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How works multiindex?
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Can anyone explain me how works multiindex?
For example here:
$|overliner| = r_0 + r_1 + cdots + r_n $
$sum_j=1^msum_a_ijleft(fracpartial taupartialx_0 right)^r_0 left(fracpartial taupartialx_1 right)^r_1 cdots left(fracpartial taupartialx_n right)^r_n $
And sometimes I see this note: $|overliner| leq mu_j$
Thank you in advance for explaining!
summation
asked Jul 15 at 9:35
Ilqur
12110
add a comment |Â
up vote
1
down vote
favorite
Can anyone explain me how works multiindex?
For example here:
$|overliner| = r_0 + r_1 + cdots + r_n $
$sum_j=1^msum_a_ijleft(fracpartial taupartialx_0 right)^r_0 left(fracpartial taupartialx_1 right)^r_1 cdots left(fracpartial taupartialx_n right)^r_n $
And sometimes I see this note: $|overliner| leq mu_j$
Thank you in advance for explaining!
summation
asked Jul 15 at 9:35
Ilqur
12110
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Can anyone explain me how works multiindex?
For example here:
$|overliner| = r_0 + r_1 + cdots + r_n $
$sum_j=1^msum_a_ijleft(fracpartial taupartialx_0 right)^r_0 left(fracpartial taupartialx_1 right)^r_1 cdots left(fracpartial taupartialx_n right)^r_n $
And sometimes I see this note: $|overliner| leq mu_j$
Thank you in advance for explaining!
summation
asked Jul 15 at 9:35
Ilqur
12110
Can anyone explain me how works multiindex?
For example here:
$|overliner| = r_0 + r_1 + cdots + r_n $
$sum_j=1^msum_a_ijleft(fracpartial taupartialx_0 right)^r_0 left(fracpartial taupartialx_1 right)^r_1 cdots left(fracpartial taupartialx_n right)^r_n $
And sometimes I see this note: $|overliner| leq mu_j$
Thank you in advance for explaining!
summation
asked Jul 15 at 9:35
Ilqur
12110
edited Jul 15 at 10:15
asked Jul 15 at 9:35
Ilqur
12110
asked Jul 15 at 9:35
Ilqur
12110
asked Jul 15 at 9:35
Ilqur
12110
12110
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The multi-index is an $n$-tuple of numbers $$r = (r_1,r_2,...r_n);;;;r_i in mathbbN;;forall i$$
(so is different from a vector for the fact that all of it's components have to be positive integers) where you can define some operations such as sum and difference, absolute value, power:
- $r+s = (r_1+s_1,dots, r_n+s_n)$
- $|r| = r_1+r_2+dots+r_n$
- $x^r = x_1^r_1x_2^r_2dots x_n^r_n$
With this facts you can define, in a compact way, lots of mathematical concept like partial derivatives. I don't want to give you a list of all the things, for that there's a very good page on Wikipedia.
Let us now look at your formula. Without any context is difficult to give it a precise explanation but let me make some hypothesis: $$vecmu = (mu_0, mu_1, mu_2) = (6,10,6);;;textvector\ r = (r_0,r_1,r_2) = (2,1,3);;;textmulti-index\ mathbfx = (x_0,x_1,x_2);;textwhere;; x_n;;textis our coordinate system\a_ij;;textis a matrix$$
From your formula, the index $i$ is not summed over so it remains at the end of the calculation. Bare in mind that $|r| = 2+1+3 = 6$
beginalign
&B_i=sum_j=1^3sum_a_ijleft(fracpartial taupartialx_0 right)^r_0 left(fracpartial taupartialx_1 right)^r_1left(fracpartial taupartialx_2 right)^r_2 = \&a_i1 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3 + ;;;;;textfor j=1: mu_1 = 6 = |r|
\&a_i2 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3 + ;;;;;textfor j=2: mu_2 = 10 neq |r| text so we don't count it
\&a_i3 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3 ;;;;;;;;;textfor j=2: mu_3 = 6 = |r|
endalign
So in the end we have
$$
B_i = a_i1 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3 + a_i3 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3
$$
so if, for example $a_ij = delta_ij$, the identity matrix, we get that
$$ B_i = left{beginmatrix 2left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)left(fracpartial taupartialx_2 right)^3&textif i=1,3 \ 0 &textif i=2endmatrixright.$$
I hope this will help you!
Edit
As @Steffen Plunder made me notice, the formula that we were trying to study is quite more complex that how I made it look like. That's because the second summation $sum_r$ requires ones to choose beforehand only the value of $|r|$ (that probably is given) and then sum ove all the multi-indices whose absolute value is $|r|$ and it's equal to $mu_j$ (which from my example is $|r|=6$ with $r=(r_0,r_1,r_2)$) so like $$(0,0,6), (1,0,5), (1,1,4), (6,0,0), (0,5,1), (2,2,2), dots$$ and so on. But I think that the concept of how does the multi-index works is still given by my previous answer, the problem now is on the formula itself. In the comment to my answer is a more precise formula for the sum over all the multi-indices whose absolute value is $|r|$ and is less or equal to some value $k$. Thanks for the correction!
answered Jul 15 at 10:23
Davide Morgante
1,904220
Thank you very much!
– Ilqur
Jul 15 at 10:32
In the example with $B_i$, after the second equality sign, you miss out all the other possible multi-indices which have absolute value $6$. Maybe it would be good to explain $sum_ < k A_bar r= sum_r_1=0^k sum_r_2=0^k-r_1 dots sum_r_n=0^k-r_1 - dots - r_n-1 A_(r_1,dots,r_n) $
– Steffen Plunder
Jul 15 at 10:33
2
Yes, that's true. I gave the example on only one multi-index as I thought that this would explain at lest how the multi-index notation works. As I said, I don't really know were that formula comes from but in a more general manner the sum has to be taken on all the possible multi-indices with absolute value of $6$. I'll edit the post to integrate your detail!
– Davide Morgante
Jul 15 at 10:43
Thanks for the edit!
– Steffen Plunder
Jul 15 at 11:06
Thank you for the correction @SteffenPlunder :)
– Davide Morgante
Jul 15 at 11:06
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
The multi-index is an $n$-tuple of numbers $$r = (r_1,r_2,...r_n);;;;r_i in mathbbN;;forall i$$
(so is different from a vector for the fact that all of it's components have to be positive integers) where you can define some operations such as sum and difference, absolute value, power:
- $r+s = (r_1+s_1,dots, r_n+s_n)$
- $|r| = r_1+r_2+dots+r_n$
- $x^r = x_1^r_1x_2^r_2dots x_n^r_n$
With this facts you can define, in a compact way, lots of mathematical concept like partial derivatives. I don't want to give you a list of all the things, for that there's a very good page on Wikipedia.
Let us now look at your formula. Without any context is difficult to give it a precise explanation but let me make some hypothesis: $$vecmu = (mu_0, mu_1, mu_2) = (6,10,6);;;textvector\ r = (r_0,r_1,r_2) = (2,1,3);;;textmulti-index\ mathbfx = (x_0,x_1,x_2);;textwhere;; x_n;;textis our coordinate system\a_ij;;textis a matrix$$
From your formula, the index $i$ is not summed over so it remains at the end of the calculation. Bare in mind that $|r| = 2+1+3 = 6$
beginalign
&B_i=sum_j=1^3sum_a_ijleft(fracpartial taupartialx_0 right)^r_0 left(fracpartial taupartialx_1 right)^r_1left(fracpartial taupartialx_2 right)^r_2 = \&a_i1 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3 + ;;;;;textfor j=1: mu_1 = 6 = |r|
\&a_i2 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3 + ;;;;;textfor j=2: mu_2 = 10 neq |r| text so we don't count it
\&a_i3 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3 ;;;;;;;;;textfor j=2: mu_3 = 6 = |r|
endalign
So in the end we have
$$
B_i = a_i1 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3 + a_i3 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3
$$
so if, for example $a_ij = delta_ij$, the identity matrix, we get that
$$ B_i = left{beginmatrix 2left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)left(fracpartial taupartialx_2 right)^3&textif i=1,3 \ 0 &textif i=2endmatrixright.$$
I hope this will help you!
Edit
As @Steffen Plunder made me notice, the formula that we were trying to study is quite more complex that how I made it look like. That's because the second summation $sum_r$ requires ones to choose beforehand only the value of $|r|$ (that probably is given) and then sum ove all the multi-indices whose absolute value is $|r|$ and it's equal to $mu_j$ (which from my example is $|r|=6$ with $r=(r_0,r_1,r_2)$) so like $$(0,0,6), (1,0,5), (1,1,4), (6,0,0), (0,5,1), (2,2,2), dots$$ and so on. But I think that the concept of how does the multi-index works is still given by my previous answer, the problem now is on the formula itself. In the comment to my answer is a more precise formula for the sum over all the multi-indices whose absolute value is $|r|$ and is less or equal to some value $k$. Thanks for the correction!
answered Jul 15 at 10:23
Davide Morgante
1,904220
Thank you very much!
– Ilqur
Jul 15 at 10:32
In the example with $B_i$, after the second equality sign, you miss out all the other possible multi-indices which have absolute value $6$. Maybe it would be good to explain $sum_ < k A_bar r= sum_r_1=0^k sum_r_2=0^k-r_1 dots sum_r_n=0^k-r_1 - dots - r_n-1 A_(r_1,dots,r_n) $
– Steffen Plunder
Jul 15 at 10:33
2
Yes, that's true. I gave the example on only one multi-index as I thought that this would explain at lest how the multi-index notation works. As I said, I don't really know were that formula comes from but in a more general manner the sum has to be taken on all the possible multi-indices with absolute value of $6$. I'll edit the post to integrate your detail!
– Davide Morgante
Jul 15 at 10:43
Thanks for the edit!
– Steffen Plunder
Jul 15 at 11:06
Thank you for the correction @SteffenPlunder :)
– Davide Morgante
Jul 15 at 11:06
add a comment |Â
up vote
1
down vote
accepted
The multi-index is an $n$-tuple of numbers $$r = (r_1,r_2,...r_n);;;;r_i in mathbbN;;forall i$$
(so is different from a vector for the fact that all of it's components have to be positive integers) where you can define some operations such as sum and difference, absolute value, power:
- $r+s = (r_1+s_1,dots, r_n+s_n)$
- $|r| = r_1+r_2+dots+r_n$
- $x^r = x_1^r_1x_2^r_2dots x_n^r_n$
With this facts you can define, in a compact way, lots of mathematical concept like partial derivatives. I don't want to give you a list of all the things, for that there's a very good page on Wikipedia.
Let us now look at your formula. Without any context is difficult to give it a precise explanation but let me make some hypothesis: $$vecmu = (mu_0, mu_1, mu_2) = (6,10,6);;;textvector\ r = (r_0,r_1,r_2) = (2,1,3);;;textmulti-index\ mathbfx = (x_0,x_1,x_2);;textwhere;; x_n;;textis our coordinate system\a_ij;;textis a matrix$$
From your formula, the index $i$ is not summed over so it remains at the end of the calculation. Bare in mind that $|r| = 2+1+3 = 6$
beginalign
&B_i=sum_j=1^3sum_a_ijleft(fracpartial taupartialx_0 right)^r_0 left(fracpartial taupartialx_1 right)^r_1left(fracpartial taupartialx_2 right)^r_2 = \&a_i1 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3 + ;;;;;textfor j=1: mu_1 = 6 = |r|
\&a_i2 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3 + ;;;;;textfor j=2: mu_2 = 10 neq |r| text so we don't count it
\&a_i3 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3 ;;;;;;;;;textfor j=2: mu_3 = 6 = |r|
endalign
So in the end we have
$$
B_i = a_i1 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3 + a_i3 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3
$$
so if, for example $a_ij = delta_ij$, the identity matrix, we get that
$$ B_i = left{beginmatrix 2left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)left(fracpartial taupartialx_2 right)^3&textif i=1,3 \ 0 &textif i=2endmatrixright.$$
I hope this will help you!
Edit
As @Steffen Plunder made me notice, the formula that we were trying to study is quite more complex that how I made it look like. That's because the second summation $sum_r$ requires ones to choose beforehand only the value of $|r|$ (that probably is given) and then sum ove all the multi-indices whose absolute value is $|r|$ and it's equal to $mu_j$ (which from my example is $|r|=6$ with $r=(r_0,r_1,r_2)$) so like $$(0,0,6), (1,0,5), (1,1,4), (6,0,0), (0,5,1), (2,2,2), dots$$ and so on. But I think that the concept of how does the multi-index works is still given by my previous answer, the problem now is on the formula itself. In the comment to my answer is a more precise formula for the sum over all the multi-indices whose absolute value is $|r|$ and is less or equal to some value $k$. Thanks for the correction!
answered Jul 15 at 10:23
Davide Morgante
1,904220
Thank you very much!
– Ilqur
Jul 15 at 10:32
In the example with $B_i$, after the second equality sign, you miss out all the other possible multi-indices which have absolute value $6$. Maybe it would be good to explain $sum_ < k A_bar r= sum_r_1=0^k sum_r_2=0^k-r_1 dots sum_r_n=0^k-r_1 - dots - r_n-1 A_(r_1,dots,r_n) $
– Steffen Plunder
Jul 15 at 10:33
2
Yes, that's true. I gave the example on only one multi-index as I thought that this would explain at lest how the multi-index notation works. As I said, I don't really know were that formula comes from but in a more general manner the sum has to be taken on all the possible multi-indices with absolute value of $6$. I'll edit the post to integrate your detail!
– Davide Morgante
Jul 15 at 10:43
Thanks for the edit!
– Steffen Plunder
Jul 15 at 11:06
Thank you for the correction @SteffenPlunder :)
– Davide Morgante
Jul 15 at 11:06
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The multi-index is an $n$-tuple of numbers $$r = (r_1,r_2,...r_n);;;;r_i in mathbbN;;forall i$$
(so is different from a vector for the fact that all of it's components have to be positive integers) where you can define some operations such as sum and difference, absolute value, power:
- $r+s = (r_1+s_1,dots, r_n+s_n)$
- $|r| = r_1+r_2+dots+r_n$
- $x^r = x_1^r_1x_2^r_2dots x_n^r_n$
With this facts you can define, in a compact way, lots of mathematical concept like partial derivatives. I don't want to give you a list of all the things, for that there's a very good page on Wikipedia.
Let us now look at your formula. Without any context is difficult to give it a precise explanation but let me make some hypothesis: $$vecmu = (mu_0, mu_1, mu_2) = (6,10,6);;;textvector\ r = (r_0,r_1,r_2) = (2,1,3);;;textmulti-index\ mathbfx = (x_0,x_1,x_2);;textwhere;; x_n;;textis our coordinate system\a_ij;;textis a matrix$$
From your formula, the index $i$ is not summed over so it remains at the end of the calculation. Bare in mind that $|r| = 2+1+3 = 6$
beginalign
&B_i=sum_j=1^3sum_a_ijleft(fracpartial taupartialx_0 right)^r_0 left(fracpartial taupartialx_1 right)^r_1left(fracpartial taupartialx_2 right)^r_2 = \&a_i1 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3 + ;;;;;textfor j=1: mu_1 = 6 = |r|
\&a_i2 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3 + ;;;;;textfor j=2: mu_2 = 10 neq |r| text so we don't count it
\&a_i3 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3 ;;;;;;;;;textfor j=2: mu_3 = 6 = |r|
endalign
So in the end we have
$$
B_i = a_i1 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3 + a_i3 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3
$$
so if, for example $a_ij = delta_ij$, the identity matrix, we get that
$$ B_i = left{beginmatrix 2left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)left(fracpartial taupartialx_2 right)^3&textif i=1,3 \ 0 &textif i=2endmatrixright.$$
I hope this will help you!
Edit
As @Steffen Plunder made me notice, the formula that we were trying to study is quite more complex that how I made it look like. That's because the second summation $sum_r$ requires ones to choose beforehand only the value of $|r|$ (that probably is given) and then sum ove all the multi-indices whose absolute value is $|r|$ and it's equal to $mu_j$ (which from my example is $|r|=6$ with $r=(r_0,r_1,r_2)$) so like $$(0,0,6), (1,0,5), (1,1,4), (6,0,0), (0,5,1), (2,2,2), dots$$ and so on. But I think that the concept of how does the multi-index works is still given by my previous answer, the problem now is on the formula itself. In the comment to my answer is a more precise formula for the sum over all the multi-indices whose absolute value is $|r|$ and is less or equal to some value $k$. Thanks for the correction!
answered Jul 15 at 10:23
Davide Morgante
1,904220
The multi-index is an $n$-tuple of numbers $$r = (r_1,r_2,...r_n);;;;r_i in mathbbN;;forall i$$
(so is different from a vector for the fact that all of it's components have to be positive integers) where you can define some operations such as sum and difference, absolute value, power:
- $r+s = (r_1+s_1,dots, r_n+s_n)$
- $|r| = r_1+r_2+dots+r_n$
- $x^r = x_1^r_1x_2^r_2dots x_n^r_n$
With this facts you can define, in a compact way, lots of mathematical concept like partial derivatives. I don't want to give you a list of all the things, for that there's a very good page on Wikipedia.
Let us now look at your formula. Without any context is difficult to give it a precise explanation but let me make some hypothesis: $$vecmu = (mu_0, mu_1, mu_2) = (6,10,6);;;textvector\ r = (r_0,r_1,r_2) = (2,1,3);;;textmulti-index\ mathbfx = (x_0,x_1,x_2);;textwhere;; x_n;;textis our coordinate system\a_ij;;textis a matrix$$
From your formula, the index $i$ is not summed over so it remains at the end of the calculation. Bare in mind that $|r| = 2+1+3 = 6$
beginalign
&B_i=sum_j=1^3sum_a_ijleft(fracpartial taupartialx_0 right)^r_0 left(fracpartial taupartialx_1 right)^r_1left(fracpartial taupartialx_2 right)^r_2 = \&a_i1 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3 + ;;;;;textfor j=1: mu_1 = 6 = |r|
\&a_i2 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3 + ;;;;;textfor j=2: mu_2 = 10 neq |r| text so we don't count it
\&a_i3 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3 ;;;;;;;;;textfor j=2: mu_3 = 6 = |r|
endalign
So in the end we have
$$
B_i = a_i1 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3 + a_i3 left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)^1left(fracpartial taupartialx_2 right)^3
$$
so if, for example $a_ij = delta_ij$, the identity matrix, we get that
$$ B_i = left{beginmatrix 2left(fracpartial taupartialx_0 right)^2 left(fracpartial taupartialx_1 right)left(fracpartial taupartialx_2 right)^3&textif i=1,3 \ 0 &textif i=2endmatrixright.$$
I hope this will help you!
Edit
As @Steffen Plunder made me notice, the formula that we were trying to study is quite more complex that how I made it look like. That's because the second summation $sum_r$ requires ones to choose beforehand only the value of $|r|$ (that probably is given) and then sum ove all the multi-indices whose absolute value is $|r|$ and it's equal to $mu_j$ (which from my example is $|r|=6$ with $r=(r_0,r_1,r_2)$) so like $$(0,0,6), (1,0,5), (1,1,4), (6,0,0), (0,5,1), (2,2,2), dots$$ and so on. But I think that the concept of how does the multi-index works is still given by my previous answer, the problem now is on the formula itself. In the comment to my answer is a more precise formula for the sum over all the multi-indices whose absolute value is $|r|$ and is less or equal to some value $k$. Thanks for the correction!
answered Jul 15 at 10:23
Davide Morgante
1,904220
edited Jul 15 at 10:59
answered Jul 15 at 10:23
Davide Morgante
1,904220
answered Jul 15 at 10:23
Davide Morgante
1,904220
answered Jul 15 at 10:23
Davide Morgante
1,904220
1,904220
Thank you very much!
– Ilqur
Jul 15 at 10:32
In the example with $B_i$, after the second equality sign, you miss out all the other possible multi-indices which have absolute value $6$. Maybe it would be good to explain $sum_ < k A_bar r= sum_r_1=0^k sum_r_2=0^k-r_1 dots sum_r_n=0^k-r_1 - dots - r_n-1 A_(r_1,dots,r_n) $
– Steffen Plunder
Jul 15 at 10:33
2
Yes, that's true. I gave the example on only one multi-index as I thought that this would explain at lest how the multi-index notation works. As I said, I don't really know were that formula comes from but in a more general manner the sum has to be taken on all the possible multi-indices with absolute value of $6$. I'll edit the post to integrate your detail!
– Davide Morgante
Jul 15 at 10:43
Thanks for the edit!
– Steffen Plunder
Jul 15 at 11:06
Thank you for the correction @SteffenPlunder :)
– Davide Morgante
Jul 15 at 11:06
add a comment |Â
Thank you very much!
– Ilqur
Jul 15 at 10:32
In the example with $B_i$, after the second equality sign, you miss out all the other possible multi-indices which have absolute value $6$. Maybe it would be good to explain $sum_ < k A_bar r= sum_r_1=0^k sum_r_2=0^k-r_1 dots sum_r_n=0^k-r_1 - dots - r_n-1 A_(r_1,dots,r_n) $
– Steffen Plunder
Jul 15 at 10:33
2
Yes, that's true. I gave the example on only one multi-index as I thought that this would explain at lest how the multi-index notation works. As I said, I don't really know were that formula comes from but in a more general manner the sum has to be taken on all the possible multi-indices with absolute value of $6$. I'll edit the post to integrate your detail!
– Davide Morgante
Jul 15 at 10:43
Thanks for the edit!
– Steffen Plunder
Jul 15 at 11:06
Thank you for the correction @SteffenPlunder :)
– Davide Morgante
Jul 15 at 11:06
Thank you very much!
– Ilqur
Jul 15 at 10:32
Thank you very much!
– Ilqur
Jul 15 at 10:32
In the example with $B_i$, after the second equality sign, you miss out all the other possible multi-indices which have absolute value $6$. Maybe it would be good to explain $sum_ < k A_bar r= sum_r_1=0^k sum_r_2=0^k-r_1 dots sum_r_n=0^k-r_1 - dots - r_n-1 A_(r_1,dots,r_n) $
– Steffen Plunder
Jul 15 at 10:33
In the example with $B_i$, after the second equality sign, you miss out all the other possible multi-indices which have absolute value $6$. Maybe it would be good to explain $sum_ < k A_bar r= sum_r_1=0^k sum_r_2=0^k-r_1 dots sum_r_n=0^k-r_1 - dots - r_n-1 A_(r_1,dots,r_n) $
– Steffen Plunder
Jul 15 at 10:33
2
2
Yes, that's true. I gave the example on only one multi-index as I thought that this would explain at lest how the multi-index notation works. As I said, I don't really know were that formula comes from but in a more general manner the sum has to be taken on all the possible multi-indices with absolute value of $6$. I'll edit the post to integrate your detail!
– Davide Morgante
Jul 15 at 10:43
Yes, that's true. I gave the example on only one multi-index as I thought that this would explain at lest how the multi-index notation works. As I said, I don't really know were that formula comes from but in a more general manner the sum has to be taken on all the possible multi-indices with absolute value of $6$. I'll edit the post to integrate your detail!
– Davide Morgante
Jul 15 at 10:43
Thanks for the edit!
– Steffen Plunder
Jul 15 at 11:06
Thanks for the edit!
– Steffen Plunder
Jul 15 at 11:06
Thank you for the correction @SteffenPlunder :)
– Davide Morgante
Jul 15 at 11:06
Thank you for the correction @SteffenPlunder :)
– Davide Morgante
Jul 15 at 11:06
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