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Tricky Series Questions

  1. Oct 23, 2007 #1
    My professor was going over some powers relating to series, and power series. He came across the problem below which he didn't even know how to solve. I am trying to figure out how to solve it, but can't get anywhere. Does anyone know how to approach this problem?

    Express this series as a function of x.
    [tex]\sum n^2x^n [/tex] from n=1 to [tex]\infty[/tex]

    So, f(x) = ?
    Last edited: Oct 24, 2007
  2. jcsd
  3. Oct 23, 2007 #2


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    come on. who's your professor, george w bush?
  4. Oct 24, 2007 #3


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    I've blanked once or twice when teaching a class before. Its embarrasing, particularly when its easy, but I've seen it happen to field medal winners
  5. Oct 24, 2007 #4
    Woops sorry guys. I forgot to post the actually question. The guestion is to express that series as a function of x.
  6. Oct 24, 2007 #5

    Gib Z

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    Quite simple really, [tex]f(x) = \sum_{n=1}^{\infty} n^2 x^n [/tex].

    Looking for a nicer (ie closed form) answer? There is no such closed form answer, unless you are looking for values of x where [itex]|x| <1[/itex], in which case there may be.
  7. Oct 24, 2007 #6


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    It's not just that there is no "closed form answer"- that series only converges for |x|< 1.
  8. Oct 24, 2007 #7
    I am not looking for the radius of convergence. I am looking how to epxress that series as a function.
  9. Oct 25, 2007 #8

    Gib Z

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    I was alluding to that point =]

    Don't you like my answer?
  10. Oct 27, 2007 #9
    Just to see if the summation converges to a single function, I Maple-ed it, and Maple gives a solution to [tex]f(x)=\sum_{n=1}^{\infty} n^2 x^n[/tex]. HINT: It is the quotient of a quadratic polynomial and a cubic polynomial.

    Perhaps this identity can be of some help (from Wikipedia):
    [tex]\sum_{i=0}^n i^2 x^i = \frac{x}{(1-x)^3} (1+x-(n+1)^2x^n+(2n^2+2n-1)x^{n+1}-n^2x^{n+2})[/tex].

    This simpler identity helps too (from here):
    [tex]\sum_{i=0}^\infty i^2 x^i = \frac{x^2+x}{(1-x)^3}[/tex]

    EDIT: Do a little manipulation to the above identity, and you'll end up with your solution.
    Last edited: Oct 27, 2007
  11. Nov 5, 2007 #10
    My word,
    1 + x + x^2 + x^3 + .. . + x^n + ... = (1-x)^(-1)

    differentiating and multiplying both sides by x we have,

    1+ 2x^2 + 3x^3 + ... + nx^n + ... = x (1-x)^(-2)
    do the same again and you'll have a result valid inside the unit disc with the origin as a centre.
  12. Nov 5, 2007 #11
    Can one go backwards from the series to the quotient of polynomials? Or must one have to know that each are equivalent from experience?
  13. Nov 5, 2007 #12


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    Here's how to derive it:

    Now, do the differentiations, and get the rational representation of f(x).
  14. Nov 6, 2007 #13
    Well we need an infinite series, S satisfying the following relation,

    S = 1 + xS

    Not all that far off, it can be more 'traditionally' derived by considering the finite geometric progression and evaluating the limit. I've provided a simple derivation that can be easily followed.
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