Question

Find the Maclaurin polynomial of degree \(n\) for the given function. $$f(x)=e^{-x}, \quad n=3$$

Short answer

The Maclaurin polynomial is \(P_3(x) = 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3\).

Step-by-step solution

Understand the Maclaurin Series

The Maclaurin series for a function \(f(x)\) is a special case of the Taylor series where it is expanded around 0. The general formula is:\[f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots + \frac{f^{(n)}(0)}{n!}x^n\]Our task is to find this series up to \(n=3\) for \(f(x) = e^{-x}\).

Compute the Derivatives

Find the first few derivatives of \(f(x) = e^{-x}\). - First derivative: \(f'(x) = -e^{-x}\)- Second derivative: \(f''(x) = e^{-x}\)- Third derivative: \(f'''(x) = -e^{-x}\)These derivatives alternate in sign and maintain the form \(\pm e^{-x}\).

Evaluate Derivatives at \(x = 0\)

Substitute \(x = 0\) into the derivatives:- \(f(0) = e^{0} = 1\)- \(f'(0) = -e^{0} = -1\)- \(f''(0) = e^{0} = 1\)- \(f'''(0) = -e^{0} = -1\)

Construct the Maclaurin Polynomial

Using the evaluated derivatives, construct the polynomial:\[P_3(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3\]Substitute their values:\[P_3(x) = 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3\]

Conclude the Maclaurin Polynomial

The Maclaurin polynomial of degree 3 for the function \(e^{-x}\) is:\[P_3(x) = 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3\]This polynomial estimates \(e^{-x}\) near \(x = 0\).

Key concepts

Taylor Series

The Taylor series is a powerful mathematical tool used to represent functions as infinite sums of terms. Each term is derived from the function's derivatives at a single point. This point is often denoted as "a." In the Maclaurin series, which is a special case of the Taylor series, the point "a" is zero. This means that the Taylor series becomes:

• Maclaurin series: the function is expanded around \(x = 0\).
• Each term involves derivatives evaluated at zero.
• The series provides a way to approximate complex functions.

For example, the function \(e^{-x}\) can be expressed through its Maclaurin series. By considering only a finite number of terms, we get a polynomial that offers a good approximation near \(x = 0\).

Polynomial Approximation

Polynomial approximation involves using polynomials to estimate more complicated functions. It means truncating an infinite series to use only a few terms, simplifying calculations while still capturing essential characteristics of the function.

With the Maclaurin polynomial, we focus on approximating the function near a specific point, often zero. In this case, the challenge was to find a polynomial of degree 3 for \(f(x) = e^{-x}\).

• This results in the polynomial: \(P_3(x) = 1 - x + \frac{1}{2}x^2 - \frac{1}{6}x^3\).
• The approximations become more accurate with an increased number of terms.
• At \(x = 0\), the polynomial perfectly replicates the behavior of \(e^{-x}\).

Thus, polynomial approximations are a valuable technique for simplifying the study of complex functions by breaking them down into more manageable, polynomial forms.

Exponential Function

The exponential function \(e^x\) is one of the most important functions in mathematics, characterized by its constant rate of growth. For every point, its derivative is identical to the function itself.

The function \(e^{-x}\) is particularly interesting because it mirrors the behavior of \(e^x\), but the roles of growth and decay are reversed. In the Maclaurin polynomial approximation for \(e^{-x}\), the terms:

• Reflect the nature of exponential decay.
• Provide a simplified way to understand the function near \(x = 0\).
• Alternate in sign due to the alternating derivatives.

This allows for an efficient way to estimate \(e^{-x}\) without complex calculations. Such understanding aids in modeling real-world scenarios, from calculating continuously compounding interest to understanding decay processes in physics and biology.