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

  • #1
RJLiberator
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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.
 

Answers and Replies

  • #2
stevendaryl
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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?
 
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  • #3
RJLiberator
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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
stevendaryl
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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
RJLiberator
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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
RJLiberator
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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
RJLiberator
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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
stevendaryl
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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.
 

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