This is a question specific to scientific writing and not about the modulo operation. The final question will be: Is this expression well-formed, or should it come with clarification?
On several occasions I read something like "indices are taken modulo $N$" in scientific publications. These usually follow a math expression like
$$ f(\vec{x},k) = x_{i+k}\,\hat{e}_i, $$
where the key is that $\vec{x}$ lives in a finite dimensional space, let's say of dimension $N$ and thus in index $i+k >N$ would not be defined.
When the authors say take the index modulo $N$, I assume most times they mean: if $i+k = m > N$, then assign $m\rightarrow m-N$, because most times the index will run from $1$ to $N$. This of course is not the usual way of applying the modulo operation, which would have a $\geq$ - symbol in the above condition. To me it is perfectly clear what is meant by the instruction to take the index modulo $N$ and I know that sometimes we rely on the good-willed nature of a reader.
My question thus is: Is this (strictly speaking) inaccurancy in the definition of mathematical expressions acceptable in the context of scientific writing style or should it be further clarified? And as a follow up: Would you recommend a different formulation? Which?
Some remarks:
- You can assume for the purpose of this question that the number that is taken modulo $N$ is supposed to take on the value $N$ and not zero if $i+k=N$. It also does not have to be an index, but they might be a special case.
- Further, the example above is simply that - an example. I am not asking about that specifically.
- If it matters, I am reading publications from an interdisciplinary field (pulse-coupled oscillators) mainly between applied mathematics and physics. Readers could be from both the Mathematics or Physics world, or possibly even Engineering, Neuroscience or Biology.
Beware that modulo has two meanings:
it is an operator that returns an integer in the range $[0,n-1]$, so that $n\bmod n=0$.
it is also an argument to the congruence relation $\equiv$; in that sense, $0\equiv n\equiv 2n\cdots\mod n$.
The second meaning is compatible with the reduction $1\to1,2\to2\cdots n-1\to n-1,n\to n,n+1\to1,n+2\to2,\cdots$ as well as with $1\to1,2\to2\cdots n-1\to n-1,n\to 0,n+1\to1,n+2\to2,\cdots$
In any case, I will always understand this modulo $n$ as the shifting by a multiple of $n$ that clamps the index back to the allowed range.