Problem 4
Question
Using the binomial series gives $$\frac{z}{(1-z)}=z\left[1+3 z+\frac{3 \cdot 4}{2 !} z+\frac{3 \cdot 4 \cdot 5}{3 !} z^{3}+\cdots\right]=z+3 z^{2}+\frac{3 \cdot 4}{2 !} z^{3}+\frac{3 \cdot 4 \cdot 5}{3 !} z^{4}+\cdots \text { where } R=1$$
Step-by-Step Solution
Verified Answer
The series for \(\frac{z}{1-z}\) is: \(z + 3z^2 + \frac{3 \cdot 4}{2!}z^3 + \frac{3 \cdot 4 \cdot 5}{3!}z^4 + \ldots\) with radius of convergence \(R = 1\).
1Step 1: Understanding the Binomial Series
The binomial series for \((1 - z)^{-n}\) is often given as \(\sum_{k=0}^{\infty} \binom{n+k-1}{k} z^k\). Here, the function \(\frac{z}{1-z}\) can be rewritten in the form of a binomial series where \(n = 1\).
2Step 2: Applying the Binomial Series
For the function \(\frac{1}{1-z}\), we have a series expansion \(1 + z + z^2 + z^3 + \ldots\). Multiplying this expansion by \(z\), we get \(z(1 + z + z^2 + z^3 + \ldots) = z + z^2 + z^3 + \ldots\).
3Step 3: Identifying Coefficients
The coefficients of the series given in the problem match the format of the binomial coefficients for integer values in a factorial expression. Specifically, the term \(\frac{3 \cdot 4 \cdot 5}{3!}\) indicates a binomial coefficient with \(n = 3\).
4Step 4: Constructing the Series
Combine your insights to write \(\frac{z}{1-z}\) as \(z[1 + 3z + \frac{3 \cdot 4}{2!}z^2 + \frac{3 \cdot 4 \cdot 5}{3!}z^3 + \ldots]\). These terms follow the pattern of general binomial expansion terms \(\frac{(3+n)!}{n!}z^n\).
5Step 5: Determine the Radius of Convergence
The radius of convergence \(R\) for the expansion of \(\frac{1}{1 - z}\) is determined by where \(|z| < 1\). Since we factor out \(z\), the condition \(|z| < 1\) remains the same for \(\frac{z}{1-z}\), confirming that \(R = 1\).
Key Concepts
Series ExpansionBinomial CoefficientsRadius of ConvergenceFactorial Expression
Series Expansion
In mathematics, a series expansion is a way of expressing a function as an infinite sum of terms. This is very useful for analyzing and approximating functions, specifically when they are too complex to deal with directly. In the case of the binomial series, we expand a fraction into a form that reveals important information about the function. For example, if we look at the function \(\frac{z}{1-z}\), its series expansion is expressed as:
- \(z + 3z^2 + \frac{3 \cdot 4}{2!} z^3 + \frac{3 \cdot 4 \cdot 5}{3!} z^4 + \cdots\)
Binomial Coefficients
Binomial coefficients are essential in algebra, primarily due to their application in expansions as observed in binomial series. They are denoted by \( \binom{n}{k} \) and defined as the number of ways to choose \(k\) elements from a set of \(n\) elements without considering the order of selection. These coefficients appear within the binomial theorem which allows a power (integer) of a binomial expression to be expanded into a sum. Within the formula:
- \( \binom{n}{k} = \frac{n!}{k!(n-k)!}\)
Radius of Convergence
Radius of convergence is an important concept in determining the region where a power series converges. For the power series expansion of a function such as \(\frac{1}{1-z}\), this means identifying the values of \(z\) where the series will effectively sum up to produce a result. Typically, the radius of convergence tells us the interval within which a series can be used to represent a function, as seen in:
- The absolute value condition \(|z| < 1\)
Factorial Expression
A factorial expression is a mathematical notation that represents the product of an integer and all of the integers below it, denoted by \(n!\). So, \(3! = 3 \times 2 \times 1 = 6\), for instance. Factorials are integral in computing binomial coefficients and appear frequently in combinatorics and series expansions. In these expansions, especially with binomial series, factorial expressions dictate the coefficients in the series:
- \(\frac{3 \cdot 4 \cdot 5}{3!}\)
Other exercises in this chapter
Problem 1
$$f(z)=\frac{1}{z}\left(1-\frac{z^{2}}{2 !}+\frac{z^{4}}{4 !}-\frac{z^{6}}{6 !}+\cdots\right)=\frac{1}{z}-\frac{z}{2 !}+\frac{z^{3}}{4 !}-\frac{z^{5}}{6 !}+\cdo
View solution Problem 4
Write \(f(z)=z^{4}-16=\left(z^{2}-4\right)\left(z^{2}+4\right)=(z-2)(z+2)(z-2 i)(z+2 i)\) to see that \(2,-2,2 i,\) and \(-2 i\) are zeros of \(f .\) Now \(f^{\
View solution Problem 5
$$f(z)=e^{-2 / z^{2}}=\sum_{k=0}^{\infty} \frac{\left(-2 / z^{2}\right)^{k}}{k !}=\cdots-\frac{2^{3}}{3 ! z^{6}}+\frac{2^{2}}{2 ! z^{4}}-\frac{2}{1 ! z^{2}}+1 ;
View solution Problem 5
Replacing \(z\) in \(e^{z}=\sum_{k=0}^{\infty} \frac{z^{k}}{k !}\) by \(-2 z\) gives \(e^{-2 z}=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k !}(2 z)^{k}\) where \(R=\i
View solution