Question

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

Short answer

The Maclaurin series for \(e^{x+y}\) is:\[e^{x+y} = \sum_{m=0}^\infty \sum_{n=0}^\infty \frac{x^m y^n}{m!n!}\]

Step-by-step solution

- Understand the Maclaurin series

The Maclaurin series is the Taylor series expansion of a function about 0. For a two-variable function, it's given by: \[ f(x,y) = \sum_{m=0}^\infty \sum_{n=0}^\infty \frac{\partial^{m+n} f(0,0)}{\partial x^m \partial y^n} \frac{x^m}{m!} \frac{y^n}{n!}. \]

- Calculate the function at (0,0)

Evaluate the function \(e^{x+y}\) at the origin (0,0). \( f(0,0) = e^{0+0} = e^0 = 1 \).

- Calculate partial derivatives

Calculate the partial derivatives of \(e^{x+y}\). Since the exponential function is simple, each partial derivative with respect to either variable will always yield \(e^{x+y}\).

- Evaluate the partial derivatives at (0,0)

Evaluate all partial derivatives at (0,0). For any partial derivative, \( \frac{\partial^{m+n} e^{x+y}}{\partial x^m \partial y^n}\bigg\vert_{x=0, y=0} = e^0 = 1 \).

- Form the series

Substitute the partial derivatives evaluated at (0,0) into the Maclaurin series formula: \[ e^{x+y} = \sum_{m=0}^\infty \sum_{n=0}^\infty \frac{1}{m!} \frac{1}{n!} x^m y^n \]

Key concepts

two-variable Maclaurin series

A two-variable Maclaurin series is simply an extension of the regular Maclaurin series to functions of two variables. The goal is to express a function of two variables as an infinite sum of its partial derivatives evaluated at a specific point—in this case, the origin (0,0). This series expansion allows us to represent more complex functions in a simpler polynomial form, which can be useful for approximations and analysis.

For a two-variable function f(x,y), the Maclaurin series is given by:
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• The sum over m and n runs from 0 to infinity.
• Each term involves the partial derivatives of the function evaluated at (0,0).
• The terms are divided by factorials of m and n and multiplied by x^m and y^n.

In the specific case of the function given, which is the exponential function e^(x+y), this simplifies the series significantly.

partial derivatives

Partial derivatives measure the rate at which a function changes as one of its input variables changes, while keeping the other variables constant. It is like a derivative, but specifically for multi-variable functions.

To find the Maclaurin series for a multi-variable function, you need to calculate the partial derivatives with respect to each variable. These calculations give us the coefficients for the terms in our series expansion.

• First-order partial derivatives: For a function f(x,y), these are denoted by f_x and f_y and represent the slope of the function along the x and y directions.
• Second-order partial derivatives: These include f_xx, f_yy, and f_xy. They provide information about the curvature of the function's surface.

In our case, since we are dealing with the exponential function e^(x+y), all partial derivatives with respect to either variable will always yield e^(x+y).
When evaluated at (0,0), these partial derivatives simplify to 1 because e^0 = 1.

exponential function

The exponential function e^(x+y) is a special type of function with unique properties. It is of the form e raised to the power of a sum of variables—in this case, x and y.

The exponential function has a fundamental property where its rate of change is equal to the value of the function itself. This self-replicating property makes it incredibly useful in various fields, including calculus.

For the function e^(x+y), every partial derivative is the same as the original function.

• This means f(x,y) = f_x(x,y) = f_y(x,y) = f_{xx}(x,y) = ... = e^(x+y).
• When evaluated at the origin (0,0), all these derivatives equal 1 because e^0 = 1.

Thus, the Maclaurin series for e^(x+y) becomes especially simple, reducing to: <

• each term in the series is given by the product of x^m and y^n divided by m! and n!.
• the coefficients are all 1 because of the nature of the exponential function.