Problem 2 Find the disk of convergence for... [FREE SOLUTION]

Chapter 2: Problem 2

Find the disk of convergence for each of the following complex power series. $$z-\frac{z^{2}}{2}+\frac{z^{3}}{3}-\frac{z^{4}}{4}+\cdots$$

Short Answer

Expert verified

The disk of convergence is $\left| z \right| < 1$.

Step by step solution

- Identify the general term

The given power series can be written in the general form. Identify the general term of the series. For the given series \[z - \frac{z^{2}}{2} + \frac{z^{3}}{3} - \frac{z^{4}}{4} + \cdots \], the general term can be expressed as: \[ a_n = \frac{(-1)^{n+1} z^n}{n} \]

- Use the Ratio Test for convergence

To find the radius of convergence, use the Ratio Test, which involves considering the limit of $|a_{n+1}/a_n|$: \[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(-1)^{(n+2)} z^{(n+1)}}{(n+1)} \cdot \frac{n}{(-1)^{(n+1)} z^n} \right| \] Simplify the expression inside the limit.

- Simplify the limit

Simplify the limit expression: \[\lim_{n \to \infty} \left| \frac{(-1)^{(n+2)} z^{(n+1)}}{(n+1)} \cdot \frac{n}{(-1)^{(n+1)} z^n} \right| = \lim_{n \to \infty} \left| \frac{z^{(n+1)}}{(n+1)} \cdot \frac{n}{z^n} \right| = \lim_{n \to \infty} \left| \frac{n \, z}{n+1} \right| = \left| z \right| \lim_{n \to \infty} \left| \frac{n}{n+1} \right| = \left| z \right| \]

- Determine the radius of convergence

The Ratio Test states that the series converges when this limit is less than 1: \[\left| z \right| < 1\]This inequality implies that the radius of convergence $R$ is 1.

- Conclusion

Conclude that the disk of convergence is given by the set of $z$ such that $\left| z \right| < 1$.

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Key Concepts

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

Ratio Test

To determine the convergence of a complex power series, one useful method is the Ratio Test. This test involves taking the limit of the ratio of successive terms in the series. Specifically, for a series with terms $a_n$, we calculate \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\.
The result of this limit helps us figure out whether the series converges or diverges:

If the limit is less than 1, the series converges.
If the limit is greater than 1, the series diverges.
If the limit equals 1, the test is inconclusive.

In the specific problem of finding the disk of convergence for the series $z - \frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} + \cdots$, the Ratio Test is applied to the terms of the series.

Radius of Convergence

The radius of convergence is a critical concept in understanding where a power series converges. This radius, denoted by \R\, defines a circle within which the series will converge.
Using the Ratio Test for the series $z - \frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} + \cdots $, we find:
\begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| &= \lim_{n \to \infty} \left| \frac{(-1)^{(n+2)} z^{(n+1)}}{(n+1)} \cdot \frac{n}{(-1)^{(n+1)} z^n} \right| \end{align*}
This cleans up to
\[ \left| z \right| < 1 \] This implies the radius of convergence $R = 1$.

Disk of Convergence

The disk of convergence is the region in the complex plane where a power series converges. For the series given in the example, after finding the radius of convergence $R = 1$, we determine that the series converges for all points inside the circle of radius 1 centered at the origin. This is written mathematically as:

\left| z \right| < 1

Meaning, the series will converge for all $z$ values that satisfy this condition. The term 'disk' comes from imagining a disk shape in the complex plane formed by these points.

Power Series

A power series is a type of series that takes the form:
\a_0 + a_1 z + a_2 z^2 + a_3 z^3 + \cdots\
Here, each term is composed of constant coefficients $a_n$ and powers of $z$. The series provided in the problem statement can be rewritten in this general form:
\sum_{n=1}^{\infty} \frac{(-1)^{n+1} z^n}{n}\br>A complex power series converges when $z$ is within a particular range, described by the radius of convergence. Outside of this range, the series may diverge.

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