# $\frac{\sum^n_{i=0} x_i} {\sum^n_{i=0} x_i^2 }$

Hello I'm not sure how to evaluate this $$\frac{\sum^n_{i=0} x_i} {\sum^n_{i=0} x_i^2 }$$ is it $$\displaystyle \frac{x_0+x_1+...+x_n}{x_0^2+x_1^2+...+.x_n^2}$$

or

$$\frac{1}{\sum^n_{i=0}x_i}$$

EDIT

Now consider this $$\frac{\sum^n_{i=0} (x_i-a)\sum^n_{i=0} (x_i-a) }{\sum^n_{i=0} (x_i-a)^2\sum^n_{i=0} (x_i-a)^2}=\frac{\sum^n_{i=0} (x_i-a)^2}{\sum^n_{i=0} (x_i-a)^2\sum^n_{i=0} (x_i-a)^2}= \frac{1}{\sum^n_{i=0} (x_i-a)^2}$$

Is it ok?

Im not sure when can we split the summation and when we can't. Where can I get more information about this. ?

It is $$\frac {x_0+x_1+x_2+...x_n}{x_0^2+x_1^2+x_2^2+...x_n^2}$$

$\sum$ starts with $i=0$.

• can you take a look at my new quesiotn? thx – Elina May 10 '17 at 19:15

It is the first one. The author would write $(\sum_{i=0}^n x_i)^2$ if they meant the second (though it looks like you missed the terms involving $x_0$).

• can you take a look at my new quesiotn? thx – Elina May 10 '17 at 19:15

$$\frac{\sum^n_{i=0} (x_i-a)\sum^n_{i=0} (x_i-a) }{\sum^n_{i=0} (x_i-a)^2\sum^n_{i=0} (x_i-a)^2}=\frac{\sum^n_{i=0} (x_i-a)^2}{\sum^n_{i=0} (x_i-a)^2\sum^n_{i=0} (x_i-a)^2}= \frac{1}{\sum^n_{i=0} (x_i-a)^2}$$

Is it ok?

No, it's not OK because $\,\sum^n_{i=0} (x_i-a)\sum^n_{i=0} (x_i-a) \,\ne\, \sum^n_{i=0} (x_i-a)^2\,$, for the same reason why, for example, $(a+b+c)(a+b+c) \ne a^2+b^2+c^2\,$ in general.

What is OK would be $\,\sum^n_{i=0} (x_i-a)\sum^n_{i=0} (x_i-a) \,=\, \big(\sum^n_{i=0} (x_i-a)\big)^2\,$.

Im not sure when can we split the summation and when we can't.

When in doubt, you can always write down the actual sums, and it becomes more apparent.

\begin{align} \sum^n_{i=0} (x_i-a)\sum^n_{i=0} (x_i-a) &= \big((x_0-a)+(x_1-a)+\cdots+(x_n-a)\big) \\[3px] &\quad\cdot \big((x_0-a)+(x_1-a)+\cdots+(x_n-a)\big) \\[3px] &\ne \big((x_0-a)^2+(x_1-a)^2+\cdots+(x_n-a)^2\big) = \sum^n_{i=0} (x_i-a)^2 \end{align}