1
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ The asterisk is multiplication? $\endgroup$ – GFauxPas Apr 23 '15 at 2:49
  • $\begingroup$ 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. $\endgroup$ – Claude Leibovici Apr 23 '15 at 3:04
  • $\begingroup$ 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)? $\endgroup$ – kathystehl Apr 25 '15 at 18:19
  • $\begingroup$ The last one is correct. $\endgroup$ – Claude Leibovici Apr 25 '15 at 18:20
  • $\begingroup$ Do you mean the edit that I just included a few seconds ago @Claude ? $\endgroup$ – kathystehl Apr 25 '15 at 18:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.