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How to change the order of triple summation?
Clash Royale CLAN TAG#URR8PPP
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How to change summation in the sum
$$
sum_i=0^n left(sum_j=1^i sum_k=j^i b_j,k right) a_i?
$$
For $n=4$ we have
$$
sum_i=0^4 left(sum_j=1^i sum_k=j^i b_j,k right) a_i=f_1,1a_1+a_2 left( f_1,1+f_1,2+f_2,2 right)
+a_3 left( f_1,1+f_1,2+f_1,3+f_2,2+f_2,3+f_
3,3 right) +a_4 left( f_1,1+f_1,2+f_1,3+f_1,4+
f_2,2+f_2,3+f_2,4+f_3,3+f_3,4+f_4,4 right)=f_1,1 left( a_1+a_2+a_3+a_4 right) +f_2,2
left( a_2+a_3+a_4 right) +f_3,3 left( a_3+a_
4 right) +f_4,4a_4+f_2,1a_2+a_3 left( f_2,1
+f_3,1+f_3,2 right) +a_4 left( f_2,1+f_3,1+f_
4,1+f_3,2+f_4,2+f_4,3 right)
$$I guess it must be something like that
$$
sum_i=0^n left(sum_j=1^i sum_k=j^i b_j,k right) a_i=sum_j=1^n sum_k=j^? left(sum_i=?^? a_i right) b_j,k
$$
but can't find true limits.
Any ideas?
combinatorics summation
asked Jul 16 at 6:41
Leox
5,0921323
add a comment |Â
up vote
1
down vote
favorite
How to change summation in the sum
$$
sum_i=0^n left(sum_j=1^i sum_k=j^i b_j,k right) a_i?
$$
For $n=4$ we have
$$
sum_i=0^4 left(sum_j=1^i sum_k=j^i b_j,k right) a_i=f_1,1a_1+a_2 left( f_1,1+f_1,2+f_2,2 right)
+a_3 left( f_1,1+f_1,2+f_1,3+f_2,2+f_2,3+f_
3,3 right) +a_4 left( f_1,1+f_1,2+f_1,3+f_1,4+
f_2,2+f_2,3+f_2,4+f_3,3+f_3,4+f_4,4 right)=f_1,1 left( a_1+a_2+a_3+a_4 right) +f_2,2
left( a_2+a_3+a_4 right) +f_3,3 left( a_3+a_
4 right) +f_4,4a_4+f_2,1a_2+a_3 left( f_2,1
+f_3,1+f_3,2 right) +a_4 left( f_2,1+f_3,1+f_
4,1+f_3,2+f_4,2+f_4,3 right)
$$I guess it must be something like that
$$
sum_i=0^n left(sum_j=1^i sum_k=j^i b_j,k right) a_i=sum_j=1^n sum_k=j^? left(sum_i=?^? a_i right) b_j,k
$$
but can't find true limits.
Any ideas?
combinatorics summation
asked Jul 16 at 6:41
Leox
5,0921323
There are six (and not two) nesting orders in which the summation can be performed. Your first displayed formula exhibits one of these. You have to tell us which is the intended final nesting.
– Christian Blatter
Jul 16 at 12:57
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
How to change summation in the sum
$$
sum_i=0^n left(sum_j=1^i sum_k=j^i b_j,k right) a_i?
$$
For $n=4$ we have
$$
sum_i=0^4 left(sum_j=1^i sum_k=j^i b_j,k right) a_i=f_1,1a_1+a_2 left( f_1,1+f_1,2+f_2,2 right)
+a_3 left( f_1,1+f_1,2+f_1,3+f_2,2+f_2,3+f_
3,3 right) +a_4 left( f_1,1+f_1,2+f_1,3+f_1,4+
f_2,2+f_2,3+f_2,4+f_3,3+f_3,4+f_4,4 right)=f_1,1 left( a_1+a_2+a_3+a_4 right) +f_2,2
left( a_2+a_3+a_4 right) +f_3,3 left( a_3+a_
4 right) +f_4,4a_4+f_2,1a_2+a_3 left( f_2,1
+f_3,1+f_3,2 right) +a_4 left( f_2,1+f_3,1+f_
4,1+f_3,2+f_4,2+f_4,3 right)
$$I guess it must be something like that
$$
sum_i=0^n left(sum_j=1^i sum_k=j^i b_j,k right) a_i=sum_j=1^n sum_k=j^? left(sum_i=?^? a_i right) b_j,k
$$
but can't find true limits.
Any ideas?
combinatorics summation
asked Jul 16 at 6:41
Leox
5,0921323
How to change summation in the sum
$$
sum_i=0^n left(sum_j=1^i sum_k=j^i b_j,k right) a_i?
$$
For $n=4$ we have
$$
sum_i=0^4 left(sum_j=1^i sum_k=j^i b_j,k right) a_i=f_1,1a_1+a_2 left( f_1,1+f_1,2+f_2,2 right)
+a_3 left( f_1,1+f_1,2+f_1,3+f_2,2+f_2,3+f_
3,3 right) +a_4 left( f_1,1+f_1,2+f_1,3+f_1,4+
f_2,2+f_2,3+f_2,4+f_3,3+f_3,4+f_4,4 right)=f_1,1 left( a_1+a_2+a_3+a_4 right) +f_2,2
left( a_2+a_3+a_4 right) +f_3,3 left( a_3+a_
4 right) +f_4,4a_4+f_2,1a_2+a_3 left( f_2,1
+f_3,1+f_3,2 right) +a_4 left( f_2,1+f_3,1+f_
4,1+f_3,2+f_4,2+f_4,3 right)
$$I guess it must be something like that
$$
sum_i=0^n left(sum_j=1^i sum_k=j^i b_j,k right) a_i=sum_j=1^n sum_k=j^? left(sum_i=?^? a_i right) b_j,k
$$
but can't find true limits.
Any ideas?
combinatorics summation
asked Jul 16 at 6:41
Leox
5,0921323
edited Jul 16 at 6:58
asked Jul 16 at 6:41
Leox
5,0921323
asked Jul 16 at 6:41
Leox
5,0921323
asked Jul 16 at 6:41
Leox
5,0921323
5,0921323
There are six (and not two) nesting orders in which the summation can be performed. Your first displayed formula exhibits one of these. You have to tell us which is the intended final nesting.
– Christian Blatter
Jul 16 at 12:57
add a comment |Â
There are six (and not two) nesting orders in which the summation can be performed. Your first displayed formula exhibits one of these. You have to tell us which is the intended final nesting.
– Christian Blatter
Jul 16 at 12:57
There are six (and not two) nesting orders in which the summation can be performed. Your first displayed formula exhibits one of these. You have to tell us which is the intended final nesting.
– Christian Blatter
Jul 16 at 12:57
There are six (and not two) nesting orders in which the summation can be performed. Your first displayed formula exhibits one of these. You have to tell us which is the intended final nesting.
– Christian Blatter
Jul 16 at 12:57
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
4
down vote
accepted
It is helpful to write the range of summation in terms of inequalities. In your case it turns out to be $$beginalign0 &leq i leq n\ 1&leq jleq i\ j&leq kleq iendaligntag1$$ and you can check that this is equivalent to $$1leq j leq k leq i leq n.tag2$$ Notice that the original order of indices is $(i,j,k)$ and take a look at $(1)$. In the first line $i$ "doesn't know about" $j$ or $k$, so it is just $1leq i leq n$ ($i = 0$ doesn't actually appear in the sum). In the second line, $j$ "doesn't know about" $k$, but it "does know" $i$, so it is $1leq j leq i$. Finally, $k$ "knows everyone", so the third line becomes $jleq kleq i$.
Let us now change the order to $(j,k,i)$. From $(2)$ we read that the first line has to be $1leq jleq n$ since $j$ "doesn't know about" $k$ and $i$. The second line is $jleq k leq n$ since $k$ "doesn't know" $i$, but "knows" $j$. Finally, the third line is $kleq ileq n$ since $i$ "knows everyone".
Thus, the change of order of summation you want is $$sum_j=1^nsum_k=j^nsum_i=k^n$$
answered Jul 16 at 7:20
Ennar
13k32343
add a comment |Â
up vote
1
down vote
$i$ from $k$ to $n$, $k$ from $j$ to $n$.
answered Jul 16 at 6:45
Kavi Rama Murthy
21k2830
add a comment |Â
up vote
1
down vote
Note: It is also common to do the transformation using the sigma-notation.
We obtain
beginalign*
colorbluesum_i=0^nleft(sum_j=1^isum_k=j^ib_j,kright)a_i
&=sum_i=0^nleft(sum_1leq jleq kleq ib_j,kright)a_i\
&=sum_1leq jleq kleq ileq n a_ib_j,k\
&=sum_j=1^nsum_jleq kleq ileq na_ib_j,k\
&=,,colorbluesum_j=1^nsum_k=j^nleft(sum_i=k^na_iright)b_j,k
endalign*
answered Jul 16 at 12:31
Markus Scheuer
56.2k450135
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
It is helpful to write the range of summation in terms of inequalities. In your case it turns out to be $$beginalign0 &leq i leq n\ 1&leq jleq i\ j&leq kleq iendaligntag1$$ and you can check that this is equivalent to $$1leq j leq k leq i leq n.tag2$$ Notice that the original order of indices is $(i,j,k)$ and take a look at $(1)$. In the first line $i$ "doesn't know about" $j$ or $k$, so it is just $1leq i leq n$ ($i = 0$ doesn't actually appear in the sum). In the second line, $j$ "doesn't know about" $k$, but it "does know" $i$, so it is $1leq j leq i$. Finally, $k$ "knows everyone", so the third line becomes $jleq kleq i$.
Let us now change the order to $(j,k,i)$. From $(2)$ we read that the first line has to be $1leq jleq n$ since $j$ "doesn't know about" $k$ and $i$. The second line is $jleq k leq n$ since $k$ "doesn't know" $i$, but "knows" $j$. Finally, the third line is $kleq ileq n$ since $i$ "knows everyone".
Thus, the change of order of summation you want is $$sum_j=1^nsum_k=j^nsum_i=k^n$$
answered Jul 16 at 7:20
Ennar
13k32343
add a comment |Â
up vote
4
down vote
accepted
It is helpful to write the range of summation in terms of inequalities. In your case it turns out to be $$beginalign0 &leq i leq n\ 1&leq jleq i\ j&leq kleq iendaligntag1$$ and you can check that this is equivalent to $$1leq j leq k leq i leq n.tag2$$ Notice that the original order of indices is $(i,j,k)$ and take a look at $(1)$. In the first line $i$ "doesn't know about" $j$ or $k$, so it is just $1leq i leq n$ ($i = 0$ doesn't actually appear in the sum). In the second line, $j$ "doesn't know about" $k$, but it "does know" $i$, so it is $1leq j leq i$. Finally, $k$ "knows everyone", so the third line becomes $jleq kleq i$.
Let us now change the order to $(j,k,i)$. From $(2)$ we read that the first line has to be $1leq jleq n$ since $j$ "doesn't know about" $k$ and $i$. The second line is $jleq k leq n$ since $k$ "doesn't know" $i$, but "knows" $j$. Finally, the third line is $kleq ileq n$ since $i$ "knows everyone".
Thus, the change of order of summation you want is $$sum_j=1^nsum_k=j^nsum_i=k^n$$
answered Jul 16 at 7:20
Ennar
13k32343
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
It is helpful to write the range of summation in terms of inequalities. In your case it turns out to be $$beginalign0 &leq i leq n\ 1&leq jleq i\ j&leq kleq iendaligntag1$$ and you can check that this is equivalent to $$1leq j leq k leq i leq n.tag2$$ Notice that the original order of indices is $(i,j,k)$ and take a look at $(1)$. In the first line $i$ "doesn't know about" $j$ or $k$, so it is just $1leq i leq n$ ($i = 0$ doesn't actually appear in the sum). In the second line, $j$ "doesn't know about" $k$, but it "does know" $i$, so it is $1leq j leq i$. Finally, $k$ "knows everyone", so the third line becomes $jleq kleq i$.
Let us now change the order to $(j,k,i)$. From $(2)$ we read that the first line has to be $1leq jleq n$ since $j$ "doesn't know about" $k$ and $i$. The second line is $jleq k leq n$ since $k$ "doesn't know" $i$, but "knows" $j$. Finally, the third line is $kleq ileq n$ since $i$ "knows everyone".
Thus, the change of order of summation you want is $$sum_j=1^nsum_k=j^nsum_i=k^n$$
answered Jul 16 at 7:20
Ennar
13k32343
It is helpful to write the range of summation in terms of inequalities. In your case it turns out to be $$beginalign0 &leq i leq n\ 1&leq jleq i\ j&leq kleq iendaligntag1$$ and you can check that this is equivalent to $$1leq j leq k leq i leq n.tag2$$ Notice that the original order of indices is $(i,j,k)$ and take a look at $(1)$. In the first line $i$ "doesn't know about" $j$ or $k$, so it is just $1leq i leq n$ ($i = 0$ doesn't actually appear in the sum). In the second line, $j$ "doesn't know about" $k$, but it "does know" $i$, so it is $1leq j leq i$. Finally, $k$ "knows everyone", so the third line becomes $jleq kleq i$.
Let us now change the order to $(j,k,i)$. From $(2)$ we read that the first line has to be $1leq jleq n$ since $j$ "doesn't know about" $k$ and $i$. The second line is $jleq k leq n$ since $k$ "doesn't know" $i$, but "knows" $j$. Finally, the third line is $kleq ileq n$ since $i$ "knows everyone".
Thus, the change of order of summation you want is $$sum_j=1^nsum_k=j^nsum_i=k^n$$
answered Jul 16 at 7:20
Ennar
13k32343
answered Jul 16 at 7:20
Ennar
13k32343
answered Jul 16 at 7:20
Ennar
13k32343
answered Jul 16 at 7:20
Ennar
13k32343
13k32343
add a comment |Â
add a comment |Â
up vote
1
down vote
$i$ from $k$ to $n$, $k$ from $j$ to $n$.
answered Jul 16 at 6:45
Kavi Rama Murthy
21k2830
add a comment |Â
up vote
1
down vote
$i$ from $k$ to $n$, $k$ from $j$ to $n$.
answered Jul 16 at 6:45
Kavi Rama Murthy
21k2830
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$i$ from $k$ to $n$, $k$ from $j$ to $n$.
answered Jul 16 at 6:45
Kavi Rama Murthy
21k2830
$i$ from $k$ to $n$, $k$ from $j$ to $n$.
answered Jul 16 at 6:45
Kavi Rama Murthy
21k2830
answered Jul 16 at 6:45
Kavi Rama Murthy
21k2830
answered Jul 16 at 6:45
Kavi Rama Murthy
21k2830
answered Jul 16 at 6:45
Kavi Rama Murthy
21k2830
21k2830
add a comment |Â
add a comment |Â
up vote
1
down vote
Note: It is also common to do the transformation using the sigma-notation.
We obtain
beginalign*
colorbluesum_i=0^nleft(sum_j=1^isum_k=j^ib_j,kright)a_i
&=sum_i=0^nleft(sum_1leq jleq kleq ib_j,kright)a_i\
&=sum_1leq jleq kleq ileq n a_ib_j,k\
&=sum_j=1^nsum_jleq kleq ileq na_ib_j,k\
&=,,colorbluesum_j=1^nsum_k=j^nleft(sum_i=k^na_iright)b_j,k
endalign*
answered Jul 16 at 12:31
Markus Scheuer
56.2k450135
add a comment |Â
up vote
1
down vote
Note: It is also common to do the transformation using the sigma-notation.
We obtain
beginalign*
colorbluesum_i=0^nleft(sum_j=1^isum_k=j^ib_j,kright)a_i
&=sum_i=0^nleft(sum_1leq jleq kleq ib_j,kright)a_i\
&=sum_1leq jleq kleq ileq n a_ib_j,k\
&=sum_j=1^nsum_jleq kleq ileq na_ib_j,k\
&=,,colorbluesum_j=1^nsum_k=j^nleft(sum_i=k^na_iright)b_j,k
endalign*
answered Jul 16 at 12:31
Markus Scheuer
56.2k450135
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Note: It is also common to do the transformation using the sigma-notation.
We obtain
beginalign*
colorbluesum_i=0^nleft(sum_j=1^isum_k=j^ib_j,kright)a_i
&=sum_i=0^nleft(sum_1leq jleq kleq ib_j,kright)a_i\
&=sum_1leq jleq kleq ileq n a_ib_j,k\
&=sum_j=1^nsum_jleq kleq ileq na_ib_j,k\
&=,,colorbluesum_j=1^nsum_k=j^nleft(sum_i=k^na_iright)b_j,k
endalign*
answered Jul 16 at 12:31
Markus Scheuer
56.2k450135
Note: It is also common to do the transformation using the sigma-notation.
We obtain
beginalign*
colorbluesum_i=0^nleft(sum_j=1^isum_k=j^ib_j,kright)a_i
&=sum_i=0^nleft(sum_1leq jleq kleq ib_j,kright)a_i\
&=sum_1leq jleq kleq ileq n a_ib_j,k\
&=sum_j=1^nsum_jleq kleq ileq na_ib_j,k\
&=,,colorbluesum_j=1^nsum_k=j^nleft(sum_i=k^na_iright)b_j,k
endalign*
answered Jul 16 at 12:31
Markus Scheuer
56.2k450135
edited Jul 16 at 12:55
answered Jul 16 at 12:31
Markus Scheuer
56.2k450135
answered Jul 16 at 12:31
Markus Scheuer
56.2k450135
answered Jul 16 at 12:31
Markus Scheuer
56.2k450135
56.2k450135
add a comment |Â
add a comment |Â
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There are six (and not two) nesting orders in which the summation can be performed. Your first displayed formula exhibits one of these. You have to tell us which is the intended final nesting.
– Christian Blatter
Jul 16 at 12:57