Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
It might help if you gave a citation where you came across this concept. –  Gerry Myerson Jan 16 '13 at 22:58
    
I came across it during a high school summer math program called NY Math Circle. –  joejacobz Jan 16 '13 at 23:15
    
Good --- what was the context? Was it used in a complete sentence? –  Gerry Myerson Jan 16 '13 at 23:32
1  
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. –  amWhy Jan 17 '13 at 1:49
    
Hi. I hope my answer does not prove entirely useless. –  000 Jan 17 '13 at 2:36

3 Answers 3

up vote 5 down vote accepted

Definition

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.

Example

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.

Addendum

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.

share|improve this answer
    
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? –  Calvin Lin Jan 17 '13 at 2:58
    
@CalvinLin It's confusing to say the least. I'll edit it in an attempt to explain momentarily. –  000 Jan 17 '13 at 3:29
    
@CalvinLin Does this edit address your concerns properly? –  000 Jan 17 '13 at 3:48
    
Wow! Thank you so much for spending so much time to answer this question! I really appreciate it. –  joejacobz Jan 18 '13 at 0:11
    
@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 $r-t$? –  Calvin Lin Jan 18 '13 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.

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

This kind of sums 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 cand find a lot in inequalities exercises.

share|improve this answer
    
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? –  joejacobz Jan 16 '13 at 23:07
    
@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 $a$, $b$, $c$. –  Iuli Jan 16 '13 at 23:15
    
Thanks for the links! –  joejacobz Jan 17 '13 at 2:26
1  
@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. –  N. S. Jan 17 '13 at 4:22

Your Answer

 
discard

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.