Transforming a summation into an anlytical expression

[leonardo2887]

Hello everybody,

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

$$\sum_{j=0}^{N}(N-j)e^{-ijy}\cdot\sum_{k=0}^{N}(N-k)e^{iky}$$.

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


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:

$$2\sum_{j=0}^{N}\sum_{k=0}^{N-j}(N-k)(N-j-k)\cos(jy) - \sum_{j=0}^{N}j^2$$ which becomes

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

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

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

Or eventually if you think it is easier to find an expression for $$\sum_{j=0}^{N}(N-j)e^{-ijy}\cdot\sum_{k=0}^{N}(N-k)e^{iky} = \sum_{j=0}^{N}\sum_{k=0}^{N}(N-j)(N-k)e^{i(k-j)y}$$, 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

 

[JJacquelin]

I am not sur to well understand the wording of the question and what you did.
My joint answer might be to easy :

 

[mathman]

Hello everybody,

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

$$\sum_{j=0}^{N}(N-j)e^{-ijy}\cdot\sum_{k=0}^{N}(N-k)e^{iky}$$.

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


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:

$$2\sum_{j=0}^{N}\sum_{k=0}^{N-j}(N-k)(N-j-k)\cos(jy) - \sum_{j=0}^{N}j^2$$ which becomes

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

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

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

Or eventually if you think it is easier to find an expression for $$\sum_{j=0}^{N}(N-j)e^{-ijy}\cdot\sum_{k=0}^{N}(N-k)e^{iky} = \sum_{j=0}^{N}\sum_{k=0}^{N}(N-j)(N-k)e^{i(k-j)y}$$, 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 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 Σkx^k = xd/dx(Σx^k) = xd/dx(1-x^N)/(1-x))
Now evaluate the derivative and substitute e^±iy for x in the appropriate expression.

 

[leonardo2887]

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 $j = 0, j\cdot e^{ijx} = 0$

 

[mathman]

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 $j = 0, j\cdot e^{ijx} = 0$

You are right.

 

[leonardo2887]

Maybe the solution can be useful to somebody:

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

which leads to a final result of:

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