Chapter 9: Problem 5
Find the first four nonzero terms of the Taylor series for the functions. $$\left(1+\frac{x}{2}\right)^{-2}$$
Short Answer
Expert verified
The Taylor series is approximately \(1 - x + \frac{3}{4}x^2 - \frac{x^3}{2}\).
Step by step solution
01
Identify the Function and its Form
We want to find the Taylor series expansion for the function \( f(x) = \left(1 + \frac{x}{2} \right)^{-2} \). This resembles the binomial series expansion \((1 + u)^n\) where \( u = \frac{x}{2} \) and \( n = -2 \).
02
Write the Binomial Series Formula
The binomial series expansion for \((1 + u)^n\) is given by:\[ (1 + u)^n = \sum_{k=0}^{\infty} \binom{n}{k} u^k \]where \( \binom{n}{k} = \frac{n(n-1)(n-2)...(n-k+1)}{k!} \) is the binomial coefficient.
03
Calculate the Binomial Coefficients
For each term, calculate the binomial coefficients for \( n = -2 \):- For \( k = 0 \): \( \binom{-2}{0} = 1 \)- For \( k = 1 \): \( \binom{-2}{1} = -2 \)- For \( k = 2 \): \( \binom{-2}{2} = 3 \)- For \( k = 3 \): \( \binom{-2}{3} = -4 \)
04
Substitute and Simplify
Substitute \( u = \frac{x}{2} \) and expand to find the first four terms:- For \( k = 0 \): \( 1 \cdot \left(\frac{x}{2}\right)^0 = 1 \)- For \( k = 1 \): \( -2 \cdot \left(\frac{x}{2}\right)^1 = -x \)- For \( k = 2 \): \( 3 \cdot \left(\frac{x}{2}\right)^2 = \frac{3}{4}x^2 \)- For \( k = 3 \): \( -4 \cdot \left(\frac{x}{2}\right)^3 = -\frac{x^3}{2} \)
05
Write Out the Series
Compile the results from the previous step to write the first four non-zero terms of the series:\[ (1 + \frac{x}{2})^{-2} \approx 1 - x + \frac{3}{4}x^2 - \frac{x^3}{2} \]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Binomial Series Expansion
The binomial series expansion is a powerful tool used to express a power of a binomial expression in an infinite series form. Specifically, for any real or complex number \( n \), the expansion of \((1+u)^n \) is expressed as a sum of terms involving powers of \( u \). This is particularly useful when dealing with negative powers or non-integer exponents.For example, the expansion is given by:
- \[ (1+u)^n = \sum_{k=0}^{\infty} \binom{n}{k} u^k \]
Binomial Coefficients
Binomial coefficients are the integral part of the binomial series expansion formula, represented as \( \binom{n}{k} \). These coefficients indicate how the terms in a binomial expansion are weighted and can be calculated in several ways. One of the formulas to compute a binomial coefficient is:
- \[ \binom{n}{k} = \frac{n(n-1)(n-2)...(n-k+1)}{k!} \]
- For \( k=0 \), \( \binom{-2}{0} = 1 \)
- For \( k=1 \), \( \binom{-2}{1} = -2 \)
- For \( k=2 \), \( \binom{-2}{2} = 3 \)
- For \( k=3 \), \( \binom{-2}{3} = -4 \)
Nonzero Terms
The concept of finding the first nonzero terms in a series is essential, especially when approximating functions. For functions represented by Taylor or binomial series, only a finite number of starting terms are needed to accurately estimate values of the function near a particular point.In our example, we found the first four nonzero terms of the given function's expansion:
- \( 1 \), for \( k=0 \), representing no input variation
- \( -x \), for \( k=1 \), showing how the linear term adjusts the output
- \( \frac{3}{4}x^2 \), for \( k=2 \), emphasizing quadratic influence
- \( -\frac{x^3}{2} \), for \( k=3 \), reflecting cubic effects