Help me decipher what this problem is asking? (Power Series)

In summary, the power series centered at a=0 with interval of convergence (-1, 1) can be represented as 1/(1-x^2). By factoring out x^2, the series can be written as x^2(1+2x+3x^2+...). To find the sum of this series, one can take derivatives and multiply the numerator and denominator by x^2 to make it identical to the series 1/(1-x).
  • #1
RJLiberator
Gold Member
1,095
63

Homework Statement



Consider the power series centered at
a= 0:
Σkx^(k+1)
From 1 to infinity

(a) Find its radius of convergence, R, and its interval of convergence. = DONE

(b) For x in the interval (-R,R) find the sum of the power series.
Help?

Homework Equations


N/a

The Attempt at a Solution



So here's what I have already. For Part a, I calculated the radius of convergence to be 1 and the interval to be from -1<x<1. So we know that R = 1 and it's asking us the sum of the power series (-1, 1).

The problem I am having conceptually is, how do you get the sum when x is -1 and 1?

My only idea is to find the sum of the power series when x = 1 and the sum when x=-1 and then add them?
Is this correct?

If I am not thinking this correctly, please help me decipher what the question is asking. Thank you.
 
Physics news on Phys.org
  • #2
I don't think that there is a uniform way to solve such problems as evaluating an infinite series. The only way I know how to do it is to play around with the series and see if you can get it into a form that you know.

Here's a hint: Do you know how to evaluate the following series: [itex]S(x) = \sum_k x^k[/itex]? Do you know how to do something to [itex]S(x)[/itex] to make it more like your series?
 
  • Like
Likes RJLiberator
  • #3
Well, if you expand out the exponent. x^(k+1) you get x^k*x which is more similar.

It would be the sum of k[x^k*x] which is not identical, but closer.
 
  • #4
RJLiberator said:
Well, if you expand out the exponent. x^(k+1) you get x^k*x which is more similar.

It would be the sum of k[x^k*x] which is not identical, but closer.

Two questions:
  1. Do you know what the infinite sum S(x) = 1 + x + x2 + x3 + ... adds up to?
  2. What happens when you take a derivative of S(x)?
 
  • #5
Well, the derivative of x^k is very similar to my function of k*x^(k-1).

To answer your questions
1) The sum adds up to (1/1-x)
2) The derivative of this is 1/(x-1)^2

I am starting to sort of understand what is going on here. We need to use the x^n series to calculate this power series. I'm almost here, but may need another hint or two.

I guess, I don't understand how the problem can ask for the sum when it is with x in an interval. (-1, 1). Should I try to find the sum at -1 and the sum at 1 and then add them?

EDIT: Well, actually, the series DIVERGES on x=-1, 1, but anything in between it converges. :/
 
  • #6
I may have just made a breakthrough!

So we know that 1/(1-x) series is very similar. If we replace x with kx^(k+1) then we have our series represented, correct?

The representation for this series is 1/(1-kx^(k+1))

Now:
1) Am I correct about this?
2) How do I evaluate it at -R, R when they both diverge? :/
Do I just plug in -1 and 1 into this as x and receive the answers of :

-1 = 1/(1-k(-1)^(k+1))
1 = 1/(1-k)

Hm...
 
  • #7
I've learned that this problem is NOT asking for an exact number, but merely a general 'formula' representation of if.

Use the series 1/(1-x) for the interval of convergence.

Take the derivative and multiple numerator and denominator by x^2 to make the series identical .

try taking some derivatives.

Differentiate term by term.

x^2/(1-x)^2 may be the answer
 
  • #8
RJLiberator said:
I may have just made a breakthrough!

So we know that 1/(1-x) series is very similar. If we replace x with kx^(k+1) then we have our series represented, correct?

The representation for this series is 1/(1-kx^(k+1))

Now:
1) Am I correct about this?
2) How do I evaluate it at -R, R when they both diverge? :/
Do I just plug in -1 and 1 into this as x and receive the answers of :

-1 = 1/(1-k(-1)^(k+1))
1 = 1/(1-k)

Hm...

FIrst of all, the series in question does NOT converge when [itex]x=\pm 1[/itex]. I think that you should interpret the question as asking what it converges to when [itex] -1 < x < +1[/itex].

Second, you can't replace [itex]x[/itex] by [itex]k x^{k+1}[/itex], because [itex]k[/itex] is not a fixed number.

Third: Your series is [itex]\sum_k k x^{k+1} = 0 + x^2 + 2 x^3 + ...[/itex]. Note that you can factor out [itex]x^2[/itex]. Try writing it in a factored form.
 

What is a power series?

A power series is an infinite series of the form ∑(n=0 to ∞) an(x-c)n, where x is a variable, c is a constant, and an is a sequence of constants. It is used to represent functions as an infinite polynomial.

What is the purpose of using a power series?

Power series are used to approximate functions that are difficult to work with, such as trigonometric functions or exponential functions. They allow us to break down a complex function into simpler, polynomial terms.

How do you determine the radius of convergence for a power series?

The radius of convergence, denoted as R, can be determined by using the ratio test. If the absolute value of the ratio of consecutive terms in the series approaches a finite value as n approaches infinity, then the series is convergent within a radius of R.

What is the difference between the interval of convergence and the radius of convergence?

The interval of convergence is the range of x values for which the power series converges. The radius of convergence is the distance from the center of the series, denoted as c, to the edge of the interval of convergence.

Can a power series be used to solve differential equations?

Yes, power series can be used to approximate solutions to differential equations. By substituting the power series into the differential equation and solving for the coefficients, we can obtain a solution in the form of an infinite polynomial.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
189
  • Calculus and Beyond Homework Help
Replies
2
Views
711
  • Calculus and Beyond Homework Help
Replies
1
Views
259
  • Calculus and Beyond Homework Help
Replies
7
Views
709
  • Calculus and Beyond Homework Help
Replies
3
Views
287
  • Calculus and Beyond Homework Help
Replies
22
Views
3K
  • Calculus and Beyond Homework Help
Replies
8
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
Back
Top