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I have difficulties in evaluating the multi-index notation in the following context:

Let $x \in R^n$ and let $i$ be a multi-index, $i=(i_1, \dots, i_n)$. Now I want to know the bound of the sum (knowing that each $x_i \leq a$, $\frac{1}{x_i} \leq b$) $$ \sum_{|i|=3} \frac{1}{i!} \frac{x^i}{x_l} \quad (\ast) $$

What I thought so far:

There are three types of summands, (i) there is one $i_j=3$ (and the rest $i_k$, $k \neq j$ is 0) in $i$, then $i!$=6; (ii) there is one index $i_{j_1}=2$ and one $i_{j_2}$=1 (the others 0), then $i!=2$; and (iii) there are three components of $i$ 1, the rest 0, the faculty would be 1. (the types would be $(0, \dots, 3, \dots, 0)$, $(0, \dots, 0 , 2, 0, \dots, 0, 1, 0, \dots, 0)$ and $(0, \dots, 0 , 1, 0, \dots, 0, 1, 0, \dots, 0, 1, 0, \dots, 0)$).

Now if I look at the first case, I would have $(n-1)$ times the expression $\frac{x_j^3}{x_l}$, $l \neq j$ and once $x_l^2$, so $$ (\ast) = \frac{1}{6} ( \underbrace{\sum_{j \neq l} \frac{x_j^3}{x_l}}_{ \leq (n-1) a^3 b} \quad + \underbrace{x_l^2}_{\leq a^2}) + \frac{1}{2} (\text{???}) + \frac{1}{1} (\text{????}) $$

Q: How can I obtain the whole bound for $(\ast)$ ?

Is this right so far? How can I continue for the second case (ii)? Is there a way to let it do mechanically be mathematica? (Does someone know how to put multi-index notation in Mathematica)

Note: I write $x_i \leq a$, $\frac{1}{x_i} \leq b$ because in the application I look at $E(\frac{x^i}{x_l})$ and have bounds for the expectation values.

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In your original expression, what is the $l$ in the index in the denominator? –  Willie Wong Sep 2 '11 at 10:04
    
a fixed (normal) index –  Johannes L Sep 2 '11 at 10:06
1  
How sharp a bound do you need? If $l$ is fixed, you can take it outside the sum, and get trivially a bound $b a^3 \sum_{|i| = 3} (i!)^{-1}$. Whatever you do you will get $O(a^3 b)$, the only issue is whether you care about the coefficients in front. Do you? –  Willie Wong Sep 2 '11 at 10:13
3  
Two remarks (in addition to @Willie's question above): (1) The obvious upper bound $n^3a^3b$ might be essentially optimal. If one thinks it is not, one should give a reason why. (2) If indeed the original problem involves expectations, using uniform upper/lower bounds to estimate expectations (possibly of independent random variables?) might lead to a substantial overestimation of the result. Thus it might be more efficient to ask the full question. –  Did Sep 2 '11 at 10:14
    
@Willie I think it should be enough to assume $x$ of the Form $h E(Y)$, and $(\ast) \leq h^2 c$, $c$ constant (that means $c$ should only depend ont he dimension, on $a$ and on $b$) And herein, $a$ and $b$ in the sense $E(Y) \leq a$, $E(1/Y) \leq b$. –  Johannes L Sep 2 '11 at 11:06
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1 Answer

up vote 1 down vote accepted

To provide an answer, it is basically just what @Willie Wong already said in his comments.

The boundedness conditions are $|x_i| \leq a$, $|\frac{1}{x_i}| \leq b$.

By definition, $x^i= x_1^{i_1} \cdots x_n^{i_n}$. Note that for $|i|=3$, we can write $x^i=x_{j_1} x_{j_2} x_{j_3}$ (it is clear that there are at most three $j_k\neq 0$, but we can also write it as three always - for example, if $i_j=3$, $x^{i_j}_j=x_{j_1} x_{j_2} x_{j_3}$ where $j_1=j_2=j_3:=j$ (and $i_{j_1}=i_{j_2}=i_{j_3}=1$)).

Therefore $x^i \leq a^3$ , and each summand $\leq a^3 b$, so that $$(\ast) \leq a^3 b C \text{, } C:= \sum_{|j|=3} \frac{1}{j!} \text{.}$$

As I said in my comments for the task behind it was important for me to see that, $\frac{(hx)^i}{hx_l} = \frac{h^{|i|}}{h} \frac{x^i}{x_l} = h^2 \frac{x^i}{x_l}$ for $|i|=3$. Thanks a lot for all your help

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