Chapter 11: Problem 9
Find the Maclaurin series for the functions in Exercises \(9-20 .\) $$ e^{-x} $$
Short Answer
Expert verified
The Maclaurin series for \( e^{-x} \) is \( \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} x^n \).
Step by step solution
01
Understand Maclaurin Series
The Maclaurin series is a type of Taylor series expansion of a function about 0. It is given by the formula \[ f(x) = f(0) + \frac{f'(0)}{1!} x + \frac{f''(0)}{2!} x^2 + \frac{f'''(0)}{3!} x^3 + \cdots \] To find the Maclaurin series of \( e^{-x} \), we need to find the derivatives of \( e^{-x} \) and evaluate them at \( x = 0 \).
02
Compute Derivatives
Calculate the first few derivatives of \( e^{-x} \):- First derivative, \( f'(x) = -e^{-x} \).- Second derivative, \( f''(x) = e^{-x} \).- Third derivative, \( f'''(x) = -e^{-x} \).- Fourth derivative, \( f^{(4)}(x) = e^{-x} \).Observe the pattern that the derivatives alternate in sign and are either \( e^{-x} \) or \( -e^{-x} \).
03
Evaluate Derivatives at 0
Evaluate the function and its derivatives at \( x = 0 \).- \( f(0) = e^{0} = 1 \).- \( f'(0) = -e^{0} = -1 \).- \( f''(0) = e^{0} = 1 \).- \( f'''(0) = -e^{0} = -1 \).- Pattern continues as noted previously.
04
Write Maclaurin Series
Substitute the values into the Maclaurin series formula:\[e^{-x} = 1 + \frac{-1}{1!}x + \frac{1}{2!}x^2 + \frac{-1}{3!}x^3 + \frac{1}{4!}x^4 + \cdots\]This can be rewritten as:\[e^{-x} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} x^n \]where \( (-1)^n \) denotes alternating signs.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Taylor series
The Taylor series is a powerful tool used in mathematics to represent functions as infinite sums of terms calculated from the values of their derivatives at a single point. It's a way to approximate complex functions with simpler polynomial expressions. The basic form of a Taylor series for a function \( f(x) \) centered around a point \( a \) is:
- \( f(a) \)
- \( \frac{f'(a)}{1!}(x-a) \)
- \( \frac{f''(a)}{2!}(x-a)^2 \)
- \( \cdots \)
Derivatives
Understanding derivatives is crucial to forming the Taylor or Maclaurin series of a function. In essence, derivatives measure how a function changes as its input changes, essentially capturing the rate of change or the slope of a function at any given point.
- First derivative \( f'(x) \): Measures the rate of change of the function.
- Second derivative \( f''(x) \): Shows how the rate of change itself changes; think of it as the acceleration or curvature.
- Higher-order derivatives \( f^{(n)}(x) \) continue to capture these changes.
Function expansion
Function expansion is the process of expressing a complex function as a series of simpler terms. This concept underlies the Taylor and Maclaurin series, providing insights into how a given function behaves around a certain point. By expanding functions using derivatives, mathematicians can:
- Create simpler models of complex functions (e.g., approximating \( e^{-x} \) with a polynomial).
- Analyze the behavior of functions closely around a point, offering solutions in calculus and differential equations.
- Enhance computational efficiency, allowing computers to approximate functions rapidly.
Infinite series
Infinite series consist of an ongoing sum of terms that can be extended without limit. They play a fundamental role in calculus and mathematical analysis by providing ways to perform calculations and understand functions beyond simple algebraic expressions.
- Convergence: An infinite series converges if the sum of the terms approaches a finite value as more terms are added.
- Divergence: If an infinite series does not converge, it diverges, meaning it doesn't settle on a finite number.