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Sum of n-th roots

  1. Nov 15, 2007 #1
    Does anyone know how to sum a*r^(1/n) for all n?
     
  2. jcsd
  3. Nov 15, 2007 #2
    Well, if you mean the roots of the equation X^n-r = 0, you should look up symmetric functions.
     
  4. Nov 15, 2007 #3
    I can't see how symmetric functions would help. I'm looking for a closed form solution for the given sum, in the sense that the infinite sum of a*r^n = a*(1-r^(n+1))/(1-r), I'm looking for the infinite sum of a*r^(1/n).
     
  5. Nov 15, 2007 #4

    Hurkyl

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    Could you write what you mean, rather than abbreviating it? I can't tell precisely what you mean, and my best guesses for what you mean are very obviously not convergent sums.
     
  6. Nov 15, 2007 #5
    What does \sum_{k=0}^{n} a*r^(1/k) equal? Given that |r| < 1, a and r are constants.
    In the sense that the geometric progression \sum_{k=0}^{n} a*r^k equals a*(1-r^(n+1))/(1-r).

    Cheers
     
  7. Nov 15, 2007 #6
    The sum of roots of unity is zero.
     
  8. Nov 15, 2007 #7
    I know that, what about sums of roots of other numbers?
     
  9. Nov 15, 2007 #8
    The n'th roots of any real number, say r, is [itex]r^{\frac{1}{n}} \zeta_n^k[/itex] where [itex]\zeta_n[/itex] is the primitive nth root of unity. So what will happen when you sum them?

    Edit: [tex] 0\leq k \leq n-1 [/tex]
     
  10. Nov 15, 2007 #9
    Thanks Kreizhn! should have noticed that myself =)
     
  11. Nov 15, 2007 #10
    Wait a second, what exactly should I have noticed? I'm summing over n not k.
     
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