I am trying t find an expression for the partial sum of a p-like-series. The problem comes from trying to sum the elements of a matrix whose entries are inversely related to their distance from the diagonal.
$$ \begin{matrix} k^{-\alpha} & (k+1)^{-\alpha} & (k+2)^{-\alpha} & (k+3)^{-\alpha} & (k+4)^{-\alpha} & (k+5)^{-\alpha} & (k+6)^{-\alpha} & (k+7)^{-\alpha}\\ & k^{-\alpha} & (k+1)^{-\alpha} & (k+2)^{-\alpha} & (k+3)^{-\alpha} & (k+4)^{-\alpha} & (k+5)^{-\alpha} & (k+6)^{-\alpha} \\ & & k^{-\alpha} & (k+1)^{-\alpha} & (k+2)^{-\alpha} & (k+3)^{-\alpha} & (k+4)^{-\alpha} & (k+5)^{-\alpha} \\ & & & k^{-\alpha} & (k+1)^{-\alpha} & (k+2)^{-\alpha} & (k+3)^{-\alpha} & (k+4)^{-\alpha} & \\ & & & & k^{-\alpha} & (k+1)^{-\alpha} & (k+2)^{-\alpha} & (k+3)^{-\alpha} \\ & & & & & k^{-\alpha} & (k+1)^{-\alpha} & (k+2)^{-\alpha} \\ & & & & & & k^{-\alpha} & (k+1)^{-\alpha} \\ & & & & & & & k^{-\alpha} \end{matrix} $$
Now, if I divide this matrix into 2 $\times$ 2 squares and sum the elements, I get
$$ \tiny \begin{matrix} 2k^{-\alpha} + (k+1)^{-\alpha} & (k+1)^{-\alpha} + 2(k+2)^{-\alpha} + (k+3)^{-\alpha} & (k+3)^{-\alpha} + 2(k+4)^{-\alpha} + (k+5)^{-\alpha} & (k+5)^{-\alpha} + 2(k+6)^{-\alpha} + (k+7)^{-\alpha} \\ & 2k^{-\alpha} + (k+1)^{-\alpha} & (k+1)^{-\alpha} + 2(k+2)^{-\alpha} + (k+3)^{-\alpha} & (k+3)^{-\alpha} + 2(k+4)^{-\alpha} + (k+5)^{-\alpha} \\ & & 2k^{-\alpha} + (k+1)^{-\alpha} & (k+1)^{-\alpha} + 2(k+2)^{-\alpha} + (k+3)^{-\alpha} \\ & & & 2k^{-\alpha} + (k+1)^{-\alpha} \end{matrix} \normalsize $$
If I repeat the process one more time we get:
$$ \tiny \begin{matrix} 4k^{-\alpha} + 3(k+1)^{-\alpha} + 2(k+2)^{-\alpha} + (k+3)^{-\alpha} & (k+1)^{-\alpha} + 2(k+2)^{-\alpha} + 3(k+3)^{-\alpha} + 4(k+4)^{-\alpha} + 3(k+5)^{-\alpha} + 2(k+6)^{-\alpha} + (k+7)^{-\alpha} \\ & 4k^{-\alpha} + 3(k+1)^{-\alpha} + 2(k+2)^{-\alpha} + (k+3)^{-\alpha} \end{matrix} \normalsize $$
The pattern which emerges is some sort of weighted p-series with strictly deceasing coefficients on the diagonal, and coefficients that increase and decrease on the off diagonals, starting at 1, then increasing to $\frac{n(n+1)}{2}$, then decreasing and returning to 1 again.
MY QUESTIONS:
- Is there a name for such a series?
- Is there an approximation for such a sum? (I found this post about approximating p-series sums, but the addition of the above coefficients complicate the issue)
Things to note:
$k$ is likely to be set to $0$
$\alpha \gt 1$
The above process is likely to be repeated up to 18 times on a very large matrix, so approximations to a large sum are welcome.
TO SUMMARISE
- Let's set k = 0.
- We have a square matrix with dimension $\propto 2^J$.
- Let $r$ and $c$ represent the row and column, respectively.
- Each cell equals $(c - r)^{-\alpha} * I(r < c)$
- I have a square, upper triangular Toeplitz matrix
$$ \begin{matrix} 0 & 1^{-\alpha} & 2^{-\alpha} & 3^{-\alpha} & 4^{-\alpha} & 5^{-\alpha} & 6^{-\alpha} & 7^{-\alpha} & ...\\ & 0 & 1^{-\alpha} & 2^{-\alpha} & 3^{-\alpha} & 4^{-\alpha} & 5^{-\alpha} & 6^{-\alpha} & ...\\ & & 0 & 1^{-\alpha} & 2^{-\alpha} & 3^{-\alpha} & 4^{-\alpha} & 5^{-\alpha} & ... \\ & & & 0 & 1^{-\alpha} & 2^{-\alpha} & 3^{-\alpha} & 4^{-\alpha} & ...\\ & & & & 0 & 1^{-\alpha} & 2^{-\alpha} & 3^{-\alpha} & ...\\ & & & & & 0 & 1^{-\alpha} & 2^{-\alpha} & ...\\ & & & & & & 0 & 1^{-\alpha} & ...\\ & & & & & & & 0 & ... \\ & & & & & & & & \ddots \end{matrix} $$
- I break the matrix into square bins of size $2^j$, $j \in {1,2,3,...., J}$
- The number of diagonals within each square bin (and therefore the number of unique terms) is 2 x dimension - 1 = $2^{j+1} - 1$.
- Since the coefficients start at 1, ascend to the midpoint, and then descend back to 1, the middle (and therefore maximum) coefficient is $\frac{(2^{j+1} - 1)+1}{2} = 2^j$.
- The summands within each square bin change as they move further away from the diagonal. If we consider the $b^{th}$ bin from the diagonal, $b \in {1,2,3,..., 2^{J-j}}$ then the maximum summand is $(b2^j - 1)$
- The minimum summand is the maximum summand - # of diagonal elements + 1 = $(b2^j - 1) - (2^{j+1} - 1) + 1 = b2^j - 2^{j+1} + 1$
- Therefore, the midpoint summand will be $\frac{max + min}{2} = \frac{(b2^j - 1)+(b2^j - 2^{j+1} + 1)}{2} = \frac{b2^{j+1} - 2^{j+1}}{2} = (b-1)*2^j$
- Bringing this all together, we get: (diagonal term) + (lower triangle terms) + (Upper triangle terms).
$$2^j((b-1)2^j)^{-\alpha} + \sum_{i=1}^{2^j - 1}(2^j -i)((b-1)2^j - i)^{-\alpha} + \sum_{i=1}^{2^j - 1}(2^j -i)((b-1)2^j + i)^{-\alpha}$$
Note that if we are on one of the "triangular" diagonal bins, the first two terms evaluate to ZERO, since $r \ge c$
$$= 2^j((b-1)2^j)^{-\alpha} + \sum_{i=1}^{2^j - 1}(2^j -i)((b-1)2^j - i)^{-\alpha} + \sum_{i=1}^{2^j - 1}(2^j -i)((b-1)2^j + i)^{-\alpha}$$
In order to make the above expression conform with Dr. Wolfgang Hintze's solution below, I simply need to change the index.
$\ell = (2^j - i) \rightarrow (i = 2^j - k)$
$$ \begin{align} & 2^j((b-1)2^j)^{-\alpha} + \sum_{i=1}^{2^j - 1}(2^j -i)((b-1)2^j - i)^{-\alpha} + \sum_{i=1}^{2^j - 1}(2^j -i)((b-1)2^j + i)^{-\alpha} \\ & = 2^j((b-1)2^j)^{-\alpha} + \sum_{\ell=1}^{2^j - 1}\ell((b-1)2^j - (2^j - \ell))^{-\alpha} + \sum_{\ell=1}^{2^j - 1}\ell((b-1)2^j + (2^j - \ell))^{-\alpha} \\ & = 2^j((b-1)2^j)^{-\alpha} + \sum_{\ell=1}^{2^j - 1}\ell((b-2)2^j + \ell)^{-\alpha} + \sum_{\ell=1}^{2^j - 1}\ell(b2^j - \ell)^{-\alpha} \\ & = 2^j((b-1)2^j)^{-\alpha} + \sum_{\ell=1}^{2^j - 1}\ell(\ell + (b-2)2^j)^{-\alpha} + \sum_{\ell=1}^{2^j - 1}\ell(- \ell + b2^j)^{-\alpha} \\ \end{align} $$
As a final pass, I also came up with the following expression by not changing the term in the final sum, instead simply distributing the power summand across the coefficient term:
$$ \begin{align} 2^j((b-1)2^j)^{-\alpha} &+ \sum_{\ell=1}^{2^j - 1}\ell(\ell + (b-2)2^j)^{-\alpha} \\ & + 2^j\sum_{i=1}^{2^j - 1}(i + (b-1)2^j)^{-\alpha} \\ & - \sum_{i=1}^{2^j - 1}i(i + (b-1)2^j)^{-\alpha} \end{align} $$
$$ \begin{align} 2^j((b-1)2^j)^{-\alpha} &+ s\left((b-2)2^j, \alpha, 1, 2^j - 1\right) \\ & + 2^j\left\{\zeta(\alpha, (b-1)2^j +1) - \zeta(\alpha, (b-2)2^j -1)\right\} \\ & - s\left((b-1)2^j, \alpha, 1, 2^j - 1\right) \end{align} $$
I'm not sure if I understand your problem correctly but it seems that you are interested in sums of the type
$$s(k,\alpha,m,n)=\sum _{i=0}^n i^m (i+k)^{-\alpha }\tag{1}$$
For $m=0$ we have
$$s(k,\alpha,0,n)=\zeta (-a,k)-\zeta (-a,k+n+1)\tag{2}$$
where $\zeta (r,n)$ is the Hurwitz Zeta function.
For general integer $m$ it is useful to start from the genrating function
$$\begin{align}g(z)= & \sum _{i=0}^n z^i (i+k)^{-\alpha }\\=& \Phi (z,\alpha ,k)-z^{n+1} \Phi (z,\alpha ,k+n+1) \end{align}\tag{3}$$
where $\Phi (z,\alpha ,k)$ is the Lerch $\Phi$ function.
You can generate the $m$-th power of $i$ by applying the operation $(z \frac{\partial}{\partial z})^m$ to $g(z)$ and letting $z\to 1$.
Since the derivative
$$z \frac{\partial \Phi (z,\alpha ,k)}{\partial z} = \Phi (z,\alpha -1,k)-k \Phi (z,\alpha ,k)\tag{4}$$
can also be expressed in terms of the Lerch $\Phi$-funcition we conclude that, the sum $(1)$ can be expressed in terms of the Lerch function $\Phi (z,\alpha ,k)$.