural logarithm, denoted as \(\ln\), which essentially reverses the operation of raising \(e\), the base of natural logarithms and an irrational mathematical constant roughly equal to 2.71828. The natural logarithm of a number \(x\) answers the question: to what power must we raise \(e\) to obtain \(x\)?

Mathematically speaking, if \(e^y = x\), then \(\(\ln x = y\).\) The natural logarithm has a crucial property that the derivative of \(\ln x\) is \(1/x\), making it a widely utilized function in calculus. The function \(f(x) = \ln \frac{1+x}{1-x}\) explored in our exercise is particularly noteworthy because its Maclaurin series provides a powerful tool for approximating values of \(\ln\) for arguments close to 1. Such series reveal the fascinating interconnectedness between calculus and logarithms.

Series Approximation

Series approximations form the backbone of numerical analysis, giving us a pragmatic method to crunch complex functions down to manageable expressions. By using the sum of sequences of numbers, typically infinite in potential length, series can approximate the values of functions by successively adding terms together to get increasingly accurate results.

For example, the function \(f(x) = \ln \frac{1+x}{1-x}\) can be approximated by the Maclaurin series to avoid the laborious and often impossible task of finding the exact value of a natural logarithm for every possible argument. Using series approximation, it's possible to estimate values such as \(\ln 3\) with a high degree of precision by merely employing the first few terms of the series, as shown in our exercise. This technique of approximation doesn't just save time; it lays the groundwork for computation and analysis in numerous applications across science and engineering domains.