Chapter 3: Problem 8
Use the Taylor series representation of \(1 /(1-z)\) around \(z=0\) for \(|z|<1\) to find a series representation of \(1 /(1-z)\) for \(|z|>1\). (Hint: use \(1 /(1-z)=-1 /(z(1-1 / z))\)
Short Answer
Expert verified
\[ \frac{1}{1-z} = -\sum_{n=0}^{\infty} z^{-(n+1)} \ for \ |z| > 1 \]
Step by step solution
01
Rewrite the Function
First, consider the given hint: \[\frac{1}{1-z} = -\frac{1}{z(1-\frac{1}{z})}\] This form will help in transforming the series.
02
Use the Taylor Series Representation
Recall the Taylor series representation around \(|z|<1\) for \[\frac{1}{1-u} = \sum_{n=0}^{\infty} u^n \ for \ |u| < 1\] Substitute \ u = \frac{1}{z}\, which we can do because we are reformulating the series for \ |z| > 1\.
03
Substitute into the Series
Substitute \ u = \frac{1}{z}\ in the series:\[\frac{1}{1-\frac{1}{z}} = \sum_{n=0}^{\infty} \left(\frac{1}{z}\right)^n = \sum_{n=0}^{\infty} z^{-n} \ for \ |z| > 1 \ and \ |\frac{1}{z}| < 1\]
04
Complete the Substitution
Now, substitute this series back into the rewritten function:\[\frac{1}{1-z} = -\frac{1}{z} \left(\sum_{n=0}^{\infty} z^{-n}\right) = -\sum_{n=0}^{\infty} z^{-(n+1)} \ for \ |z|> 1\]
05
Final Series Representation
Therefore, the series representation of \ frac{1}{1-z} \ for \ |z| > 1\ is:\[\frac{1}{1-z} = -\sum_{n=0}^{\infty} z^{-(n+1)}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
complex variables
In complex analysis, a **complex variable** is a variable whose values are complex numbers. A complex number has the form \(z = x + iy\), where \(x\) and \(y\) are real numbers, and \(i\) is the imaginary unit with the property \(i^2=-1\). Complex variables are fundamental in many areas of mathematics and physics.
This exercise involves the function \(\frac{1}{1-z}\), which is analyzed for both \(|z| < 1\) and \(|z| > 1\). These regions refer to the magnitude of the complex variable \(z\) in the complex plane. When dealing with complex variables, it's important to consider:
This exercise involves the function \(\frac{1}{1-z}\), which is analyzed for both \(|z| < 1\) and \(|z| > 1\). These regions refer to the magnitude of the complex variable \(z\) in the complex plane. When dealing with complex variables, it's important to consider:
- The real part and imaginary part of the variable
- The magnitude (or modulus), given by \(|z| = \sqrt{x^2 + y^2}\)
- The argument (or angle), denoted as \(\arg(z)\), which specifies the direction of the complex number
series expansion
A **series expansion** is the representation of a function as an infinite sum of terms. The Taylor series is a specific type of series expansion centered around a point (often zero). For instance, the Taylor series for \(\frac{1}{1-z}\) around \(z=0\) is:
\[\frac{1}{1-z} = \sum_{n=0}^{\infty} z^n \ \text{for} \ |z| < 1\]
This series converges when the magnitude of \(z\) is less than 1. In contrast, the challenge in our exercise is to find a suitable series expansion for \(\frac{1}{1-z}\) when \(|z| > 1\).
The steps are:
\[\frac{1}{1-z} = \sum_{n=0}^{\infty} z^n \ \text{for} \ |z| < 1\]
This series converges when the magnitude of \(z\) is less than 1. In contrast, the challenge in our exercise is to find a suitable series expansion for \(\frac{1}{1-z}\) when \(|z| > 1\).
The steps are:
- Rewriting the function with a useful substitution
- Using a known series expansion
- Substituting and simplifying expressions
analytic continuation
**Analytic continuation** is a method to extend the domain of a given analytic (complex differentiable) function. It involves finding a new function that matches the original one within its initial domain.
For our series, the original Taylor series \(\frac{1}{1-z}\) expands around \(z=0\) for \(|z| < 1\). To continue this analytically for \(|z| > 1\), we:
For our series, the original Taylor series \(\frac{1}{1-z}\) expands around \(z=0\) for \(|z| < 1\). To continue this analytically for \(|z| > 1\), we:
- Reformulate the function in a manner suitable for the new domain
- Apply the series expansion to the transformed function
- Combine and simplify terms
radius of convergence
The **radius of convergence** is the range within which a power series converges. For a series \(\sum_{n=0}^{\infty} a_n z^n\), the radius of convergence \(R\) is defined by:
\[\frac{1}{R} = \limsup_{n \rightarrow \infty} \sqrt[n]{|a_n|}\]
In simpler terms, it tells us the distance from the center (in this case, \(z=0\)) within which the series provides valid, convergent values.
In the case of \(\frac{1}{1-z}\) and its Taylor series around \(z=0\), the radius of convergence is 1, applicable for \(|z| < 1\).
Our task extends representation beyond this circle. For \(|z| > 1\), we transform the function using substitutions. The new series obtained, \(\sum_{n=0}^{\infty} z^{-(n+1)}\), converges for \(|z| > 1\), effectively shifting the radius of convergence to this region.
Understanding the radius of convergence is vital as it ensures the correctness and applicability of the series expansion.
\[\frac{1}{R} = \limsup_{n \rightarrow \infty} \sqrt[n]{|a_n|}\]
In simpler terms, it tells us the distance from the center (in this case, \(z=0\)) within which the series provides valid, convergent values.
In the case of \(\frac{1}{1-z}\) and its Taylor series around \(z=0\), the radius of convergence is 1, applicable for \(|z| < 1\).
Our task extends representation beyond this circle. For \(|z| > 1\), we transform the function using substitutions. The new series obtained, \(\sum_{n=0}^{\infty} z^{-(n+1)}\), converges for \(|z| > 1\), effectively shifting the radius of convergence to this region.
Understanding the radius of convergence is vital as it ensures the correctness and applicability of the series expansion.