Dealing with constants in summation notation

Asked Apr 23 '15 at 01:43

Active Apr 25 '15 at 18:31

Viewed 597 times

I haven't worked with summation notation in a while, and am unsure how to approach the following:

$\sum_1^n [-\frac 12 * \frac{(x_i - \alpha)^2}{\alpha}]$ where $\alpha \in R$

What would be the best/correct way to simplify this, i.e. pull the summation through to $x_i$? Would you leave the constants in the brackets, such that

$= -\frac n2 * \frac{(x_i - \alpha)^2}{\alpha}$

Or, should you pull the constants out so that we have

$-\frac{1}{2\alpha} \sum_1^n (x_i - \alpha)^2$ ?

Or should I multiple out the $(x_i - \alpha)^2$ term so that we have

$-\frac{1}{2\alpha} \sum_1^n (x_i)^2 + \sum_1^n (x_i) - \frac{n\alpha}{2}$ ?

edited Apr 25 '15 at 18:31

asked Apr 23 '15 at 01:43

kathystehl

The asterisk is multiplication? – GFauxPas Apr 23 '15 at 02:49
Since there are multiplicative terms which are independent on $i$, just put them on the front of the $\Sigma$. Try writing the sum with three terms and the answer would become clear. – Claude Leibovici Apr 23 '15 at 03:04
Yes, the asterisk is multiplication. @Claude, do you mean that I should multiply out the $(x_i - alpha)^2$ term and end up with the last thing that I have added in the question section (just added)? – kathystehl Apr 25 '15 at 18:19
The last one is correct. – Claude Leibovici Apr 25 '15 at 18:20
Do you mean the edit that I just included a few seconds ago @Claude ? – kathystehl Apr 25 '15 at 18:22
The last last one has a sign error for the last term. The one before was OK. – Claude Leibovici Apr 25 '15 at 18:23
Okay, thank you. To others, I have corrected the sign error that Claude noted. The last term should now be correct. Thanks for your help. – kathystehl Apr 25 '15 at 18:32

Dealing with constants in summation notation

0 Answers0