Rearrange the index of a summation to get an infinite sum

114 Views Asked by At

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

1

There are 1 best solutions below

2
On BEST ANSWER

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:$$\cdots,-3,-2,-1,0,1,2,3,\cdots$$