Find the power series for representatino for the function

In summary: My suggestion to write out a few terms isn't meant to be a proof. It's meant to help you get an idea of what kind of series you are dealing with. Then you prove it.
  • #1
vande060
186
0

Homework Statement



Ive been able to do every single problem in my homework to the point where I have to test the edges of the interval of convergence. I have not been able to figure out a single one of the problems at the point of testing the edge of convergence, and I am to the point of pulling my hair out, so I really need an explanation of what to do at that point. Ill give an example of a problem.find the power series for representation for the function

f(x) = x/(9+x^2)

Homework Equations


The Attempt at a Solution



f(x) = x/(9+x2)

f(x) = x/(9 -(-x2)

f(x) = x/9(1-(-x2/9)

(∞, n=0) (x/9)∑ (-1)n [(x2)/(9)] n

(∞, n=0) ∑ (-1)n (x2n+1)/ (9n+1)

then i do a quicktest to find the radius of convergence

lim n--> ∞ |cn/cn+1|

lim n--> ∞ |[(-1)n/(9n+1)] * [(9n+2)/(-1)n+1] = 9

|x2| < |9|

-3 <x < 3

this is where i start to feel bad, i know the radius of convergence is 3 and the interval of convergence is (-3,3), but I have to test the edges of convergence, and I can't seem to do it. I know i should plug -3 and 3 back into the power series and test for convergence, but i have no idea how to test these two series for convergence

(∞, n=0) ∑ (-1)n ((-3)2n+1)/ (9n+1)

(∞, n=0) ∑ (-1)n ((3)2n+1)/ (9n+1)
 
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  • #2
What are 3^(2n) and (-3)^(2n) expressed as powers of 9?
 
  • #3
Dick said:
What are 3^(2n) and (-3)^(2n) expressed as powers of 9?

9^n and 9^n(∞, n=0) ∑ (-1)n ((9)n+1/2)/ (9n+1)

which diverges by alternating series test because bn+1< bn and limit runs to 0, right?

so interval of convergence is (-3,3)
 
  • #4
vande060 said:
9^n and 9^n(∞, n=0) ∑ (-1)n ((9)n+1/2)/ (9n+1)

which diverges by alternating series test because bn+1< bn and limit runs to 0, right?

so interval of convergence is (-3,3)

Why don't you explicitly write out the first three or four terms in each series and then think about that again? You don't need an alternating series test to conclude that they diverge.
 
  • #5
Dick said:
Why don't you explicitly write out the first three or four terms in each series and then think about that again? You don't need an alternating series test to conclude that they diverge.

unfortunately I do :( Professor said that writing out the first few terms of a series is not enough to draw conclusions. We have to use a test, and cite what test we used, for every problem. I get where you are coming from though, and thanks for the help, I cracked out a few more problems on my homework after working through this one :D
 
  • #6
Don't forget that for some problems you can differentiate or integrate the power series of a simpler function.
 
  • #7
vande060 said:
unfortunately I do :( Professor said that writing out the first few terms of a series is not enough to draw conclusions. We have to use a test, and cite what test we used, for every problem. I get where you are coming from though, and thanks for the help, I cracked out a few more problems on my homework after working through this one :D

My suggestion to write out a few terms isn't meant to be a proof. It's meant to help you get an idea of what kind of series you are dealing with. Then you prove it.
 

1. What is a power series representation for a function?

A power series representation for a function is an infinite polynomial expression that can be used to approximate the value of the function at any point within its radius of convergence. It is written in the form of a summation, with the variable raised to different powers and multiplied by coefficients.

2. How do you find the power series representation for a function?

To find the power series representation for a function, you need to first determine the function's radius of convergence. This can be done using tests such as the ratio test or the root test. Once the radius of convergence is known, the coefficients of the power series can be found by repeatedly differentiating the function and evaluating at the center of the series.

3. What is the radius of convergence and why is it important?

The radius of convergence represents the interval of values for which the power series will converge to the actual value of the function. It is important because it determines the range of values for which the power series representation can be used to approximate the function accurately.

4. Can a function always be represented by a power series?

No, not all functions can be represented by a power series. The function must be analytic, meaning it is infinitely differentiable at every point within its domain. If a function is not analytic, it cannot be represented by a power series.

5. How does the power series representation differ from the Taylor series representation?

The power series representation is a special case of the Taylor series representation, where the center of the series is located at x=0. The Taylor series can have a center at any point within the domain of the function, while the power series is centered at the origin. Additionally, the coefficients in the power series are found by evaluating derivatives of the function at the center, while the Taylor series coefficients are found by evaluating derivatives at any chosen point within the domain.

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