Chapter 10: Problem 1
Using known Taylor series, find the first four nonzero terms of the Taylor series about 0 for the function. $$e^{-x}$$
Short Answer
Expert verified
The first four nonzero terms are: \( 1 - x + \frac{x^2}{2} - \frac{x^3}{6} \).
Step by step solution
01
Recall the Taylor Series Formula
The Taylor series for a function \( f(x) \) about 0 (also known as the Maclaurin series) is given by: \[ f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots \] We will use this formula to find the first four nonzero terms of the Taylor series for \( e^{-x} \).
02
Find the Derivatives of the Function
Compute the first few derivatives of the function \( e^{-x} \): - \( f(x) = e^{-x} \) - \( f'(x) = -e^{-x} \) - \( f''(x) = e^{-x} \) - \( f'''(x) = -e^{-x} \) - \( f^{(4)}(x) = e^{-x} \) Observe the pattern that the magnitude of each derivative remains \( e^{-x} \), alternating signs.
03
Evaluate Derivatives at Zero
Find the value of these derivatives at \( x = 0 \): - \( f(0) = e^{0} = 1 \) - \( f'(0) = -e^{0} = -1 \) - \( f''(0) = e^{0} = 1 \) - \( f'''(0) = -e^{0} = -1 \) - \( f^{(4)}(0) = e^{0} = 1 \)
04
Write the Terms of the Series
Using the derivatives, write out the first four nonzero terms of the series: \[ f(x) = 1 - \frac{1}{1!}x + \frac{1}{2!}x^2 - \frac{1}{3!}x^3 + \cdots \] Simplifying, we have: \[ 1 - x + \frac{x^2}{2} - \frac{x^3}{6} \]
05
Conclusion
The first four nonzero terms of the Taylor series for \( e^{-x} \) about \( 0 \) are \( 1 - x + \frac{x^2}{2} - \frac{x^3}{6} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Maclaurin Series
A Maclaurin series is a specific case of the Taylor series, which is expanded around the point zero. It's a way to represent functions as infinite sums of terms calculated from the function's derivatives at a single point. This helps in approximating functions by a polynomial, especially when the function itself might be complicated to deal with directly. For a given function like \( f(x) \), the Maclaurin series expansion can be represented as:
- \( f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \cdots \)
Derivative Evaluation
Calculating derivatives is a crucial step in forming the Maclaurin series. Derivatives measure how a function changes as its input changes. When you take the derivatives of a function, you explore how sensitive the function is to changes in its input.For our function \( e^{-x} \), we observe the derivatives are:
- \( f(x) = e^{-x} \)
- \( f'(x) = -e^{-x} \)
- \( f''(x) = e^{-x} \)
- \( f'''(x) = -e^{-x} \)
- \( f^{(4)}(x) = e^{-x} \)
Function Expansion
Function expansion is the process of expressing a function in terms of basic components, like power series. It's essentially breaking a complex function down into an infinite sum of simpler terms.By using the derivatives calculated at zero, you can write the Maclaurin series for \( e^{-x} \) as:
- \( f(x) = 1 - \frac{1}{1!}x + \frac{1}{2!}x^2 - \frac{1}{3!}x^3 + \cdots \)
- \( 1 - x + \frac{x^2}{2} - \frac{x^3}{6} \)
Exponential Functions
Exponential functions, like \( e^x \) or \( e^{-x} \), are functions where a constant base is raised to the power of the variable \( x \). They are ubiquitous in mathematics, especially in growth, finance, and natural phenomena.The function \( e^{-x} \) decreases rapidly as \( x \) increases, reflecting exponential decay. Expanding it into a series form helps to estimate its values without a calculator or further simplifies its integration or differentiation.Such functions have derivative patterns that are straightforward but recurrent, a feature that aids series expansion. When dealing with exponential functions, recognizing inherent properties, such as constant proportional derivatives, assists in forming accurate series representations.