/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 Use the binomial series to expan... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use the binomial series to expand the function as a power series. State the radius of convergence. \(\frac{1}{(2+x)^{3}}\)

Short Answer

Expert verified
Power series: \(\sum_{n=0}^{\infty} \binom{-3}{n} \left(\frac{x}{2}\right)^n\). Radius of convergence: 2.

Step by step solution

01

Identify the Binomial Series Pattern

The binomial series expansion for \((1+x)^m\) is given by \(1 + \binom{m}{1}x + \binom{m}{2}x^2 + \cdots\). We need our function in the form of \((1+x)^m\). Start by rewriting \(\frac{1}{(2+x)^{3}}\) as \((\frac{1}{2})^3 (1 + \frac{x}{2})^{-3}\). This identifies that we need to expand \((1+u)^{-3}\) where \(u = \frac{x}{2}\).
02

Apply the Binomial Series Expansion

The binomial series expansion for \((1+u)^{m}\) is: \(\sum_{n=0}^{\infty} \binom{m}{n} u^n\). Here \(m = -3\), and \(u = \frac{x}{2}\), so it becomes \(\sum_{n=0}^{\infty} \binom{-3}{n} \left(\frac{x}{2}\right)^n\).
03

Compute Binomial Coefficients

The binomial coefficient for negative integers can be calculated using \(\binom{m}{n} = \frac{m(m-1)(m-2)...(m-n+1)}{n!}\). For \(\binom{-3}{n}\), it simplifies to: \(\frac{(-3)(-4)(-5)...(-3-n+1)}{n!}\).
04

Write the Power Series Expression

Putting it all together, we write the power series: \[ \frac{1}{(2+x)^3} = \left(\frac{1}{8}\right) \sum_{n=0}^{\infty} \binom{-3}{n} \left(\frac{x}{2}\right)^n = \left(\frac{1}{8}\right) \left(1 - \frac{3x}{2} + \frac{3 \times 4 x^2}{8} - \frac{3 \times 4 \times 5 x^3}{48} + \cdots \right).\]
05

Determine the Radius of Convergence

The binomial series \((1+u)^{-3}\) converges when \(|u| < 1\), here \(u = \frac{x}{2}\), so the condition is \(|\frac{x}{2}| < 1\) which implies \(|x| < 2\). Thus, the radius of convergence is 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.

Power Series
A power series is a type of infinite series that can be represented in the form: \[\sum_{n=0}^{fty} a_n (x - c)^n\]where:
  • \(a_n\) are the coefficients of the series, which can be constants or expressions.
  • \(x\) is the variable.
  • \(c\) is the center of the series, which is the point at which the series is expanded.
Power series are incredibly useful in mathematics, particularly in calculus and analysis, as they allow complex functions to be expressed as a sum of simpler functions. This form is particularly handy when solving differential equations or evaluating functions at specific points.
In the context of the binomial series, we wrote \[ \frac{1}{(2+x)^3} \] as a power series by expressing it in the form \[ (1 + u)^{-3}, \] where \(u = \frac{x}{2}\). This transformation allowed us to use the power series expansion technique to simplify the expression into an infinite series.
The power series form enabled us to represent this function as a series involving powers of \(x\). This series could then be manipulated and used for further mathematical analysis.
Radius of Convergence
The radius of convergence is a key concept when dealing with power series. It indicates the region within which the series converges to a value. For a power series centered at \(c\), it converges for values of \(x\) such that \[|x - c| < R \]where \(R\) is the radius of convergence.
Determining the radius of convergence is crucial, as it tells us within what input range the power series can faithfully represent the function.
In our case, we were dealing with the transformed function \[\left(1 + \frac{x}{2}\right)^{-3}\]. To find its radius of convergence, we needed to determine where this binomial series would converge. Since the condition is \[|u| < 1\]for convergence, and \(u = \frac{x}{2}\), we had \[|\frac{x}{2}| < 1\] which simplifies to \[|x| < 2\]. Thus, the radius of convergence is 2, meaning the power series representation converges for all \(x\) values within this range.
Understanding the radius of convergence allows mathematicians and students to know the limitations of their series expansions and ensure calculations remain accurate within specified bounds.
Binomial Coefficients
Binomial coefficients are a fundamental part of the binomial series expansion used in our problem. They are typically written as \(\binom{m}{n}\) and are used to express the coefficients in a binomial expansion. The formula for binomial coefficients is: \[\binom{m}{n} = \frac{m(m-1)(m-2)...(m-n+1)}{n!}\]These are crucial for determining the terms of a binomial series, especially when dealing with series involving powers of functions.
In the context of the exercise, we were working with a negative integer, \(m = -3\). This requires calculating terms in a slightly different manner since traditional binomial coefficients are usually applied to positive integers. For negative \(m\), the coefficients still provide important information for the expansion.
  • Start with \(m = -3\) and apply the formula to find each \(\binom{-3}{n}\).
  • These coefficients are then used to determine each term in the series for powers of \((\frac{x}{2})^n\).
Understanding binomial coefficients allows us to effectively expand and manipulate series, providing a structured approach to handle complex algebraic expressions efficiently. This systematic approach is vital in working with infinite series.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the Taylor polynomial \(T_{n}(x)\) for the function \(f\) at the number a. Graph \(f\) and \(T_{3}\) on the same screen. $$f(x)=\frac{\ln x}{x}, \quad a=1$$

(a) Approximate \(f\) by a Taylor polynomial with degree \(n\) at the number a. (b) Use Taylor's Inequality to estimate the accuracy of the approximation \(f(x) \approx T_{n}(x)\) when \(x\) lies in the given interval. (c) Check your result in part (b) by graphing \(\left|R_{n}(x)\right|\) $$f(x)=\sqrt{x}, \quad a=4, \quad n=2, \quad 4 \leqslant x \leqslant 4.2$$

(a) Find the Taylor polynomials up to degree 6 for \(f(x)=\cos x\) centered at \(a=0 .\) Graph \(f\) and these polynomials on a common screen. (b) Evaluate \(f\) and these polynomials at \(x=\pi / 4, \pi / 2\) and \(\pi .\) (c) Comment on how the Taylor polynomials converge to \(f(x) .\)

Test the series for convergence or divergence. $$\sum_{n=1}^{\infty}(\sqrt[n]{2}-1)^{n}$$

The Cantor set, named after the German mathematician Georg Cantor \((1845-1918),\) is constructed as follows. We start with the closed interval \([0,1]\) and remove the open interval \(\left(\frac{1}{3}, \frac{2}{3}\right) .\) That leaves the two intervals \(\left[0, \frac{1}{3}\right]\) and \(\left[\frac{2}{3}, 1\right]\) and we remove the open middle third of each. Four intervals remain and again we remove the open middle third of each of them. We continue this procedure indefinitely, at each step removing the open middle third of every interval that remains from the preceding step. The Cantor set consists of the numbers that remain in \([0,1]\) after all those intervals have been removed. (a) Show that the total length of all the intervals that are removed is \(1 .\) Despite that, the Cantor set contains infinitely many numbers. Give examples of some numbers in the Cantor set. (b) The Sierpinski carpet is a two-dimensional counterpart of the Cantor set. It is constructed by removing the center one-ninth of a square of side \(1,\) then removing the centers of the eight smaller remaining squares, and so on. (The figure shows the first three steps of the construction.) Show that the sum of the areas of the removed squares is \(1 .\) This implies that the Sierpinski carpet has area \(0 .\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.