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Let $j \in \mathbb{R}^n$ be a multi-index with $|j|=3$. Fix an index $l$, then there are how many $X^j$ with $j_l=0$?, more precise: I want to know what $ \text{#} \{ X^j : |j|=3, j_l \neq 0\}$ is.

So did the following: If $j$, $|j|=3$, then $X^j=X_{i_1} X_{i_2} X_{i_3}$ (for example $j_i=3$, then choose $i_1=i_2=i_3=i$), and $j_l=1$ means that $i_3=l$ already fixed, $X^j=X_{i_1} X_{i_2} X_l$. So if we disjunctly seperate it in the cases $j_l=1,2$ and $3$, we would have

  1. $\text{#} \{ X^j : |j|=3, j_l =1 \}= \text{#}\{ X_{i_1} X_{i_2} X_l : i_1,i_2 \neq l \}=(n-1)^2$ (place 2 stones $i_1, i_2$ on a line with $n-1$ free places (place l is "occupied") )
  2. $\text{#} \{ X^j : |j|=3, j_l =2 \}= \text{#}\{ X_{i_1} X_l^2 : i_1 \neq l \}=n-1$
  3. $\text{#} \{ X^j : |j|=3, j_l =3 \}= \text{#}\{ X_l^3 \}=1$

So in sum $n^2-n-1$. The problem is that, if I check this, it should equal $\text{#} \{X^j : |j|=3\} - \text{#} \{ X^j : |j|=3, j_l=0 \} = n^3 - \text{#} \{ X_{i_1} X_{i_2} X_{i_3} : i_1, i_2, i_3 \neq l \}=$ $=n^3-(n-1)^3$

But $n^3-(n-1)^3=1-3n+3n^2$, so I did something wrong somewhere.. Does somebody see the mistake(s)? (Or has another way to assure the result)

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    yes, as you said $|j|=j_1+ \dots + j_n$ for a multi-index $j$.2011-09-08

1 Answers 1

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You're oscillating between distinguishing by order and not. For instance, to get $(n-1)^2$, you're treating $i_1$ and $i_2$ as distinguishable from each other, but not from $l$. This should be either $3(n-1)^2$ if you want to distinguish by order, or $n(n-1)/2$ if you don't. For multi-indices, it would make sense not to; then with the correct value for 1. you have

$\frac{n(n-1)}2+(n-1)+1=\frac{n(n+1)}2\;,$

and the check yields

$ \begin{eqnarray} \binom {n+2}3-\binom {n+1}3 &=& \frac{(n+2)(n+1)n}6-\frac{(n+1)n(n-1)}6 \\ &=& ((n+2)-(n-1))\frac{(n+1)n}6 \\ &=& \frac{n(n+1)}2 \end{eqnarray} $

as expected.

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    @joriki yes, thanks for the help with your answer!2019-04-02