Mixing
Reducing an algebraic summation expression
$begingroup$
Consider the following elements of an algebraic series
$$sum_{i=1}^Nc(theta_i)$$
$$frac{1}{N}sum_{j=1}^Nsum_{i=1,ineq j}^Nc(theta_i)$$
$$frac{1}{N-1}frac{1}{N}sum_{k=1}^Nsum_{array{j=1 \jneq k}}^Nsum_{array{i=1\ineq j\ineq k}}^Nc(theta_i)$$
$$frac{1}{N-2}frac{1}{N-1}frac{1}{N}sum_{l=1}^Nsum_{array{k=1\lneq k}}^Nsum_{array{j=1 \jneq k\jneq l}}^Nsum_{array{i=1\ineq j\ineq k \i neq l}}^Nc(theta_i)$$
where all the other elements run until $(N-1)$
How can I write a general expression for this series, and can the summation and average of all the elements in the series can be written?
sequences-and-series summation
asked Jan 16 at 9:18
jarheadjarhead
1308
$endgroup$
add a comment |
$begingroup$
Consider the following elements of an algebraic series
$$sum_{i=1}^Nc(theta_i)$$
$$frac{1}{N}sum_{j=1}^Nsum_{i=1,ineq j}^Nc(theta_i)$$
$$frac{1}{N-1}frac{1}{N}sum_{k=1}^Nsum_{array{j=1 \jneq k}}^Nsum_{array{i=1\ineq j\ineq k}}^Nc(theta_i)$$
$$frac{1}{N-2}frac{1}{N-1}frac{1}{N}sum_{l=1}^Nsum_{array{k=1\lneq k}}^Nsum_{array{j=1 \jneq k\jneq l}}^Nsum_{array{i=1\ineq j\ineq k \i neq l}}^Nc(theta_i)$$
where all the other elements run until $(N-1)$
How can I write a general expression for this series, and can the summation and average of all the elements in the series can be written?
sequences-and-series summation
asked Jan 16 at 9:18
jarheadjarhead
1308
$endgroup$
$begingroup$
The argument of the sum only depends on $i$. Are you sure it is meant this way, shouldn't it also depend on $j, k, l$?
$endgroup$
– Florian
Jan 16 at 9:23
$begingroup$
@Florian, yes each expression is actuly an average of all the possible configurations of $c(theta_i)$ when we count $N,N-1,N-2...$ elements.
$endgroup$
– jarhead
Jan 16 at 9:25
add a comment |
$begingroup$
Consider the following elements of an algebraic series
$$sum_{i=1}^Nc(theta_i)$$
$$frac{1}{N}sum_{j=1}^Nsum_{i=1,ineq j}^Nc(theta_i)$$
$$frac{1}{N-1}frac{1}{N}sum_{k=1}^Nsum_{array{j=1 \jneq k}}^Nsum_{array{i=1\ineq j\ineq k}}^Nc(theta_i)$$
$$frac{1}{N-2}frac{1}{N-1}frac{1}{N}sum_{l=1}^Nsum_{array{k=1\lneq k}}^Nsum_{array{j=1 \jneq k\jneq l}}^Nsum_{array{i=1\ineq j\ineq k \i neq l}}^Nc(theta_i)$$
where all the other elements run until $(N-1)$
How can I write a general expression for this series, and can the summation and average of all the elements in the series can be written?
sequences-and-series summation
asked Jan 16 at 9:18
jarheadjarhead
1308
$endgroup$
Consider the following elements of an algebraic series
$$sum_{i=1}^Nc(theta_i)$$
$$frac{1}{N}sum_{j=1}^Nsum_{i=1,ineq j}^Nc(theta_i)$$
$$frac{1}{N-1}frac{1}{N}sum_{k=1}^Nsum_{array{j=1 \jneq k}}^Nsum_{array{i=1\ineq j\ineq k}}^Nc(theta_i)$$
$$frac{1}{N-2}frac{1}{N-1}frac{1}{N}sum_{l=1}^Nsum_{array{k=1\lneq k}}^Nsum_{array{j=1 \jneq k\jneq l}}^Nsum_{array{i=1\ineq j\ineq k \i neq l}}^Nc(theta_i)$$
where all the other elements run until $(N-1)$
How can I write a general expression for this series, and can the summation and average of all the elements in the series can be written?
sequences-and-series summation
sequences-and-series summation
asked Jan 16 at 9:18
jarheadjarhead
1308
asked Jan 16 at 9:18
jarheadjarhead
1308
asked Jan 16 at 9:18
jarheadjarhead
1308
asked Jan 16 at 9:18
jarheadjarhead
1308
asked Jan 16 at 9:18
jarheadjarhead
1308
1308
$begingroup$
The argument of the sum only depends on $i$. Are you sure it is meant this way, shouldn't it also depend on $j, k, l$?
$endgroup$
– Florian
Jan 16 at 9:23
$begingroup$
@Florian, yes each expression is actuly an average of all the possible configurations of $c(theta_i)$ when we count $N,N-1,N-2...$ elements.
$endgroup$
– jarhead
Jan 16 at 9:25
add a comment |
$begingroup$
The argument of the sum only depends on $i$. Are you sure it is meant this way, shouldn't it also depend on $j, k, l$?
$endgroup$
– Florian
Jan 16 at 9:23
$begingroup$
@Florian, yes each expression is actuly an average of all the possible configurations of $c(theta_i)$ when we count $N,N-1,N-2...$ elements.
$endgroup$
– jarhead
Jan 16 at 9:25
$begingroup$
The argument of the sum only depends on $i$. Are you sure it is meant this way, shouldn't it also depend on $j, k, l$?
$endgroup$
– Florian
Jan 16 at 9:23
$begingroup$
The argument of the sum only depends on $i$. Are you sure it is meant this way, shouldn't it also depend on $j, k, l$?
$endgroup$
– Florian
Jan 16 at 9:23
$begingroup$
@Florian, yes each expression is actuly an average of all the possible configurations of $c(theta_i)$ when we count $N,N-1,N-2...$ elements.
$endgroup$
– jarhead
Jan 16 at 9:25
$begingroup$
@Florian, yes each expression is actuly an average of all the possible configurations of $c(theta_i)$ when we count $N,N-1,N-2...$ elements.
$endgroup$
– jarhead
Jan 16 at 9:25
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We just need to look at two instances in order to see what's going on.
We obtain
begin{align*}
color{blue}{sum_{j=1}^Nsum_{{i=1}atop{ineq j}}^Nc(theta_i)}
&=sum_{i=1}^Nc(theta_i)sum_{{j=1}atop{jneq i}}^N1\
&,,color{blue}{=(N-1)sum_{i=1}^Nc(theta_i)}
end{align*}
and
begin{align*}
color{blue}{sum_{k=1}^Nsum_{{j=1}atop{jneq k}}^Nsum_{{i=1}atop{ineq j,ineq k}}^Nc(theta_i)}
&=sum_{i=1}^Nc(theta_i)sum_{{j=1}atop{jneq i}}^Nsum_{{i=1}atop{ineq j,ineq k}}^N1\
&=(N-2)sum_{i=1}^Nc(theta_i)sum_{{j=1}atop{jneq i}}^N1\
&,,color{blue}{=(N-1)(N-2)sum_{i=1}^Nc(theta_i)}
end{align*}
We deduce for positive integers $m<N$:
begin{align*}
frac{(N-m)!}{N!}sum_{k_0=1}^N
sum_{{k_1=1}atop{k_1neq k_j, 0leq j< 1}}^Ncdots
sum_{{k_m=1}atop{k_mneq k_j, 0leq j<m}}^Nc(theta_{k_m})
=frac{N-m}{N}sum_{k_m=1}^Nc(theta_{k_m})tag{1}
end{align*}
which can be shown rigorously for instance by using induction by $mgeq 1$.
The general part corresponds to the examples above by
begin{align*}
&(j,i)to(k_0,k_1)quad &(m=1)\
&(k,j,i)to (k_0,k_1,k_2)quad &(m=2)
end{align*}
We obtain for $m=3$ from (1):
begin{align*}
color{blue}{frac{(N-3)!}{N!}}&color{blue}{sum_{k_0=1}^N
sum_{{k_1=1}atop{k_1neq k_j, 0leq j< 1}}^N
sum_{{k_2=1}atop{k_2neq k_j, 0leq j<2}}^N
sum_{{k_3=1}atop{k_3neq k_j, 0leq j<3}}^N
c(theta_{k_3})}\
&=frac{1}{N(N-1)(N-2)}sum_{k_3=1}^Nc(theta_{k_3})
sum_{{k_2=1}atop{k_2neq k_{3-j}, 0leq j< 1}}^N
sum_{{k_1=1}atop{k_1neq k_{3-j}, 0leq j<2}}^N
sum_{{k_0=1}atop{k_0neq k_{3-j}, 0leq j<3}}^N1\
&=frac{N-3}{N(N-1)(N-2)}sum_{k_3=1}^Nc(theta_{k_3})
sum_{{k_2=1}atop{k_2neq k_{3-j}, 0leq j< 1}}^N
sum_{{k_1=1}atop{k_1neq k_{3-j}, 0leq j<2}}^N1\
&=frac{N-3}{N(N-1)}sum_{k_3=1}^Nc(theta_{k_3})
sum_{{k_2=1}atop{k_2neq k_{3-j}, 0leq j< 1}}^N1\
&,,color{blue}{=frac{N-3}{N}sum_{k_3=1}^Nc(theta_{k_3})}
end{align*}
answered Jan 16 at 14:53
Markus ScheuerMarkus Scheuer
61.9k457149
$endgroup$
$begingroup$
in the final result, the counting is not over the $i$'s. what does the counting of $k_0$ represent?
$endgroup$
– jarhead
Jan 17 at 12:00
1
$begingroup$
@jarhead: Typos corrected. Thanks for pointing at it.
$endgroup$
– Markus Scheuer
Jan 17 at 12:16
1
$begingroup$
@jarhead: I've added some info which might help to clarify open aspects.
$endgroup$
– Markus Scheuer
Jan 17 at 13:45
1
$begingroup$
@jarhead: When considering the special cases $m=1$ and $m=2$, we have $(N-1)!/N!=1/N$ and $(N-2)!/N!=1/(Ncdot (N-1))$ in accordance with the factors at the left-hand side of your second and third example.
$endgroup$
– Markus Scheuer
Jan 17 at 15:15
1
$begingroup$
@jarhead: Example added which should clarify the open points. $m$ is just a fixed number indicating the number of sums we sum over.
$endgroup$
– Markus Scheuer
Jan 17 at 16:11
|
show 6 more comments
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1 Answer
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1 Answer
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oldest
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$begingroup$
We just need to look at two instances in order to see what's going on.
We obtain
begin{align*}
color{blue}{sum_{j=1}^Nsum_{{i=1}atop{ineq j}}^Nc(theta_i)}
&=sum_{i=1}^Nc(theta_i)sum_{{j=1}atop{jneq i}}^N1\
&,,color{blue}{=(N-1)sum_{i=1}^Nc(theta_i)}
end{align*}
and
begin{align*}
color{blue}{sum_{k=1}^Nsum_{{j=1}atop{jneq k}}^Nsum_{{i=1}atop{ineq j,ineq k}}^Nc(theta_i)}
&=sum_{i=1}^Nc(theta_i)sum_{{j=1}atop{jneq i}}^Nsum_{{i=1}atop{ineq j,ineq k}}^N1\
&=(N-2)sum_{i=1}^Nc(theta_i)sum_{{j=1}atop{jneq i}}^N1\
&,,color{blue}{=(N-1)(N-2)sum_{i=1}^Nc(theta_i)}
end{align*}
We deduce for positive integers $m<N$:
begin{align*}
frac{(N-m)!}{N!}sum_{k_0=1}^N
sum_{{k_1=1}atop{k_1neq k_j, 0leq j< 1}}^Ncdots
sum_{{k_m=1}atop{k_mneq k_j, 0leq j<m}}^Nc(theta_{k_m})
=frac{N-m}{N}sum_{k_m=1}^Nc(theta_{k_m})tag{1}
end{align*}
which can be shown rigorously for instance by using induction by $mgeq 1$.
The general part corresponds to the examples above by
begin{align*}
&(j,i)to(k_0,k_1)quad &(m=1)\
&(k,j,i)to (k_0,k_1,k_2)quad &(m=2)
end{align*}
We obtain for $m=3$ from (1):
begin{align*}
color{blue}{frac{(N-3)!}{N!}}&color{blue}{sum_{k_0=1}^N
sum_{{k_1=1}atop{k_1neq k_j, 0leq j< 1}}^N
sum_{{k_2=1}atop{k_2neq k_j, 0leq j<2}}^N
sum_{{k_3=1}atop{k_3neq k_j, 0leq j<3}}^N
c(theta_{k_3})}\
&=frac{1}{N(N-1)(N-2)}sum_{k_3=1}^Nc(theta_{k_3})
sum_{{k_2=1}atop{k_2neq k_{3-j}, 0leq j< 1}}^N
sum_{{k_1=1}atop{k_1neq k_{3-j}, 0leq j<2}}^N
sum_{{k_0=1}atop{k_0neq k_{3-j}, 0leq j<3}}^N1\
&=frac{N-3}{N(N-1)(N-2)}sum_{k_3=1}^Nc(theta_{k_3})
sum_{{k_2=1}atop{k_2neq k_{3-j}, 0leq j< 1}}^N
sum_{{k_1=1}atop{k_1neq k_{3-j}, 0leq j<2}}^N1\
&=frac{N-3}{N(N-1)}sum_{k_3=1}^Nc(theta_{k_3})
sum_{{k_2=1}atop{k_2neq k_{3-j}, 0leq j< 1}}^N1\
&,,color{blue}{=frac{N-3}{N}sum_{k_3=1}^Nc(theta_{k_3})}
end{align*}
answered Jan 16 at 14:53
Markus ScheuerMarkus Scheuer
61.9k457149
$endgroup$
$begingroup$
in the final result, the counting is not over the $i$'s. what does the counting of $k_0$ represent?
$endgroup$
– jarhead
Jan 17 at 12:00
1
$begingroup$
@jarhead: Typos corrected. Thanks for pointing at it.
$endgroup$
– Markus Scheuer
Jan 17 at 12:16
1
$begingroup$
@jarhead: I've added some info which might help to clarify open aspects.
$endgroup$
– Markus Scheuer
Jan 17 at 13:45
1
$begingroup$
@jarhead: When considering the special cases $m=1$ and $m=2$, we have $(N-1)!/N!=1/N$ and $(N-2)!/N!=1/(Ncdot (N-1))$ in accordance with the factors at the left-hand side of your second and third example.
$endgroup$
– Markus Scheuer
Jan 17 at 15:15
1
$begingroup$
@jarhead: Example added which should clarify the open points. $m$ is just a fixed number indicating the number of sums we sum over.
$endgroup$
– Markus Scheuer
Jan 17 at 16:11
|
show 6 more comments
$begingroup$
We just need to look at two instances in order to see what's going on.
We obtain
begin{align*}
color{blue}{sum_{j=1}^Nsum_{{i=1}atop{ineq j}}^Nc(theta_i)}
&=sum_{i=1}^Nc(theta_i)sum_{{j=1}atop{jneq i}}^N1\
&,,color{blue}{=(N-1)sum_{i=1}^Nc(theta_i)}
end{align*}
and
begin{align*}
color{blue}{sum_{k=1}^Nsum_{{j=1}atop{jneq k}}^Nsum_{{i=1}atop{ineq j,ineq k}}^Nc(theta_i)}
&=sum_{i=1}^Nc(theta_i)sum_{{j=1}atop{jneq i}}^Nsum_{{i=1}atop{ineq j,ineq k}}^N1\
&=(N-2)sum_{i=1}^Nc(theta_i)sum_{{j=1}atop{jneq i}}^N1\
&,,color{blue}{=(N-1)(N-2)sum_{i=1}^Nc(theta_i)}
end{align*}
We deduce for positive integers $m<N$:
begin{align*}
frac{(N-m)!}{N!}sum_{k_0=1}^N
sum_{{k_1=1}atop{k_1neq k_j, 0leq j< 1}}^Ncdots
sum_{{k_m=1}atop{k_mneq k_j, 0leq j<m}}^Nc(theta_{k_m})
=frac{N-m}{N}sum_{k_m=1}^Nc(theta_{k_m})tag{1}
end{align*}
which can be shown rigorously for instance by using induction by $mgeq 1$.
The general part corresponds to the examples above by
begin{align*}
&(j,i)to(k_0,k_1)quad &(m=1)\
&(k,j,i)to (k_0,k_1,k_2)quad &(m=2)
end{align*}
We obtain for $m=3$ from (1):
begin{align*}
color{blue}{frac{(N-3)!}{N!}}&color{blue}{sum_{k_0=1}^N
sum_{{k_1=1}atop{k_1neq k_j, 0leq j< 1}}^N
sum_{{k_2=1}atop{k_2neq k_j, 0leq j<2}}^N
sum_{{k_3=1}atop{k_3neq k_j, 0leq j<3}}^N
c(theta_{k_3})}\
&=frac{1}{N(N-1)(N-2)}sum_{k_3=1}^Nc(theta_{k_3})
sum_{{k_2=1}atop{k_2neq k_{3-j}, 0leq j< 1}}^N
sum_{{k_1=1}atop{k_1neq k_{3-j}, 0leq j<2}}^N
sum_{{k_0=1}atop{k_0neq k_{3-j}, 0leq j<3}}^N1\
&=frac{N-3}{N(N-1)(N-2)}sum_{k_3=1}^Nc(theta_{k_3})
sum_{{k_2=1}atop{k_2neq k_{3-j}, 0leq j< 1}}^N
sum_{{k_1=1}atop{k_1neq k_{3-j}, 0leq j<2}}^N1\
&=frac{N-3}{N(N-1)}sum_{k_3=1}^Nc(theta_{k_3})
sum_{{k_2=1}atop{k_2neq k_{3-j}, 0leq j< 1}}^N1\
&,,color{blue}{=frac{N-3}{N}sum_{k_3=1}^Nc(theta_{k_3})}
end{align*}
answered Jan 16 at 14:53
Markus ScheuerMarkus Scheuer
61.9k457149
$endgroup$
$begingroup$
in the final result, the counting is not over the $i$'s. what does the counting of $k_0$ represent?
$endgroup$
– jarhead
Jan 17 at 12:00
1
$begingroup$
@jarhead: Typos corrected. Thanks for pointing at it.
$endgroup$
– Markus Scheuer
Jan 17 at 12:16
1
$begingroup$
@jarhead: I've added some info which might help to clarify open aspects.
$endgroup$
– Markus Scheuer
Jan 17 at 13:45
1
$begingroup$
@jarhead: When considering the special cases $m=1$ and $m=2$, we have $(N-1)!/N!=1/N$ and $(N-2)!/N!=1/(Ncdot (N-1))$ in accordance with the factors at the left-hand side of your second and third example.
$endgroup$
– Markus Scheuer
Jan 17 at 15:15
1
$begingroup$
@jarhead: Example added which should clarify the open points. $m$ is just a fixed number indicating the number of sums we sum over.
$endgroup$
– Markus Scheuer
Jan 17 at 16:11
|
show 6 more comments
$begingroup$
We just need to look at two instances in order to see what's going on.
We obtain
begin{align*}
color{blue}{sum_{j=1}^Nsum_{{i=1}atop{ineq j}}^Nc(theta_i)}
&=sum_{i=1}^Nc(theta_i)sum_{{j=1}atop{jneq i}}^N1\
&,,color{blue}{=(N-1)sum_{i=1}^Nc(theta_i)}
end{align*}
and
begin{align*}
color{blue}{sum_{k=1}^Nsum_{{j=1}atop{jneq k}}^Nsum_{{i=1}atop{ineq j,ineq k}}^Nc(theta_i)}
&=sum_{i=1}^Nc(theta_i)sum_{{j=1}atop{jneq i}}^Nsum_{{i=1}atop{ineq j,ineq k}}^N1\
&=(N-2)sum_{i=1}^Nc(theta_i)sum_{{j=1}atop{jneq i}}^N1\
&,,color{blue}{=(N-1)(N-2)sum_{i=1}^Nc(theta_i)}
end{align*}
We deduce for positive integers $m<N$:
begin{align*}
frac{(N-m)!}{N!}sum_{k_0=1}^N
sum_{{k_1=1}atop{k_1neq k_j, 0leq j< 1}}^Ncdots
sum_{{k_m=1}atop{k_mneq k_j, 0leq j<m}}^Nc(theta_{k_m})
=frac{N-m}{N}sum_{k_m=1}^Nc(theta_{k_m})tag{1}
end{align*}
which can be shown rigorously for instance by using induction by $mgeq 1$.
The general part corresponds to the examples above by
begin{align*}
&(j,i)to(k_0,k_1)quad &(m=1)\
&(k,j,i)to (k_0,k_1,k_2)quad &(m=2)
end{align*}
We obtain for $m=3$ from (1):
begin{align*}
color{blue}{frac{(N-3)!}{N!}}&color{blue}{sum_{k_0=1}^N
sum_{{k_1=1}atop{k_1neq k_j, 0leq j< 1}}^N
sum_{{k_2=1}atop{k_2neq k_j, 0leq j<2}}^N
sum_{{k_3=1}atop{k_3neq k_j, 0leq j<3}}^N
c(theta_{k_3})}\
&=frac{1}{N(N-1)(N-2)}sum_{k_3=1}^Nc(theta_{k_3})
sum_{{k_2=1}atop{k_2neq k_{3-j}, 0leq j< 1}}^N
sum_{{k_1=1}atop{k_1neq k_{3-j}, 0leq j<2}}^N
sum_{{k_0=1}atop{k_0neq k_{3-j}, 0leq j<3}}^N1\
&=frac{N-3}{N(N-1)(N-2)}sum_{k_3=1}^Nc(theta_{k_3})
sum_{{k_2=1}atop{k_2neq k_{3-j}, 0leq j< 1}}^N
sum_{{k_1=1}atop{k_1neq k_{3-j}, 0leq j<2}}^N1\
&=frac{N-3}{N(N-1)}sum_{k_3=1}^Nc(theta_{k_3})
sum_{{k_2=1}atop{k_2neq k_{3-j}, 0leq j< 1}}^N1\
&,,color{blue}{=frac{N-3}{N}sum_{k_3=1}^Nc(theta_{k_3})}
end{align*}
answered Jan 16 at 14:53
Markus ScheuerMarkus Scheuer
61.9k457149
$endgroup$
We just need to look at two instances in order to see what's going on.
We obtain
begin{align*}
color{blue}{sum_{j=1}^Nsum_{{i=1}atop{ineq j}}^Nc(theta_i)}
&=sum_{i=1}^Nc(theta_i)sum_{{j=1}atop{jneq i}}^N1\
&,,color{blue}{=(N-1)sum_{i=1}^Nc(theta_i)}
end{align*}
and
begin{align*}
color{blue}{sum_{k=1}^Nsum_{{j=1}atop{jneq k}}^Nsum_{{i=1}atop{ineq j,ineq k}}^Nc(theta_i)}
&=sum_{i=1}^Nc(theta_i)sum_{{j=1}atop{jneq i}}^Nsum_{{i=1}atop{ineq j,ineq k}}^N1\
&=(N-2)sum_{i=1}^Nc(theta_i)sum_{{j=1}atop{jneq i}}^N1\
&,,color{blue}{=(N-1)(N-2)sum_{i=1}^Nc(theta_i)}
end{align*}
We deduce for positive integers $m<N$:
begin{align*}
frac{(N-m)!}{N!}sum_{k_0=1}^N
sum_{{k_1=1}atop{k_1neq k_j, 0leq j< 1}}^Ncdots
sum_{{k_m=1}atop{k_mneq k_j, 0leq j<m}}^Nc(theta_{k_m})
=frac{N-m}{N}sum_{k_m=1}^Nc(theta_{k_m})tag{1}
end{align*}
which can be shown rigorously for instance by using induction by $mgeq 1$.
The general part corresponds to the examples above by
begin{align*}
&(j,i)to(k_0,k_1)quad &(m=1)\
&(k,j,i)to (k_0,k_1,k_2)quad &(m=2)
end{align*}
We obtain for $m=3$ from (1):
begin{align*}
color{blue}{frac{(N-3)!}{N!}}&color{blue}{sum_{k_0=1}^N
sum_{{k_1=1}atop{k_1neq k_j, 0leq j< 1}}^N
sum_{{k_2=1}atop{k_2neq k_j, 0leq j<2}}^N
sum_{{k_3=1}atop{k_3neq k_j, 0leq j<3}}^N
c(theta_{k_3})}\
&=frac{1}{N(N-1)(N-2)}sum_{k_3=1}^Nc(theta_{k_3})
sum_{{k_2=1}atop{k_2neq k_{3-j}, 0leq j< 1}}^N
sum_{{k_1=1}atop{k_1neq k_{3-j}, 0leq j<2}}^N
sum_{{k_0=1}atop{k_0neq k_{3-j}, 0leq j<3}}^N1\
&=frac{N-3}{N(N-1)(N-2)}sum_{k_3=1}^Nc(theta_{k_3})
sum_{{k_2=1}atop{k_2neq k_{3-j}, 0leq j< 1}}^N
sum_{{k_1=1}atop{k_1neq k_{3-j}, 0leq j<2}}^N1\
&=frac{N-3}{N(N-1)}sum_{k_3=1}^Nc(theta_{k_3})
sum_{{k_2=1}atop{k_2neq k_{3-j}, 0leq j< 1}}^N1\
&,,color{blue}{=frac{N-3}{N}sum_{k_3=1}^Nc(theta_{k_3})}
end{align*}
answered Jan 16 at 14:53
Markus ScheuerMarkus Scheuer
61.9k457149
edited Jan 17 at 16:08
answered Jan 16 at 14:53
Markus ScheuerMarkus Scheuer
61.9k457149
answered Jan 16 at 14:53
Markus ScheuerMarkus Scheuer
61.9k457149
answered Jan 16 at 14:53
Markus ScheuerMarkus Scheuer
61.9k457149
61.9k457149
$begingroup$
in the final result, the counting is not over the $i$'s. what does the counting of $k_0$ represent?
$endgroup$
– jarhead
Jan 17 at 12:00
1
$begingroup$
@jarhead: Typos corrected. Thanks for pointing at it.
$endgroup$
– Markus Scheuer
Jan 17 at 12:16
1
$begingroup$
@jarhead: I've added some info which might help to clarify open aspects.
$endgroup$
– Markus Scheuer
Jan 17 at 13:45
1
$begingroup$
@jarhead: When considering the special cases $m=1$ and $m=2$, we have $(N-1)!/N!=1/N$ and $(N-2)!/N!=1/(Ncdot (N-1))$ in accordance with the factors at the left-hand side of your second and third example.
$endgroup$
– Markus Scheuer
Jan 17 at 15:15
1
$begingroup$
@jarhead: Example added which should clarify the open points. $m$ is just a fixed number indicating the number of sums we sum over.
$endgroup$
– Markus Scheuer
Jan 17 at 16:11
|
show 6 more comments
$begingroup$
in the final result, the counting is not over the $i$'s. what does the counting of $k_0$ represent?
$endgroup$
– jarhead
Jan 17 at 12:00
1
$begingroup$
@jarhead: Typos corrected. Thanks for pointing at it.
$endgroup$
– Markus Scheuer
Jan 17 at 12:16
1
$begingroup$
@jarhead: I've added some info which might help to clarify open aspects.
$endgroup$
– Markus Scheuer
Jan 17 at 13:45
1
$begingroup$
@jarhead: When considering the special cases $m=1$ and $m=2$, we have $(N-1)!/N!=1/N$ and $(N-2)!/N!=1/(Ncdot (N-1))$ in accordance with the factors at the left-hand side of your second and third example.
$endgroup$
– Markus Scheuer
Jan 17 at 15:15
1
$begingroup$
@jarhead: Example added which should clarify the open points. $m$ is just a fixed number indicating the number of sums we sum over.
$endgroup$
– Markus Scheuer
Jan 17 at 16:11
$begingroup$
in the final result, the counting is not over the $i$'s. what does the counting of $k_0$ represent?
$endgroup$
– jarhead
Jan 17 at 12:00
$begingroup$
in the final result, the counting is not over the $i$'s. what does the counting of $k_0$ represent?
$endgroup$
– jarhead
Jan 17 at 12:00
1
1
$begingroup$
@jarhead: Typos corrected. Thanks for pointing at it.
$endgroup$
– Markus Scheuer
Jan 17 at 12:16
$begingroup$
@jarhead: Typos corrected. Thanks for pointing at it.
$endgroup$
– Markus Scheuer
Jan 17 at 12:16
1
1
$begingroup$
@jarhead: I've added some info which might help to clarify open aspects.
$endgroup$
– Markus Scheuer
Jan 17 at 13:45
$begingroup$
@jarhead: I've added some info which might help to clarify open aspects.
$endgroup$
– Markus Scheuer
Jan 17 at 13:45
1
1
$begingroup$
@jarhead: When considering the special cases $m=1$ and $m=2$, we have $(N-1)!/N!=1/N$ and $(N-2)!/N!=1/(Ncdot (N-1))$ in accordance with the factors at the left-hand side of your second and third example.
$endgroup$
– Markus Scheuer
Jan 17 at 15:15
$begingroup$
@jarhead: When considering the special cases $m=1$ and $m=2$, we have $(N-1)!/N!=1/N$ and $(N-2)!/N!=1/(Ncdot (N-1))$ in accordance with the factors at the left-hand side of your second and third example.
$endgroup$
– Markus Scheuer
Jan 17 at 15:15
1
1
$begingroup$
@jarhead: Example added which should clarify the open points. $m$ is just a fixed number indicating the number of sums we sum over.
$endgroup$
– Markus Scheuer
Jan 17 at 16:11
$begingroup$
@jarhead: Example added which should clarify the open points. $m$ is just a fixed number indicating the number of sums we sum over.
$endgroup$
– Markus Scheuer
Jan 17 at 16:11
|
show 6 more comments
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$begingroup$
The argument of the sum only depends on $i$. Are you sure it is meant this way, shouldn't it also depend on $j, k, l$?
$endgroup$
– Florian
Jan 16 at 9:23
$begingroup$
@Florian, yes each expression is actuly an average of all the possible configurations of $c(theta_i)$ when we count $N,N-1,N-2...$ elements.
$endgroup$
– jarhead
Jan 16 at 9:25