- #1
leonardo2887
- 3
- 0
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
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