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Distance between n-th Roots of Unity

  1. Mar 25, 2014 #1
    1. The problem statement, all variables and given/known data

    *Find the distance between 1 and the various n-th roots of unity - denoted d(k)

    *Find a formula for the sum of distances between 1 and each of the n-th roots of unity - denoted S(n)

    *Find the limit as n->infinity of (1/n).S(n)

    2. Relevant equations

    *The n-th roots of unity are z[0], z[1], z[2],...,z[n-1] - ie. there are 5 5th roots of unity. z[0] is always 1.

    *d(k) = abs{z(k)-1}

    *S(n) = sum(from k=1 to k=n-1) of d(k) - so n=5 has four terms in the sum

    3. The attempt at a solution

    By vector subtraction and finding the magnitude, I calculated:

    d(k)=2sin( k.pi / n )

    Hence:

    S(n) = 2.sum(from k=1 to k=n-1) of sin( k.pi / n )

    But now I'm stuck on the limit question. I know the answer is about 1.27 from calculating the first dozen term by calculato, however I can't see how to prove this.

    What I have is (where limit is as n goes to infinity):

    lim(1/n)S[n] = (2/n){sin(pi/n)+sin(2pi/n)+sin(3pi/n)+sin(4pi/n)+...+sin((n-1)pi/n)}

    There are n-1 terms in the braces. Every time I try solving it I end up with zero, which I know is not the case, since when I do a spreadsheet S[n]/n starts at 1 (the n=2 case) and increases with increasing n. I've tried decomposing sine into x-x^3/3!+x^5/5!+... but it doesn't seem to help.

    Is anyone able to steer me in the right direction?
     
  2. jcsd
  3. Mar 25, 2014 #2
    Hi bank! Welcome to PF!

    Your work looks good to me. For the third part, you have the following:

    $$\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{r=1}^{n-1}\left(2\sin\left(\frac{k\pi}{n}\right)\right)$$

    Can you see a way to convert the above to a definite integral? ;)
     
  4. Mar 25, 2014 #3
    Cheers, I'll have to have a look later, do you mean the whole thing, or just the part after the summation? I'll have to check my notes as I haven't done any analysis in a year.

    Until you mentioned that I was going to look into the sums for:

    1+2+3+4+5+...+(n-1) = ?
    1^3+2^3+3^3+...+(n-1)^3 = ?

    As if I can find those I'm on the way to an answer through the Taylor expansion of sine - or is the way you've mentioned much easier?
     
  5. Mar 25, 2014 #4
    Have you got any examples of using definite integrals to solve this sort of thing?
     
  6. Mar 25, 2014 #5
    I seem to be getting somewhere now.

    I'd still like to know how to do it the way you've suggested using definite integrals, but by using the strategy I was thinking of, that is expanding the sines as series and substituting the values for the integer power summations, I arrive at an answer for the limit that is equal to:

    SUM(from k=1 to infinity) -> [(-1)^n+1][pi^(2k-1)] / [k*((2k-1)!)]

    I've plugged this into Excel and it gives the correct answer, I just need to arrive at the correct derivation for the limit - how would I arrive at this (1.2734)?
     
  7. Mar 25, 2014 #6
    Sorry to keep you waiting, I had to leave for the tuition.

    I mean the whole thing.

    I am not sure if the Taylor expansion would help but the summation you have is pretty easy to solve by converting it into a definite integral. Its something called converting a Riemann sum to a definite integral. It should be somewhere in your notes. For example, look here: http://in.answers.yahoo.com/question/index?qid=20101015152613AAJJ3c0

    I hope that helps.
     
  8. Mar 25, 2014 #7
    I found my required answer simple enough.

    I could manipulate my series to (2/pi).SUM(from n=1 to infinity) of (-1)^n+1 . pi^2n / (2n)!

    This is thus after a change of sign:

    -(2/pi) [pi^2/2! - pi^4/4! + pi^6/6! -...]

    =-(2/pi) . cos(pi) - 1

    = 4/pi as required

    I've had a look next week, and the integration stuff seems to be there - so I have I feeling I was meant to solve it the way i did, though the results needed for 1^k+2^k+3^3k+...+(n-1)^k etc. are a little worrying in terms of proving proof for the general form in terms of n^k / k + O(n^k-1).

    Are you able to go through how I could have done it with the integration method?
     
  9. Mar 25, 2014 #8
    Nice!

    For the integration method, the sum can be written as the following definite integral:
    $$\int_0^1 2\sin(\pi x)\,dx=\frac{4}{\pi}$$
    You will learn this when you deal with integrals. :)
     
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