Problem 7
Question
In Exercises \(7-18\), find the Maclaurin polynomial of degree \(n\) for the function. $$ f(x)=e^{-x}, \quad n=3 $$
Step-by-Step Solution
Verified Answer
The Maclaurin polynomial of degree \(3\) for the function \(f(x) = e^{-x}\) is given by \(f(x) = 1 - x + \frac{x^2}{2} - \frac{x^3}{6}\)
1Step 1: 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 \(f'''(0) = -1\), \(f''''(0) = 1\)
2Step 2: 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.
3Step 3: 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 FunctionMaclaurin SeriesDerivatives in Calculus
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.
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.
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.
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.
Other exercises in this chapter
Problem 7
Write the first five terms of the sequence. \(a_{n}=\frac{(-1)^{n(n+1) / 2}}{n^{2}}\)
View solution Problem 7
Use the Integral Test to determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{1}{\sqrt{n+1}} $$
View solution Problem 7
Use the definition to find the Taylor series (centered at \(c\) ) for the function. $$ f(x)=\sin 2 x, \quad c=0 $$
View solution Problem 7
Determine the convergence or divergence of the series. $$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(n+1)}{\ln (n+1)} $$
View solution