Question

Find the two-variable Maclaurin series for the following functions. $$e^{x+y}$$

Short answer

The Maclaurin series for \( e^{x+y} \) is: \[ 1 + x + y + \frac{x^2}{2!} + \frac{y^2}{2!} + xy + ...\]

Step-by-step solution

Recall the Maclaurin series definition

The Maclaurin series for a function of two variables, \(f(x, y)\), can be expressed as an infinite sum of its partial derivatives at the point \((0,0)\). The general form for the Maclaurin series is: \[ f(x, y) = \frac{\frac{\text{∂}^0 }{\text{∂}x^0 \text{∂}y^0} f(0,0)}{0!0!} + \frac{\text{∂} f(0,0)}{\text{∂}x}x + \frac{\text{∂} f(0,0)}{\text{∂}y}y + \frac{\text{∂}^2 f(0,0)}{\text{∂}x^2} \frac{x^2}{2!} + \frac{\text{∂}^2 f(0,0)}{\text{∂}y^2} \frac{y^2}{2!} + \frac{\text{∂}^2 f(0,0)}{\text{∂}x \text{∂}y}xy + \frac{\text{∂}^3 f(0,0)}{\text{∂}x^3} \frac{x^3}{3!} + ... \]

Determine the function and its partial derivatives

Given the function \(f(x, y) = e^{x+y}\), calculate its partial derivatives at \((0,0)\). The function value: \[ f(0,0) = e^{0+0} = 1 \]

Calculate first-order partial derivatives

Calculate the first-order partial derivatives: \( \frac{\text{∂} f}{\text{∂} x} = e^{x+y} \) and \( \frac{\text{∂} f}{\text{∂} y} = e^{x+y} \), both evaluated at (0,0) yield: \( \frac{\text{∂} f}{\text{∂} x} \bigg|_{(0,0)} = e^{0+0} = 1 \) and \( \frac{\text{∂} f}{\text{∂} y} \bigg|_{(0,0)} = e^{0+0} = 1 \)

Calculate higher-order partial derivatives

Determine the second-order partial derivatives: \( \frac{\text{∂}^2 f}{\text{∂} x^2} = e^{x+y}\), \( \frac{\text{∂}^2 f}{\text{∂} y^2} = e^{x+y} \), and \( \frac{\text{∂}^2 f}{\text{∂} x \text{∂} y} = e^{x+y} \). Evaluating at (0,0) gives: \( \frac{\text{∂}^2 f}{\text{∂} x^2} \bigg|_{(0,0)} = 1 \), \( \frac{\text{∂}^2 f}{\text{∂} y^2} \bigg|_{(0,0)} = 1 \), and \( \frac{\text{∂}^2 f}{\text{∂} x \text{∂} y} \bigg|_{(0,0)} = 1 \)

Combine the terms to form the series

The Maclaurin series includes the terms up to the desired order. For this function: \[ f(x, y) = 1 + x + y + \frac{x^2}{2!} + \frac{y^2}{2!} + xy + \frac{x^3}{3!} + \frac{x^2 y}{2!} + \frac{xy^2}{2!} + \frac{y^3}{3!} + ... \]Or more compactly: \[ e^{x+y} = \bigg( \frac{e^x e^y}{x! y!} \bigg) \]

Key concepts

Partial Derivatives

Understanding partial derivatives is essential in grasping the Maclaurin series in two variables. Unlike single-variable calculus, partial derivatives deal with functions of more than one variable. When taking a partial derivative of a function, you differentiate with respect to one variable while keeping the other(s) constant. For instance, given the function \( f(x, y) = e^{x+y} \), the partial derivatives with respect to \( x \) and \( y \) are calculated separately. To find \( \frac{\text{∂} f}{\text{∂} x} \), we treat \( y \) as a constant and differentiate \( e^{x+y} \) with respect to \( x \), yielding \( e^{x+y} \). Similarly, the partial derivative \( \frac{\text{∂} f}{\text{∂} y} = e^{x+y} \) treats \( x \) as a constant. These derivatives play a crucial role in forming the terms of the Maclaurin series.

Multivariable Calculus

Multivariable calculus is an extension of single-variable calculus to functions of two or more variables. It includes operations like taking partial derivatives, integrating over multiple dimensions, and forming series expansions. In the context of the two-variable Maclaurin series, we use partial derivatives at the point \((0,0)\) to represent the function as an infinite sum. For example, the Maclaurin series for \( f(x, y) = e^{x+y} \) is built by evaluating its partial derivatives at \( (0,0) \) and combining them into a series. The series can be written as: \[ f(x, y) = 1 + x + y + \frac{x^2}{2!} + \frac{y^2}{2!} + xy + ... \] This representation helps in approximating the value of the function around the origin.

Exponential Function

The exponential function \( e^z \) is one of the most important functions in calculus and mathematical analysis. When dealing with multivariable functions, the exponential function can be extended to forms like \( e^{x+y} \). This function has unique properties, such as its derivative being the function itself, which simplifies the process of taking partial derivatives. For instance, the partial derivatives of \( f(x, y) = e^{x+y} \) are straightforward: \( \frac{\text{∂} f}{\text{∂} x} = e^{x+y} \) and \( \frac{\text{∂} f}{\text{∂} y} = e^{x+y} \). This property makes the exponential function an ideal candidate for series expansions, as seen with the Maclaurin series: \[ e^{x+y} = 1 + x + y + \frac{x^2}{2!} + \frac{y^2}{2!} + xy + ... \] This series helps in approximating the function near the origin, providing insights into its behavior.