Convergence on the unit circle

In summary: This means that the root test is also inconclusive. However, we can still use other methods to determine the behavior of convergence.One approach is to look at the terms of the series and see if they approach 0 as n approaches infinity. In this case, we can see that as n gets larger, the term | \frac{z^{n}}{n^{
  • #1
Trevor Vadas
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Homework Statement



Determine the behavior of convergence on the unit circle, ie |z| = 1 of:

Ʃ [itex]\frac{z^{n}}{n^{2}(1 - z^{n})}[/itex]

Homework Equations



Obviously this is divergent then z is a root of unity. The question is what happens when z is not a root of unity.

The Attempt at a Solution


My thought was originally that it would converge. Now I think it may diverge. To show this I look at
| [itex]\frac{z^{n}}{n^{2}(1 - z^{n})}[/itex] | = [itex]\frac{1}{n^{2}|1 - z^{n}|}[/itex], since |z| = 1, and show this does not converge to 0 and hence the series must diverge.

now for ε = 1. I want to show that z[itex]^{n}[/itex] will land close enough to 1(for infinitely many n) so that
[itex]\frac{1}{n^{2}|1 - z^{n}|}[/itex] > 1.

or there exists infinitely many n such that [itex]\frac{1}{(|1 - z^{n}|}[/itex] [itex]\geq[/itex] n[itex]^{2}[/itex]

Homework Statement



We know that with |z|=1, z[itex]^{n}[/itex] is dense on the unit circle. Hence for any n there exsists an m that would make
[itex]\frac{1}{(|1 - z^{m}|}[/itex] [itex]\geq[/itex] n[itex]^{2}[/itex] true.
Now can we show the stars will allign and get m to equal n infinitely many times.

Thanks for any help.
 
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  • #2


Thank you for your question. When considering the behavior of convergence on the unit circle, we must first determine the values of z for which the series will converge. As you correctly noted, the series will diverge when z is a root of unity. This is because when z is a root of unity, z^n will cycle through the same values and the denominator, 1-z^n, will approach 0, making the series divergent.

However, when z is not a root of unity, the behavior of the series becomes more complicated. We can use the ratio test to determine the behavior of convergence for this series. The ratio test states that if the limit of |a_{n+1}/a_n| as n approaches infinity is less than 1, then the series will converge. If the limit is greater than 1, then the series will diverge. If the limit is equal to 1, the test is inconclusive and we must use other methods to determine the behavior of convergence.

Applying this test to our series, we have:

| \frac{z^{n+1}}{(n+1)^{2}(1 - z^{n+1})} | / | \frac{z^{n}}{n^{2}(1 - z^{n})} | = | \frac{z}{(n+1)^{2}}| * | \frac{1 - z^{n}}{1 - z^{n+1}}|

Since |z| = 1, we can simplify this to:

| \frac{1 - z^{n}}{1 - z^{n+1}}| = | \frac{1 - z^{n}}{1 - z^{n}z}| = 1

This means that the limit of |a_{n+1}/a_n| is equal to 1, and the ratio test is inconclusive. Therefore, we must use other methods to determine the behavior of convergence.

One method we can use is the root test, which states that if the limit of the nth root of |a_n| as n approaches infinity is less than 1, then the series will converge. If the limit is greater than 1, then the series will diverge. If the limit is equal to 1, the test is inconclusive.

Applying this test to our series, we have:

lim_{n->∞} (|
 

FAQ: Convergence on the unit circle

What is "Convergence on the unit circle"?

Convergence on the unit circle is a mathematical concept that refers to the behavior of a sequence of points that approach a specific point on the unit circle, which has a radius of 1. The sequence of points can approach the point from any direction on the circle.

How is "Convergence on the unit circle" different from other types of convergence?

Unlike other types of convergence, which focus on the behavior of a sequence of points approaching a specific value, convergence on the unit circle focuses on the behavior of a sequence of points approaching a specific point on the unit circle.

What is the significance of "Convergence on the unit circle" in mathematics?

"Convergence on the unit circle" is significant in mathematics because it is a fundamental concept for understanding the behavior and limits of complex numbers. It is also used in various applications, such as signal processing and circuit analysis.

How is "Convergence on the unit circle" related to trigonometric functions?

Convergence on the unit circle is closely related to trigonometric functions, as the points on the unit circle can be represented using trigonometric functions such as sine and cosine. The behavior of these functions as a sequence of points approach a specific point on the unit circle is essential in understanding convergence on the unit circle.

Are there any real-world applications of "Convergence on the unit circle"?

Yes, convergence on the unit circle has various real-world applications, such as in the analysis of oscillating systems, electrical engineering, and physics. It is also used in digital signal processing to remove noise from signals and improve their accuracy.

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