Sigma Notations

Question

I have troubles understanding the sigma notation. If for example we have $c_i$ as $$c_i=\frac {x_i-x}{\sum(x_i-x)^2}$$ $$\sum c_i=\sum\frac{x_i-x}{\sum(x_i-x)^2}$$ Do we distribute the sigma to both top and bottom? But then the bottom would have double sigmas which dont make sense?

Answer 1

Your notation is confusing. Assuming $i$ goes from $0$ to $n$ and $c_i$ is actually $$c_i=\frac{x_i-x}{\sum^n_{k=0}(x_k-x)^2}.$$ Then, we have \begin{align} \sum_{i=0}^n c_i &= c_0 + c_1 + \dots + c_n\\ &=\frac{x_0-x}{\sum^n_{k=0}(x_k-x)^2} + \frac{x_1-x}{\sum^n_{k=0}(x_k-x)^2} + \dots +\frac{x_n-x}{\sum^n_{k=0}(x_k-x)^2}\\ &=\frac{\sum_{i=0}^n x_i-x}{\sum_{k=0}^n (x_k-x)^2}. \end{align}

On the other hand, if $c_i$ is actually: $$c_i=\frac{x_i-x}{\sum^i_{k=0}(x_k-x)^2}$$ Then, we have $$ \begin{align} \sum c_i&= c_0 + c_1 + \dots + c_n\\ &=\frac{x_0-x}{\sum^0_{k=0}(x_k-x)^2} + \frac{x_1-x}{\sum^1_{k=0}(x_k-x)^2} + \dots +\frac{x_n-x}{\sum^n_{k=0}(x_k-x)^2}\\ &=\frac{x_0-x}{(x_0-x)^2} + \frac{x_1-x}{(x_0-x)^2 + (x_1-x)^2} + \dots +\frac{x_n-x}{(x_0-x)^2 + (x_1-x)^2 + \dots + (x_n-x)^2} \end{align} $$ We see here that the denominators of the terms of the sum are not equal so you can't distribute the sigma