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Chapter 6: Problem 3

Use known results to expand the given function in a Maclaurin series. Give the radius of convergence \(R\) of each series. $$ f(z)=\frac{1}{(1+2 z)^{2}} $$

Short Answer

Expert verified
The Maclaurin series is \(1 - 4z + 12z^2 - 32z^3 + \ldots\) with radius of convergence \(R = \frac{1}{2}\).

Step by step solution

01

Recognize the Basic Series

The function can be expanded using the binomial series. The binomial series is given by \[(1+x)^{-n} = \sum_{k=0}^{\infty} \binom{-n}{k} x^k = 1 - nx + \frac{n(n+1)}{2!}x^2 - \frac{n(n+1)(n+2)}{3!}x^3 + \ldots\]In our case, we need to match this formula to \[(1+2z)^{-2}.\]
02

Identify the Parameters

We observe that in our function \((1+2z)^{-2}\), we have \(x = 2z\) and \(n = 2\). This fits the binomial series with \(-n = -2\).
03

Write the Series Expansion

We substitute \(x = 2z\) and \(n = 2\) into the binomial series formula:\[(1+2z)^{-2} = \sum_{k=0}^{\infty} \binom{-2}{k} (2z)^k.\]We use the binomial coefficient \(\binom{-2}{k} = \frac{(-2)(-3)(-4)\ldots(-2-k+1)}{k!}\).
04

Simplify the Binomial Coefficients

Calculate the initial terms:- For \(k=0\): \(\binom{-2}{0} = 1\).- For \(k=1\): \(\binom{-2}{1} = -2\).- For \(k=2\): \(\binom{-2}{2} = 3\).- etc.Now, plug these into the series:\[1 - 4z + 12z^2 - 32z^3 + \ldots,\]where each term matches the expansion coefficient from the binomial series formula.
05

Determine the Radius of Convergence

The binomial series \((1+2z)^{-2}\) has a radius of convergence \(R = \frac{1}{|2|} = \frac{1}{2}\), since the series converges for \(|x| < 1\) and \(x = 2z\) here implies \(|2z| < 1\) or \(|z| < \frac{1}{2}\).

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Key Concepts

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

Radius of Convergence
The radius of convergence is a crucial parameter when working with power series. It tells us the interval within which the series converges to the function it represents.
  • For a power series centered at zero, the radius of convergence is the largest value of \(r\) such that the series converges for all \(|z| < r\).
  • In our example, we used the binomial series with \(x = 2z\). For the series \((1+2z)^{-2}\) to converge, \(|2z| < 1\) must be true, simplifying to \(|z| < \frac{1}{2}\).
This radius of convergence, \(R = rac{1}{2}\), indicates the region of the complex plane in which the series \((1+2z)^{-2}\) uniformly converges. Remember, outside this circle, the series may diverge or fail to represent the function accurately.
Binomial Series
The binomial series is a versatile tool for expanding expressions of the form \( (1+x)^{-n} \). This formula allows us to deal with fractional and negative exponents in elegant series form.
  • The general form of the binomial series is \( (1+x)^{-n} = \sum_{k=0}^{\infty} inom{-n}{k} x^k\).
  • The coefficients are calculated using binomial coefficients, such as \( \binom{-n}{k} = rac{(-n)(-n-1)(-n-2)...(-n-k+1)}{k!} \).
Applying this to the function \( (1+2z)^{-2} \), with \( x=2z \) and \( n=2 \, \ -n=-2 \), provides a series expansion that expresses the function in an infinite sum of simpler terms. This makes it easier to analyze and solve problems involving complex functions.
Complex Analysis
Complex analysis is a fascinating area of mathematics that extends the real number system to the complex numbers. This field examines functions of complex numbers, offering powerful results for real-world applications in physics, engineering, and beyond. Some core concepts include:
  • **Holomorphic Functions**: These are functions that are complex differentiable in a neighborhood of every point in their domain. They exhibit many beautiful properties, such as being infinitely differentiable and having power series expansions.
  • **Contour Integrals**: Integrals of complex functions that are computed along a path or contour in the complex plane. They are integral to many theorems in complex analysis.
  • **Residue Theorem**: A key tool for evaluating complex integrals, it relates contour integrals to the sum of residues of a function's singularities within the contour.
In the context of our exercise, understanding complex functions and their properties allows us to expand into the Maclaurin series effectively, gaining insights into the behavior of functions beyond what real analysis alone would offer.

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Most popular questions from this chapter

In Problems 11-14, the indicated number is a zero of the given function. Use a Maclaurin or Taylor series to determine the order of the zero. $$ f(z)=z\left(1-\cos ^{2} z\right) ; z=0 $$

In Problems 1-6, use an appropriate Laurent series to find the indicated residue. $$ f(z)=\frac{2}{(z-1)(z+4)} ; \operatorname{Res}(f(z), 1) $$

Determine the order of the poles for the given function. $$ f(z)=5-\frac{6}{z^{2}} $$

In Problems 1-4, show that \(z=0\) is a removable singularity of the given function. Supply a definition of \(f(0)\) so that \(f\) is analytic at \(z=0\). $$ f(z)=\frac{e^{2 z}-1}{z} $$

Consider the function \(f(z)=\frac{1}{(z-5)^{3}}\). What is the Laurent series expansion of \(f\) about \(z_{0}=5\) that is valid on the annulus \(0<|z-5|<\infty\) ?
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