# Is there such a thing as backwards sigma?

Is there a math function, similar to sigma, that can run down? For example instead of $\sum\limits_{i=1}^{10}i$ ,
something that adds from 10 to 1 (like a backwards run)...

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Umm... addition is associative and commutative so what would the difference be? – Derek Elkins Mar 26 at 17:05
In Mathematica it would look like Sum[i,{i,10,1,-1}] (I think). – Steven Gregory Mar 27 at 0:30
Summation isn't a for-loop. – David Richerby Mar 27 at 5:32

The problem with your example is that addition is commutative, so it is not really useful to have a distinction for a sum from $1$ to $10$ or from $10$ to $1$.

However, your question makes sense in a noncommutative setting. Suppose for instance you have 10 matrices $A_1$, ..., $A_{10}$. Since the product of matrices is not in general commutative, the product $A_1 \dotsm A_{10}$ is in general different from $A_{10} \dotsm A_1$. In this case, you may consider writing $\prod_{i=1}^{10}A_i$ in the first case and $\prod_{i=10}^{1}A_i$ in the second case, but this is probably not the most satisfying solution.

A better solution is to consider a totally ordered finite set $(I, \leqslant)$ and to write $\prod_{i \in I}A_i$. The first case of my example can now be obtained by considering the set $\{1, \ldots, 10\}$ ordered by $1 \leqslant 2 \dotsm \leqslant 10$ and the second one by considering the set $\{1, \ldots, 10\}$ ordered by $10 \leqslant 9 \dotsm \leqslant 1$.

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I think that rather than resorting to unusual orderings, most people would simply write $\prod_{i=1}^{10} A_{11-i}$ as suggested by other answers. – Nate Eldredge Mar 27 at 4:32
Or just consider a list / tuple of indices $(10,\ldots,1)$. I think the notation $\prod_{i\in (10,\ldots,1)}A_i$ should be clear enough, though it may not quite be standard. – leftaroundabout Mar 27 at 13:34

Sure:

$$\sum_{i=0}^{9}(10-i)=10+9+8+\ldots+1$$

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More generally, what OP wants is $\sum_{i=0}^n f(i)=\sum_{i=0}^n f(n-i)$. – YoTengoUnLCD Mar 26 at 17:08

Try $$\sum_{i=1}^{10} (11 - i)$$ There is no reason to introduce another symbol when a simple subtraction can do the work.

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Doing subtraction like this isn't necessary the best notation. If $(11 - i)$ appears twenty times in your summand, it's probably best to think of an alternative. – Tanner Swett Mar 26 at 18:49
@TannerSwett Actually, in cases such as J.-E. Pin mentions where it actually makes a difference which order you do something in, This is exactly the sort of notation I would expect to see: $$\prod_{i=1}^{10}A_{11-i}$$. It is quite common. For example, the binomial theorem:$$(x + y)^m = \sum_{n=0}^m {m\choose n}x^ny^{\color{red}{m-n}}$$. – Paul Sinclair Mar 27 at 2:05
But that's why I said "if it appears twenty times in your summand". In your two examples, the subtraction only appears once. If it appeared twenty times, it would start to get cumbersome. – Tanner Swett Apr 3 at 5:36

If $a, b > 0$ then you can say

$$\sum_{k=a}^b a_k = \sum_{k=0}^b a_k - \sum_{k=0}^{a-1} a_k.$$

For example:

\begin{align} \sum_{k=3}^7 k &= \sum_{k=0}^7 k - \sum_{k=0}^2 k\\[0.3cm] &= (0+1+2+3+4+5+6+7) - (0+1+2)\\[0.3cm] &= 3 + 4 + 5 + 6 + 7\\[0.3cm] &= 25 \end{align}

And: \begin{align} \sum_{k=7}^3 k &= \sum_{k=0}^3 k - \sum_{k=0}^6 k\\[0.3cm] &= (0+1+2+3) - (0+1+2+3+4+5+6)\\[0.3cm] &= -(4 + 5 + 6)\\[0.3cm] &= -15 \end{align}

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There is nothing wrong with writing $$\sum_{i=10}^1 i$$

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In many cases, there's a convention that a sum where the upper limit is smaller is interpreted as 0. This is sometimes convenient because, for instance, one can write "The sum of the first $n$ positive integers is $\sum_{i=1}^n i$" and still be correct when $n=0$. So your suggestion could be misleading to someone who's used to this convention. – Nate Eldredge Mar 27 at 4:35
It depends. One often sees the convention that the empty sum is assumed when the upper limit is less than the lower one. – J. M. Mar 27 at 4:38