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Does a closed form exist for the following product of two summations with the same indices?

Assuming $a_i$, $b_i \in (-\infty, \infty)$, $i = \{1, 2, ..., n\}$, we have: $$ \sum_{i=1}^{n} \frac{a_i^2}{(b_i - a_i)^2} \sum_{i=1}^{n} \frac{(b_i - a_i)^2}{2a_i^2} $$

What I have so far, is that since there are cross-product terms, I will need to rewrite the summation as the following :

$$ \sum_{i=1}^{n} \sum_{j=1}^{n} \frac{a_i^2}{(b_i - a_i)^2} \frac{(b_j - a_j)^2}{2a_j^2} $$

It is evident that the portion of the summation, whenever $i = j$, would sum up to $\frac{n}{2}$. What confuses me are the cross-terms, which do not seem to converge to some closed form or a constant.

However, since the indices are not the same in the two fractions, is there nothing I can do to the summation to reduce it further into a closed form? (particularly, in the context of the question I am working on, this summation is supposed to converge to some sort of a constant over 2).

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1 Answer 1

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The question is stated in an overcomplicated way. Setting $$x_i = \frac{a_i^2}{(b_i-a_i)^2},$$ You are asking for

$$\frac12 \sum_{i=1}^n x_i \sum_{i=1}^n \frac1{x_i}=\frac12 (n + \sum_{i\neq j} \frac{x_i}{x_j})$$

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  • $\begingroup$ Re-writing in that form makes sense. However, is there a closed form (or convergence to a constant) for $\sum_{i \ne j} \frac{x_i}{x_j}$ ? The cross-terms are the ones that stumped me. $\endgroup$
    – NicTam
    Commented Mar 11, 2021 at 20:19
  • $\begingroup$ @NicTam To answer your questions: no, there is no closed form, and I don't know about convergence, since the $x_$ are arbitrary (positive). $\endgroup$
    – Igor Rivin
    Commented Mar 11, 2021 at 20:21
  • $\begingroup$ For arbitrary $x_i$ we can not teally hope for convergence. For example, let $a_i = i$ and $b_i = 0$. Clearly $x_i = 1$ and the cross term sum diverges as $n \to \infty$ $\endgroup$ Commented Mar 11, 2021 at 20:29

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