Summation Notation for Current Index

Question

I have a question on how to properly present a summation with the use of variables. I am not so familiar with how to properly use summation notation, and am looking for some advice on how to do so properly.

Right now, I have built the following summation:$$t_\text{pal}=\sum_{i=1}^{N_R}\left(n_it_i-\sum_{j=\text{current index}}^{N_R}n_jt_j\right)$$

Within the outer summation, I have an inner summation that I want to start at the index of the current term (represented as j = current index) . How should this be represented in the inner summation, would it just be j = i?

Also, I want to also have a condition where if the value of the term is less than 0, the term takes the value of 0 (if term < 0, term = 0). How should this be properly represented in the summation?

Thank you.

Answer 1

Here is a derivation with some intermediate steps. We obtain

\begin{align*} \color{blue}{t_{\text{pal}}}&=\sum_{i=1}^{N_R}\left(n_it_i-\sum_{j=i}^{N_R}n_jt_j\right)\tag{1}\\ &=\sum_{i=1}^{N_R}\left(n_it_i-\left(n_it_i+\sum_{j={i+1}}^{N_R}n_jt_j\right)\right)\tag{2}\\ &=\sum_{i=1}^{N_R}\left(-\sum_{j={i+1}}^{N_R}n_jt_j\right)\tag{3}\\ &\,\,\color{blue}{=-\sum_{i=1}^{N_R}\sum_{j={i+1}}^{N_R}n_jt_j}\tag{4}\\ \end{align*}

Comment:

• In (1) we start with $j=i$ in the inner sum.

• In (2) we separate the first summand from the inner sum.

• In (3) we cancel the terms $n_it_i$.

• In (4) we factor out $-1$.

Note, we can the sum (4) also write as \begin{align*} -\sum_{i=1}^{N_R}\sum_{j={i+1}}^{N_R}n_jt_j=\sum_{i=1}^{N_R}\sum_{j={i+1}}^{N_R}\left(-n_jt_j\right) \end{align*} If depending on $n_j$ and $t_j$ the innermost terms $(-n_jt_j)$ is positive or should be set to $0$ otherwise we can write for instance \begin{align*} \color{blue}{\sum_{i=1}^{N_R}\sum_{j={i+1}}^{N_R}\max\{-n_jt_j,0\}}\qquad\text{or}\qquad\color{blue}{\sum_{i=1}^{N_R}\sum_{j={i+1}}^{N_R}\left(-n_jt_j\right)^{+}} \end{align*} as already indicated in the comment section.