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I was reading about double summation of series when the variable are dependent.

My book derived the the formula by creating a matrix to identify a pattern in the series:

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Then they derive the formula:

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What I am having trouble understanding is the range in the red box.

How can the range be $0 \leq i < j \leq n$ when the initial values for both $i$ and $j$ is $1$?

Is it a typo in the book? If we had some functions in place of $i$ and $j$, would the discrepancy in initial values matter? If it is not an error, then what is it that I am not getting?

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2 Answers 2

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The first summation handle the case where $i \ne j$.

The second part of the summation handle the case when $i=j$.

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  • $\begingroup$ I understand that. My problem is shouldn't the first part begin from $i=1$ (instead of $i=0$ as given in the book)? $\endgroup$ May 5, 2021 at 5:59
  • $\begingroup$ I see, fortunately, if you multiply with $0$, it doesn't add anything to the sum. so ya, you could have started from $1$. $\endgroup$ May 5, 2021 at 6:29
  • $\begingroup$ So it means the book is wrong then? What if instead of just $i$ and $j$, there were functions? Then would not the entire sum be wrong just cause I took the wrong initial values? $\endgroup$ May 5, 2021 at 6:44
  • $\begingroup$ if $i=0$, $ij=0$, so it wouldn't affect the sum. $0+1=1$. $\endgroup$ May 5, 2021 at 6:46
  • $\begingroup$ Ok. But what if we were to replace $i$ and $j$ by $f(i)$ and $f(j)$? Then would this still be valid? $\endgroup$ May 5, 2021 at 7:20
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Indexes in pair $(i,j)$ are equal, $i=j$, only on so called "main" diagonal from left upper corner to right down one.

This diagonal splits square in two triangles: left down triangle, where indexes starts from $i=2$(row) and $j=1$(column). In all pairs here we have $i \gt j$; right upper triangle, where indexes starts from $i=1$(row) and $j=2$(column). In all pairs here we have $i \lt j$.

Generally there are three sums accordingly to "main" diagonal, left down triangle and right upper triangle. When in triangles we have same elements, for example, when matrix is symmetric, then we obtain doubled sum for $i \ne j$, which is first in your formula.

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  • $\begingroup$ I get that. I am asking about the starting value of $i$ for the triangles, which according to part of the formula I have highlighted is $0$. How can they take it to be $0$ when the series begin from $1$ for both $i$ and $j$. $\endgroup$ May 5, 2021 at 5:54
  • $\begingroup$ @Alpha Delta. If you add summand which contains factor with $i=0$ or $j=0$, then this doesn't change sum, right? $\endgroup$
    – zkutch
    May 5, 2021 at 11:09
  • $\begingroup$ The sum changes if instead of just $i$ and $j$, $f(i)$ or $f(j)$ is used $\endgroup$ May 5, 2021 at 11:32
  • $\begingroup$ @Alpha Delta. Of course if/when you write other sum, then it changes. No? If you change summand $i\cdot j$ on something another, $f(i)\cdot f(j)$ for example, then, generally, it will not be same for zero indexes. $\endgroup$
    – zkutch
    May 5, 2021 at 12:45

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