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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?

Anthony S.
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2 Answers2

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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!

Mathphys meister
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  • Yes if you individually consider each summation then easily you can find that the each half of the summation will give you those values in the series. I think a further clarification on the question would be, why does j = i ?

    If you extended the table of values to include another set of values for the variable j, would that change the answer or would the original values of i still be used for j?

     – Anthony S. Aug 12 '18 at 22:48
  • I understand your problem. I completely missed that point. If you write $\sum^5_{i=1}x_i \sum^5_{i=1} y_i = \sum^5_{j=1} x_j \sum^5_{i=1} y_i$ then this is completely fine because i and j are dummy indices (you sum over them). But it is not true that $\sum^5_{i=1} \sum^5_{i=1} x_i y_i$ is a valid thing to do, then you should use a index j for one summation since otherwise you won’t recover the cross terms you are looking for. The last question of you is about $\sum^5_{i=1} x_i y_i = x_1 y_1 + ... + x_5 y_5$ so no cross terms at all so this does not work (first sum had cross terms). – Mathphys meister Aug 12 '18 at 23:06
  • The moral of the story is that you need two indices to get cross terms if you write the sums both left and otherwise you need to do the two sums separately and then multiply them giving the same result. – Mathphys meister Aug 12 '18 at 23:09
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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.

Deepesh Meena
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