Product of Summation for a single table of values.

Question

I've been looking around but can't get a exactly clear answer on my question.

I'm provided a table of values of $x_i$ and $y_i$ for $i = 1$ to $i = 5$.

I'm then asked to evaluate $$\sum_{i=1}^5\sum_{j=1}^5x_jy_i$$

I know that they're recognised as similar notation so that $\sum_{i=1}^5y_i\cdot\sum_{i=1}^5x_i$ is the solution, but why are they ($j$ and $i$) recognised to be similar [see edit]? Why not $\sum_{i=1}^5\sum_{i=1}^5x_iy_i$ ?

EDIT: my simplified version: Why does $j = i$? Or why are the values for $i$ used for $j$?

Is $\sum_{i=1}^5\sum_{i=1}^5x_iy_i$ invalid and would $\sum_{i=1}^5x_iy_i$ be the same as the expression for evaluation?

Answer 1

Okay, let us start with your expression. I will give you a hint and you can work it out to see what I meant.

$\sum^5_{i=1} \sum^5_{j=1} x_i y_j = x_1 (y_1 + ... + y_5) + ... + x_5 (y_1 + ... + y_5)$

The other summation will be the following.

$\sum^5_{i=1} x_i \sum^5_{j=1} y_j = (x_1 + ... + x_5) \times (y_1 + ... y_5) $

Now, your job is to show that these will end up being the same. You can already see terms like $x_1 y_1$ occuring in both expressions. This holds also for the cross terms and therefore they are the same. That is way people write summations most of the time as left as possible, it is really all the same!

Answer 2

why are they recognized as similar expression here is why

$$\sum_{i=1}^5\sum_{j=1}^5x_jy_i$$

lets first expand the inner sum we can write it as $$\sum_{i=1}^5(x_1y_i+x_2y_i+x_3y_i+x_4y_i+x_5y_i)$$ as you can $y_i$ is not changing becuase it depends on i thus it is common and we can take it out now our expression becomes $$\sum_{i=1}^5y_i\cdot(x_1+x_2+x_3+x_4+x_5)$$ and we can write $x_1+x_2+x_3+x_4+x_5$ it as $\sum_{j=1}^{5}x_j$ thus our expression becomes $$\sum_{i=1}^5y_i\cdot \sum_{j=1}^{5}x_j$$ thus $$\sum_{i=1}^5\sum_{j=1}^5x_jy_i=\sum_{i=1}^5y_i\cdot \sum_{j=1}^{5}x_j$$

you couldn't that if you had an expression like this $$\sum_{i=1}^5\sum_{j=1}^5(x_jy_i+1)$$ now you can not take $y_i$ as common

I hope you understand what I am trying to say.