Question

Find the two-variable Maclaurin series for the following functions. \(e^{x y}\)

Short answer

The Maclaurin series for \(e^{xy}\) is \(1 + xy + \cdots\)

Step-by-step solution

- Understand the Maclaurin Series for a Single Variable

Recall the Maclaurin series expansion for any function. For a single variable function, it's given by \[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \]

- Generalize to Two Variables

In the two-variable case, the Maclaurin series expansion for a function \(f(x, y)\) is \[ f(x, y) = f(0, 0) + f_x(0, 0)x + f_y(0, 0)y + \frac{1}{2!}(f_{xx}(0, 0)x^2 + 2f_{xy}(0, 0)xy + f_{yy}(0, 0)y^2) + \cdots \]

- Identify the Function and Its Derivatives

For the function \(f(x, y) = e^{xy}\), we need to find its partial derivatives at \((0, 0)\).\[f(0, 0) = e^{0} = 1\]\[f_x = y e^{xy}\] so \[f_x(0, 0) = 0\]\[f_y = x e^{xy}\] so \[f_y(0, 0) = 0\]

- Compute Second Order Derivatives

Find second order partial derivatives:\[f_{xx} = y^2 e^{xy}\] so \[f_{xx}(0, 0) = 0\]\[f_{yy} = x^2 e^{xy}\] so \[f_{yy}(0, 0) = 0\]\[f_{xy} = (1 + xy) e^{xy}\] so \[f_{xy}(0, 0) = 1\]

- Form the Maclaurin Series Expansion

Substitute all the values into the Maclaurin series formula:\[f(x, y) = 1 + 0 \cdot x + 0 \cdot y + \frac{1}{2!}(0 \cdot x^2 + 2 \cdot 1 \cdot xy + 0 \cdot y^2) + \cdots\]Simplify it to: \[f(x, y) = 1 + xy + \cdots \]

Key concepts

Taylor series expansion

The Taylor series expansion is a powerful tool in calculus that allows us to express a function as an infinite sum of terms calculated from the values of its derivatives at a single point. For a single-variable function, the Taylor series expansion around a point (often zero, leading to the Maclaurin series) is given by the formula:

For a multivariable function like in our exercise, the series includes terms for each variable and their combinations.
Understanding the Taylor series helps us approximate complex functions using polynomials, which are easier to work with.

partial derivatives

Partial derivatives measure how a function changes as its variables change, one at a time. In our example, we need the partial derivatives of the function \(e^{xy}\) with respect to both \(x\) and \(y\).

To find the partial derivative with respect to \(x\), we treat \(y\) as a constant and differentiate. Similarly, for the partial derivative with respect to \(y\), we treat \(x\) as constant and differentiate.

Finding these helps us understand the behavior of the function in response to changes in each variable, which is crucial for forming the Maclaurin series in multivariable functions.

multivariable calculus

Multivariable calculus extends calculus concepts to functions of more than one variable. It involves partial derivatives, multiple integrals, and series expansions like the Maclaurin series for two variables.

For the function \(e^{xy}\), we need to compute partial derivatives and mixed partial derivatives to form its series expansion.

This is essential for analyzing the function's behavior in higher dimensions, as it allows us to approximate the function near a point and understand its local behavior. The Maclaurin series expansion in two variables looks at combinations of up to \(n\) derivatives of \(x\) and \(y\).

exponential function

The exponential function \(e^{xy}\) is a fundamental mathematical function with unique properties. In our problem, this function forms the basis for the Maclaurin series expansion we are working on.

The exponential function \(e^x\) is known for its property that the derivative of \(e^x\) with respect to \(x\) is \(e^x\) itself. This self-replicating property simplifies the process of finding derivatives, as seen in the steps.

Knowing these properties helps in computing the necessary partial derivatives for our series expansion, making the process more straightforward and manageable.