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Problem Given N 3-dimensional points which are {$p_1,p_2,..,p_n$} where $p_i = (x_i,y_i,z_i) $ . I have to find the value of the formula

$$ \sum \limits_{i=0}^n \sum \limits_{j=i+1}^n \frac{ \mid \alpha*(x_i-x_j+\beta*(y_i-y_j)+\gamma*(z_i-z_j)+\delta) \mid }{ \sqrt{ (x_i-x_j)^4+(y_i-y_j)^4+(z_i-z_j)^4} } $$

for some given $ \alpha, \beta, \gamma, \delta $. all numbers are between 1 and M ( = 100).

I need an efficient method for the reduction of this formula to something which is easy to calculate.

So far, I have recognized that the above expression is the normal component of vector joining i and j from the the plane formed by $ \alpha, \beta, \gamma, \delta $ divided by the square of the distance between i and j in $l_4$ norm.

Please give any idea about how to reduce it to simpler standard form which can be calculated efficiently.

EDIT: By transforming and rotating the coordinate system such that the plane formed by $\alpha , \beta ,\gamma$ and $ \delta $ is represented by $z=0$, problem reduces to finding

$$ \sum \limits_{i=0}^n \sum \limits_{j=i+1}^n \frac{ \mid z_i-z_j \mid }{ \sqrt{ (x_i-x_j)^4+(y_i-y_j)^4+(z_i-z_j)^4} } $$

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  • $\begingroup$ Are you sure it is $4$ in exponent in denominator? $\endgroup$
    – user76479
    Sep 7, 2015 at 19:54
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    $\begingroup$ Where does this problem come from, and why do you expect a "simpler standard form" to exist? $\endgroup$
    – Stefan Kohl
    Sep 7, 2015 at 19:59
  • $\begingroup$ @Arul, yes it is 4. $\endgroup$
    – yejc
    Sep 8, 2015 at 3:24
  • $\begingroup$ @StefanKohl, I need to compute the expression but for large n, brute force method is very efficient, so I asked you guys to suggest some modifications to the expression to calculate it efficiently. $\endgroup$
    – yejc
    Sep 8, 2015 at 3:32
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    $\begingroup$ Is this the same question as mathoverflow.net/questions/217713/… ? $\endgroup$ Sep 8, 2015 at 5:41

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