Chapter 11: Problem 6
Use the definition of a Taylor series to find the first four nonzero terms of the series for \(f(x)\) centered at the given value of \(a .\) $$ f(x)=\frac{1}{1+x}, \quad a=2 $$
Short Answer
Expert verified
The first four nonzero terms are \( \frac{1}{3} - \frac{1}{9}(x-2) + \frac{1}{27}(x-2)^2 - \frac{1}{81}(x-2)^3 \).
Step by step solution
01
Identify the Function and Point of Expansion
We are given the function \( f(x) = \frac{1}{1+x} \) and we need to find the Taylor series centered at \( a = 2 \). The Taylor series formula is: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n \] where \( f^{(n)}(a) \) is the nth derivative of \( f \) evaluated at \( a \).
02
Calculate Derivatives
Compute the derivatives of \( f(x) = \frac{1}{1+x} \):- \( f(x) \) is \( \frac{1}{1+x} \).- The first derivative \( f'(x) = -\frac{1}{(1+x)^2} \).- The second derivative \( f''(x) = \frac{2}{(1+x)^3} \).- The third derivative \( f'''(x) = -\frac{6}{(1+x)^4} \).- The fourth derivative \( f^{(4)}(x) = \frac{24}{(1+x)^5} \).
03
Evaluate Derivatives at a = 2
Evaluate each derivative at \( x = a = 2 \):- \( f(2) = \frac{1}{1+2} = \frac{1}{3} \).- \( f'(2) = -\frac{1}{9} \).- \( f''(2) = \frac{2}{27} \).- \( f'''(2) = -\frac{6}{81} = -\frac{2}{27} \).- \( f^{(4)}(2) = \frac{24}{243} = \frac{8}{81} \).
04
Apply the Taylor Series Formula
Substitute the derivatives evaluated at \( a = 2 \) into the Taylor series formula up to the fourth term:- First term: \( \frac{f(2)}{0!}(x-2)^0 = \frac{1}{3} \)- Second term: \( \frac{f'(2)}{1!}(x-2) = -\frac{1}{9}(x-2) \)- Third term: \( \frac{f''(2)}{2!}(x-2)^2 = \frac{2}{27 \times 2}(x-2)^2 = \frac{1}{27}(x-2)^2 \)- Fourth term: \( \frac{f'''(2)}{3!}(x-2)^3 = -\frac{2}{27 \times 6}(x-2)^3 = -\frac{1}{81}(x-2)^3 \)
05
Write the First Four Nonzero Terms
Combine the terms to write the first four nonzero terms of the Taylor series:\[\frac{1}{3} - \frac{1}{9}(x-2) + \frac{1}{27}(x-2)^2 - \frac{1}{81}(x-2)^3\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivatives
Derivatives are fundamental to understanding Taylor series. They measure how a function changes as its input changes. In simpler terms, the derivative represents the slope of the function's graph.
- The first derivative, denoted as \( f'(x) \), gives the rate of change of the function.
- The second derivative, \( f''(x) \), provides information on how the rate of change itself is changing, which can tell us whether the slope is increasing or decreasing.
- The third and higher-order derivatives keep adding layers of how much the previous derivative is changing.
- The first derivative: \( f'(x) = -\frac{1}{(1+x)^2} \).
- The second derivative: \( f''(x) = \frac{2}{(1+x)^3} \).
Power Series
A power series is an infinite series of the form \( \sum_{n=0}^{\infty} a_n (x-c)^n \), where \( a_n \) are coefficients and \( c \) is the center of the series. Power series can represent functions as an infinite sum of terms.With Taylor series, the power series becomes a practical tool for approximation.
- Each term in the power series captures a different level of the function's behavior.
- These series can be very beneficial in approximating functions that are difficult to compute directly.
Function Expansion
Function expansion transforms a complex function into a series that is easier to work with. Taylor series is one of the most popular methods of function expansion in calculus.Using derivatives, a function \( f(x) \) can be expanded about a point \( a \) as a sum of polynomial terms:
- Constant term: represents value of the function at \( a \).
- Linear and higher terms: help to better approximate the function by incorporating slopes and curvatures.
- \( \frac{1}{3} - \frac{1}{9}(x-2) + \frac{1}{27}(x-2)^2 - \frac{1}{81}(x-2)^3 \).
Calculus
Calculus is a branch of mathematics focused on change and motion, playing a key role in fields like physics and engineering. It consists of two main areas: differential calculus and integral calculus.
When discussing Taylor series, we primarily engage with:
- Differential calculus: studies the concept of derivatives and the rates of change.
- Integral calculus: not directly involved here, but important if considering the summation of series.