Question

In Exercises \(7-18\), find the Maclaurin polynomial of degree \(n\) for the function. $$ f(x)=e^{-x}, \quad n=3 $$

Short answer

The Maclaurin polynomial of degree \(3\) for the function \(f(x) = e^{-x}\) is given by \(f(x) = 1 - x + \frac{x^2}{2} - \frac{x^3}{6}\)

Step-by-step solution

Getting the function's derivatives

First, we need to find the first four (as \(n\) is 3, we go one step further to \(n+1\)) derivatives of the function \(f(x) = e^{-x}\) at \(x = 0\). The differentiation rule for an exponential function is clear, we differentiate the exponent and multiply it. This means the \(n\)th derivative of \(f(x)\) will be \((-1)^n e^{-x}\). Now, we substitute \(0\) for \(x\) in each derivative to get the coefficients of the Maclaurin series. This gives us \(f(0) = 1\), \(f'(0) = -1\), \(f''(0) = 1\), and \(f'''(0) = -1\), \(f''''(0) = 1\)

Constructing the Maclaurin polynomial

Now, the Maclaurin series for \(f(x)\) can be written as:\[f(x) = f(0) + f'(0)x + f''(0)\frac{x^2}{2!} + f'''(0)\frac{x^3}{3!}\]Substituting our coefficients from Step 1:\[f(x) = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!}\]Finally, where \(x = -x\), the appearance of minus signs alternates between terms, an important property for the exponential function.

Simplifying the polynomial

By simplifying the result from Step 2, we obtain the final polynomial: \[f(x) = 1 - x + \frac{x^2}{2} - \frac{x^3}{6}\]

Key concepts

Exponential Function

In calculus, exponential functions are pivotal in understanding growth and decay phenomena across various disciplines. An exponential function can be written in the form of f(x) = a^x, where a is a constant base and x represents the exponent. One special case is the natural exponential function, denoted as e^x, where e is Euler's number approximately equal to 2.71828.

Exponential functions have unique properties that make them essential in calculus, like the fact that their rate of growth is proportional to their current value, which leads to the function's own derivative being a multiple of itself. That is, the derivative of e^x is e^x again. This self-replicating feature is evident in the exercise provided, where the function to be explored is f(x) = e^{-x}, an exponential function with decay instead of growth due to the negative exponent.

Maclaurin Series

The Maclaurin series is a special case of the Taylor series, centered at zero. It provides a powerful method for approximating functions with a polynomial called the Maclaurin polynomial. The Maclaurin series of a function f(x) is written as an infinite sum of terms of the form \(\frac{f^{(n)}(0)}{n!}x^n\), where n! denotes the factorial of n, and f^{(n)}(0) is the n-th derivative of the function evaluated at x=0.

Constructing a Maclaurin series entails finding these derivatives at zero and assembling them into an infinite sum. However, for practical purposes, we often stop at a certain degree to obtain a polynomial approximation. In the exercise, the Maclaurin polynomial of degree 3 gives a tangible approximation for the function f(x) = e^{-x} using the first four non-zero terms of the series.

Derivatives in Calculus

Derivatives represent the cornerstone of calculus, involving the study of how functions change—that is, rates of change and slopes of curves. Fundamentally, a derivative measures how a function value changes as its input changes. The act of finding a derivative is called differentiation.

In the context of the exercise, differentiation plays a critical role in finding the coefficients of the Maclaurin polynomial. Remembering that the n-th derivative of an exponential function f(x) = e^x is the function itself, we can apply the rule for differentiation to find higher-order derivatives that will be needed for constructing the Maclaurin series, allowing us to understand the local behavior of the function through its polynomial approximation.