Can someone give me an example of how cyclic sums are used? I don't really understand how they're used in problem-solving. For example, $$\sum_{a,b,c}a^2$$ Any help would be appreciated, and I'm not sure if that's the correct notation, so forgive me. Thanks for the help!

  • $\begingroup$ It might help if you gave a citation where you came across this concept. $\endgroup$ Jan 16, 2013 at 22:58
  • $\begingroup$ I came across it during a high school summer math program called NY Math Circle. $\endgroup$
    – joejacobz
    Jan 16, 2013 at 23:15
  • $\begingroup$ Good --- what was the context? Was it used in a complete sentence? $\endgroup$ Jan 16, 2013 at 23:32
  • 1
    $\begingroup$ joe - when you find answers helpful, you can upvote them by clicking on the "upwards" arrow (greyed-out) above the vote count next to the answer. You can also accept one answer per question by clicking on the "greyed-out" check-mark to the left of the answer you want to accept. $\endgroup$
    – amWhy
    Jan 17, 2013 at 1:49
  • $\begingroup$ Hi. I hope my answer does not prove entirely useless. $\endgroup$
    – 000
    Jan 17, 2013 at 2:36

3 Answers 3



Cyclic sums are used to denote summations over permutations. Consider the permutation $p=(a\, b\, c)$. The cyclic sum $\sum_{p}a$ is the sum ran through the entire permutation one cycle: $$ \sum_p a=a+b+c. $$

The first term is derived from the fact that the first term of a cyclic sum is always just the term being permutated. In this case, we are permutating $a$. The second term is derived from the operation of the permutation once: $a\mapsto b$. The third term is derived from the operation of the permutation twice: $a\mapsto b$ and $b\mapsto c$. In other words, each term is the permutation iterated $n-1$ times where $n$ is the position of the term in the permutation.

More rigorously, we have:

$$ \sum_p f(x_1,\ldots, x_n)=\sum_{0\le k\le n-1}p^k\left( f(x_1,\ldots, x_n)\right), $$

for some permutation $p$ wherein $p^k$ denotes $p$ iterated $k$ times.


With $p=(x_1\, \ldots \, x_n)$, the following inequality is true for even $n\le 12$ and odd $n\le 23$:

$$ \sum_p \frac{x_1}{x_2+x_3}\ge \frac{1}{2}n, $$

where $x_i$ is nonnegative for all $x_i$ in $\{x_1,\ldots, x_n\}$ and the denominators are positive. This is referred to as Shapiro's Cyclic Sum Constant.

Also, the following paper may be of interest to you: Combinatorial Remarks on the Cyclic Sum Formula for Multiple Zeta Values.


A personal thing I looked into was the following: Given the function $f(x)=ax^2+bx+c$ and the representation of $(s-t)$, $(r-t)$, and $(r-s)$ as $p_1$, $p_2$, and $p_3$, respectively, we can state the leading coefficient of $f$ as follows:

$$ a=\sum_{\substack{p\\ 1\le i\le 3}}f(r)\prod_{\substack{1\le j\le 3\\ j\ne i}}\frac{1}{p_i}, $$

where $p=(r\, s\, t)$ and $r$, $s$, and $t$ are $x$-coordinates of points on the parabola $f(x)$. I don't know if this is correct, as it was solely some calculation in my spare time.

For Mr. Lin:

The summation expands as follows:

\begin{align} a&=\sum_{\substack{p\\ 1\le i\le 3}}f(r)\prod_{\substack{1\le j\le 3\\ j\ne i}}\frac{1}{p_i}\\ &=\sum_{\substack{(r\, s\, t)\\ 1\le i\le 3}}f(r)\prod_{\substack{1\le j\le 3\\ j\ne i}}\frac{1}{p_i}\\ &=f(r)\prod_{\substack{1\le j\le 3\\ j\ne 1}}\frac{1}{p_i}+f(s)\prod_{\substack{1\le j\le 3\\ j\ne 2}}\frac{1}{p_i}+f(t)\prod_{\substack{1\le j\le 3\\ j\ne 3}}\frac{1}{p_i}. \end{align}

I am pretty sure there is a typo with the $\frac{1}{p_i}$. I believe it should be $\frac{1}{p_j}$. With that in mind:

\begin{align} f(r)\prod_{\substack{1\le j\le 3\\ j\ne 1}}\frac{1}{p_i}+f(s)\prod_{\substack{1\le j\le 3\\ j\ne 2}}\frac{1}{p_i}+f(t)\prod_{\substack{1\le j\le 3\\ j\ne 3}}\frac{1}{p_i}&=\frac{f(r)}{p_2p_3}+\frac{f(s)}{p_1p_3}+\frac{f(t)}{p_1p_2}. \end{align}

This notation is surely not that fun to enjoy: One must perform the permutation and the expansion upon the index set $1\le i\le 3$ simultaneously.

  • $\begingroup$ Can you explain the notation of the Addendum? In particular, what is $\sum_p_{1\leq i \leq 3} f(r)$, since I do not see $p$ nor $i$. Also, the product doesn't have a $j$ involved? $\endgroup$
    – Calvin Lin
    Jan 17, 2013 at 2:58
  • $\begingroup$ @CalvinLin It's confusing to say the least. I'll edit it in an attempt to explain momentarily. $\endgroup$
    – 000
    Jan 17, 2013 at 3:29
  • $\begingroup$ @CalvinLin Does this edit address your concerns properly? $\endgroup$
    – 000
    Jan 17, 2013 at 3:48
  • $\begingroup$ Wow! Thank you so much for spending so much time to answer this question! I really appreciate it. $\endgroup$
    – joejacobz
    Jan 18, 2013 at 0:11
  • $\begingroup$ @Limitless Thanks! Now I understand what you mean. Wasn't used to the notation so got confused. When you say representation, do you mean "Set $p_1 = s-t$"? Also, shouldn't it be cyclic, aka $t-r$ instead or $r-t$? $\endgroup$
    – Calvin Lin
    Jan 18, 2013 at 0:46

Since Iuli explained what it means, just some comments to explain why we use this type of notation.

Cyclic (and symmetric) sums occur very often in inequalities. The cyclic sum is simply the sum obtained by permuting the letters cyclical, i.e. moving the first letter to last. Same way the symmetric sum is the sum over all permutations.

They are actually extremely helpful in inequalities involving many letters, as long as you can handle them easily.

For example, lets say that you need to use in some problem the so called Muirhead inequality, in the particular case $[7,3,2,1] > [5,4,3,1]$. Then all you write is
$$\sum_{sym} a^7b^3c^2d \geq \sum_{sym} a^5b^4c^3d $$

If you write them explicitly, each side has 24 terms and the exact inequality is:

$$a^7b^3c^2d + a^7b^3d^2c+ a^7c^3b^2d + a^7c^3d^2b+a^7d^3c^2b+a^7d^3c^2b+ b^7a^3c^2d + b^7a^3d^2c+ b^7c^3a^2d + b^7c^3d^2a+b^7d^3c^2a+b^7d^3c^2a +c^7b^3a^2d + c^7b^3d^2a+ c^7a^3b^2d + c^7a^3d^2b+c^7d^3a^2b+c^7d^3a^2b c^7b^3a^2d + c^7b^3d^2a+ c^7a^3b^2d + c^7a^3d^2b+c^7d^3a^2b+c^7d^3a^2b \geq a^5b^4c^3d + a^5b^4d^3c+ a^5c^4b^3d + a^5c^4d^3b+a^5d^4c^3b+a^5d^4c^3b+ b^5a^4c^3d + b^5a^4d^3c+ b^5c^4a^3d + b^5c^4d^3a+b^5d^4c^3a+b^5d^4c^3a +c^5b^4a^3d + c^5b^4d^3a+ c^5a^4b^3d + c^5a^4d^3b+c^5d^4a^3b+c^5d^4a^3b c^5b^4a^3d + c^5b^4d^3a+ c^5a^4b^3d + c^5a^4d^3b+c^5d^4a^3b+c^5d^4a^3b $$

Which form would rather use, esspecially if it is part of a larger exercise? Note than in general, with $n$ variables, the Muirhead inequality has n! terms on each side. So 5 variables means 240 terms, 6 variables means 1680 terms, wouldn't you rather use the half line form of it?

There are actually few inequalities which use the cyclic notation. And even if cyclic have usually much less terms than symmetric, still they are much simpler to use in the short form.

  • $\begingroup$ Muirhead's inequality looks exceptionally powerful. $\endgroup$
    – 000
    Jan 17, 2013 at 18:08
  • $\begingroup$ @Limitless It is :) many of the commonly used inequalities are particular cases of it . $\endgroup$
    – N. S.
    Jan 17, 2013 at 22:29
  • $\begingroup$ Thanks for the hard work you put into this post; it really means a lot. This also really helped out. Thanks again! $\endgroup$
    – joejacobz
    Jan 18, 2013 at 0:13

This kind of sum can be used in inequalities.

An example of a cyclic sum is:

$$\sum_{a,b,c}{\frac{a^3}{3}}=\frac{a^3}{3}+\frac{b^3}{3}+\frac{c^3}{3}\geq \sqrt[3]{\frac{(abc)^3}{3^3}}=\frac{abc}{3}.$$

This is a simple example but you can find a lot of them in inequality exercises.

  • $\begingroup$ Hm, okay. So, would I ever be in a position where it would be easier to make terms into a cyclic sum to simplify a problem, or is it usually the other way around? $\endgroup$
    – joejacobz
    Jan 16, 2013 at 23:07
  • $\begingroup$ @joejacobz please look at the section "First year at AoPS-MathLinks!" from this link :artofproblemsolving.com/Forum/blog.php?u=22804&cat=97 and then here: h artofproblemsolving.com/Forum/viewtopic.php?t=161062. It is useful to use the notation of $\sum_{a}$ or $\sum_{cyc}$ to describe all the terms of the sum; -it's a circle drawing trough 3 points $a$, $b$, $c$. $\endgroup$
    – Iuli
    Jan 16, 2013 at 23:15
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    $\begingroup$ @Iuli I think that $\sum_{cyc}$ is much better as notation in general, since $\sum_{a,b,c}$ could be confused with the symmetric sum. $\endgroup$
    – N. S.
    Jan 17, 2013 at 4:22

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