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Closed Form of Power Series

  • Thread starter Yagoda
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  • #1
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Homework Statement


Using that [itex]\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n [/itex] for |x|<1 and that
[itex] f'(x) =\sum_{n=0}^{\infty} (n+1)a_{n+1}(x-x_0)^n [/itex], write [itex] \sum_{n=0}^{\infty} n^2x^n [/itex] in closed form.


Homework Equations





The Attempt at a Solution

In this series, [itex]a_n = n^2 [/itex] and [itex]x_0 = 0 [/itex]. Applying the theorem I get that [itex] f'(x) =\sum_{n=0}^{\infty} (n+1)(n+1)^2(x)^n = \sum_{n=0}^{\infty} (n+1)^3(x)^n[/itex]. I know I want to try to apply the sum of the geometric series and then integrate to get f(x) (or maybe those things in reverse order), but the [itex](n+1)^3[/itex] is giving me trouble.
 

Answers and Replies

  • #2
CompuChip
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I think you failed to define f(x) - is that 1/(1 - x) or a general function ##f(x) = \sum_{n = 0}^\infty a_n x^n## ?

Consider this:
$$\sum_{n = 0}^\infty n^2 x^n = \sum_{n = 1}^\infty n^2 x^n = \sum_{n = 0}^\infty (n + 1)^2 x^{n + 1} = x \sum_{n = 0}^\infty (n + 1) (n + 1) x^n$$

and try taking it from there.
 
  • #3
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I am considering f(x) to be the closed form of the series that I'm looking for so I use the theorem to find its derivative f'(x) and then hopefully find f itself. So what I've got so far is that given f'(x), [itex]f(x) =\sum_{n=0}^{\infty} (n+1)^2(x)^n [/itex], which tell us that [itex]f(x) =\sum_{n=0}^{\infty} (n+1)^2(x)^n =\sum_{n=0}^{\infty} n^2x^n [/itex] since that was the original series, if that makes sense.
Maybe this is an elementary question, but is there a way to apply the geometric series identity to your expression to get a closed form or does some other type manipulation that I'm not seeing have to be done?
 
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  • #4
Ray Vickson
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I am considering f(x) to be the closed form of the series that I'm looking for so I use the theorem to find its derivative f'(x) and then hopefully find f itself. So what I've got so far is that given f'(x), [itex]f(x) =\sum_{n=0}^{\infty} (n+1)^2(x)^n [/itex], which tell us that [itex]f(x) =\sum_{n=0}^{\infty} (n+1)^2(x)^n =\sum_{n=0}^{\infty} n^2x^n [/itex] since that was the original series, if that makes sense.
Maybe this is an elementary question, but is there a way to apply the geometric series identity to your expression to get a closed form or does some other type manipulation that I'm not seeing have to be done?
[tex] n^2 x^n = x \frac{d}{dx}\left( x \frac{d}{dx} x^n \right). [/tex]
Alternatively, if ##g(x) = \sum x^n,## look at the series for ##g'(x)## and ##g''(x).## Can you see how to get ##\sum n^2 x^n## from those two series?
 

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