Question

Find the two-variable Maclaurin series for the following functions. $$\frac{\ln (1+x)}{1+y}$$

Short answer

\(\frac{\text{ln}(1+x)}{1+y} = (x - \frac{x^2}{2} + \frac{x^3}{3})(1 - y + y^2)\)

Step-by-step solution

- Understanding the Function

The given function is \(\frac{\text{ln}(1+x)}{1+y}\). The goal is to find the Maclaurin series for this function, which means to expand it around \(x = 0\) and \(y = 0\).

- Maclaurin Series for \(\text{ln}(1+x)\)

The Maclaurin series for \(\text{ln}(1+x)\) is: \(\text{ln}(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\)

- Series Expansion for \(\frac{1}{1+y}\)

The Maclaurin series for \(\frac{1}{1+y}\) is: \(\frac{1}{1+y} = 1 - y + y^2 - y^3 + \cdots\)

- Combine Both Series

Now, substitute the series expansions into the original function: \[\frac{\text{ln}(1+x)}{1+y} = (x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots)(1 - y + y^2 - y^3 + \cdots)\]

- Multiply the Series Terms

Multiply the series terms to get the combined Maclaurin series. Consider terms up to \(x^3\) and \(y^3\) for simplicity: \[ (x - \frac{x^2}{2} + \frac{x^3}{3})(1 - y + y^2) = x - xy + xy^2 - \frac{x^2}{2} + \frac{x^2}{2}y + \dots\]

Key concepts

Two-Variable Series Expansion

Two-variable series expansions help us understand how a function behaves around a particular point, typically \(0,0\). This method decomposes a function into a sum of simpler terms that are easier to manage. In this exercise, we expand \(\frac{\ln (1+x)}{1+y}\), around \(x = 0\) and \(y = 0\). This approach makes complex functions more accessible by breaking them down into their component parts. Note that higher-order terms are often included to ensure accuracy.

ln(1+x) Series

The natural logarithm function, \(\ln(1+x)\), can be expanded using the Maclaurin series. For \(\ln(1+x)\), this series is represented as: \[\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\] This expansion gives us a simpler way to work with the logarithmic function by expressing it as an infinite series. Each term corresponds to a power of \(x\), making it easier to understand and manipulate.

Series Multiplication

To solve the given exercise, we need to multiply the series for \(\ln(1+x)\) and \(\frac{1}{1+y}\). Each series represents a different aspect of the function we are working with, and multiplying them combines these aspects into a single expression. Let's substitute the Maclaurin series expansions into our function and perform the multiplication:
\[\frac{\ln(1+x)}{1+y} = (x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots)(1 - y + y^2 - y^3 + \cdots)\] By multiplying the terms step-by-step, up to a reasonable cutoff (e.g., \(x^3\) and \(y^3\)), we obtain:
\[ (x - \frac{x^2}{2} + \frac{x^3}{3})(1 - y + y^2) = x - xy + xy^2 - \frac{x^2}{2} + \frac{x^2}{2}y + \cdots\] This combined series is the simplified form of the original function.

Maclaurin Series for Rational Functions

The Maclaurin series is a special case of the Taylor series, expanded around zero. For rational functions like \(\frac{1}{1+y}\), it simplifies expressions significantly. The Maclaurin series for \(\frac{1}{1+y}\) is: \[ \frac{1}{1+y} = 1 - y + y^2 - y^3 + \cdots\] This series expresses a fraction as an infinite sum, making it easier to handle in mathematical calculations. By combining this with the series for \(\ln(1+x)\), it allows us to decompose and recompose functions in manageable steps.