Formula for analytic functions?
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In here (third under double infinite series) they list the following formula.
$$displaystyle sum_{k=0}^{infty} sum_{j=0}^{infty} a_{k,j} = sum_{j=0}^{infty} sum_{k=0}^{j} a_{k, j-k}$$
Is this true and how does one prove it?
real-analysis sequences-and-series combinatorics
asked Dec 10 '18 at 4:56
Art VandelayArt Vandelay
133
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add a comment |
$begingroup$
In here (third under double infinite series) they list the following formula.
$$displaystyle sum_{k=0}^{infty} sum_{j=0}^{infty} a_{k,j} = sum_{j=0}^{infty} sum_{k=0}^{j} a_{k, j-k}$$
Is this true and how does one prove it?
real-analysis sequences-and-series combinatorics
asked Dec 10 '18 at 4:56
Art VandelayArt Vandelay
133
$endgroup$
$begingroup$
The same terms appear on both sides: of course if these are real numbers, one needs the series to be absolutely convergent.
$endgroup$
– Lord Shark the Unknown
Dec 10 '18 at 5:00
$begingroup$
$(a_{0,0} + a_{0,1} + a_{0,2}) + (a_{1,0} + a_{1,1} + a_{1,2}) + (a_{2,0} + a_{2,1} + a_{2,2}) = (a_{0,0}) + (a_{0,1} + a_{1,0}) + (a_{0,2} + a_{1,2} + a_{2,0})$
$endgroup$
– rsadhvika
Dec 10 '18 at 5:12
$begingroup$
Notice that on the right hand side the indices add up to a fixed number in each group. $(k + j - k = j)$
$endgroup$
– rsadhvika
Dec 10 '18 at 5:16
add a comment |
$begingroup$
In here (third under double infinite series) they list the following formula.
$$displaystyle sum_{k=0}^{infty} sum_{j=0}^{infty} a_{k,j} = sum_{j=0}^{infty} sum_{k=0}^{j} a_{k, j-k}$$
Is this true and how does one prove it?
real-analysis sequences-and-series combinatorics
asked Dec 10 '18 at 4:56
Art VandelayArt Vandelay
133
$endgroup$
In here (third under double infinite series) they list the following formula.
$$displaystyle sum_{k=0}^{infty} sum_{j=0}^{infty} a_{k,j} = sum_{j=0}^{infty} sum_{k=0}^{j} a_{k, j-k}$$
Is this true and how does one prove it?
real-analysis sequences-and-series combinatorics
real-analysis sequences-and-series combinatorics
asked Dec 10 '18 at 4:56
Art VandelayArt Vandelay
133
asked Dec 10 '18 at 4:56
Art VandelayArt Vandelay
133
asked Dec 10 '18 at 4:56
Art VandelayArt Vandelay
133
asked Dec 10 '18 at 4:56
Art VandelayArt Vandelay
133
asked Dec 10 '18 at 4:56
Art VandelayArt Vandelay
133
133
$begingroup$
The same terms appear on both sides: of course if these are real numbers, one needs the series to be absolutely convergent.
$endgroup$
– Lord Shark the Unknown
Dec 10 '18 at 5:00
$begingroup$
$(a_{0,0} + a_{0,1} + a_{0,2}) + (a_{1,0} + a_{1,1} + a_{1,2}) + (a_{2,0} + a_{2,1} + a_{2,2}) = (a_{0,0}) + (a_{0,1} + a_{1,0}) + (a_{0,2} + a_{1,2} + a_{2,0})$
$endgroup$
– rsadhvika
Dec 10 '18 at 5:12
$begingroup$
Notice that on the right hand side the indices add up to a fixed number in each group. $(k + j - k = j)$
$endgroup$
– rsadhvika
Dec 10 '18 at 5:16
add a comment |
$begingroup$
The same terms appear on both sides: of course if these are real numbers, one needs the series to be absolutely convergent.
$endgroup$
– Lord Shark the Unknown
Dec 10 '18 at 5:00
$begingroup$
$(a_{0,0} + a_{0,1} + a_{0,2}) + (a_{1,0} + a_{1,1} + a_{1,2}) + (a_{2,0} + a_{2,1} + a_{2,2}) = (a_{0,0}) + (a_{0,1} + a_{1,0}) + (a_{0,2} + a_{1,2} + a_{2,0})$
$endgroup$
– rsadhvika
Dec 10 '18 at 5:12
$begingroup$
Notice that on the right hand side the indices add up to a fixed number in each group. $(k + j - k = j)$
$endgroup$
– rsadhvika
Dec 10 '18 at 5:16
$begingroup$
The same terms appear on both sides: of course if these are real numbers, one needs the series to be absolutely convergent.
$endgroup$
– Lord Shark the Unknown
Dec 10 '18 at 5:00
$begingroup$
The same terms appear on both sides: of course if these are real numbers, one needs the series to be absolutely convergent.
$endgroup$
– Lord Shark the Unknown
Dec 10 '18 at 5:00
$begingroup$
$(a_{0,0} + a_{0,1} + a_{0,2}) + (a_{1,0} + a_{1,1} + a_{1,2}) + (a_{2,0} + a_{2,1} + a_{2,2}) = (a_{0,0}) + (a_{0,1} + a_{1,0}) + (a_{0,2} + a_{1,2} + a_{2,0})$
$endgroup$
– rsadhvika
Dec 10 '18 at 5:12
$begingroup$
$(a_{0,0} + a_{0,1} + a_{0,2}) + (a_{1,0} + a_{1,1} + a_{1,2}) + (a_{2,0} + a_{2,1} + a_{2,2}) = (a_{0,0}) + (a_{0,1} + a_{1,0}) + (a_{0,2} + a_{1,2} + a_{2,0})$
$endgroup$
– rsadhvika
Dec 10 '18 at 5:12
$begingroup$
Notice that on the right hand side the indices add up to a fixed number in each group. $(k + j - k = j)$
$endgroup$
– rsadhvika
Dec 10 '18 at 5:16
$begingroup$
Notice that on the right hand side the indices add up to a fixed number in each group. $(k + j - k = j)$
$endgroup$
– rsadhvika
Dec 10 '18 at 5:16
add a comment |
1 Answer
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I'll give an analogy here.
Imagine painting a rectangular wall. You can cover the wall in many ways.
One way is to paint it column by column :
On $xy$ plane, this represents a situation when you keep $x$ constant and vary $y$ in each column.
Alternatively you could also go like this :
On $xy$ plane, this represents a situation when you keep $x+y$ constant in each slanted line.
answered Dec 10 '18 at 5:24
rsadhvikarsadhvika
1,7101228
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
I'll give an analogy here.
Imagine painting a rectangular wall. You can cover the wall in many ways.
One way is to paint it column by column :
On $xy$ plane, this represents a situation when you keep $x$ constant and vary $y$ in each column.
Alternatively you could also go like this :
On $xy$ plane, this represents a situation when you keep $x+y$ constant in each slanted line.
answered Dec 10 '18 at 5:24
rsadhvikarsadhvika
1,7101228
$endgroup$
add a comment |
$begingroup$
I'll give an analogy here.
Imagine painting a rectangular wall. You can cover the wall in many ways.
One way is to paint it column by column :
On $xy$ plane, this represents a situation when you keep $x$ constant and vary $y$ in each column.
Alternatively you could also go like this :
On $xy$ plane, this represents a situation when you keep $x+y$ constant in each slanted line.
answered Dec 10 '18 at 5:24
rsadhvikarsadhvika
1,7101228
$endgroup$
add a comment |
$begingroup$
I'll give an analogy here.
Imagine painting a rectangular wall. You can cover the wall in many ways.
One way is to paint it column by column :
On $xy$ plane, this represents a situation when you keep $x$ constant and vary $y$ in each column.
Alternatively you could also go like this :
On $xy$ plane, this represents a situation when you keep $x+y$ constant in each slanted line.
answered Dec 10 '18 at 5:24
rsadhvikarsadhvika
1,7101228
$endgroup$
I'll give an analogy here.
Imagine painting a rectangular wall. You can cover the wall in many ways.
One way is to paint it column by column :
On $xy$ plane, this represents a situation when you keep $x$ constant and vary $y$ in each column.
Alternatively you could also go like this :
On $xy$ plane, this represents a situation when you keep $x+y$ constant in each slanted line.
answered Dec 10 '18 at 5:24
rsadhvikarsadhvika
1,7101228
answered Dec 10 '18 at 5:24
rsadhvikarsadhvika
1,7101228
answered Dec 10 '18 at 5:24
rsadhvikarsadhvika
1,7101228
answered Dec 10 '18 at 5:24
rsadhvikarsadhvika
1,7101228
1,7101228
add a comment |
add a comment |
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$begingroup$
The same terms appear on both sides: of course if these are real numbers, one needs the series to be absolutely convergent.
$endgroup$
– Lord Shark the Unknown
Dec 10 '18 at 5:00
$begingroup$
$(a_{0,0} + a_{0,1} + a_{0,2}) + (a_{1,0} + a_{1,1} + a_{1,2}) + (a_{2,0} + a_{2,1} + a_{2,2}) = (a_{0,0}) + (a_{0,1} + a_{1,0}) + (a_{0,2} + a_{1,2} + a_{2,0})$
$endgroup$
– rsadhvika
Dec 10 '18 at 5:12
$begingroup$
Notice that on the right hand side the indices add up to a fixed number in each group. $(k + j - k = j)$
$endgroup$
– rsadhvika
Dec 10 '18 at 5:16