# summation question

• Oct 8th 2013, 09:26 AM
kingsolomonsgrave
summation question
Maple says that

Attachment 29407

why is each of the|Xi-Xj| multiplied by two?

Is it that |Xi - Xj| =|Xj - Xi|?

I had not realized that double summation multiplies sums every combination of Xi and Xj
• Oct 8th 2013, 10:05 AM
emakarov
Re: summation question
Quote:

Originally Posted by kingsolomonsgrave
Is it that |Xi - Xj| =|Xj - Xi|?

Yes. There are nine ordered pairs $\displaystyle (i, j)$, and three of them (with equal elements) give rise to $\displaystyle |X_i - X_i| = 0$. The other six are split into three pairs because $\displaystyle |X_i - X_j| =|X_j - X_i|$.

If you want just $\displaystyle |X_1-X_2|+|X_1-X_3|+|X_2-X_3|$, you can denote it by $\displaystyle \sum_{1\le i<j\le 3}|X_i-X_j|$.
• Oct 8th 2013, 08:44 PM
ibdutt
Re: summation question
• Oct 9th 2013, 04:27 AM
SlipEternal
Re: summation question
On line 4, you keep i=1 for all three terms. On line 5, you keep i=2 for all three terms. Then on line 6, you keep i=3 for all three terms.

It should read (beginning on line 4):
\displaystyle \begin{align*}= & \hspace{1em}\hspace{3pt} |x_1-x_1| + |x_2-x_1| + |x_3-x_1| \\ & + |x_1-x_2| + |x_2-x_2| + |x_3-x_2| \\ & + |x_1-x_3| + |x_2-x_3| + |x_3-x_3| \\ = & \hspace{1em}\hspace{3pt} 0 + |x_2-x_1| + |x_3-x_1| \\ & + |x_1-x_2| + 0 + |x_3-x_2| \\ & + |x_1-x_3| + |x_2-x_3| + 0 \\ = & \, \left(|x_1-x_2| + |x_2-x_1|\right) + \left(|x_1-x_3| + |x_3-x_1|\right) + \left(|x_2-x_3| + |x_3-x_2|\right) \\ = & \, 2\left(|x_1-x_2| + |x_1-x_3| + |x_2-x_3|\right)\end{align*}
• Oct 9th 2013, 08:39 PM
ibdutt
Re: summation question
Quote:

Originally Posted by ibdutt

Sorry i made some typing mistake, it shd be as under
Attachment 29432
• Oct 10th 2013, 06:44 AM
SlipEternal
Re: summation question
Yes, that is correct.