Power Series (Which test can i use to determine divergence at the end points)

In summary, the power series representation of the function f(-4x)= 1/(1+4x) was found using the geometric series and the limit of the series was determined using the ratio test. The endpoints for the series were tested and it was determined that they were divergent by the divergence test. This answer is also acceptable and another possible answer is that the endpoints were divergent by the geometric series because r is equal to 1.
  • #1
yeny
7
0
Hello,

I was given f(-4x)= 1/(1+4x), and I used the geometric series to find the power series representation of this function. I then took the limit of (-4x)^k by using ratio test. The answer is abs. value of x. So -1/4<x<1/4

I then plugged in those end points to the series going from k=0 to infinity of (-4x)^k

here's where I'm stuck. How do I determine convergence/divergence of the endpoints?

When I tested x=-1/4, my series was k=0 to infinity of (1)^k, for that series, I wrote " Divergent by divergence test because lim as k --> infinity does not equal zero.

Is that an acceptabe answer? I also had another possible answer which was, Divergent by geometric series because r is less than or equal to 1"

Thank you so much for taking the time to look at this. Hope you all have a wonderful weekend =)
 
Physics news on Phys.org
  • #2
"When I tested x=-1/4, my series was k=0 to infinity of (1)^k, for that series, I wrote " Divergent by divergence test because lim as k --> infinity does not equal zero."
Yes, that is completely valid

"Is that an acceptabe answer? I also had another possible answer which was, Divergent by geometric series because r is less than or equal to 1"
First, a geometric series is convergent for r< 1. Did you mean "divergent because r is larger than or equal to 1"? I would see no reason to include the "larger than". You are specifically talking about r= 1.

Or, simply, the partial sums are [tex]S_n= \sum_{k= 1}^n 1^k= n[/tex]. What is the limit of that as n goes to infinity.
 
Last edited:

FAQ: Power Series (Which test can i use to determine divergence at the end points)

1. What is a power series?

A power series is an infinite series of the form ∑n=0^∞ cn(x-a)n, where cn are coefficients, x is a variable, and a is a constant called the center of the series. It is a type of mathematical series commonly used in calculus and other branches of mathematics.

2. How can I determine the convergence or divergence of a power series?

There are several tests that can be used to determine the convergence or divergence of a power series. These include the ratio test, the root test, the integral test, and the comparison test. Each test has its own criteria and conditions for determining convergence or divergence, so it is important to understand and apply them correctly.

3. Can I use these tests to determine divergence at the end points of a power series?

Yes, these tests can also be used to determine the convergence or divergence at the end points of a power series. However, it is important to note that some tests may only be applicable at the interior points of a series, so it is necessary to check the conditions of each test before applying it to the end points.

4. Is there a specific test that is best for determining divergence at the end points?

No, there is no specific test that is best for determining divergence at the end points of a power series. It is recommended to try different tests and see which one gives the most conclusive result. In some cases, it may be necessary to use multiple tests to fully determine the convergence or divergence of a power series.

5. What should I do if the tests for determining divergence at the end points give conflicting results?

If the tests for determining divergence at the end points of a power series give conflicting results, it is best to use other methods such as the Cauchy-Hadamard theorem or the radius of convergence formula. It is also important to carefully check the conditions and assumptions of each test and make sure they are being applied correctly.

Similar threads

Replies
5
Views
541
Replies
17
Views
3K
Replies
1
Views
1K
Replies
3
Views
1K
Replies
10
Views
4K
Replies
3
Views
1K
Replies
6
Views
2K
Back
Top