Question

Find the two-variable Maclaurin series for the following functions. $$\cos (x+y)$$

Short answer

The two-variable Maclaurin series for cos(x + y) is 1 - (1/2) x² - (1/2) y² - (1/2) xy + ...

Step-by-step solution

Understand the Maclaurin Series Expansion

The Maclaurin series for a function is a Taylor series expansion of the function about 0. For a function of two variables, f(x, y), the Maclaurin series is a sum of partial derivatives of f evaluated at (0, 0), each multiplied by powers of the variables divided by factorial terms.

Write the General Form

The general form of the Maclaurin series for a function of two variables, f(x, y), is given by: f(x, y) = f(0, 0) + ( ∂f/∂x )(0, 0) x + ( ∂f/∂y )(0, 0) y + ( 1/2! ) ( ∂²f/∂x² )(0, 0) x² + ( 1/2! ) ( ∂²f/∂y² )(0, 0) y² + ( 1/2! )( ∂²f/ ∂x∂y )(0, 0) xy + ...

Compute the Necessary Partial Derivatives

First, let's compute the necessary partial derivatives at (0, 0) for the function cos(x + y): f(x, y) = cos(x + y). - f(0, 0) = cos(0) = 1. - ( ∂f/ ∂x )(x, y) = - sin(x + y), so ( ∂f/ ∂x )(0, 0) = 0. - ( ∂f/ ∂y )(x, y) = - sin(x + y), so ( ∂f/ ∂y )(0,0) = 0. - ( ∂²f/ ∂x² )(x,y) = - cos(x + y), so ( ∂²f/ ∂x² )(0,0) = -1. - ( ∂²f/ ∂y² )(x,y) = - cos(x + y), so ( ∂²f/ ∂y² )(0,0) = -1. - ( ∂²f/ ∂x ∂y )(x,y) = - cos(x + y), so ( ∂²f/ ∂x ∂y )(0,0) = -1.

Substitute Values into the Series

Now, substitute the partial derivatives into the general form of Maclaurin series up to the second order terms. f(x, y) = 1 + 0 x + 0 y + ( -1/2! ) x² + ( -1/2! ) y² + ( -1/2! ) xy + ... Thus, the series is: f(x, y) = 1 - (1/2) x² - (1/2) y² - ( 1/2) xy + ...

Key concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. They measure how a function changes as one specific variable changes while keeping other variables constant.

For a function of two variables, say \(f(x, y)\), the partial derivative with respect to \(x\) is denoted as \(\frac{\partial f}{\partial x}\) and with respect to \(y\) as \(\frac{\partial f}{\partial y}\). These partial derivatives are used to understand the rate of change or the slope of the function along the \(x\) or \(y\) direction respectively.

To compute a partial derivative of a function like \(\cos(x + y)\), differentiate the function with respect to one variable while treating the other variable as a constant. For example:

• The partial derivative with respect to \(x\) is \(\frac{\partial}{\partial x} \cos(x+y) = -\sin(x+y)\)
• The partial derivative with respect to \(y\) is \(\frac{\partial}{\partial y} \cos(x+y) = -\sin(x+y)\)

In the Maclaurin series, partial derivatives of a function at 0 help form each term in the series expansion.

Taylor Series

The Taylor series is a mathematical concept used to approximate functions. It represents a function as an infinite sum of terms calculated from the values of its derivatives at a single point.

For a function of one variable, say \(f(x)\), the general form is:

• \(f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots\)

For a function of two variables, \(f(x, y)\), the series expansion becomes more complex and involves mixed partial derivatives:

• \(f(x, y) = f(0, 0) + \left(\frac{\partial f}{\partial x}\right)(0,0)x + \left(\frac{\partial f}{\partial y}\right)(0,0)y + \frac{1}{2!}\left(\frac{\partial^2 f}{\partial x^2}\right)(0,0)x^2 + \frac{1}{2!}\left(\frac{\partial^2 f}{\partial y^2}\right)(0,0)y^2 + \frac{1}{2!}\left(\frac{\partial^2 f}{\partial x \partial y}\right)(0,0)xy + \dots\)

The Maclaurin series is a special case of the Taylor series, where the expansion is about 0. This means we evaluate all derivatives at the point \((0,0)\).

Cosine Function

The cosine function, denoted as \(\cos(x)\), is a fundamental trigonometric function that describes the horizontal coordinate of a point on the unit circle as the angle \(x\) varies. Its important properties include periodicity, even symmetry, and the specific values it takes at key angles.

For instance, \(\cos(0) = 1\), which is used when evaluating the function at \((0,0)\) for our Maclaurin series expansion. Its derivatives show how the function behaves as the angle changes:

• \(\cos'(x) = -\sin(x)\)
• \(\cos''(x) = -\cos(x)\)
• \(\cos'''(x) = \sin(x)\)
• \(\cos^{(4)}(x) = \cos(x)\)

When dealing with \(\cos(x + y)\), we apply similar rules but as it is a composition, the resulting derivatives impact both \(x\) and \(y\). Calculating higher-order partial derivatives for series expansions leverages these trigonometric identities.

Two-Variable Functions

Functions of two variables, like \(f(x, y)\), depend on both \(x\) and \(y\) inputs to produce outputs. This makes them more complex than single-variable functions since they change in more than one direction.

Understanding such functions often involves plotting surfaces to visualize how outputs vary with changes in both inputs. In calculus, it includes computing partial derivatives, higher-order derivatives, and using them in series expansions such as the Maclaurin series.

For example, in the function \(\cos(x + y)\), every change in either \(x\) or \(y\) affects the output due to the combined argument \((x + y)\). Constructing the Maclaurin series involves deriving partial derivatives of various orders and evaluating them at \((0,0)\). This approach helps approximate the function near that point using polynomial terms derived from these partial derivatives.