Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)...

share|cite|improve this question
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
up vote 11 down vote accepted

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 $10 \leqslant 9 \dotsm \leqslant 1$.

share|cite|improve this answer
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



share|cite|improve this answer
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.

share|cite|improve this answer
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}

share|cite|improve this answer

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.