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I need represent this expression $L_{u_3}u(k-1)+L_{u_2}u(k-2)+L_{u_1}u(k-3)$ by using a summatory, $L_{u}$ is a vector that contains d elements $L_u=\begin{pmatrix}L_{u_1} L_{u_2} L_{u_3}....L_{u_d}\end{pmatrix}$ and $u(\cdot)$ also depends on d as is seen in the expression, where d in that case is 3, but it can be bigger, i was thinking to solve it by $\sum_{j=1}^{d}L_{u_j}u(k-(?))$ but how is the index for $u(k-(?)$? I don't know how to represent it while some terms go upwards (L_{u_j}) and the other goes downwards? If you have some advice please let me know. Thanks

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$$\sum_{j=1}^3L_{u_j}u(k-(4-j))$$ should do the trick.

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Almost. You want $u(k-(d+1-j))$, or $u(k-(d+1)+j)$, not $u(k-(d+1)-j)$. –  Gerry Myerson Mar 8 '13 at 12:53
    
yes that is it! Thank you so much! –  Gina Torres Mar 8 '13 at 14:10
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