Rearrange the index of a summation to get an infinite sum
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I'd like to know the basis for the following transformation:
$$sum_i,j:i+j=ka_ib_j quad k=0, dots n+m$$
Let $j = k-i$ then:
$$sum_i=-infty^inftya_ib_k-i quad k=0, dots n+m$$
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 $i$.
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summation
asked Sep 1 at 7:15
Dianne
31
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up vote
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down vote
favorite
I'd like to know the basis for the following transformation:
$$sum_i,j:i+j=ka_ib_j quad k=0, dots n+m$$
Let $j = k-i$ then:
$$sum_i=-infty^inftya_ib_k-i quad k=0, dots n+m$$
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
summation
asked Sep 1 at 7:15
Dianne
31
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'd like to know the basis for the following transformation:
$$sum_i,j:i+j=ka_ib_j quad k=0, dots n+m$$
Let $j = k-i$ then:
$$sum_i=-infty^inftya_ib_k-i quad k=0, dots n+m$$
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
summation
asked Sep 1 at 7:15
Dianne
31
I'd like to know the basis for the following transformation:
$$sum_i,j:i+j=ka_ib_j quad k=0, dots n+m$$
Let $j = k-i$ then:
$$sum_i=-infty^inftya_ib_k-i quad k=0, dots n+m$$
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
summation
summation
asked Sep 1 at 7:15
Dianne
31
asked Sep 1 at 7:15
Dianne
31
asked Sep 1 at 7:15
Dianne
31
asked Sep 1 at 7:15
Dianne
31
asked Sep 1 at 7:15
Dianne
31
31
add a comment |Â
add a comment |Â
1 Answer
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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:$$cdots,-3,-2,-1,0,1,2,3,cdots$$
answered Sep 1 at 7:28
drhab
88.9k541122
1
... 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 1 at 7:39
@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 1 at 7:42
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
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:$$cdots,-3,-2,-1,0,1,2,3,cdots$$
answered Sep 1 at 7:28
drhab
88.9k541122
1
... 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 1 at 7:39
@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 1 at 7:42
add a comment |Â
up vote
1
down vote
accepted
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:$$cdots,-3,-2,-1,0,1,2,3,cdots$$
answered Sep 1 at 7:28
drhab
88.9k541122
1
... 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 1 at 7:39
@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 1 at 7:42
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
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:$$cdots,-3,-2,-1,0,1,2,3,cdots$$
answered Sep 1 at 7:28
drhab
88.9k541122
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:$$cdots,-3,-2,-1,0,1,2,3,cdots$$
answered Sep 1 at 7:28
drhab
88.9k541122
answered Sep 1 at 7:28
drhab
88.9k541122
answered Sep 1 at 7:28
drhab
88.9k541122
answered Sep 1 at 7:28
drhab
88.9k541122
88.9k541122
1
... 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 1 at 7:39
@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 1 at 7:42
add a comment |Â
1
... 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 1 at 7:39
@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 1 at 7:42
1
1
... 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 1 at 7:39
... 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 1 at 7:39
@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 1 at 7:42
@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 1 at 7:42
add a comment |Â
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