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How would I interpret the following summation (where $r_{k}$ is a function that produces a scalar):
$f^{int}_i = \frac{(\kappa \mathbf{b})_{i} + (\kappa \mathbf{b})_{i + 1}}{2} \displaystyle\sum_{k=i+1}^{N-2} [r_{k} - \frac{(\kappa \mathbf{b})_{i - 1} + (\kappa \mathbf{b})_{i}}{2}\displaystyle\sum_{k=i}^{N-2}r_{k}]$

Is using k for both summations just out of laziness? Is this an equivalent function:

$f^{int}_i = \frac{(\kappa \mathbf{b})_{i} + (\kappa \mathbf{b})_{i + 1}}{2}\displaystyle\sum_{j=i+1}^{N-2} [r_{j} - \frac{(\kappa \mathbf{b})_{i-1} + (\kappa \mathbf{b})_{i}}{2}\displaystyle\sum_{k=i}^{N-2}r_{k}]$

where $r_{k} = (\boldsymbol{\omega_{i}}-\boldsymbol{\bar{\omega_{i}}})^TB_{i}R_{\pi/2}\boldsymbol{\omega_{i}}$

The original continuous form of the equation is as follows. This quantity is being calculated along a curve. There are two other terms for calculating $f_{int}^{i}$ but I am confident about their translation from the continuous case to the discrete case.

$f^{int}(s) = [(\int_{s}^{L} (\boldsymbol{\omega} - \boldsymbol{\bar{\omega}})^TB(s)R_{\pi/2}\boldsymbol{\omega}dt) (\kappa\mathbf{b})^T]'(s)$

stagermane
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    In my opinion it's not just laziness but actually an incorrect usage as well. How would you know which $k$ matches with which sum? – wltrup Aug 11 '15 at 20:22
  • What wltrup said. Except I'll say it stronger: The notation is simply wrong. Demand your money back... – David C. Ullrich Aug 11 '15 at 22:24
  • Yes after looking at it more closely, I can tell there are problems with the notation just based on ending up with scalars and vectors for the inner summation (the final quantity should be a vector). I've posted the continuous solution for context. – stagermane Aug 12 '15 at 06:53

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To me, the only interpretation which makes sense is to interpret this as equivalent to

$$f^{int}_i = \frac{(\kappa \mathbf{b})_{i-1} + (\kappa \mathbf{b})_{i}}{2} \displaystyle\sum_{k=i+1}^{N-2} [r_{k} - \frac{(\kappa \mathbf{b})_{i-1} + (\kappa \mathbf{b})_{i}}{2} \displaystyle\sum_{j=i}^{N-2}r_{j}] $$

I would factor things in this way:

$s_i =\sum_{j=i}^{N-2}r_{j} $, $t_i = \frac{(\kappa \mathbf{b})_{i-1} + (\kappa \mathbf{b})_{i}}{2} $, so that

$\begin{array}\\ f^{int}_i &= t_i \sum_{k=i+1}^{N-2} [r_{k} - t_is_i]\\ &= t_i \sum_{k=i+1}^{N-2} r_{k} -t_i \sum_{k=i+1}^{N-2} t_is_i\\ &= t_i s_{i+1} -t_i^2s_i (N-2-i)\\ \end{array} $

This also shows me places where the formula seems to have a chance of being in error. For example, both the inside sum and its multiplier are independent of $k$; I would look carefully at this.

marty cohen
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