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Transforming a summation into an anlytical expression

  1. Jan 16, 2012 #1
    Hello everybody,

    I have some problems in finding an analytical expression for this product:

    [tex] \sum_{j=0}^{N}(N-j)e^{-ijy}\cdot\sum_{k=0}^{N}(N-k)e^{iky} [/tex].

    I have solved the problem for several Ns, applying the Euler rule [itex]2\cos(x) = e^{ix} + e^{-ix}[/itex]


    Now, I'm trying to express the product with some function (which would be valid for any N). I found that I can express the product as:

    [tex] 2\sum_{j=0}^{N}\sum_{k=0}^{N-j}(N-k)(N-j-k)\cos(jy) - \sum_{j=0}^{N}j^2 [/tex] which becomes

    [tex] 2\sum_{j=0}^{N}\sum_{k=0}^{N-j}(N^2 - 2Nk - Nj - jk + k^2)\cos(jy) - \sum_{j=0}^{N}j^2 [/tex]

    I've found expressions for [itex]\sum j^2 = f(N) [/itex], for [itex]\sum\cos(ky) = f(N,y)[/itex], for [itex]\sum k\cos(ky) = f(N,y)[/itex] and I'm quite sure that I'll find also for [itex]\sum k^2\cos(ky)[/itex].

    My question now is, are there some easy ways to develop the mixed terms [itex]\sum_{j=0}^{N}\sum_{k=0}^{N-j}k\cos(jz)[/itex] and [itex]\sum_{j=0}^{N}\sum_{k=0}^{N-j}jk\cos(jz)[/itex].

    Or eventually if you think it is easier to find an expression for [tex] \sum_{j=0}^{N}(N-j)e^{-ijy}\cdot\sum_{k=0}^{N}(N-k)e^{iky} = \sum_{j=0}^{N}\sum_{k=0}^{N}(N-j)(N-k)e^{i(k-j)y}[/tex], which would also be a more elegant way to solve the problem.

    Thank you in advance for any help and please apologise my bad english/math...

    Leonardo
     
  2. jcsd
  3. Jan 16, 2012 #2
    I am not sur to well understand the wording of the question and what you did.
    My joint answer might be to easy :
     

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  4. Jan 16, 2012 #3

    mathman

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    I would start by replacing j by j'=N-j and k by k'=N-k, also note that sums in both cases go from 1 to N.

    Next note that Σkxk = xd/dx(Σxk) = xd/dx(1-xN)/(1-x))
    Now evaluate the derivative and substitute e±iy for x in the appropriate expression.
     
  5. Jan 17, 2012 #4
    Thank you very much, this seems to be the easiest, and also most elegant, solution.

    Just the last question, the sum can also go from 0 to N? It should make no difference since for [itex] j = 0, j\cdot e^{ijx} = 0 [/itex]
     
    Last edited: Jan 17, 2012
  6. Jan 17, 2012 #5

    mathman

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    You are right.
     
  7. Feb 6, 2012 #6
    Maybe the solution can be useful to somebody:

    the expression [tex] f(N,z) =\left(\sum_{j'=0}^{N}j'exp(-ij'z)\right)\left(\sum_{k'=0}^{N}j'exp(ik'z)\right) [/tex] was solved recalling that:
    [tex]
    \sum_{l=0}^Nle^{ly} = \sum_{l=0}^N\frac{\partial e^{ly}}{\partial y} = \frac{\partial{\sum_{l=0}^Ne^{ly}}}{\partial y}
    [/tex] and that:
    [tex]
    \sum_{l=0}^Ne^{ly} = \frac{\sin\left(\frac{N+1}{2}y\right)} {\sin\left(\frac{y}{2}\right)}e^{\frac{N}{2}y}
    [/tex]

    which leads to a final result of:

    [tex]
    f(N,z) = \frac{N\cos\left((N+1)z\right)-(N+1)\cos(Nz)-N(N+1)\cos{z} + 1 + N(N+1)}{2(\cos{z}-1)^2}
    [/tex]
     
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