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I'd like to know the basis for the following transformation:

$$\sum_{i,j:i+j=k}a_ib_j \quad k=0, \dots n+m$$

Let $j = k-i$ then:

$$\sum_{i=-\infty}^{\infty}a_ib_{k-i} \quad k=0, \dots n+m$$

I understand that the $j$ index is eliminated by the substitution, I just don't get how it suddenly becomes an infinite sum across all $i$.

Source

Dianne
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1 Answers1

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In the first sum you have pairs of indices $(i,j)$ all with property $i+j=k$.

So actually for e.g. $k=2$ like this :$$\cdots,(-3,5),(-2,4),(-1,3),(0,2),(1,3),(2,4),(3,5),\cdots$$

In the second sum only the $i$ remains and get the corresponding sequence:$$\cdots,-3,-2,-1,0,1,2,3,\cdots$$

drhab
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    ... and since $i+j=k$ we have $j=k-i$ I.e. once we know the first index, we can calculate the second. Although the first sum has two indices, the index values only have one degree of freedom. – gandalf61 Sep 01 '18 at 07:39
  • @gandalf61 Yep. I left that out because the OP said "I understand that the index $j$ is eliminated by the substitution". But it does not harm to mention that too. – drhab Sep 01 '18 at 07:42