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Limit of Complex summation

  1. May 4, 2010 #1
    I tried integration then applying limit as n tends to infinity, for k = 1, it becomes a circle, but as k increases, points decrease hence it should be wrong.

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  3. May 4, 2010 #2
    The summand is of the form [F(k) - F(k-1)], which is the sum of
    F(k) minus the sum of F(k-1). The second summation contains almost all the terms of the first summation.
  4. May 5, 2010 #3
    tried in this way, I am getting 0, but that shouldn't be...
  5. May 5, 2010 #4
    is it 2??
  6. May 5, 2010 #5
    What are the two terms that do not cancel?
  7. May 5, 2010 #6
    did lil. bit manipulation but it seems weird now.
  8. May 5, 2010 #7
    f(x) =sum | e^k - e^k/e| where e = e^2pi i/n n roots of unity.

    f(x) =sum |e^k|/|e| *[|e-1|] = n*[|e-1|] = 2n sin pi/n ; which diverges...
  9. May 5, 2010 #8
    Put f(k) = exp(2 pi i k/n)

    Then the summation is the difference of:

    f(1) + f(2) + ...f(n-1) +f(n)


    f(0) + f(1) + f(2)+....+f(n-1)

    If you subtract these two summations from each other, what are you left with?
  10. May 5, 2010 #9
    No, it isn't. There's an absolute sign, so it's really of the form |F(k) - F(k - 1)|. You need to take a common factor from both terms and use the properties of absolute values (|z1*z2| = |z1|*|z2| and so on). You should get an indeterminate expression of the form 0.[itex]\infty[/itex] and use L'Hospital's rule to obtain the limit.
  11. May 5, 2010 #10
    Ah, I see that I really do need new glasses!
  12. May 5, 2010 #11
    but as its modulus, common factor will 1; absolute value of n roots of equity will be 1.

  13. May 5, 2010 #12
    That's right, so the summand does not depend on k anymore, making it very easy to compute the sum.
  14. May 5, 2010 #13
    There is also a geometric way to see what the answer is: If you look at where the terms exp(2 pi i k/n) and exp(2 pi i (k-1)/n) on the complex plane are, you see that they are located on the unit circle. Then you can say something about the distance between the two points in terms of the polar angle difference between the two points when n is very large.
  15. May 5, 2010 #14
    No. What remains in the absolute value is a complex number whose absolute value is not equal to one. But, as a poster above me had noticed, it is independent of k, so it is just a sum of n identical terms. After you evaluate this finite sum, only then do you take the limit [itex]n \rightarrow \infty [/itex].
  16. May 5, 2010 #15
    its easier to solve this logically,

    the term is distance between two consecutive roots of unity, so, for n = 3 its equi triangle, 4 its square, so as n tends to infinity it becomes a circle, and sum of distance between all points will be its perimeter on unit circle.

    so the limit tends to 2 pi.
  17. May 6, 2010 #16
    geometrically the difference |f(k)-f(k-1)| is a particular edge of a polygon with n edges connecting unity roots.

    n=3 case:

    for n->∞ it approach a circle. because we add up the edges which are going to be infinitely short.

    sum |f k - f k-1| >= sum |f k| - sum |f k-1| if i'm not mistaken.
    and for unity roots z_k we have 0=p(z)= z^n-1, if one takes the linear factors it is obvious that the sum of unity roots z_k from k=0 to n-1 is zero and the product is (-1)^(n-1).
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