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At 11:15 of this video, Michael Penn takes this sum,

$$ \sum_{i=1}^{n} \sum_{j=1}^{i} \frac{1}{ \alpha_j \alpha_{i-j} }$$

And splits into three sums,

$j=i-j$ , $ j<i-j$ and $j>i-j$

but I don't understand, what's the intuition behind this? Like, what exactly is the procedure behind putting constraints on indexs to split the sums into sub sums?

Looking for explanations based on using a concrete example how inserting restriction on it helps us make subsums?

tryst with freedom
  • 11,538

2 Answers2

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Let's consider a simpler case:

You have $$a^2+b^2+c^2+2ab+2bc+2ca-2ab-2bc-2ca$$

You may write it as $$\sum_{i\in\{a,b,c\}}\sum_{j\in\{a,b,c\}}{ij}-2\sum_{i,j\in\{a,b,c\}, i<j}{ij}$$.

This is just a fancy way of writing the top equation but it makes the thing less obvious to see the difference is just the $i=j$ cases. What the video is saying, is that the left summation and the right summation has difference for those $i=j$ cases.

cr001
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If I have the sum $$\text{apple} +\text{lion} +\text{banana} +\text{England} +\text{Brazil} +\text{donkey} +\text{Australia} +\text{China} +\text{bear} $$

I can rewrite it like $$(\text{apple} +\text{banana}) +(\text{England} +\text{Brazil} +\text{Australia} +\text{China})+(\text{donkey} +\text{bear} +\text{lion})$$

Rather than summing among all these strange things together, I prefer to put the fruit-things together, the country-things together, and the animal-things together (hoping maybe that the three sub-sums will be easier to handle).

TheSilverDoe
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  • hmm, nice, but could you explain using a concrete example how inserting restriction on it helps us make the subsums? – tryst with freedom Sep 02 '20 at 14:19
  • Well, this is a stupid example, but let's say that you want to calculate $\sum_{k=1}^{10} (-1)^k$. Because $(-1)^k$ is equal to $1$ if $k$ is even and $-1$ if $k$ is odd, you can split the sum in two parts : the one for the even $k$'s, and the one for the odd $k$'s. You have then that $\sum_{k=1}^{10} (-1)^k = \sum_{k=1 \ k \text{ even}}^{10} (-1)^k + \sum_{k=1 \ k \text{ odd}}^{10} (-1)^k = \sum_{k=1 \ k \text{ even}}^{10} 1 + \sum_{k=1 \ k \text{ odd}}^{10} (-1)$. So it is equal to the number of even numbers between $1$ and $10$ minus the numbers of odd ones. So it is $0$. – TheSilverDoe Sep 02 '20 at 14:27