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Chapter 30: Problem 76

Show that the radius of convergence of the binomial series is 1 .

Short Answer

Expert verified
The radius of convergence of the binomial series is 1.

Step by step solution

01

- Define the Binomial series

The general binomial series can be described as: \[ (1+x)^{\alpha} = 1 + \alpha x + \frac{\alpha (\alpha - 1)x^{2}}{2!} + \frac{\alpha (\alpha - 1)(\alpha - 2)x^{3}}{3!} + \dots\] where the coefficient of the nth term is \[a_{n} = \frac{\alpha(\alpha - 1) \dots (\alpha - n + 1)}{n!} x^{n}\]
02

- Apply the Ratio Test

To find the radius of convergence, apply the Ratio Test. The Ratio Test requires solving for the absolute value of the limit of the ratio of the (n+1)th term to the nth term as n approaches infinity. So the ratio \(|a_{n+1}/a_{n}|\) becomes \( \left| \frac{ \frac{\alpha(\alpha - 1) \dots (\alpha - n + 1)}{n!} x^{n} }{ \frac{\alpha(\alpha - 1) \dots (\alpha - n)}{(n+1)!} x^{n+1} } \right|\). Simplify this to \( |x| \cdot \left| \frac{\alpha - n}{n+1} \right|\).
03

- Take the limit

Find the limit of the absolute value of the ratio \( |x| \cdot \frac{\alpha - n}{n+1} \) as n approaches infinity. This limit gives \( |x| \cdot 0 \) which simplifies to 0. Therefore according to the ratio test, since the limit is less than 1, the series converges.
04

- Find the radius of convergence

The radius of convergence is the reciprocal of the limit result, in this case 0. Since anything divided by 0 tends towards infinity, the interval of convergence is (-1, 1). The radius of convergence, being half the length of this interval, is r = 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Series
The binomial series is a way of expanding expressions raised to any power, not just integers. It's written as:
  • \[(1+x)^{\alpha} = 1 + \alpha x + \frac{\alpha (\alpha - 1)x^{2}}{2!} + \cdots\]
In this series, each term increases in power by 1 and involves a coefficient based on \(\alpha\), which can be any real number. This makes it flexible and widely used in mathematical calculations. The nth-term coefficient is given by:
  • \[\frac{\alpha(\alpha - 1) \cdots (\alpha - n + 1)}{n!} x^{n}\]
Understanding the terms' coefficient is crucial, as they determine how rapidly each term grows or shrinks, impacting convergence.
Ratio Test
The Ratio Test is a method used to determine the convergence of a series. It's useful for series where each term is a function of the previous one, like the binomial series. The basic idea is to look at the ratio of consecutive terms to check their behavior as they move towards infinity.To apply the test:
  • Calculate \(|a_{n+1}/a_{n}|\) for the series terms.
  • Find the limit as n approaches infinity of this ratio.
For convergence, the limit must be less than 1.In our example, the ratio simplifies to:
  • \(|x| \cdot \left| \frac{\alpha - n}{n+1} \right|\)
As \(n\) grows, \(\alpha - n\) becomes very negative, leading the limit to zero. Since 0 is less than 1, the series converges.
Convergence
Convergence in the context of series means that as we add more terms, the series approaches a finite value. It's important to understand whether a series converges, as it tells us about the stability and applicability of the series for different values.In the binomial series example:
  • The limit \(|x| \cdot \frac{\alpha - n}{n+1}\) as \(n\) approaches infinity is 0.
  • This is less than 1, indicating convergence within the interval of \(-1 < x < 1\).
Convergence ensures the series behaves well within this range, making calculations reliable.
Power Series
A power series is an infinite series of the form:
  • \[\sum_{n=0}^{\infty} c_n x^n\]
Each term involves a power of \(x\), and these series are extremely useful in approximating functions. The convergence of a power series depends on \(x\) and the series coefficients.In the context of the binomial series:
  • The series takes a power series form when expanded around 0.
  • The convergence radius tells us the interval where the series gives accurate results.
By examining the power series, we can understand how well it approximates functions like \((1+x)^{\alpha}\) within a specific range.

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Most popular questions from this chapter

Let \(\sum_{k=1}^{\infty} a_{k}\) be a series and \(S_{n}=\sum_{k=1}^{n} a_{k}\) its \(n\) th partial sum, where \(n=1,2,3, \ldots\). For each of the following, decide whether or not enough information is given to assure that \(\sum_{k=1}^{\infty} a_{k}\) converges. \(M\) and \(m\) are constants. Explain your reasoning. (a) \(a_{k}>0\) for all \(k\) and \(S_{n}>m\) for all \(n\). (b) \(a_{k}>0\) for all \(k\) and \(S_{n}m\) for all \(n\). (d) \(m

Determine whether the series converges or diverges. It is possible to solve Problems 4 through 19 without the Limit Comparison, Ratio, and Root Tests. \(\sum_{k=10}^{\infty} \frac{10}{k \sqrt{k}}\)

Determine whether the series converges or diverges. In this set of problems knowledge of the Limit Comparison Test is assumed. \(\sum_{k=3}^{\infty} \frac{k}{2 k^{3}-2}\)

According to Einstein s theory of special relativity, the mass of an object moving with velocity \(v \mathrm{~m} / \mathrm{s}\) is given by $$ m=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}} $$ where \(m_{0}\) is the mass of the object at rest and \(c\) is the speed of light, \(c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\). (a) Use the rst degree Taylor polynomial for \(\frac{1}{\sqrt{1+x}}\) to arrive at the estimate $$ m \approx m_{0}+\frac{m_{0}}{2} \frac{v^{2}}{c^{2}} $$ (b) If an object is moving at \(100 \mathrm{~m} / \mathrm{s}\), nd an upper bound for the error involved in using the approximation given in part (a).

Determine whether the series converges or diverges. In this set of problems knowledge of all the convergence tests from the chapter is assumed. \(\sum_{k=1}^{\infty} \frac{k^{3}}{k !}\)
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