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I am a novice in this type of sums and I can't even understand the meaning of the three sigmas. Somehow, I am guessing that the answer might be $0$ but I am not sure. I need a well-explained answer(with examples) explaining the meaning and the method to solve it. Lots and lots of thanks beforehand. Sum is written below :

Suppose that $x_1,x_2,...,x_n(n>2)$ are real numbers such that $x_i=-x_{n-i+1}$ for $1\le i\le n$. Consider the sum $S=\Sigma\Sigma\Sigma x_ix_jx_k$, where the summations are taken over all $i,j,k:1\le i,j,k\le n$ and $i,j,k$ are all distinct. Then $S$ equals ____ ?

User Not Found
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2 Answers2

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By problem $S=$ $\displaystyle \sum_{k=1}^n$ $\displaystyle \sum_{j=1}^n$ $\displaystyle \sum_{i=1}^n$ $x_k x_j x_i$. Now notice that,

$$\displaystyle \sum_{k=1}^n \displaystyle \sum_{j=1}^n \displaystyle \sum_{i=1}^n x_k x_j x_i =S=\displaystyle \sum_{k=1}^n \displaystyle \sum_{j=1}^n \displaystyle \sum_{i=1}^n x_k x_j x_{n+i-1}=-\displaystyle \sum_{k=1}^n \displaystyle \sum_{j=1}^n \displaystyle \sum_{i=1}^n x_k x_j x_i$$

The rest part is easy.

William Hilbert
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  • Thanks but what is the meaning of three sigmas. Will $x_i$ come first, or $x_j$ or $x_k$ or are they equivalent. Also what should one write below the first sigma : $i=1$,$j=1$ or $k=1$ or are they equivalent – User Not Found Jun 25 '14 at 05:07
  • They are equivalent. By the way, what do you mean by-"Also what should one write below the first sigma : $i=1$,$j=1$ or $k=1$ or are they equivalent."? – William Hilbert Jun 25 '14 at 05:10
  • Just like you have written k=1 below the first sigma, j=1 below the second and i=1 below the third. – User Not Found Jun 25 '14 at 05:13
  • When $\sum$ sign is used, its range in understood implicitly. In the answer I have just written this range (as given by your question) explicitly. – William Hilbert Jun 25 '14 at 05:16
  • Is $\displaystyle \sum_{k=1}^n \displaystyle \sum_{j=1}^n \displaystyle \sum_{i=1}^n$ equivalent to $\displaystyle \sum_{i=1}^n \displaystyle \sum_{j=1}^n \displaystyle \sum_{k=1}^n$ – User Not Found Jun 25 '14 at 05:20
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    I think that you should try exploring the properties of multiple summation yourself. In that way you will learn better. – William Hilbert Jun 25 '14 at 05:23
  • @WilliamHilbert how could you write the last line....clearly u had used $x_{i} = x_{n+i-1}$ which is not given in OP –  Jul 23 '20 at 05:03
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The $\sum\sum\sum$ here really means sum over the 3 indices $i,j,k$. However, it is a bad notation and misleading. Nowhere does it capture the requirement $i,j,k$ are distinct. Literally, you should interpret it as

$$\sum\sum\sum\quad\equiv\quad \sum_{i=1}^n\sum_{\substack{j=1\\j\ne i}}^n\sum_{\substack{k=1\\k \ne i,j}}^n$$ A less misleading notation could be something like $\displaystyle\;\sum_{i\ne j\ne k}\;$ instead.

For the question of the sum. Yes, it does vanishes. Notice

$$i,j,k\;\;\text{ all distinct } \quad\iff\quad n-i+1, n-j+1, n-k+1\;\;\text{ all distinct }$$

By a relabeling of indices $x_\alpha \mapsto x_{n-\alpha+1}$, we have

$$\sum_{i\ne j \ne k} x_i x_j x_k = \sum_{i\ne j \ne k} x_{n-i+1} x_{n-j+1} x_{n-k+1}$$

Apply the given condition $x_{n-\alpha+1} = -x_\alpha$, you can rewrite RHS as

$$RHS = \sum_{i\ne j \ne k} (-x_i)(-x_j)(-x_k) = -\sum_{i\ne j \ne k} x_i x_j x_k = -LHS$$

From this, you can conclude $S = LHS = RHS = 0$.

achille hui
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  • In the question it is only told that $x_i=-x_{n+i-1}$ why did you assume it to be true for $j$ and $k$ also? – William Hilbert Jun 25 '14 at 05:28
  • @WilliamHilbert $i,j,k$ are dummy indices. As long as $j$ or $k$ are taking from same set of values as $i$ (i.e form the set ${;1,\ldots,n;}$, it will continue to work. – achille hui Jun 25 '14 at 05:31
  • I don't think so. The question only says that it works on $i$. It will be illogical to assume that it will also work for $j$ and $k$ only due to the reason that their range of values are the same set. – William Hilbert Jun 25 '14 at 05:33
  • @William: I would agree with you if it were $\sum\sum\sum x_iy_jz_k$. But it's not: $x_3$ is $x_3$ regardless of whether $i=3$ or $j=3$ or $k=3$. – Eric Stucky Jun 25 '14 at 05:34
  • @WilliamHilbert This is rule of substitution in predicate calculus. If you have a statement $P(x)$ that depends on a variable $x$. If $y, z$ are any two variables that doesn't appear in $P$, then $(\forall y, P(z)) \iff (\forall z,P(z))$. How you name the variable (index in this case) doesn't matter. – achille hui Jun 25 '14 at 05:59
  • @achillehui though it was long year's back I have a question how did u write $\sum_{i\ne j \ne k} x_i x_j x_k = \sum_{i\ne j \ne k} x_{n-i+1} x_{n-j+1} x_{n-k+1}$.....moreover I have come across this problem(https://math.stackexchange.com/questions/3765705/evaluating-s-depending-upon-following-condition-calculate-the-sum-s-sigma-si) and got your solution to related this topic –  Jul 23 '20 at 04:52